instuctors-solutions-manual-for-statistics-for-business-and-economics
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INSTRUCTOR’S SOLUTIONS
MANUAL
NANCY S. BOUDREAU
Bowling Green State University
S TATISTICS FOR B USINESS
AND E CONOMICS
TWELFTH EDITION
James T. McClave
Info Tech, Inc.
University of Florida
P. George Benson
College of Charleston
Terry Sincich
University of South Florida
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author.
Copyright © 2014, 2011, 2008 Pearson Education, Inc.
Publishing as Pearson, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-83681-6
ISBN-10: 0-321-83681-2
www.pearsonhighered.com
Chapter 1
Statistics, Data, and Statistical Thinking
1.1
Statistics is a science that deals with the collection, classification, analysis, and interpretation of
information or data. It is a meaningful, useful science with a broad, almost limitless scope of applications
to business, government, and the physical and social sciences.
1.2
Descriptive statistics utilizes numerical and graphical methods to look for patterns, to summarize, and to
present the information in a set of data. Inferential statistics utilizes sample data to make estimates,
decisions, predictions, or other generalizations about a larger set of data.
1.3
The four elements of a descriptive statistics problem are:
1.
2.
3.
4.
1.4
The population or sample of interest. This is the collection of all the units upon which the variable is
measured.
One or more variables that are to be investigated. These are the types of data that are to be collected.
Tables, graphs, or numerical summary tools. These are tools used to display the characteristic of the
sample or population.
Identification of patterns in the data. These are conclusions drawn from what the summary tools
revealed about the population or sample.
The five elements of an inferential statistical analysis are:
1.
2.
3.
4.
5.
The population of interest. The population is a set of existing units.
One or more variables that are to be investigated. A variable is a characteristic or property of an
individual population unit.
The sample of population units. A sample is a subset of the units of a population.
The inference about the population based on information contained in the sample. A statistical
inference is an estimate, prediction, or generalization about a population based on information
contained in a sample.
A measure of reliability for the inference. The reliability of an inference is how confident one is that
the inference is correct.
1.5
The first major method of collecting data is from a published source. These data have already been
collected by someone else and are available in a published source. The second method of collecting data is
from a designed experiment. These data are collected by a researcher who exerts strict control over the
experimental units in a study. These data are measured directly from the experimental units. The final
method of collecting data is observational. These data are collected directly from experimental units by
simply observing the experimental units in their natural environment and recording the values of the
desired characteristics. The most common type of observational study is a survey.
1.6
Quantitative data are measurements that are recorded on a meaningful numerical scale. Qualitative data are
measurements that are not numerical in nature; they can only be classified into one of a group of categories.
1.7
A population is a set of existing units such as people, objects, transactions, or events. A variable is a
characteristic or property of an individual population unit such as height of a person, time of a reflex,
amount of a transaction, etc.
1
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2
Chapter 1
1.8
A population is a set of existing units such as people, objects, transactions, or events. A sample is a subset
of the units of a population.
1.9
A representative sample is a sample that exhibits characteristics similar to those possessed by the target
population. A representative sample is essential if inferential statistics is to be applied. If a sample does
not possess the same characteristics as the target population, then any inferences made using the sample
will be unreliable.
1.10
An inference without a measure of reliability is nothing more than a guess. A measure of reliability
separates statistical inference from fortune telling or guessing. Reliability gives a measure of how
confident one is that the inference is correct.
1.11
A population is a set of existing units such as people, objects, transactions, or events. A process is a series
of actions or operations that transform inputs to outputs. A process produces or generates output over time.
Examples of processes are assembly lines, oil refineries, and stock prices.
1.12
Statistical thinking involves applying rational thought processes to critically assess data and inferences
made from the data. It involves not taking all data and inferences presented at face value, but rather
making sure the inferences and data are valid.
1.13
The data consisting of the classifications A, B, C, and D are qualitative. These data are nominal and thus
are qualitative. After the data are input as 1, 2, 3, and 4, they are still nominal and thus qualitative. The
only differences between the two data sets are the names of the categories. The numbers associated with
the four groups are meaningless.
1.14
Answers will vary. First, number the elements of the population from 1 to 200,000. Using MINITAB,
generate 10 numbers on the interval from 1 to 200,000, eliminating any duplicates.
The 10 numbers selected for the random sample are:
135075
89127
189226
83899
112367
191496
110021
44853
42091
198461
Elements with the above numbers are selected for the sample.
1.15
a.
The experimental unit for this study is a single-family residential property in Arlington, Texas.
b.
The variables measured are the sale price and the Zillow estimated value. Both of these variables are
quantitative.
c.
If these 2,045 properties were all the properties sold in Arlington, Texas in the past 6 months, then
this would be considered the population.
d.
If these 2,045 properties represent a sample, then the population would be all the single-family
residential properties sold in the last 6 months in Arlington, Texas.
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Statistics, Data, and Statistical Thinking
1.16
1.17
1.18
1.19
3
e.
No. The real estate market across the United States varies greatly. The prices of single-family
residential properties in this small area are probably not representative of all properties across the
United States.
a.
The experimental unit for this study is an NFL quarterback.
b.
The variables measured in this study include draft position, NFL winning ratio, and QB production
score. Since the draft position was put into 3 categories, it is a qualitative variable. The NFL winning
ratio and the QB production score are both quantitative.
c.
Since we want to project the performance of future NFL QBs, this would be an application of
inferential statistics.
a.
The population of interest is all citizens of the United States.
b.
The variable of interest is the view of each citizen as to whether the president is doing a good or bad
job. It is qualitative.
c.
The sample is the 2000 individuals selected for the poll.
d.
The inference of interest is to estimate the proportion of all U.S. citizens who believe the president is
doing a good job.
e.
The method of data collection is a survey.
f.
It is not very likely that the sample will be representative of the population of all citizens of the
United States. By selecting phone numbers at random, the sample will be limited to only those
people who have telephones. Also, many people share the same phone number, so each person
would not have an equal chance of being contacted. Another possible problem is the time of day the
calls are made. If the calls are made in the evening, those people who work in the evening would not
be represented.
a.
High school GPA is a number usually between 0.0 and 4.0. Therefore, it is quantitative.
b.
Honors/awards would have responses that name things. Therefore, it would be qualitative.
c.
The scores on the SAT's are numbers between 200 and 800. Therefore, it is quantitative.
d.
Gender is either male or female. Therefore, it is qualitative.
e.
Parent's income is a number: $25,000, $45,000, etc. Therefore, it is quantitative.
f.
Age is a number: 17, 18, etc. Therefore, it is quantitative.
I.
Qualitative; the possible responses are "yes" or "no," which are non-numerical.
II.
Quantitative; age is measured on a numerical scale, such as 15, 32, etc.
III.
Qualitative; the possible responses are “yes” or “no,” which are non-numerical.
IV.
Qualitative; the possible responses are "laser printer" or "another type of printer," which are nonnumerical.
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1.20
Chapter 1
V.
Qualitative; the speeds can be classified as "slower," "unchanged," or "faster," which are nonnumerical.
VI.
Quantitative; the number of people in a household who have used Windows 95 at least once is
measured on a numerical scale, such as 0, 1, 2, etc.
a.
For question 1, the data collected would be qualitative. The possible response would be “yes” or
“no”.
For question 2, the data collected would be quantitative. The responses would be numbers such as 0,
1, 2, etc.
For question 3, the data collected would be qualitative. The possible responses would be “yes” or
“no”.
1.21
1.22
1.23
b.
The data collected from the 1,066 adults would be a sample. These adults would only be a part of all
adults in the United States.
a.
Whether the data collected on the chief executive officers at the 500 largest U. S. companies is a
population or a sample depends on what one is interested in. If one is only interested in the
information from the CEO’s of the 500 largest U.S. companies, then these data form a population. If
one is interested in the information on CEO’s from all U.S. firms, then these data would form a
sample.
b.
1.
The industry type of the CEO’s company is a qualitative variable. The industry type is a name.
2.
The CEO’s total compensation is a meaningful number. Thus, it is a quantitative variable.
3.
The CEO’s total compensation over the previous five years is a meaningful number. Thus, it is
a quantitative variable.
4.
The number of company stock shares (millions) held is a meaningful number. Thus, it is a
quantitative variable.
5.
The CEO’s age is a meaningful number. Thus, it is a quantitative variable.
6.
The CEO’s efficiency rating is a meaningful number. Thus, it is a quantitative variable.
a.
The population of interest is the status of computer crime at all United States businesses and
government agencies.
b.
The method of data collection was a survey. Since not all of those who were sent a survey
responded, the sample was self-selected. The results are probably not representative of the
population. Usually, those who respond to surveys have very strong opinions, either positive or
negative.
c.
The variable of interest is whether or not the firm or agency had unauthorized use of its computer
systems during the year. Since the response would be either yes or no, the variable would be
qualitative.
d.
If the sample was representative, we could infer that approximately 41% of all U. S. corporations and
government agencies experienced unauthorized use of their computer systems during the year.
Since the data collected consist of the entire population, this would represent a descriptive study. Flaherty
used the data to help describe the condition of the U.S. Treasury in 1861.
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Statistics, Data, and Statistical Thinking
5
1.24
This study would be an example of inferential statistics. The researchers collected data over 2 years.
Using this information, the researchers are projecting or making inferences about what will happen in the
future.
1.25
a.
The population of interest is all individuals who earned MBA degrees since January 1990.
b.
The method of data collection was a survey.
c.
This is probably not a representative sample. The sample was self-selected. Not all of those who
were selected for the study responded to all four surveys. Those who did respond to all 4 surveys
probably have very strong opinions, either positive or negative, which may not be representative of
all of those in the population.
a.
The population of interest is all CPA firms.
b.
A survey was used to collect the data.
c.
This sample was probably not representative. Not all of those selected to be in the sample responded.
In fact, only 992 of the 23,500 people who were sent the survey responded. Generally, those who do
respond to surveys have very strong opinions, either positive or negative. These may not be the
opinions of all CPA firms.
d.
Since the sample may not be representative, the inferences drawn in the study may not be valid.
a.
Length of maximum span can take on values such as 15 feet, 50 feet, 75 feet, etc. Therefore, it is
quantitative.
b.
The number of vehicle lanes can take on values such as 2, 4, etc. Therefore, it is quantitative.
c.
The answer to this item is "yes" or "no," which is not numeric. Therefore, it is qualitative.
d.
Average daily traffic could take on values such as 150 vehicles, 3,579 vehicles, 53,295 vehicles, etc.
Therefore, it is quantitative.
e.
Condition can take on values "good," "fair," or "poor," which are not numeric. Therefore, it is
qualitative.
f.
The length of the bypass or detour could take on values such as 1 mile, 4 miles, etc. Therefore, it is
quantitative.
g.
Route type can take on values "interstate," U.S.," "state," "county," or "city," which are not numeric.
Therefore, it is qualitative.
a.
The variable of interest to the researchers is the rating of highway bridges.
b.
Since the rating of a bridge can be categorized as one of three possible values, it is qualitative.
c.
The data set analyzed is a population since all highway bridges in the U.S. were categorized.
d.
The data were collected observationally. Each bridge was observed in its natural setting.
1.26
1.27
1.28
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6
Chapter 1
1.29
a.
The process being studied is the distribution of pipes, valves, and fittings to the refining, chemical,
and petrochemical industries by the Wallace Company of Houston.
b.
The variables of interest are the speed of the deliveries, the accuracy of the invoices, and the quality
of the packaging of the products.
c.
The sampling plan was to monitor a subset of current customers by sending out a questionnaire twice
a year and asking the customers to rate the speed of the deliveries, the accuracy of the invoices, and
the quality of the packaging minutes. The sample is the total numbers of questionnaires received.
d.
The Wallace Company's immediate interest is learning about the delivery process of its distribution
of pipes, valves, and fittings. To do this, it is measuring the speed of deliveries, the accuracy of the
invoices, and the quality of its packaging from the sample of its customers to make an inference
about the delivery process to all customers. In particular, it might use the mean speed of its deliveries
to the sampled customers to estimate the mean speed of its deliveries to all its customers. It might
use the mean accuracy of its invoices from the sampled customers to estimate the mean accuracy of
its invoices of all its customers. It might use the mean rating of the quality of its packaging from the
sampled customers to estimate the mean rating of the quality of its packaging of all its customers.
e.
Several factors might affect the reliability of the inferences. One factor is the set of customers
selected to receive the survey. If this set is not representative of all the customers, the wrong
inferences could be made. Also, the set of customers returning the surveys may not be representative
of all its customers. Again, this could influence the reliability of the inferences made.
a.
The population of interest would be the set of all students. The sample of interest would be the
students participating in the experiment. The variable measured in this study is whether the student
would spend money on repairing a very old car or not.
b.
The data-collection method used was a designed experiment. The students participating in the
experiment were randomly assigned to one of three emotional states and then asked a question.
c.
The researcher could estimate the proportion of all students in each of the three emotional states who
would spend money to repair a very old car.
d.
One factor that might affect the reliability of the inference drawn is whether the students in the
experiment were representative of all students. It is stated that the sample was made up of volunteer
students. Chances are that these volunteer students were not representative of all students. In
addition, if these students were all from the same school, they probably would not be representative
of the population of students either.
a.
The population of interest would be all accounting alumni of a large southwestern university.
b.
Age would produce quantitative data – the responses would be numbers.
1.30
1.31
Gender would produce qualitative data – the responses would be ‘male’ or ‘female’.
Level of education would produce qualitative data – the responses could be categories such college
degree, master’s degree, or PhD degree.
Income would produce quantitative data – the responses would be numbers.
Job satisfaction score would produce quantitative data. We would assume that a satisfaction score
would be a number, where the higher the number, the higher the job satisfaction.
Machiavellian rating score would produce quantitative data. We would assume that a rating score
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Statistics, Data, and Statistical Thinking
7
would be a number, where the higher the score, the higher the Machiavellian traits.
c.
The sample is the 198 people who returned the useable questionnaires.
d.
The data collection method used was a survey.
e.
The inference made by the researcher is that Machiavellian behavior is not required to achieve
success in the accounting profession.
f.
Generally, those who respond to surveys are those with strong feelings (in either direction) toward the
subject matter. Those who do not have strong feelings for the subject matter tend not to answer
surveys. Those who did not respond might be those who are not real happy with their jobs or those
who are not real unhappy with their jobs. Thus, we might have no idea what type of scores these
people would have on the Machiavellian rating score.
1.32
a.
Give each stock in the NYSE-Composite Transactions table of the Wall Street Journal a number (1 to
m). Using a random number table or a computer program, select n different numbers on the interval
from 1 to m. The stocks with the same numbers as the n chosen numbers will be selected for the
sample.
1.33
a.
The experimental units for this study are engaged couples who used a particular website.
b.
There are two variables of interest – the price of the engagement ring and the level of appreciation.
Price of the engagement ring is a quantitative variable because it is measured on a numerical scale.
Level of appreciation is a qualitative variable. There are 7 different categories for this variable that
are then assigned numbers.
c.
The population of interest would be all engaged couples.
d.
No, the sample is probably not representative. Only engaged couples who used a particular web site
were eligible to be in the sample. Then, only those with “average” American names were invited to
be in the sample.
e.
Answers will vary. First, we will number the individuals from 1 to 50. Using MINITAB, 25 random
numbers were generated on the interval from 1 to 50. The random numbers are:
1, 4, 5, 8, 12, 13, 17, 18, 19, 20, 22, 26, 27, 30, 31, 33, 34, 35, 38, 39, 40, 42, 43, 46, 49
The individuals who were assigned the numbers corresponding to the above numbers would be
assigned to one role and the remaining individuals would be assigned to the other role.
1.34
Answers will vary. Using MINITAB, the 5 seven-digit phone numbers generated with area code 373 were:
373-639-0598
373-411-9164
373-502-7699
373-782-2719
373-930-3231
1.35
a.
Some possible questions are:
1.
In your opinion, why has the banking industry consolidated in the past few years? Check all
that apply.
a.
Too many small banks with not enough capital.
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8
Chapter 1
b.
c.
d.
e.
f.
2.
A result of the Savings and Loan scandals.
To eliminate duplicated resources in the upper management positions.
To provide more efficient service to the customers.
To provide a more complete list of financial opportunities for the customers.
Other. Please list.
Using a scale from 1 to 5, where 1 means strongly disagree and 5 means strongly agree,
indicate your agreement to the following statement: "The trend of consolidation in the banking
industry will continue in the next five years."
1 strongly disagree
1.36
1.37
1.38
2 disagree
3 no opinion
4 agree
5 strongly agree
b.
The population of interest is the set of all bank presidents in the United States.
c.
It would be extremely difficult and costly to obtain information from all bank presidents. Thus, it
would be more efficient to sample just 200 bank presidents. However, by sending the questionnaires
to only 200 bank presidents, one risks getting the results from a sample which is not representative of
the population. The sample must be chosen in such a way that the results will be representative of the
entire population of bank presidents in order to be of any use.
a.
The process being studied is the process of filling beverage cans with soft drink at CCSB's Wakefield
plant.
b.
The variable of interest is the amount of carbon dioxide added to each can of beverage.
c.
The sampling plan was to monitor five filled cans every 15 minutes. The sample is the total number
of cans selected.
d.
The company's immediate interest is learning about the process of filling beverage cans with soft
drink at CCSB's Wakefield plant. To do this, they are measuring the amount of carbon dioxide added
to a can of beverage to make an inference about the process of filling beverage cans. In particular,
they might use the mean amount of carbon dioxide added to the sampled cans of beverage to estimate
the mean amount of carbon dioxide added to all the cans on the process line.
e.
The technician would then be dealing with a population. The cans of beverage have already been
processed. He/she is now interested in the outputs.
a.
The population of interest is the set of all people in the United States over 14 years of age.
b.
The variable being measured is the employment status of each person. This variable is qualitative.
Each person is either employed or not.
c.
The problem of interest to the Census Bureau is inferential. Based on the information contained in
the sample, the Census Bureau wants to estimate the percentage of all people in the labor force who
are unemployed.
Suppose we want to select 900 intersections by numbering the intersections from 1 to 500,000. We would
then use a random number table or a random number generator from a software program to select 900
distinct intersection points. These would then be the sampled markets.
Now, suppose we want to select the 900 intersections by selecting a row from the 500 and a column from
the 1,000. We would first number the rows from 1 to 500 and number the columns from 1 to 1,000. Using
a random number generator, we would generate a sample of 900 from the 500 rows. Obviously, many rows
will be selected more than once. At the same time, we use a random number generator to select 900
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Statistics, Data, and Statistical Thinking
9
columns from the 1,000 columns. Again, some of the columns could be selected more than once. Placing
these two sets of random numbers side-by-side, we would use the row-column combinations to select the
intersections. For example, suppose the first row selected was 453 and the first column selected was 731.
The first intersection selected would be row 453, column 731. This process would be continued until 900
unique intersections were selected.
1.39
Answers will vary.
a.
The results as stated indicate that by eating oat bran, one can improve his/her health. However, the
only way to get the stated benefit is to eat only oat bran with limited results. People may change their
eating habits expecting an outcome that is almost impossible.
b.
To investigate the impact of domestic violence on birth defects, one would need to collect data on all
kinds of birth defects and whether the mother suffered any domestic violence or not during her
pregnancy. One could use an observational study survey to collect the data.
c.
Very few people are always happy with the way they are. However, many people are happy with
themselves most of the time. One might want to ask a series of questions to measure self-esteem
rather than just one. One question might ask what percent of the time the high school girl is happy
with the way she is.
d.
The results of the study are probably misleading because of the fact that if someone relied on a
limited number of foods to feed her children it does not imply that the children are hungry. In
addition, one might cut the size of a meal because the children were overweight, not because there
was not enough food. One might get better information about the proportion of hungry American
children by actually recording what a large, representative sample of children eat in a week.
e.
A leading question gives information that seems to be true, but may not be complete. Based on the
incomplete information, the respondent may come to a different decision than if the information was
not provided.
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Chapter 2
Methods for Describing Sets of Data
2.1
First, we find the frequency of the grade A. The sum of the frequencies for all five grades must be 200.
Therefore, subtract the sum of the frequencies of the other four grades from 200. The frequency for grade
A is:
200 (36 + 90 + 30 + 28) = 200 184 = 16
To find the relative frequency for each grade, divide the frequency by the total sample size, 200. The
relative frequency for the grade B is 36/200 = .18. The rest of the relative frequencies are found in a
similar manner and appear in the table:
Grade on Statistics Exam
A: 90 100
B: 80 89
C: 65 79
D: 50 64
F: Below 50
Total
2.2
a.
Relative Frequency
.08
.18
.45
.15
.14
1.00
To find the frequency for each class, count the number of times each letter occurs. The frequencies
for the three classes are:
Class
X
Y
Z
Total
b.
Frequency
16
36
90
30
28
200
Frequency
8
9
3
20
The relative frequency for each class is found by dividing the frequency by the total sample size. The
relative frequency for the class X is 8/20 = .40. The relative frequency for the class Y is 9/20 = .45.
The relative frequency for the class Z is 3/20 = .15.
Class
X
Y
Z
Total
Frequency
8
9
3
20
Relative Frequency
.40
.45
.15
1.00
10
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Methods for Describing Sets of Data
c.
11
The frequency bar chart is:
9
8
Frequency
7
6
5
4
3
2
1
0
X
d.
Y
C la s s
Z
The pie chart for the frequency distribution is:
Pie Chart of Class
Category
X
Z
15.0%
Y
Z
X
40.0%
Y
45.0%
2.3
a.
The type of graph is a bar graph.
b.
The variable measured for each of the robots is type of robotic limbs.
c.
From the graph, the design used the most is the “legs only” design.
d.
The relative frequencies are computed by dividing the frequencies by the total sample size. The total
sample size is n = 106. The relative frequencies for each of the categories are:
Type of Limbs
None
Both
Legs ONLY
Wheels ONLY
Total
Frequency
15
8
63
20
106
Relative Frequency
15/106 = .142
8 / 106 = .075
63/106 = .594
20/106 = .189
1.000
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12
Chapter 2
e.
Using MINITAB, the Pareto diagram is:
.60
Relative Frequency
.50
.40
.30
.20
.10
0
Legs
Wheels
None
Both
Type
Percent within all data.
2.4
a.
From the pie chart, 50.4% or .504 of the sampled adults living in the U.S. use the internet and pay to
download music. From the data, 506 out of 1,003 adults or 506/1,003 = .504 of sampled adults in the
U.S. use the internet and pay to download music. These two results agree.
b.
Using MINITAB, a pie chart of the data is:
Pie Chart of Download-Music
Category
Pay
No Pay
No Pay
33.0%
Pay
67.0%
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Methods for Describing Sets of Data
2.5
13
Using MINITAB, the Pareto diagram for the data is:
Chart of Tenants
50
Percent
40
30
20
10
0
Small
SmallStandard
Large
Tenants
Major
Anchor
Percent within all data.
Most of the tenants in UK shopping malls are small or small standard. They account for approximately
84% of all tenants ([711 + 819]/1,821 = .84). Very few (less than 1%) of the tenants are anchors.
2.6
a.
The relative frequency for each response category is found by dividing the frequency by the
total sample size. The relative frequency for the category “Insurance Companies” is
869/2119 = .410. The rest of the relative frequencies are found in a similar manner and are reported
in the table.
Most responsible for rising
health-care costs
Insurance companies
Pharmaceutical companies
Government
Hospitals
Physicians
Other
Not at all sure
TOTAL
Number responding
869
339
338
127
85
128
233
2,119
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Relative Frequencies
869/2119 = .410
339/2119 = .160
338/2119 = .160
127/2119 = .060
85/2119 = .040
128/2119 = .060
233/2119 = .110
1.000
14
Chapter 2
b.
Using MINITAB, the relative frequency bar chart is:
Chart of Category
40
Count
30
20
10
0
Insurance Co
c.
Pharm
Government Hospitals Physicians
Category
Other
Not sure
O ther
Phy sicians
Using MINITAB, the Pareto diagram is:
Chart of Category
Relative Frequency
.40
.30
.20
.10
0
Insurance Co. Gov ernment
Pharm
Not sure
Hospitals
Category
Most American adults in the sample (41%) believe that the Insurance companies are the most
responsible for the rising costs of health care. The next highest categories are Government and
Pharmaceutical companies with about 16% each. Only 4% of American adults in the sample believe
physicians are the most responsible for the rising health care costs.
2.7
a.
Since the variable measured is manufacturer, the data type is qualitative.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
15
Using MINITAB, a frequency bar chart for the data is:
Number Shipped
120000
Frequency
100000
80000
60000
40000
Pax Tech
Provenco
SZZT
Toshiba
Urmet
Toshiba
Pax Tech
Glintt
EIntelligent
Urmet
Omron
KwangWoo
EIntelligent
Glintt
Fujian Landi
Bitel
0
CyberNet
20000
Manufacturer
Using MINITAB, the Pareto diagram is:
Number Shipped
120000
100000
Frequency
80000
60000
40000
Bitel
CyberNet
Provenco
Omron
KwangWoo
0
SZZT
20000
Fujian Landi
c.
Manufacturer
Most PIN pads shipped in 2007 were manufactured by either Fujian Landi or SZZT Electronics.
These two categories make up (119,000 + 67,300)/334,039= 186,300/334,039 = .558 of all PIN pads
shipped in 2007. Urmet shipped the fewest number of PIN pads among these 12 manufacturers.
Copyright © 2014 Pearson Education, Inc.
16
2.8
Chapter 2
Using MINITAB, the bar graphs of the 2 waves is:
Sch
NoWorkGrad
NoWorkBusSch
WorkMBA
Sch
2
NoWorkGrad
WorkMBA
NoWorkBusSch
1
90
80
70
60
50
40
30
20
10
0
WorkNoMBA
Percent
WorkNoMBA
Chart of Job Status
Job Status
Panel variable: Wave; Percent within all data.
In wave 1, most of those taking the GMAT were working (2657/3244 =.819) and none had MBA’s. About
20% were not working but were in either a 4-year institution or other graduate school ([36 + 551]/3244 =
.181). In wave 2, almost all were now working ([1787 + 1372]/3244 = .974). Of those working, more than
half had MBA’s (1787/[1787 + 1372] = .566). Of those not working, most were in another graduate
school.
2.9
Using MINITAB, the pie chart is:
Pie Chart of Percent vs Blog/Forum
C ategory
C ompany
Employ ees
Third Party
Not Identified
Not Identified
15.4%
Company
38.5%
Third Party
11.5%
Employ ees
34.6%
Companies and Employees represent (38.5 + 34.6 = 73.1) slightly more than 73% of the entities creating
blogs/forums. Third parties are the least common entity.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.10
17
Using MINITAB, a bar chart of the data is:
Chart of INDUSTRY
14
12
Count
10
8
6
4
2
Aerospace & Defense
Banking
Business Services &
Capital Goods
Chemicals
Conglomerates
Construction
Consumer Durables
Diversified Financia
Drugs & biotechnolog
Food Drink & tobacco
Health care equipmen
Hotels, Restaurants
Household & personal
Insurance
Materials
Media
Oil & Gas Operations
Retailing
Semiconductors
Software & Services
Technology Hardware
Telecommunications s
Transportation
Utilities
0
INDUSTRY
Industries with the highest frequencies include Oil & Gas Operations, Retailing, Drugs & biotechnologies,
and Health care equipment. Industries with the smallest frequencies include Business Services,
Construction, Banking, and Consumer Durables.
2.11
a.
Using MINITAB, a pie chart of the data is:
Pie Chart of PREVUSE
Category
NEVER
USED
USED
28.8%
NEVER
71.2%
From the chart, 71.2% or .712 of the sampled physicians have never used ethics consultation.
Copyright © 2014 Pearson Education, Inc.
18
Chapter 2
b.
Using MINITAB, a pie chart of the data is:
Pie Chart of FUTUREUSE
C ategory
NO
YES
NO
19.5%
YES
80.5%
From the chart, 19.5% or .195 of the sampled physicians state that they will not use the services in the
future.
c.
Using MINITAB, the side-by-side pie charts are:
Pie Chart of PREVUSE
MED
SURG
C ategory
NEVER
USED
USED
27.9%
USED
29.3%
NEVER
70.7%
NEVER
72.1%
Panel variable: SPEC
The proportion of medical practitioners who have never used ethics consultation is .707. The
proportion of surgical practitioners who have never used ethics consultation is .721. These two
proportions are almost the same.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
d.
19
Using MINITAB, the side-by-side pie charts are:
Pie Chart of FUTUREUSE
MED
SURG
Category
NO
YES
NO
17.3%
NO
23.3%
YES
76.7%
YES
82.7%
Panel variable: SPEC
The proportion of medical practitioners who will not use ethics consultation in the future is .173. The
proportion of surgical practitioners who will not use ethics consultation in the future is .233. The
proportion of surgical practitioners who will not use ethics consultation in the future is greater than
that of the medical practitioners.
Using MINITAB, the side-by-side bar graphs are:
Chart of Acquisitions
No
1980
Yes
1990
100
75
50
Percent
2.12
25
0
2000
100
75
50
25
0
No
Yes
Acquisitions
Panel variable: Year; Percent within all data.
In 1980, very few firms had acquisitions 18 / 1,963 .009 . By 1990, the proportion of firms having
acquisitions increased to 350 / 2,197 .159 . By 2000, the proportion of firms having acquisitions increased
to 748 / 2,778 .269 .
Copyright © 2014 Pearson Education, Inc.
20
2.13
Chapter 2
Using MINITAB, the side-by-side bar graphs are:
Chart of Dive
Left
Middle
Ahead
Right
Behind
80
60
Percent
40
20
0
Tied
80
60
40
20
0
Left
Middle
Right
Dive
Panel variable: Situation; Percent within all data.
From the graphs, it appears that if the team is either tied or ahead, the goal-keepers tend to dive either right
or left with equal probability, with very few diving in the middle. However, if the team is behind, then the
majority of goal-keepers tend to dive right (71%).
2.14
Using MINITAB, a pie chart of the data is:
Pie Chart of Measure
Big Shows
20.0%
Total visitors
26.7%
Category
Big Shows
Funds Raised
Members
Paying visitors
Total visitors
Funds Raised
23.3%
Paying visitors
16.7%
Members
13.3%
Since the sizes of the slices are close to each other, it appears that the researcher is correct. There is a large
amount of variation within the museum community with regard to performance measurement and
evaluation.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.15
21
a.
The variable measured by Performark is the length of time it took for each advertiser to respond back.
b.
The pie chart is:
Pie Chart of Response Time
C ategory
> 120 days
13-59 day s
60-120 days
Never responded
> 120 days
12.0%
Never responded
21.0%
13-59 days
33.0%
60-120 day s
34.0%
Twenty-one percent or .21 17,000 3,570 of the advertisers never respond to the sales lead.
d.
The information from the pie chart does not indicate how effective the "bingo cards" are. It just
indicates how long it takes advertisers to respond, if at all.
a.
Using MINITAB, the side-by-side graphs are:
Chart of Frequency vs Stars
5
Content
4
3
2
1
Exposure
16
12
8
Frequency
2.16
c.
4
Faculty
Opportunity
0
16
12
8
4
0
5
4
3
2
1
Stars
Panel variable: Criteria
From these graphs, one can see that very few of the top 30 MBA programs got 5-stars in any criteria.
In addition, about the same number of programs got 4 stars in each of the 4 criteria. The biggest
difference in ratings among the 4 criteria was in the number of programs receiving 3-stars. More
programs received 3-stars in Course Content than in any of the other criteria. Consequently, fewer
programs received 2-stars in Course Content than in any of the other criteria.
Copyright © 2014 Pearson Education, Inc.
b.
Since this chart lists the rankings of only the top 30 MBA programs in the world, it is reasonable that
none of these best programs would be rated as 1-star on any criteria.
a.
Using MINITAB, bar charts for the 3 variables are:
Chart of Well Class
120
100
Count
80
60
40
20
0
Private
Public
Well Class
Chart of Aquifer
200
150
Count
2.17
Chapter 2
100
50
0
Bedrock
Unconsolidated
Aquifer
Chart of Detection
160
140
120
100
Count
22
80
60
40
20
0
Below Limit
Detect
Detection
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
23
Using MINITAB, the side-by-side bar chart is:
Chart of Detection
Below Limit
Private
Detect
Public
80
70
Percent
60
50
40
30
20
10
0
Below Limit
Detect
Detection
Panel variable: Well Class; Percent within all data.
c.
Using MINITAB, the side-by-side bar chart is:
Chart of Detection
Below Limit
Bedrock
Detect
Unconsoli
70
60
Percent
50
40
30
20
10
0
Below Limit
Detect
Detection
Panel variable: Aquifer; Percent within all data.
d.
From the bar charts in parts a-c, one can infer that most aquifers are bedrock and most levels of
MTBE were below the limit ( 2 / 3) . Also the percentages of public wells verses private wells are
relatively close. Approximately 80% of private wells are not contaminated, while only about 60% of
public wells are not contaminated. The percentage of contaminated wells is about the same for both
types of aquifers ( 30%) .
Copyright © 2014 Pearson Education, Inc.
24
2.18
Chapter 2
Using MINITAB, the relative frequency histogram is:
.25
Relative Frequency
.20
.15
.10
.05
0
0
2.5
4.5
6.5
8.5
Class
10.5
12.5
14.5
16.5
To find the number of measurements for each measurement class, multiply the relative frequency by the
total number of observations, n = 500. The frequency table is:
Measurement Class
Relative Frequency
.10
.5 2.5
.15
2.5 4.5
.25
4.5 6.5
.20
6.5 8.5
.05
8.5 10.5
.10
10.5 12.5
.10
12.5 14.5
.05
14.5 16.5
Frequency
500(.10) = 50
500(.15) = 75
500(.25) = 125
500(.20) = 100
500(.05) = 25
500(.10) = 50
500(.10) = 50
500(.05) = 25
500
Using MINITAB, the frequency histogram is:
140
120
100
Frequency
2.19
.5
80
60
40
20
0
0
.5
2.5
4.5
6.5
8.5
Class
10.5
12.5
14.6
16.5
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.20
a.
The original data set has 1 + 3 + 5 + 7 + 4 + 3 = 23 observations.
b.
For the bottom row of the stem-and-leaf display:
25
The stem is 0.
The leaves are 0, 1, 2.
Assuming that the data are up to two digits, rounded off to the nearest whole number, the
numbers in the original data set are 0, 1, and 2.
2.21
2.22
2.23
c.
Again, assuming that the data are up to two digits, rounded off to the nearest whole number, the dot
plot corresponding to all the data points is:
a.
This is a frequency histogram because the number of observations is graphed for each interval rather
than the relative frequency.
b.
There are 14 measurement classes.
c.
There are 49 measurements in the data set.
a.
The measurement class 10 – 20 has the highest proportion of respondents.
b.
The approximate proportion of the 144 organizations that reported a percentage monetary loss from
malicious insider actions less than 20% is .30 + .38 = .68.
c.
The approximate proportion of the 144 organizations that reported a percentage monetary loss from
malicious insider actions greater than 60% is .07 + .03 + .04 + .05 = .19.
d.
The approximate proportion of the 144 organizations that reported a percentage monetary loss from
malicious insider actions between 20% and 30% is .11. Therefore about .11(144) = 15.84 or 16 of
the 144 organizations reported a percentage monetary loss from malicious insider actions between
20% and 30%.
a.
Since the label on the vertical axis is Percent, this is a relative frequency histogram. We can divide
the percents by 100% to get the relative frequencies.
b.
Summing the percents represented by all of the bars above 100, we get approximately 12%.
Copyright © 2014 Pearson Education, Inc.
26
Chapter 2
2.24
a.
Using MINITAB, the stem-and-leaf display and histogram are:
Stem-and-Leaf Display: Score
Stem-and-leaf of Score
Leaf Unit = 1.0
1
1
2
3
3
4
4
5
7
11
17
24
41
62
(37)
87
27
6
7
7
7
7
7
8
8
8
8
8
9
9
9
9
9
10
N
= 186
9
3
4
8
3
44
6667
888999
0001111
22222222222333333
444444555555555555555
6666666666666667777777777777777777777
888888888888888888888888888899999999999999999999999999999999
000000000000000000000000000
Histogram of Score
60
Frequency
50
40
30
20
10
0
72
76
80
84
Score
88
92
96
100
b.
From the stem-and-leaf display, there are only 7 observations with sanitation scores less than 86. The
proportion of ships with accepted sanitation standards is (186 7) / 186 179 / 186 .962 .
c.
The score of 69 is highlighted in the stem-and-leaf display.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.25
a.
Using MINITAB, a dot plot of the data is:
Dotplot of Acquisitions
0
240
360
480
Acquisitions
600
720
840
b.
By looking at the dot plot, one can conclude that the years 1996-2000 had the highest number of
firms with at least one acquisition. The lowest number of acquisitions in that time frame (748) is
almost 100 higher than the highest value from the remaining years.
a.
Using MINITAB, a histogram of the current values of the 32 NFL teams is:
Histogram of Value ($mil)
14
12
10
Frequency
2.26
120
8
6
4
2
0
750
900
1050
1200
1350
Value ($mil)
1500
1650
1800
Copyright © 2014 Pearson Education, Inc.
27
Chapter 2
b.
Using MINITAB, a histogram of the 1-year change in current value for the 32 NFL teams is:
Histogram of Chang1Yr (%)
10
Frequency
8
6
4
2
0
-4
c.
-2
0
2
4
Chang1Yr (%)
6
8
10
Using MINITAB, a histogram of the debt-to-value ratios for the 32 NFL teams is:
Histogram of Debt/Value (%)
20
15
Frequency
28
10
5
0
0
16
32
Debt/Value (%)
48
64
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
d.
29
Using MINITAB, a histogram of the annual revenues for the 32 NFL teams is:
Histogram of Revenue ($mil)
16
14
Frequency
12
10
8
6
4
2
0
225
e.
250
275
300
325
Revenue ($mil)
350
375
400
Using MINITAB, a histogram of the operating incomes for the 32 NFL teams is:
Histogram of Income ($mil)
10
Frequency
8
6
4
2
0
0
f.
20
40
60
Income ($mil)
80
100
120
For all of the histograms, there is 1 team that has a very high score. The Dallas Cowboys have the
largest values for current value, annual revenues, and operating income. However, the New York
Giants have the highest 1-year change, while the New York Jets have the highest debt-to-value ratio.
All of the graphs except the one showing the 1-Yr Value Changes are skewed to the right.
Copyright © 2014 Pearson Education, Inc.
30
Chapter 2
2.27
a.
Using MINITAB, the frequency histograms for 2011 and 2010 SAT mathematics scores are:
His togr a m of M ATH 2 0 1 1 , M A TH 2 0 1 0
480
520
Frequency
M A T H2011
560
600
M A T H2010
14
14
12
12
10
10
8
8
6
6
4
4
2
2
0
0
480
520
560
600
It appears that the scores have not changed very much at all. The graphs are very similar.
Using MINITAB, the frequency histograms for 2011 and 2001 SAT mathematics scores are:
His togr am of M ATH2 0 1 1 , M ATH2 0 0 1
480
M A T H2011
510
540
570
600
M A T H2001
14
12
12
10
10
Frequency
b.
8
8
6
6
4
4
2
2
0
0
480
520
560
600
It appears that the scores have shifted to the right. The scores in 2011 appear to be somewhat better
than the scores in 2011.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
31
Using MINITAB, the frequency histogram of the differences is:
H is togr a m of D iffM a th
16
14
Frequency
12
10
8
6
4
2
0
-32
-16
0
DiffM a t h
16
32
From this graph of the differences, we can see that there are more observations to the right of 0 than
to the left of 0. This indicates that, in general, the scores have improved since 2001.
d.
2.28
From the graph, the largest improvement score is in the neighborhood of 32. The actual largest score
is 32 and it is associated with Michigan.
Using MINITAB, the two dot plots are:
Dotplot of Arrive, Depart
Arrive
Depart
108
120
132
144
156
168
Data
Yes. Most of the numbers of items arriving at the work center per hour are in the 135 to 165 area. Most of
the numbers of items departing the work center per hour are in the 110 to 140 area. Because the number of
items arriving is larger than the number of items departing, there will probably be some sort of bottleneck.
Copyright © 2014 Pearson Education, Inc.
32
2.29
Chapter 2
Using MINITAB, the stem-and-leaf display is:
Stem-and-Leaf Display: Dioxide
Stem-and-leaf of Dioxide
Leaf Unit = 0.10
5
7
(2)
7
7
5
5
4
4
0
0
1
1
2
2
3
3
4
N
= 16
12234
55
34
44
3
0000
The highlighted values are values that correspond to water specimens that contain oil. There is a tendency
for crude oil to be present in water with lower levels of dioxide as 6 of the lowest 8 specimens with the
lowest levels of dioxide contain oil.
2.30
Yes, we would agree with the statement that honey may be the preferable treatment for the cough and sleep
difficulty associated with childhood upper respiratory tract infection. For those receiving the honey
dosage, 14 of the 35 children (or 40%) had improvement scores of 12 or higher. For those receiving the
DM dosage, only 9 of the 33 (or 24%) children had improvement scores of 12 or higher. For those
receiving no dosage, only 2 of the 37 children (or 5%) had improvement scores of 12 or higher. In
addition, the median improvement score for those receiving the honey dosage was 11, the median for those
receiving the DM dosage was 9 and the median for those receiving no dosage was 7.
2.31
Using MINITAB, the relative frequency histograms of the years in practice for the two groups of doctors
are:
Histogram of YRSPRAC
0.0
NO
25
7.5
15.0
22.5
30.0
37.5
YES
Percent
20
15
10
5
0
0.0
7.5
15.0
22.5
30.0
37.5
YRSPRAC
Panel variable: FUTUREUSE
The researchers hypothesized that older, more experienced physicians will be less likely to use ethics
consultation in the future. From the histograms, approximately 38% of the doctors that said “no” have
more than 20 years of experience. Only about 19% of the doctors that said “yes” had more than 20 years of
experience. This supports the researchers’ assertion.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
a.
Using MINITAB, the stem-and-leaf display is as follows, where the stems are the units place and the
leaves are the decimal places:
Stem-and-Leaf Display: Time
Stem-and-leaf of Time
Leaf Unit = 0.10
(26)
23
15
9
4
2
2
1
1
1
1
2
3
4
5
6
7
8
9
10
N
= 49
00001122222344444445555679
11446799
002899
11125
24
8
1
b.
A little more than half (26/49 = .53) of all companies spent less than 2 months in bankruptcy. Only
two of the 49 companies spent more than 6 months in bankruptcy. It appears that, in general, the
length of time in bankruptcy for firms using "prepacks" is less than that of firms not using prepacks."
c.
A dot diagram will be used to compare the time in bankruptcy for the three types of "prepack" firms:
Dotplot of Time vs Votes
Votes
2.32
33
Joint
None
Prepack
1.2
2.4
3.6
4.8
6.0
7.2
8.4
9.6
Time
d.
The highlighted times in part a correspond to companies that were reorganized through a leverage
buyout. There does not appear to be any pattern to these points. They appear to be scattered about
evenly throughout the distribution of all times.
Copyright © 2014 Pearson Education, Inc.
34
2.33
Chapter 2
Using MINITAB, the histogram of the data is:
Histogram of INTTIME
60
50
Frequency
40
30
20
10
0
0
75
150
225
300
INTTIME
375
450
525
This histogram looks very similar to the one shown in the problem. Thus, there appears that there was
minimal or no collaboration or collusion from within the company. We could conclude that the phishing
attack against the organization was not an inside job.
2.34
Using MINITAB, the stem-and-leaf display for the data is:
Stem-and-Leaf Display: Time
Stem-and-leaf of Time
Leaf Unit = 1.0
3
7
(7)
11
6
4
2
1
N
= 25
3 239
4 3499
5 0011469
6 34458
7 13
8 26
9 5
10 2
The numbers in bold represent delivery times associated with customers who subsequently did not place
additional orders with the firm. Since there were only 2 customers with delivery times of 68 days or longer
that placed additional orders, I would say the maximum tolerable delivery time is about 65 to 67 days.
Everyone with delivery times less than 67 days placed additional orders.
2.35
x 3.2 2.5 2.1 3.7 2.8 2.0 16.3 2.717
Assume the data are a sample. The sample mean is:
x
n
6
6
The median is the average of the middle two numbers when the data are arranged in order (since n = 6 is
even). The data arranged in order are: 2.0, 2.1, 2.5, 2.8, 3.2, 3.7. The middle two numbers are 2.5 and 2.8.
The median is:
2.5 2.8 5.3
2.65
2
2
Copyright © 2014 Pearson Education, Inc.
2.36
Methods for Describing Sets of Data
x 85 8.5
a.
x
b.
x
400
25
16
c.
x
35
.778
45
d.
x
242
13.44
18
n
35
10
2.37
The mean and median of a symmetric data set are equal to each other. The mean is larger than the median
when the data set is skewed to the right. The mean is less than the median when the data set is skewed to
the left. Thus, by comparing the mean and median, one can determine whether the data set is symmetric,
skewed right, or skewed left.
2.38
The median is the middle number once the data have been arranged in order. If n is even, there is not a
single middle number. Thus, to compute the median, we take the average of the middle two numbers. If n
is odd, there is a single middle number. The median is this middle number.
A data set with five measurements arranged in order is 1, 3, 5, 6, 8. The median is the middle number,
which is 5.
A data set with six measurements arranged in order is 1, 3, 5, 5, 6, 8. The median is the average of the
5 5 10
middle two numbers which is
5.
2
2
2.39
Assume the data are a sample. The mode is the observation that occurs most frequently. For this sample,
the mode is 15, which occurs three times.
x 18 10 15 13 17 15 12 15 18 16 11 160 14.545
The sample mean is:
x
11
n
11
The median is the middle number when the data are arranged in order. The data arranged in order are: 10,
11, 12, 13, 15, 15, 15, 16, 17, 18, 18. The middle number is the 6th number, which is 15.
2.40
a.
b.
x
x 7 4 15 2.5
x
x 2 4 40 3.08
6
6
33
Median =
3 (mean of 3rd and 4th numbers, after ordering)
2
Mode = 3
n
13
13
n
Median = 3 (7th number, after ordering)
Mode = 3
Copyright © 2014 Pearson Education, Inc.
36
Chapter 2
c.
2.41
2.42
x
x 51 37 496 49.6
10
10
48 50
Median =
49 (mean of 5th and 6th numbers, after ordering)
2
Mode = 50
n
a.
For a distribution that is skewed to the left, the mean is less than the median.
b.
For a distribution that is skewed to the right, the mean is greater than the median.
c.
For a symmetric distribution, the mean and median are equal.
a.
b.
The mean is
x 9 (.1) (1.6) 14.6 16.0 7.7 19.9 9.8 3.2 24.8 17.6 10.7 9.1 140.7 10.82
x
n
13
13
The average annualized percentage return on investment for 13 randomly selected stock screeners is
10.82.
Since the number of observations is odd, the median is the middle number once the data have been
arranged in order. The data arranged in order are:
-1.6 -.1 3.2 7.7 9.0 9.1 9.8 10.7 14.6 16.0 17.6 19.9 24.8
The middle number is 9.8 which is the median. Half of the annualized percentage returns on
investment are below 9.8 and half are above 9.8.
2.43
2.44
a.
The mean amount exported on the printout is 653. This means that the average amount of money per
market from exporting sparkling wine was $653,000.
b.
The median amount exported on the printout is 231. Since the median is the middle value, this means
that half of the 30 sparkling wine export values were above $231,000 and half of the sparkling wine
export values were below $231,000.
c.
The mean 3-year percentage change on the printout is 481. This means that in the last three years, the
average change is 481%, which indicates a large increase.
d.
The median 3-year percentage change on the printout is 156. Since the median is the middle value,
this means that half, or 15 of the 30 countries’ 3-year percentage change values were above 156% and
half, or 15 of the 30 countries’ 3-year percentage change values were below 156%.
a.
The sample mean is:
x
n
x i 1
n
i
1.72 2.50 2.16 1.95 37.62
1.881
20
20
The sample average surface roughness of the 20 observations is 1.881.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
37
The median is found as the average of the 10th and 11th observations, once the data have been
ordered. The ordered data are:
1.06 1.09 1.19 1.26 1.27 1.40 1.51 1.72 1.95 2.03 2.05 2.13 2.13 2.16 2.24 2.31 2.41 2.50 2.57 2.64
The 10th and 11th observations are 2.03 and 2.05. The median is:
2.03 2.05 4.08
2.04
2
2
The middle surface roughness measurement is 2.04. Half of the sample measurements were less than
2.04 and half were greater than 2.04.
2.45
c.
The data are somewhat skewed to the left. Thus, the median might be a better measure of central
tendency than the mean. The few small values in the data tend to make the mean smaller than the
median.
a.
The mean is x
b.
x 1,680,927 885,182 881, 777 563,967 15,192, 021 759, 601.05 .
n
20
20
The average research expenditures for the top 20 ranked universities is 759,601.05 thousand dollars.
Since the number of observations is even, the median is the average of the middle 2 numbers once the
data have been arranged in order. Since the data are already arranged in order, the median is
702,592 688, 225
695, 408.5 .
2
Half of the institutions have a research expenditure less than 695,408.5 thousand dollars and half
have research expenditures greater than 695,408.5 thousand dollars.
2.46
2.47
c.
No, the mean from part a would not be a good measure for the center of the distribution for all
American universities. The data in part a come from only the top 20 universities. These universities
would not be representative of all American universities.
a.
The mean is 67.755. The statement is accurate.
b.
The median is 68.000. The statement is accurate.
c.
The mode is 64. The statement is not accurate. A better statement would be: “The most common
reported level of support for corporate sustainability for the 992 senior managers was 64.
d.
Since the mean and median are almost the same, the distribution of the 992 support levels should be
fairly symmetric. The histogram in Exercise 2.23 is almost symmetric.
a.
The median is the middle number (18th) once the data have been arranged in order because n = 35 is
odd. The honey dosage data arranged in order are:
4,5,6,8,8,8,8,9,9,9,9,10,10,10,10,10,10,11,11,11,11,12,12,12,12,12,12,13,13,14,15,15,15,15,16
The 18th number is the median = 11.
Copyright © 2014 Pearson Education, Inc.
38
Chapter 2
b.
The median is the middle number (17th) once the data have been arranged in order because n = 33 is
odd. The DM dosage data arranged in order are:
3,4,4,4,4,4,4,6,6,6,7,7,7,7,7,8,9,9,9,9,9,10,10,10,11,12,12,12,12,12,13,13,15
The 17th number is the median = 9.
c.
The median is the middle number (19th) once the data have been arranged in order because n = 37 is
odd. The No dosage data arranged in order are:
0,1,1,1,3,3,4,4,5,5,5,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,8,8,9,9,9,9,10,11,12,12
The 19th number is the median = 7.
2.48
d.
Since the median for the Honey dosage is larger than the other two, it appears that the honey dosage
leads to more improvement than the other two treatments.
a.
The mean dioxide level is x
3.3 0.5 1.3 4.0 29
1.81 . The average dioxide amount is
16
16
1.81.
b.
Since the number of observations is even, the median is the average of the middle 2 numbers once the
data are arranged in order. The data arranged in order are:
0.1 0.2 0.2 0.3 0.4 0.5 0.5 1.3 1.4 2.4 2.4 3.3 4.0 4.0 4.0 4.0
The median is
1.3 1.4 2.7
1.35 . Half of the dioxide levels are below 1.35 and half are above
2
2
1.35.
c.
The mode is the number that occurs the most. For this data set the mode is 4.0. The most frequent
level of dioxide is 4.0.
d.
Since the number of observations is even, the median is the average of the middle 2 numbers once the
data are arranged in order. The data arranged in order are:
0.1 0.3 1.4 2.4 2.4 3.3 4.0 4.0 4.0 4.0
The median is
e.
2.4 3.3 5.7
2.85 .
2
2
Since the number of observations is even, the median is the average of the middle 2 numbers once the
data are arranged in order. The data arranged in order are:
0.2 0.2 0.4 0.5 0.5 1.3
The median is
f.
0.4 0.5 0.9
0.45 .
2
2
The median level of dioxide when crude oil is present is 0.45. The median level of dioxide when
crude oil is not present is 2.85. It is apparent that the level of dioxide is much higher when crude oil
is not present.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.49
2.50
39
a.
Skewed to the right. There will be a few people with very high salaries such as the president and
football coach.
b.
Skewed to the left. On an easy test, most students will have high scores with only a few low scores.
c.
Skewed to the right. On a difficult test, most students will have low scores with only a few high
scores.
d.
Skewed to the right. Most students will have a moderate amount of time studying while a few students
might study a long time.
e.
Skewed to the left. Most cars will be relatively new with a few much older.
f.
Skewed to the left. Most students will take the entire time to take the exam while a few might leave
early.
a.
The sample means is:
x
x 3.58 3.48 3.27 1.17 77.07 1.927
40
n
40
The median is found as the 20th and 21st observations, once the data have been ordered. The 20th and
21st observations are 1.75 and 1.76. The median is:
1.75 1.76 3.51
1.755
2
2
The mode is the number that occurs the most and is 1.4, which occurs 3 times.
b.
The sample average driving performance index is 1.927. The median driving performance index is
1.755. Half of all driving performance indexes are less than 1.755 and half are higher. The most
common driving performance index value is 1.4.
c.
Since the mean is larger than the median, the data are skewed to the right. Using MINITAB, a
histogram of the driving performance index values is:
Histogram of INDEX
10
Frequency
8
6
4
2
0
1.5
2.51
2.0
2.5
INDEX
3.0
3.5
The mean is 141.31 hours. This means that the average number of semester hours per candidate for the
CPA exam is 141.31 hours. The median is 140 hours. This means that 50% of the candidates had more
than 140 semester hours of credit and 50% had less than 140 semester hours of credit. Since the mean and
median are so close in value, the data are probably not skewed, but close to symmetric.
Copyright © 2014 Pearson Education, Inc.
40
Chapter 2
2.52
a.
Using MINITAB, the output is:
Descriptive Statistics: YRSPRAC
Variable
YRSPRAC
N
112
N*
6
Mean
14.598
Minimum
1.000
Median
14.000
Maximum
40.000
Mode
14, 20, 25
N for
Mode
9
The mean is 14.598. The average length of time in practice for this sample is 14.598 years. The
median is 14. Half of the physicians have been in practice less than 14 years and half have been in
practice longer than 14 years. There are 3 modes: 14, 20, and 25. The most frequent years in practice
are 14, 20, and 25 years.
b.
Using MINITAB, the results are:
Descriptive Statistics: YRSPRAC
Variable
YRSPRAC
FUTUREUSE
NO
YES
N
21
91
N*
2
4
Mean
16.43
14.176
Minimum
1.00
1.000
Median
18.00
14.000
Maximum
35.00
40.000
Mode
25
14, 20
N for
Mode
5
8
The mean for the physicians who would refuse to use ethics consultation in the future is 16.43. The
average time in practice for these physicians is 16.43 years. The median is 18. Half of the physicians
who would refuse ethics consultation in the future have been in practice less than 18 years and half
have been in practice more than 18 years. The mode is 25. The most frequent years in practice for
these physicians is 25 years.
2.53
c.
From the results in part b, the mean for the physicians who would use ethics consultation in the future
is 14.176. The average time in practice for these physicians is 14.176 years. The median is 14. Half
of the physicians who would use ethics consultation in the future have been in practice less than 14
years and half have been in practice more than 14 years. There are 2 modes: 14 and 20. The most
frequent years in practice for these physicians are 14 and 20 years.
d.
The results in parts b and c confirm the researchers’ theory. The mean, median and mode of years in
practice are larger for the physicians who would refuse to use ethics consultation in the future than
those who would use ethics consultation in the future.
For the "Joint exchange offer with prepack" firms, the mean time is 2.6545 months, and the median is 1.5
months. Thus, the average time spent in bankruptcy for "Joint" firms is 2.6545 months, while half of the
firms spend 1.5 months or less in bankruptcy.
For the "No prefiling vote held" firms, the mean time is 4.2364 months, and the median is 3.2 months.
Thus, the average time spent in bankruptcy for "No prefiling vote held" firms is 4.2364 months, while half
of the firms spend 3.2 months or less in bankruptcy.
For the "Prepack solicitation only" firms, the mean time is 1.8185 months, and the median is 1.4 months.
Thus, the average time spent in bankruptcy for "Prepack solicitation only" firms is 1.8185 months, while
half of the firms spend 1.4 months or less in bankruptcy.
Since the means and medians for the three groups of firms differ quite a bit, it would be unreasonable to use
a single number to locate the center of the time in bankruptcy. Three different "centers" should be used.
2.54
a.
The sample mean is:
x
n
x i 1
n
i
5 2 4 ... 3 78
3.90
20
20
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
41
The sample median is found by finding the average of the 10th and 11th observations once the data are
arranged in order. The data arranged in order are:
1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9 11
The 10th and 11th observations are 3 and 4. The average of these two numbers (median) is:
median
3 4 7
3.5
2
2
The mode is the observation appearing the most. For this data set, the mode is 1, which appears 5
times.
b.
Eliminating the largest number which is 11 results in the following:
The sample mean is:
x
n
i
x i 1
n
5 2 4 ... 3 67
3.53
19
19
The sample median is found by finding the middle observation once the data are arranged in order.
The data arranged in order are:
1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9
The 10th observation is 3. The median is 3
The mode is the observations appearing the most. For this data set, the mode is 1, which appears 5
times.
By dropping the largest number, the mean is reduced from 4.05 to 3.68. The median is reduced from
3.5 to 3. There is no effect on the mode.
c.
The data arranged in order are:
1 1 1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7 9 11
If we drop the lowest 2 and largest 2 observations we are left with:
1 1 1 2 2 3 3 3 4 4 4 5 5 5 6 7
The sample 10% trimmed mean is:
x
n
x i 1
n
i
1 1 2 ... 7 56
3.5
16
16
The advantage of the trimmed mean over the regular mean is that very large and very small numbers
that could greatly affect the mean have been eliminated.
2.55
2.56
a.
Due to the "elite" superstars, the salary distribution is skewed to the right. Since this implies that the
median is less than the mean, the players' association would want to use the median.
b.
The owners, by the logic of part a, would want to use the mean.
a.
The primary disadvantage of using the range to compare variability of data sets is that the two data
sets can have the same range and be vastly different with respect to data variation. Also, the range is
greatly affected by extreme measures.
Copyright © 2014 Pearson Education, Inc.
42
Chapter 2
b.
The sample variance is the sum of the squared deviations of the observations from the sample mean
divided by the sample size minus 1. The population variance is the sum of the squared deviations of
the values from the population mean divided by the population size.
c.
The variance of a data set can never be negative. The variance of a sample is the sum of the squared
deviations from the mean divided by n 1. The square of any number, positive or negative, is
always positive. Thus, the variance will be positive.
The variance is usually greater than the standard deviation. However, it is possible for the variance to
be smaller than the standard deviation. If the data are between 0 and 1, the variance will be smaller
than the standard deviation. For example, suppose the data set is .8, .7, .9, .5, and .3. The sample
mean is:
x
x .8 .7 .9 .5 .3 3.2 .64
.5
n
5
The sample variance is: s 2
x
x
2
2
n 1
n
3.22
13 .232 .058
5 1
4
2.28
The standard deviation is s .058 .241
2.57
a.
s2
b.
a.
b.
n
x
s 2.3 1.52
x2
n 1
2
n
x
x
17 2
7 3.619
7 1
s 3.619 1.9
302
10 7.111
10 1
s 7.111 2.67
63
2
2
n 1
n
x
154
Range = 1 (3) = 4
s2
2.58
n 1
82
5 2.3
5 1
22
Range = 8 (2) = 10
s2
d.
2
2
Range = 6 0 = 6
s2
c.
x
x
Range = 4 0 = 4
s2
s2
x2
n 1
n
x
x
2
2
n 1
2
n
x
x
202
10 4.8889
10 1
84
2
2
n 1
n
(6.8) 2
17 1.395
17 1
25.04
1002
40 3.3333
40 1
380
s 1.395 1.18
s 4.8889 2.211
s 3.3333 1.826
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
2.59
a.
s2
2
n 1
17 2
20 .1868
20 1
18
n
x 3 1 10 10 4 28
x
s2
b.
x
x
2
x 28 5.6
s2
s .1868 .432
x 3 1 10 10 4 226
2
2
2
2
2
2
5
n
x
x
2
2
n 1
n
282
5 69.2 17.3
5 1
4
226
x 8 10 32 5 55
x
43
s 17.3 4.1593
x 8 10 32 5 1213
2
x 55 13.75 feet
2
2
2
2
4
n
x2
x
n 1
2
n
552
4 456.75 152.25 square feet
4 1
3
1213
s 152.25 12.339 feet
c.
x 1 (4) (3) 1 (4) (4) 15 x (1) (4) (3) 1 (4) (4) 59
2
x
s2
d.
s2
2
2
2
2
2
x 15 2.5
6
n
x
x
2
2
n 1
x
x
2
n
(15) 2
6 21.5 4.3
6 1
5
59
1 1 1 2 1 4 10
2
5 5 5 5 5 5 5
x 2 1 .33 ounce
n
6
24
1 1 1 2 1 4
x 2 5 5 5 5 5 5 25 .96
2
2
2
2
2
2
3
x
x
2
n 1
s 4.3 2.0736
n
2
24 22
.2933
25 6
.0587 square ounce
6 1
5
s .0587 .2422 ounce
Copyright © 2014 Pearson Education, Inc.
44
Chapter 2
2.60
a.
s2
b.
n 1
n
x
1992
5 3.7
5 1
s 3.7 1.92
3032
9 1,949.25
9 1
s 1,949.25 44.15
2952
8 1,307.84
8 1
s 1,307.84 36.16
7935
x2
n 1
2
n
x
x
25, 795
Range = 100 2 = 98
s2
2.61
2
2
Range = 100 1 = 99
s2
c.
x
x
Range = 42 37 = 5
2
2
n 1
n
20, 033
This is one possibility for the two data sets.
Data Set 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Data Set 2: 0, 0, 1, 1, 2, 2, 3, 3, 9, 9
The two sets of data above have the same range = largest measurement smallest measurement = 9 0 = 9.
The means for the two data sets are:
x1
x2
x 0 1 2 3 4 5 6 7 8 9 45 4.5
n
10
10
n
10
10
x 0 0 1 1 2 2 3 3 9 9 30 3
The dot diagrams for the two data sets are shown below.
Dotplot of x1, x2
x1
0
2
4
x
6
8
6
8
x2
0
2
x
4
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.62
This is one possibility for the two data sets.
Data Set 1: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5
Data Set 2: 1, 1, 1, 1, 1, 5, 5, 5, 5, 5
x1
x 1 1 2 2 3 3 4 4 5 5 30 3
x2
x 1 1 1 1 1 5 5 5 5 5 30 3
n
10
10
n
10
10
Therefore, the two data sets have the same mean. The variances for the two data sets are:
s12
s22
x
x
n 1
2
2
n
x
x
2
2
n 1
302
10 20 2.2222
9
9
110
n
302
10 40 4.4444
9
9
130
The dot diagrams for the two data sets are shown below.
Dotplot of x1, x2
x1
x
1
2
3
x2
1
2
3
4
5
4
5
x
2.63
a.
s2
b.
x
x
Range = 3 0 = 3
2
n 1
n
2
72
5 1.3
5 1
15
s 1.3 1.14
After adding 3 to each of the data points,
Range = 6 3 = 3
Copyright © 2014 Pearson Education, Inc.
45
46
Chapter 2
s2
c.
x
x
2
2
n
n 1
222
5 1.3
5 1
102
s 1.3 1.14
After subtracting 4 from each of the data points,
x
x
Range = 1 (4) = 3
s2
2.64
2
2
n
n 1
(13) 2
5 1.3
5 1
39
s 1.3 1.14
d.
The range, variance, and standard deviation remain the same when any number is added to or
subtracted from each measurement in the data set.
a.
The range is the difference between the maximum and minimum values. The range
24.8 – 1.6 26.4 . The units of measurement are percents.
b.
The variance is
s2
x2
x
n 1
n
2
140.7 2
13 2236.41 1522.8069 713.6031 59.4669
13 1
12
12
2236.41
The units are square percents.
2.65
c.
The standard deviation is s 59.4669 7.7115 . The units are percents.
a.
The range is the difference between the largest observation and the smallest observation. From the
printout, the largest observation is $4,852 thousand and the smallest observation is $70 thousand. The
range is:
R $4,852 $70 $4,882 thousand
b.
From the printout, the standard deviation is s = $1,113 thousand.
c.
The variance is the standard deviation squared. The variance is:
s 2 1,1132 1, 238, 769 million dollars squared
2.66
a.
The sample variance of the honey dosage group is:
s2
x
x
2
2
n 1
n
3752
35 277.142857 8.1512605
35-1
34
4295-
The standard deviation is: s 8.1512605 2.855
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
47
The sample variance of the DM dosage group is:
s2
x
x
2
2
n 1
n
2752
33 339.33333 10.604167
33-1
32
2631-
The standard deviation is: s 10.604167 3.256
c.
The sample variance of the control group is:
s2
x
x
2
2
n 1
n
2412
37 311.243243 8.6456456
37-1
36
1881-
The standard deviation is: s 8.6456456 2.940
2.67
2.68
d.
The group with the most variability is the group with the largest standard deviation, which is the DM
group. The group with the least variability is the group with the smallest standard deviation, which is
the honey group.
a.
The range is 155. The statement is accurate.
b.
The variance is 722.036. The statement is not accurate. A more accurate statement would be: “The
variance of the levels of supports for corporate sustainability for the 992 senior managers is 722.036.”
c.
The standard deviation is 26.871. If the units of measure for the two distributions are the same, then
the distribution of support levels for the 992 senior managers has less variation than a distribution
with a standard deviation of 50. If the units of measure for the second distribution is not known, then
we cannot compare the variation in the two distributions by looking at the standard deviations alone.
d.
The standard deviation best describes the variation in the distribution. The range can be greatly
affected by extreme measures. The variance is measured in square units, which is hard to interpret.
Thus, the standard deviation is the best measure to describe the variation.
a.
Using MINITAB, the results are:
Descriptive Statistics: YRSPRAC
Variable
YRSPRAC
N
112
N*
6
Mean
14.598
StDev
9.161
Variance
83.918
Range
39.000
The range is 39. The difference between the largest years in practice and the smallest years in
practice is 39 years. The variance is 83.918 square years. The standard deviation is 9.161 years.
b.
Using MINITAB, the results are:
Descriptive Statistics: YRSPRAC
Variable
YRSPRAC
FUTUREUSE
NO
YES
N
21
91
N*
2
4
Mean
16.43
14.176
StDev
10.05
8.950
Variance
100.96
80.102
Range
34.00
39.000
For the physicians who would refuse to use ethics consultation in the future, the standard deviation is
10.05 years.
Copyright © 2014 Pearson Education, Inc.
48
2.69
Chapter 2
c.
For the physicians who would use ethics consultation in the future, the standard deviation is 8.95
years.
d.
The variation in the length of time in practice for the physicians who would refuse to use ethics
consultation in the future is greater than that for the physicians who would use ethics consultation in
the future.
a.
The range is the largest observation minus the smallest observation or 11 – 1 = 10.
xi
782
xi2 i
450
n
20 7.6737
The variance is: s 2 i
n 1
20 1
2
The standard deviation is: s s 2 7.6737 2.77
b.
The largest observation is 11. It is deleted from the data set. The new range is: 9 – 1 = 8.
xi
67 2
xi2 i
329
n
19 5.1520
The variance is: s 2 i
n 1
19 1
2
The standard deviation is: s s 2 5.1520 2.27
When the largest observation is deleted, the range, variance and standard deviation decrease.
c.
The largest observation is 11 and the smallest is 1. When these two observations are deleted from the
data set, the new range is: 9 – 1 = 8.
xi
i
662
xi2
328
n
18 5.0588
The variance is: s 2 i
n 1
18 1
2
The standard deviation is: s s 2 5.0588 2.25
When the largest and smallest observations are deleted, the range, variance and standard deviation
decrease.
2.70
a.
A worker's overall time to complete the operation under study is determined by adding the subtasktime averages.
Worker A
The average for subtask 1 is: x
The average for subtask 2 is: x
x 211 30.14
x 21 3
n
7
n
7
Worker A's overall time is 30.14 + 3 = 33.14.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
Worker B
The average for subtask 1 is: x
The average for subtask 2 is: x
49
x 213 30.43
x 29 4.14
n
7
n
7
Worker B's overall time is 30.43 + 4.14 = 34.57.
b.
x
x
Worker A
s
n 1
Worker B
s
2
2
x2
n
x
n 1
2
n
2112
7 15.8095 3.98
7 1
6455
2132
7 .9524 .98
7 1
6487
c.
The standard deviations represent the amount of variability in the time it takes the worker to complete
subtask 1.
d.
Worker A
s
x
x
2
2
n 1
n
x
x
212
7 .6667 .82
7 1
67
Worker B
s
e.
2
2
n 1
n
292
7 4.4762 2.12
7 1
147
I would choose workers similar to worker B to perform subtask 1. Worker B has a slightly higher
average time on subtask 1 (A: x 30.14 , B: x 30.43 ). However, Worker B has a smaller
variability in the time it takes to complete subtask 1 (part b). He or she is more consistent in the time
needed to complete the task.
I would choose workers similar to Worker A to perform subtask 2. Worker A has a smaller average
time on subtask 2 (A: x 3 , B: x 4.14 ). Worker A also has a smaller variability in the time
needed to complete subtask 2 (part d).
2.71
a.
The unit of measurement of the variable of interest is dollars (the same as the mean and standard
deviation). Based on this, the data are quantitative.
b.
Since no information is given about the shape of the data set, we can only use Chebyshev's Rule.
$900 is 2 standard deviations below the mean, and $2100 is 2 standard deviations above the mean.
Using Chebyshev's Rule, at least 3/4 of the measurements (or 3/4 200 = 150 measurements) will
fall between $900 and $2100.
Copyright © 2014 Pearson Education, Inc.
50
Chapter 2
$600 is 3 standard deviations below the mean and $2400 is 3 standard deviations above the mean.
Using Chebyshev's Rule, at least 8/9 of the measurements (or 8/9 200 178 measurements) will
fall between $600 and $2400.
$1200 is 1 standard deviation below the mean and $1800 is 1 standard deviation above the mean.
Using Chebyshev's Rule, nothing can be said about the number of measurements that will fall
between $1200 and $1800.
$1500 is equal to the mean and $2100 is 2 standard deviations above the mean. Using Chebyshev's
Rule, at least 3/4 of the measurements (or 3/4 200 = 150 measurements) will fall between $900 and
$2100. It is possible that all of the 150 measurements will be between $900 and $1500. Thus,
nothing can be said about the number of measurements between $1500 and $2100.
2.72
2.73
2.74
Since no information is given about the data set, we can only use Chebyshev's Rule.
a.
Nothing can be said about the percentage of measurements which will fall between
x s and x s .
b.
At least 3/4 or 75% of the measurements will fall between x 2 s and x 2 s .
c.
At least 8/9 or 89% of the measurements will fall between x 3s and x 3s .
According to the Empirical Rule:
a.
Approximately 68% of the measurements will be contained in the interval x s to x s .
b.
Approximately 95% of the measurements will be contained in the interval x 2 s to x 2 s .
c.
Essentially all the measurements will be contained in the interval x 3s to x 3s .
a.
x
s2
x 206 8.24
n
25
x
x
2
2
n 1
n
2062
25 3.357
25 1
1778
s 3.357 1.83
b.
Number of Measurements
in Interval
Interval
c.
Percentage
x s , or (6.41, 10.07)
18
18 / 25 .72 or 72%
x 2 s , or (4.58, 11.90)
24
24 / 25 .96 or 96%
x 3s , or (2.75, 13.73)
25
25 / 25 1.00 or 100%
The percentages in part b are in agreement with Chebyshev's Rule and agree fairly well with the
percentages given by the Empirical Rule.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
d.
Range 12 5 7 and s
51
Range 7
1.75
4
4
The range approximation provides a satisfactory estimate of s 1.83 from part a.
2.75
Using Chebyshev's Rule, at least 8/9 of the measurements will fall within 3 standard deviations of the
mean. Thus, the range of the data would be around 6 standard deviations. Using the Empirical Rule,
approximately 95% of the observations are within 2 standard deviations of the mean. Thus, the range of
the data would be around 4 standard deviations. We would expect the standard deviation to be somewhere
between Range/6 and Range/4.
For our data, the range 760 135 625 .
The
Range 625
Range 625
156.25 .
104.17 and
6
6
4
4
Therefore, I would estimate that the standard deviation of the data set is between 104.17 and 156.25.
It would not be feasible to have a standard deviation of 25. If the standard deviation were 25, the data
would span 625/25 = 25 standard deviations. This would be extremely unlikely.
a.
Using MINITAB, the histogram of the data is:
Histogram of Wheels
12
10
8
Frequency
2.76
6
4
2
0
1
2
3
4
5
6
7
8
Wheels
Since the distribution is skewed to the right, it is not mound-shaped and it is not symmetric.
b.
Using MINITAB, the results are:
Descriptive Statistics: Wheels
Variable
Wheels
N
28
Mean
3.214
StDev
1.371
Minimum
1.000
Q1
2.000
Median
3.000
Q3
4.000
Maximum
8.000
The mean is 3.214 and the standard deviation is 1.371.
c.
d.
The interval is: x 2 s 3.214 2(1.371) 3.214 2.742 (0.472, 5.956) .
According to Chebyshev’s rule, at least 75% of the observations will fall within 2 standard deviations
of the mean.
Copyright © 2014 Pearson Education, Inc.
52
2.77
2.78
Chapter 2
e.
According to the Empirical Rule, approximately 95% of the observations will fall within 2 standard
deviations of the mean.
f.
Actually, 26 of the 28 or 26/28 = .929 of the observations fall within the interval. This value is close
to the 95% that we would expect with the Empirical Rule.
a.
The interval x 2 s will contain at least 75% of the observations. This interval is
x 2 s 3.11 2(.66) 3.11 1.32 (1.79, 4.43) .
b.
No. The value 1.25 does not fall in the interval x 2 s . We know that at least 75% of all
observations will fall within 2 standard deviations of the mean. Since 1.25 falls more than 2 standard
deviations from the mean, it would not be a likely value to observe.
a.
Using Chebyshev’s Rule, at least 75% of the observations will fall within 2 standard deviations of the
mean.
x 2 s 4.25 2(12.02) 4.25 24.04 ( 19.79, 28.29) or (0, 28.29) since we cannot have a
negative number blogs.
2.79
2.80
b.
We would expect the distribution to be skewed to the right. We know that we cannot have a negative
number of blogs/forums. Even 1 standard deviation below the mean is a negative number. We would
assume that there are a few very large observations because the standard deviation is so big compared
to the mean.
a.
The 2 standard deviation interval around the mean is:
b.
Using Chebyshev’s Theorem, at least ¾ of the observations will fall within 2 standard
deviations of the mean. Thus, at least ¾ of first-time candidates for the CPA exam have total credit
hours between 105.77 and 176.85.
c.
In order for the above statement to be true, nothing needs to be known about the shape of the
distribution of total semester hours.
a.
Since the data are mound-shaped and symmetric, we know from the Empirical Rule that
approximately 95% of the observations will fall within 2 standard deviations of the mean. This
interval will be: x 2 s 39 2(6) 39 12 (27, 51) .
b.
We know that approximately .05 of the observations will fall outside the range 27 to 51. Since the
distribution of scores is symmetric, we know that half of the .05 or .025 will fall above 51.
c.
We know from the Empirical Rule that approximately 99.7% (essentially all) of the observations will
fall within 3 standard deviations of the mean. This interval is:
x 3s 39 3(6) 39 18 (21, 57) .
x 2 s 141.31 2(17.77) 141.31 35.54 (105.77, 176.85)
x
n
2.81
a.
The sample mean is: x i 1
n
i
17,800
95.699
186
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Methods for Describing Sets of Data
53
n
xi
n
2
17,8002
x i 1
1,707,998
n
186 24.6332
The sample variance is: s 2 i 1
n 1
186 1
2
The standard deviation is: s s 2 24.6332 4.9632
x s 95.699 4.963 (90.736, 100.662)
b.
x 2 s 95.699 2(4.963) 95.699 9.926 (85.773, 105.625)
x 3s 95.699 3(4.963) 95.699 14.889 (80.810, 110.558)
c.
There are 166 out of 186 observations in the first interval. This is (166 / 186) 100% 89.2% . There
are 179 out of 186 observations in the second interval. This is (179 / 186) 100% 96.2% . There
are 182 out of 186 observations in the second interval. This is (182 / 186) 100% 97.8% .
The percentages for the first 2 intervals are much larger than we would expect using the Empirical
Rule. The Empirical Rule indicates that approximately 68% of the observations will fall within 1
standard deviation of the mean. It also indicates that approximately 95% of the observations will fall
within 2 standard deviations of the mean. Chebyshev’s Theorem says that at least ¾ or 75% of the
observations will fall within 2 standard deviations of the mean and at least 8/9 or 88.9% of the
observations will fall within 3 standard deviations of the mean. It appears that our observed
percentages agree with Chebyshev’s Theorem better than the Empirical Rule.
2.82
2.83
a.
The interval is: x 2 s 13.2 2(19.5) 13.2 39 ( 25.8, 52.2) or (0, 52.2) since we cannot
have negative number of minutes.
b.
Since this interval contains negative numbers, we know that the distribution cannot be symmetric.
One cannot have negative values for time spent on a laptop computer.
c.
Since we know the data are not symmetric, we must use Chebyshev’s Rule. At least ¾ or 75% of the
observations will fall between -25.8 and 52.2 or between 0 and 52.2 minutes.
x
The sample mean is:
n
x i 1
n
i
240.9 248.8 215.7 238.0 2347.4
234.74
10
10
The sample variance deviation is:
n
xi
n
2
2347.42
xi i 1
551,912.1
883.424
n
10
98.1582
s 2 i 1
9
9
n 1
2
The sample standard deviation is:
s2 98.1582 9.91
The data are fairly symmetric, so we can use the Empirical Rule. We know from the Empirical Rule that
almost all of the observations will fall within 3 standard deviations of the mean. This interval would be:
x 3s 234.74 3(9.91) 234.74 29.73 (205.01, 264.47)
Copyright © 2014 Pearson Education, Inc.
54
Chapter 2
2.84
a.
Using MINITAB, the frequency histogram for the time in bankruptcy is:
Histogram of TIME
20
Frequency
15
10
5
0
2
4
6
Time in Bankruptcy
8
10
The Empirical Rule is not applicable because the data are not mound shaped.
b.
Using MINITAB, the descriptive measures are:
Descriptive Statistics: TIME
Variable
TIME
N
49
Mean
2.549
StDev
1.828
Minimum
1.000
Q1
1.350
Median
1.700
Q3
3.500
Maximum
10.100
From Chebyshev’s Theorem, we know that at least 75% of the observations will fall within 2
standard deviations of the mean. This interval is:
x 2 s 2.549 2(1.828) 2.549 3.656 ( 1.107, 6.205) or (0, 6.205) since we cannot have
negative months.
2.85
c.
There are 47 of the 49 observations within this interval. The percentage would be
(47 / 49) 100% 95.9% . This agrees with Chebyshev’s Theorem (at least 75%). It also agrees
with the Empirical Rule (approximately 95%).
d.
From the above interval we know that about 95% of all firms filing for prepackaged bankruptcy will
be in bankruptcy between 0 and 6.2 months. Thus, we would estimate that a firm considering filing
for bankruptcy will be in bankruptcy up to 6.2 months.
a.
b.
The interval x 2 s for the flexed arm group is x 2 s 59 3(4) 59 12 (47, 71) . The interval
for the extended are group is x 2 s 43 3(2) 43 6 (37, 49) . We know that at least 8/9 or
88.9% of the observations will fall within 3 standard deviations of the mean using Chebyshev’s Rule.
Since these 2 intervals barely overlap, the information supports the researchers’ theory. The shoppers
from the flexed arm group are more likely to select vice options than the extended arm group.
The interval x 2 s for the flexed arm group is x 2 s 59 2(10) 59 20 (39, 79) . The
interval for the extended are group is x 2 s 43 2(15) 43 30 (13, 73) . Since these two
intervals overlap almost completely, the information does not support the researcher’s theory. There
does not appear to be any difference between the two groups.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.86
55
a.
Yes. The distribution of the buy-side analysts is fairly flat and skewed to the right. The distribution
of the sell-side analysts is more mound shaped and is not spread out as far as the buy-side
distribution. Since the buy-side distribution is more spread out, the variance of the buy-side
distribution will be larger than the variance of the sell-side distribution. Because the buy-side
distribution is skewed to the right, the mean will be pulled to the right. Thus, the mean of the buyside distribution will be greater than the mean of the sell-side distribution.
b.
Since the sell-side distribution is fairly mound-shaped, we can use the Empirical Rule. The Empirical
Rule says that approximately 95% of the observations will fall within 2 standard deviations of the
mean. The interval for the sell-side distribution would be:
x 2 s .05 2(.85) .05 1.7 ( 1.75, 1.65)
Since the buy-side distribution is skewed to the right, we cannot use the Empirical Rule. Thus, we
will use Chebyshev’s Rule. We know that at least (1 – 1/k2) will fall within k standard deviations of
the mean. If we choose k 4 , then (1 1/ 42 ) .9375 or 93.75%. This is very close to 95%
requested in the problem. The interval for the buy-side distribution to contain at least 93.75% of the
observations would be: x 4 s .85 4(1.93) .85 7.72 ( 6.87, 8.57)
Note: This interval will contain at least 93.75% of the observations. It may contain more than
93.75% of the observations.
2.87
Since we do not know if the distribution of the heights of the trees is mound-shaped, we need to apply
Chebyshev's Rule. We know 30 and 3 . Therefore, 3 30 3(3) 30 9 (21, 39) .
According to Chebyshev's Rule, at least 8 / 9 .89 of the tree heights on this piece of land fall within this
interval and at most 1 / 9 .11 of the tree heights will fall above the interval. However, the buyer will only
1000
purchase the land if at least
.20 of the tree heights are at least 40 feet tall. Therefore, the buyer
5000
should not buy the piece of land.
2.88
a.
Since we do not have any idea of the shape of the distribution of SAT-Math score changes, we must
use Chebyshev’s Theorem. We know that at least 8/9 of the observations will fall within 3 standard
deviations of the mean. This interval would be:
x 3s 19 3(65) 19 195 ( 176, 214)
Thus, for a randomly selected student, we could be pretty sure that this student’s score would be
anywhere from 176 points below his/her previous SAT-Math score to 214 points above his/her
previous SAT-Math score.
b.
Since we do not have any idea of the shape of the distribution of SAT-Verbal score changes, we must
use Chebyshev’s Theorem. We know that at least 8/9 of the observations will fall within 3 standard
deviations of the mean. This interval would be:
x 3s 7 3(49) 7 147 ( 140, 154)
Thus, for a randomly selected student, we could be pretty sure that this student’s score would be
anywhere from 140 points below his/her previous SAT-Verbal score to 154 points above his/her
previous SAT-Verbal score.
Copyright © 2014 Pearson Education, Inc.
56
Chapter 2
c.
2.89
A change of 140 points on the SAT-Math would be a little less than 2 standard deviations from the
mean. A change of 140 points on the SAT-Verbal would be a little less than 3 standard deviations
from the mean. Since the 140 point change for the SAT-Math is not as big a change as the 140 point
on the SAT-Verbal, it would be most likely that the score was a SAT-Math score.
We know 25 and 1 . Therefore, 2 25 2(.1) 25 .2 (24.8, 25.2)
The machine is shut down for adjustment if the contents of two consecutive bags fall more than 2 standard
deviations from the mean (i.e., outside the interval (24.8, 25.2)). Therefore, the machine was shut down
yesterday at 11:30 (25.23 and 25.25 are outside the interval) and again at 4:00 (24.71 and 25.31 are outside
the interval).
2.90
2.91
a.
z
b.
z
c.
z
d.
z
x x 40 30
2 (sample)
5
s
x
x
2 standard deviations above the mean.
90 89
.5 (population)
2
.5 standard deviations above the mean.
50 50
0 (population)
5
0 standard deviations above the mean.
x x 20 30
2.5 (sample)
4
s
2.5 standard deviations below the mean.
Using the definition of a percentile:
a.
Percentile
75th
Percentage
Above
25%
Percentage
Below
75%
b.
50th
50%
50%
c.
20th
80%
20%
d.
84th
16%
84%
2.92
QL corresponds to the 25th percentile. QM corresponds to the 50th percentile. QU corresponds to the 75th
percentile.
2.93
We first compute z-scores for each x value.
a.
z
b.
z
c.
d.
x
x
100 50
2
25
1 4
3
1
x 0 200
2
z
100
z
x
10 5
1.67
3
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
57
The above z-scores indicate that the x value in part a lies the greatest distance above the mean and the
x value of part b lies the greatest distance below the mean.
2.94
Since the element 40 has a z-score of 2 and 90 has a z-score of 3,
2
40
and
2 40
2 40
40 2
3
90
3 90
3 90
By substitution, 40 2 3 90 5 50 10 and 40 2(10) 60 .
Therefore, the population mean is 60 and the standard deviation is 10.
2.95
The mean score of U.S. eighth-graders on a mathematics assessment test is 283. This is the average score.
The 25th percentile is 259. This means that 25% of the U.S. eighth-graders score below 259 on the test and
75% score higher. The 75th percentile is 308. This means that 75% of the U.S. eighth-graders score below
308 on the test and 25% score higher. The 90th percentile is 329. This means that 90% of the U.S. eighthgraders score below 329 on the test and 10% score higher.
2.96
a.
The z-score is z
b.
Since the data are mound-shaped and symmetric and 39 is the mean, .5 of the sampled drug dealers
will have WR scores below 39.
c.
If 5% of the drug dealers have WR scores above 49, then 95% will have WR scores below 49. Thus,
49 will be the 95th percentile.
x x 30 39
1.5 . A score of 30 is 1.5 standard deviations below the mean.
6
s
2.97
A median starting salary of $41,100 indicates that half of the University of South Florida graduates had
starting salaries less than $41,100 and half had starting salaries greater than $41,100. At mid-career, half of
the University of South Florida graduates had a salary less than $71,100 and half had salaries greater than
$71,100. At mid-career, 90% of the University of South Florida graduates had salaries under $131,000 and
10% had salaries greater than $131,000.
2.98
a.
From Exercise 2.81, x 95.699 and s = 4.963. The z-score for an observation of 74 is:
z
x x 74 95.699
4.37
4.963
s
This z-score indicates that an observation of 74 is 4.37 standard deviations below the mean. Very
few observations will be lower than this one.
b.
The z-score for an observation of 98 is:
z
x x 92 95.699
0.75
4.963
s
This z-score indicates that an observation of 92 is .75 standard deviations below the mean. This score
Copyright © 2014 Pearson Education, Inc.
58
Chapter 2
is not an unusual observation in the data set.
2.99
2.100
2.101
Since the 90th percentile of the study sample in the subdivision was .00372 mg/L, which is less than the
USEPA level of .015 mg/L, the water customers in the subdivision are not at risk of drinking water with
unhealthy lead levels.
x x 155 67.755
3.25 . This score would not be
s
26.871
considered a typical level of support. It is 3.25 standard deviations above the mean. Very few observations
would be above this value.
The z-score associated with a score of 155 is z
a.
The 10th percentile is the score that has at least 10% of the observations less than it. If we arrange the
data in order from the smallest to the largest, the 10th percentile score will be the .10(75) = 7.5 or 8th
observation. When the data are arranged in order, the 8th observation is 0. Thus, the 10th percentile is
0.
b.
The 95th percentile is the score that has at least 95% of the observations less than it. If we arrange the
data in order from the smallest to the largest, the 95th percentile score will be the .95(75) = 71.25 or
72nd observation. When the data are arranged in order, the 72nd observation is 21. Thus, the 95th
percentile is 21.
x
n
c.
The sample mean is: x i 1
n
i
393
5.24
75
xi
3932
xi2 i
5943
n
75 52.482
The sample variance is: s 2 i
75 1
n 1
2
The standard deviation is: s s 2 52.482 7.244
The z-score for a county with 48 Superfund sites is: z
2.102
x x 48 5.24
5.90
7.244
s
d.
Yes. A score of 48 is almost 6 standard deviations from the mean. We know that for any data set
almost all (at least 8/9 using Chebyshev’s Theorem) of the observations are within 3 standard
deviations of the mean. To be almost 6 standard deviations from the mean is very unusual.
a.
Since the data are approximately mound-shaped, we can use the Empirical Rule.
On the blue exam, the mean is 53% and the standard deviation is 15%. We know that approximately
68% of all students will score within 1 standard deviation of the mean. This interval is:
x s 53 15 (38, 68)
About 95% of all students will score within 2 standard deviations of the mean. This interval is:
x 2 s 53 2(15) 53 30 (23, 83)
About 99.7% of all students will score within 3 standard deviations of the mean. This interval is:
x 3s 53 3(15) 53 45 (8, 98)
b.
Since the data are approximately mound-shaped, we can use the Empirical Rule.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
59
On the red exam, the mean is 39% and the standard deviation is 12%. We know that approximately
68% of all students will score within 1 standard deviation of the mean. This interval is:
x s 39 12 (27, 51)
About 95% of all students will score within 2 standard deviations of the mean. This interval is:
x 2 s 39 2(12) 39 24 (15, 63)
About 99.7% of all students will score within 3 standard deviations of the mean. This interval is:
x 3s 39 3(12) 39 36 (3, 75)
The student would have been more likely to have taken the red exam. For the blue exam, we know
that approximately 95% of all scores will be from 23% to 83%. The observed 20% score does not
fall in this range. For the red exam, we know that approximately 95% of all scores will be from 15%
to 63%. The observed 20% score does fall in this range. Thus, it is more likely that the student
would have taken the red exam.
a.
The z-score for Harvard is z = 5.08. This means that Harvard’s productivity score was 5.08 standard
deviations above the mean. This is extremely high and extremely unusual.
b.
The z-score for Howard University is z = .85. This means that Howard University’s productivity
score was .85 standard deviations below the mean. This is not an unusual z-score.
c.
Yes. Other indicators that the distribution is skewed to the right are the values of the highest and
lowest z-scores. The lowest z-score is less than 1 standard deviation below the mean while the
highest z-score is 5.08 standard deviations above the mean.
Using MINITAB, the histogram of the z-scores is:
Histogram of Z-Score
70
60
50
Frequency
2.103
c.
40
30
20
10
0
-1
0
1
2
Z-Score
3
4
5
This histogram does imply that the data are skewed to the right.
Copyright © 2014 Pearson Education, Inc.
60
Chapter 2
2.104
a.
From the problem, 2.7 and .5
z
x
z x x z
For z = 2.0, x 2.7 2.0(.5) 3.7
For z = 1.0, x 2.7 1.0(.5) 2.2
For z = .5, x 2.7 .5(.5) 2.95
For z = 2.5, x 2.7 2.5(.5) 1.45
b.
For z = 1.6, x 2.7 1.6(.5) 1.9
c.
If we assume the distribution of GPAs is approximately mound-shaped, we can use the Empirical
Rule.
From the Empirical Rule, we know that .025 or 2.5% of the students will have GPAs above 3.7
(with z = 2). Thus, the GPA corresponding to summa cum laude (top 2.5%) will be greater than 3.7
(z > 2).
We know that .16 or 16% of the students will have GPAs above 3.2 (z = 1). Thus, the limit on
GPAs for cum laude (top 16%) will be greater than 3.2 (z > 1).
We must assume the distribution is mound-shaped.
2.105
Not necessarily. Because the distribution is highly skewed to the right, the standard deviation is very large.
Remember that the z-score represents the number of standard deviations a score is from the mean. If the
standard deviation is very large, then the z-scores for observations somewhat near the mean will appear to
be fairly small. If we deleted the schools with the very high productivity scores and recomputed the mean
and standard deviation, the standard deviation would be much smaller. Thus, most of the z-scores would be
larger because we would be dividing by a much smaller standard deviation. This would imply a bigger
spread among the rest of the schools than the original distribution with the few outliers.
2.106
To determine if the measurements are outliers, compute the z-score.
a.
b.
z
x x 65 57
.727
11
s
Since the z-score is less than 3, this would not be an outlier.
x x 21 57
3.273 Since the z-score is greater than 3 in absolute value, this would be an
11
s
outlier.
z
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
d.
2.107
z
61
x x 72 57
1.364 Since the z-score is less than 3, this would not be an outlier.
11
s
x x 98 57
3.727 Since the z-score is greater than 3 in absolute value, this would be an
11
s
outlier.
z
The interquartile range is IQR QU QL 85 60 25 .
The lower inner fence = QL 1.5( IQR ) 60 1.5(25) 22.5 .
The upper inner fence = QU 1.5( IQR ) 85 1.5(25) 122.5 .
The lower outer fence = QL 3( IQR ) 60 3(25) 15 .
The upper outer fence = QU 3( IQR ) 85 3(25) 160 .
With only this information, the box plot would look something like the following:
*
────────────
──────────────────│
+
│──────
────────────
─┼────┼────┼────┼────┼────┼────┼────┼────┼────┼────┼───
10
20
30
40
50
60
70
80
90 100 110
The whiskers extend to the inner fences unless no data points are that small or that large. The upper inner
fence is 122.5. However, the largest data point is 100, so the whisker stops at 100. The lower inner fence
is 22.5. The smallest data point is 18, so the whisker extends to 22.5. Since 18 is between the inner and
outer fences, it is designated with a *. We do not know if there is any more than one data point below 22.5,
so we cannot be sure that the box plot is entirely correct.
2.108
a.
Median is approximately 4.
b.
QL is approximately 3 (Lower Quartile)
QU is approximately 6 (Upper Quartile)
c.
IQR QU QL 6 3 3
d.
The data set is skewed to the right since the right whisker is longer than the left, there is one outlier,
and there are two potential outliers.
e.
50% of the measurements are to the right of the median and 75% are to the left of the upper quartile.
f.
The upper inner fence is QU 1.5( IQR ) 6 1.5(3) 10.5 . The upper outer fence is
QU 3( IQR ) 6 3(3) 15 . Thus, there are two suspect outliers, 12 and 13. There is one highly
suspect outlier, 16.
Copyright © 2014 Pearson Education, Inc.
62
Chapter 2
2.109
a.
Using MINITAB, the box plot for sample A is given below.
Boxplot of Sample A
200
Sample A
175
150
125
100
Using MINITAB, the box plot for sample B is given below.
Boxplot of Sample B
210
200
Sample B
190
180
170
160
150
140
b.
In sample A, the measurement 84 is an outlier. This measurement falls outside the lower outer fence.
Lower outer fence = Lower hinge 3( IQR ) 150 3(172 150) 150 3(22) 84
Lower inner fence = Lower hinge 1.5( IQR ) 150 1.5(22) 117
Upper inner fence = Upper hinge 1.5( IQR ) 172 1.5(22) 205
In addition, 100 may be an outlier. It lies outside the inner fence.
In sample B, 140 and 206 may be outliers. The point 140 lies outside the inner fence while the point
206 lies right at the inner fence.
Lower outer fence = Lower hinge 3( IQR ) 168 3(184 169) 168 3(15) 123
Lower inner fence = Lower hinge 1.5( IQR ) 168 1.5(15) 145.5
Upper inner fence = Upper hinge 1.5( IQR ) 184 1.5(15) 206.5
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.110
2.111
2.112
a.
The approximate 25th percentile PASI score before treatment is 10. The approximate median before
treatment is 15. The approximate 75th percentile PASI score before treatment is 28.
b.
The approximate 25th percentile PASI score after treatment is 3. The approximate median after
treatment is 5. The approximate 75th percentile PASI score after treatment is 7.5.
c.
Since the 75th percentile after treatment is lower than the 25th percentile before treatment, it appears
that the ichthyotherapy is effective in treating psoriasis.
a.
The average expenditure per full-time employee is $6,563. The median expenditure per employee is
$6,232. Half of all expenditures per employee were less than $6,232 and half were greater than
$6,232. The lower quartile is $5,309. Twenty-five percent of all expenditures per employee were
below $5,309. The upper quartile is $7,216. Seventy-five percent of all expenditures per employee
were below $7,216.
b.
IQR QU QL $7, 216 $5, 309 $1, 907 .
c.
The interquartile range goes from the 25th percentile to the 75th percentile. Thus, .5 .75 .25 of the
1,751 army hospitals have expenses between $5,309 and $7,216.
a.
From the printout, x 52.334 and s = 9.224.
The highest salary is 75 (thousand).
The z-score is z
x x 75 52.334
2.46
9.224
s
Therefore, the highest salary is 2.46 standard deviations above the mean.
The lowest salary is 35.0 (thousand).
The z-score is z
x x 35.0 52.334
1.88
9.224
s
Therefore, the lowest salary is 1.88 standard deviations below the mean.
The mean salary offer is 52.33 (thousand).
The z-score is z
x x 52.33 52.334
0
9.224
s
The z-score for the mean salary offer is 0 standard deviations from the mean.
No, the highest salary offer is not unusually high. For any distribution, at least 8/9 of the salaries
should have z-scores between 3 and 3. A z-score of 2.46 would not be that unusual.
2.113
63
b.
Since no salaries are outside the inner fences, none of them are suspect or highly suspect outliers.
a.
The z-score is: z
x x 160 141.31
1.05
17.77
s
Since the z-score is not large, it is not considered an outlier.
Copyright © 2014 Pearson Education, Inc.
64
Chapter 2
b.
Z-scores with values greater than 3 in absolute value are considered outliers. An observation with a
z-score of 3 would have the value:
z
xx
x 141.31
3
3(17.77) x 141.31 53.31 x 141.31 x 194.62
17.77
s
An observation with a z-score of 3 would have the value:
z
xx
x 141.31
3
3(17.77) x 141.31 53.31 x 141.31 x 88.00
17.77
s
Thus any observation of semester hours that is greater than or equal to 194.62 or less than or equal to
88 would be considered an outlier.
2.114
From Exercise 2.100, x 67.755 and s 26.87 . Using MINITAB, a boxplot of the data is:
Boxplot of Support
160
140
120
Support
100
80
60
40
20
0
From the boxplot, the support level of 155 would be an outlier. From Exercise 2.100, we found the z-score
x x 155 67.755
associated with a score of 155 as z
3.25 . Since this z-score is greater than 3, the
26.871
s
observation 155 is considered an outlier.
a.
Using MINITAB, the boxplots for each type of firm are:
Boxplot of TIME vs VOTES
10
8
TIME
2.115
6
4
2
0
Joint
None
VOTES
Prepack
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.116
65
b.
The median bankruptcy time for Joint firms is about 1.5. The median bankruptcy time for None
firms is about 3.2. The median bankruptcy time for Prepack firms is about 1.4.
c.
The range of the "Prepack" firms is less than the other two, while the range of the "None" firms is the
largest. The interquartile range of the "Prepack" firms is less than the other two, while the
interquartile range of the "Joint" firms is larger than the other two.
d.
No. The interquartile range for the "Prepack" firms is the smallest which corresponds to the smallest
standard deviation. However, the second smallest interquartile range corresponds to the "None"
firms. The second smallest standard deviation corresponds to the "Joint" firms.
e.
Yes. There is evidence of two outliers in the "Prepack" firms. These are indicated by the two *'s.
There is also evidence of two outliers in the "None" firms. These are indicated by the two *'s.
a.
From Exercise 2.101, x 5.24 , s 2 52.482 , and s 7.244 .
We will use 3 standard deviations from the mean as the cutoff for outliers. Z-scores with values
greater than 3 in absolute value are considered outliers. An observation with a z-score of 3 would
have the value:
z
xx
x 5.24
3
3(7.244) x 5.24 21.732 x 5.24 x 26.972
7.244
s
An observation with a z-score of -3 would have the value:
xx
x 5.24
z
3
3(7.244) x 5.24 21.732 x 5.24 x 16.492
7.244
s
Thus, any observation that is greater than 26.972 or less than -16.492 would be considered an outlier.
In this data set there would be 1 outlier: 48.
x
n
b.
Deleting the observation 48, the sample mean is: x i 1
n
i
345
4.66
74
xi
3452
xi2 i
3639
n
74 27.8158
The sample variance is: s 2 i
74 1
n 1
2
The standard deviation is: s s 2 27.8158 5.274
The mean has decreased from 5.24 to 4.66, while the standard deviation decreased from 7.244 to
5.274.
Copyright © 2014 Pearson Education, Inc.
66
Chapter 2
2.117
a.
Using MINITAB, the boxplot is:
Boxplot of Score
100
95
Score
90
85
80
75
70
From the boxplot, there appears to be 10 outliers: 69, 73, 74, 78, 83, 84, 84, 86, 86, and 86.
b.
From Exercise 2.81, x 95.699 and s = 4.963. Since the data are skewed to the left, we will
consider observations more than 2 standard deviations from the mean to be outliers. An observation
with a z-score of 2 would have the value:
z
xx
x 95.699
2
2(4.963) x 95.699 9.926 x 95.699 x 105.625
4.963
s
An observation with a z-score of -2 would have the value:
z
xx
x 95.699
2
2(4.963) x 95.699 9.926 x 95.699 x 85.773
4.963
s
Observations greater than 105.625 or less than 85.773 would be considered outliers. Using this
criterion, the following observations would be outliers: 69, 73, 74, 78, 83, 84, and 84.
c.
No, these methods do not agree exactly. Using the boxplot, 10 observations were identified as
outliers. Using the z-score method, only 7 observations were identified as outliers. However, the 3
additional points that were not identified as outliers using the z-score method were very close to the
cutoff value.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.118
a.
67
Using MINITAB, the box plot is:
Boxplot of TIME
70
60
50
TIME
40
30
20
10
0
The median is about 18. The data appear to be skewed to the right since there are 3 suspect outliers
to the right and none to the left. The variability of the data is fairly small because the IQR is fairly
small, approximately 26 10 = 16.
b.
The customers associated with the suspected outliers are customers 268, 269, and 264.
c.
In order to find the z-scores, we must first find the mean and standard deviation.
x
x 815 20.375
n
s2
40
x
x
2
n 1
n
2
2
24129 815
40 192.90705
40 1
s 192.90705 13.89
The z-scores associated with the suspected outliers are:
Customer 268 z
49 20.375
2.06
13.89
Customer 269 z
50 20.375
2.13
13.89
Customer 264 z
64 20.375
3.14
13.89
All the z-scores are greater than 2. These are unusual values.
2.119
From the stem-and-leaf display in Exercise 2.34, the data are fairly mound-shaped, but skewed somewhat
to the right.
The sample mean is x
x 1493 59.72 .
n
25
Copyright © 2014 Pearson Education, Inc.
68
Chapter 2
The sample variance is s 2
x
x
2
2
n 1
n
14932
25 321.7933 .
25 1
96,885
The sample standard deviation is s 321.7933 17.9386 .
The z-score associated with the largest value is z
Since
x x 102 59.72
2.36 .
17.9386
s
the data are not extremely skewed to the right, this observation is probably not an outlier.
The observations associated with the one-time customers are 5 of the largest 7 observations. Thus, repeat
customers tend to have shorter delivery times than one-time customers.
2.120
For Perturbed Intrinsics, but no Perturbed Projections:
x
2
n
x i 1
n
i
n
xi
n
i 1
2
8.12
x
15.63
i
n
5 2.508 .627
s 2 i 1
n 1
5 1
4
8.1
1.62
5
The z-score corresponding to a value of 4.5 is z
s s 2 .627 .792
x x 4.5 1.62
3.63
.792
s
Since this z-score is greater than 3, we would consider this an outlier for perturbed intrinsics, but no
perturbed projections.
For Perturbed Projections, but no Perturbed Intrinsics:
x
2
n
x i 1
n
i
n
xi
n
2
125.82
xi i 1
3350.1
n
5 184.972 46.243
s 2 i 1
5 1
4
n 1
125.8
25.16
5
s s 2 46.243 6.800
The z-score corresponding to a value of 4.5 is z
x x 4.5 25.16
3.038
6.800
s
Since this z-score is less than -3, we would consider this an outlier for perturbed projections, but no
perturbed intrinsics.
Since the z-score corresponding to 4.5 for the perturbed projections, but no perturbed intrinsics is smaller in
absolute value than that for perturbed intrinsics, but no perturbed projections, it is more likely that the that
the type of camera perturbation is perturbed projections, but no perturbed intrinsics.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.121
69
Using MINITAB, the scatterplot is:
Scatterplot of Var2 vs Var1
18
16
14
Var2
12
10
8
6
4
2
0
0
1
2
3
4
5
6
8
Var1
2.122
Using MINITAB, a scatterplot of the data is:
Scatterplot of Var2 vs Var1
14
12
Var2
10
8
6
4
2
0
0
2
4
Var1
2.123.
From the scatterplot of the data, it appears that as the number of punishments increases, the average payoff
decreases. Thus, there appears to be a negative linear relationship between punishment use and average
payoff. This supports the researchers conclusion that “winners” don’t punish”.
Copyright © 2014 Pearson Education, Inc.
70
2.124
Chapter 2
Using MINITAB, the scatterplot of the data is:
Scatterplot of Catch vs Search
7000
Catch
6000
5000
4000
3000
15
20
25
Search
30
35
There is an apparent negative linear trend between the search frequency and the total catch. As the search
frequency increases, the total catch tends to decrease.
Using MINITAB, a scattergram of the data is:
Scatterplot of SLUGPCT vs ELEVATION
0.625
0.600
0.575
SLUGPCT
2.125
0.550
0.525
0.500
0.475
0.450
0
1000
2000
3000
ELEVATION
4000
5000
6000
If we include the observation from Denver, then we would say there might be a linear relationship between
slugging percentage and elevation. If we eliminated the observation from Denver, it appears that there
might not be a relationship between slugging percentage and elevation.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.126
71
Using MINITAB, the scatterplot of the data is:
Scatterplot of MATH2011 vs MATH2001
625
600
MATH2011
575
550
525
500
475
450
460
480
500
520
540
560
MATH2001
580
600
620
There appears to be a positive linear trend between the Math SAT scores in 2001 and the Math SAT scores
in 2011. As the 2001 Math SAT scores increase, the 2011 Math SAT scores also tend to increase.
a.
Using MINITAB, a scatterplot of JIF and cost is:
Scatterplot of JIF vs Cost
3.5
3.0
2.5
JIF
2.0
1.5
1.0
0.5
0.0
0
200
400
600
800
1000
Cost
1200
1400
1600
1800
There is a slight negative linear trend to the data. As cost increases, JIF tends to decrease.
b.
Using MINITAB, a scatterplot of the number of cities and cost is:
Scatterplot of Cites vs Cost
800
700
600
500
Cites
2.127
400
300
200
100
0
0
200
400
600
800
1000
Cost
1200
1400
1600
1800
Copyright © 2014 Pearson Education, Inc.
72
Chapter 2
There is a moderate positive trend to the data. As cost increases, the number of cities tends to
increase.
c.
Using MINITAB, a scatterplot of RPI and cost is:
Scatterplot of RPI vs Cost
4
RPI
3
2
1
0
0
200
400
600
800
1000
Cost
1200
1400
1600
1800
There is a slight positive trend to the data. As cost increases, RPI tends to increase.
Using MINITAB, the scatterplot of the data is:
Scatterplot of Mass vs Time
7
6
5
4
M ass
2.128
3
2
1
0
0
10
20
30
T ime
40
50
60
There is evidence to indicate that the mass of the spill tends to diminish as time
increases. As time is getting larger, the mass is decreasing.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.129
a.
73
Using MINITAB, a scatterplot of the data is:
Scatterplot of Year2 vs Y ear1
55
50
Year2
45
40
35
30
20
30
40
Year1
50
60
There is a moderate positive trend to the data. As the scores for Year1 increase, the scores for Year2
also tend to increase.
b.
Using MINITAB, the scattergram of the data is:
Scatterplot of Value ($mil) vs Income ($mil)
2000
1800
1600
Value ($mil)
2.130
From the graph, two agencies that had greater than expected PARS evaluation scores for Year2 were
USAID and State.
1400
1200
1000
800
600
0
20
40
60
Income ($mil)
80
100
120
There is a moderate positive trend to the data. As operating income increases, the 2011 value also tends to
increase. Since the trend is moderate, we would recommend that an NFL executive use operating income
to predict a team’s current value.
Copyright © 2014 Pearson Education, Inc.
74
Chapter 2
2.131
a.
Using MINITAB, the scatterplot of the data is:
Scatterplot of YRSPRAC vs EDHRS
40
YRSPRAC
30
20
10
0
0
200
400
600
800
1000
EDHRS
There does not appear to be much of a relationship between the years of experience and the amount
of exposure to ethics in medical school.
Using MINITAB, a boxplot of the amount of exposure to ethics in medical school is:
Boxplot of EDHRS
1000
800
600
EDHRS
b.
400
200
0
The one data point that is an extreme outlier is the value of 1000.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
75
After removing this data point, the scatterplot of the data is:
Scatterplot of YRSPRAC vs EDHRS
40
YRSPRAC
30
20
10
0
0
10
20
30
40
50
EDHRS
60
70
80
90
With the data point removed, there now appears to be a negative trend to the data. As the amount of
exposure to ethics in medical school increases, the years of experience decreases.
2.132
Using MINITAB, a scatterplot of the data is:
Scatterplot of ACCURACY vs DISTANCE
75
70
ACCURACY
65
60
55
50
45
280
290
300
DISTANCE
310
320
Yes, his concern is a valid one. From the scatterplot, there appears to be a fairly strong negative
relationship between accuracy and driving distance. As driving distance increases, the driving accuracy
tend to decrease.
2.133
One way the bar graph can mislead the viewer is that the vertical axis has been cut off. Instead of starting
at 0, the vertical axis starts at 12. Another way the bar graph can mislead the viewer is that as the bars get
taller, the widths of the bars also increase.
Copyright © 2014 Pearson Education, Inc.
76
Chapter 2
2.134
a.
Using MINITAB, the time series plot is:
Time Series Plot of Deaths
900
800
700
Deaths
600
500
400
300
200
100
0
2003
2004
2005
2006
Index
2.135
b.
The time series plot is misleading because the information for 2006 is incomplete – it is based on
only 2 months while all of the rest of the years are based on 12 months.
c.
In order to construct a plot that accurately reflects the trend in American casualties from the Iraq War,
we would want complete data for 2006 and information for the years 2007 through 2011.
a.
The graph might be misleading because the scales on the vertical axes are different. The left vertical
axis ranges from 0 to $120 million. The right vertical axis ranges from 0 to $20 billion.
b.
Using MINITAB, the redrawn graph is:
Time Series Plot of Craigslist, NewspaperAds
18000
Variable
C raigslist
NewspaperA ds
16000
14000
Data
12000
10000
8000
6000
4000
2000
0
2003
2004
2005
2006
Index
2007
2008
2009
Although the amount of revenue produced by Craigslist has increased dramatically from 2003 to
2009, it is still much smaller than the revenue produced by newspaper ad sales.
1.136
a.
This graph is misleading because it looks like as the days are increasing, the number of barrels
collected per day are also increasing. However, the bars are the cumulative number of barrels
collected. The cumulative value can never decrease.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
b.
77
Using MINITAB, the graph of the daily collection of oil is:
Chart of Barrells
2500
Barrells
2000
1500
1000
500
0
May-16
May-17
May-18
May-19 May-20
Day
May-21
May-22
May-23
From this graph, it shows that there has not been a steady improvement in the suctioning process.
There was an increase for 3 days, then a leveling off for 3 days, then a decrease.
2.137
The relative frequency histogram is:
Histogram of Class
Relative frequency
.20
.15
.10
.05
0
1.125
2.625
4.125
5.625
Measurement Class
7.125
8.625
2.138
The mean is sensitive to extreme values in a data set. Therefore, the median is preferred to the mean when
a data set is skewed in one direction or the other.
2.139
a.
z
b.
z
c
z
d.
z
x
x
x
x
50 60
1
10
z
70 60
1
10
z
80 60
2
10
50 50
0
5
z
70 50
4
5
z
80 50
6
5
50 40
1
10
z
70 40
3
10
z
80 40
4
10
50 40
.1
100
z
70 40
.3
100
z
80 40
.4
100
Copyright © 2014 Pearson Education, Inc.
78
Chapter 2
2.140
2.141
a.
If we assume that the data are about mound-shaped, then any observation with a z-score greater than
3 in absolute value would be considered an outlier. From Exercise 2.139, the z-score corresponding
to 50 is 1, the z-score corresponding to 70 is 1, and the z-score corresponding to 80 is 2. Since none
of these z-scores is greater than 3 in absolute value, none would be considered outliers.
b.
From Exercise 2.139, the z-score corresponding to 50 is 2, the z-score corresponding to 70 is 2, and
the z-score corresponding to 80 is 4. Since the z-score corresponding to 80 is greater than 3, 80
would be considered an outlier.
c.
From Exercise 2.139, the z-score corresponding to 50 is 1, the z-score corresponding to 70 is 3, and
the z-score corresponding to 80 is 4. Since the z-scores corresponding to 70 and 80 are greater than
or equal to 3, 70 and 80 would be considered outliers.
d.
From Exercise 2.139, the z-score corresponding to 50 is .1, the z-score corresponding to 70 is .3, and
the z-score corresponding to 80 is .4. Since none of these z-scores is greater than 3 in absolute value,
none would be considered outliers.
a.
x 13 1 10 3 3 30
x
b.
x 25 6.25
x
n
49
7
7
s2
x 12 3
n
4
2
2
n
2
2
302
5 108 27
5 1
4
288
s 27 5.20
x 13 6 6 0 241
2
x 3 3 3 3 12
x
2
x
x
n 1
s2
4
n
2
2
2
2
x
x
2
n 1
n
2
x
x
2
n
252
4 84.75 28.25
4 1
3
241
s 28.25 5.32
x 1 0 1 10 11 11 15 569 .
2
n 1
2
2
2
x 1 0 1 10 11 11 15 49
x
d.
s2
5
n
2
x 13 6 6 0 25
x
c.
x 30 6
x 13 1 10 3 3 288
2
2
2
2
492
7 226 37.67
7 1
6
569
2
2
s 37.67 6.14
x 3 3 3 3 36
2
s2
2
x
x
2
2
2
2
n 1
n
2
122
4 0 0
4 1
3
36
Copyright © 2014 Pearson Education, Inc.
2
s 0 0
2.142
a.
x 4 6 6 5 6 7 34
x
b.
Methods for Describing Sets of Data
x 34 5.67
s2
6
n
79
x 4 6 6 5 6 7 198
2
2
x
x
2
2
2
n 1
n
2
2
2
2
342
6 5.3333 1.0667
6 1
5
198
s 1.067 1.03
x 1 4 (3) 0 (3) (6) 9 x (1) 4 (3) 0 (3) (6) 71
2
x
x
n
9
$1.5
6
s2
x
x
2
2
2
n 1
n
2
2
2
2
2
(9) 2
6 57.5 11.5 dollars squared
6 1
5
71
s 11.5 $3.39
c.
x 5 5 5 5 16 2.0625
3
x
s2
4
2
1
1
x 2.0625 .4125%
n
x 5 5 5 5 16 1.2039
2
3
2
4
2
2
2
1
2
1
2
5
x
x
2
2
n 1
n
2.06252
.3531
5
.0883% squared
5 1
4
1.2039
s .0883 .30%
d.
(a)
Range = 7 4 = 3
(b)
Range = $4 ($-6) = $10
(c)
Range =
4
1
64
5
59
% %
% %
% .7375%
5
16
80
80
80
2.143
The range is found by taking the largest measurement in the data set and subtracting the smallest
measurement. Therefore, it only uses two measurements from the whole data set. The standard deviation
uses every measurement in the data set. Therefore, it takes every measurement into account—not just two.
The range is affected by extreme values more than the standard deviation.
2.144
range 20
5
4
4
Copyright © 2014 Pearson Education, Inc.
80
Chapter 2
2.145
Using MINITAB, the scatterplot is:
Scatterplot of Var2 vs Var1
30
Var2
25
20
15
10
100
a.
300
Var1
Management System
Cause Category
Engineering & Design
Procedures & Practices
Management & Oversight
Training & Communication
TOTAL
b.
400
500
To find relative frequencies, we divide the frequencies of each category by the total number of
incidents. The relative frequencies of the number of incidents for each of the cause categories are:
Number of Incidents
Relative Frequencies
27
24
22
10
83
27 / 83 = .325
24 / 83 = .289
22 / 83 = .265
10 / 83 = .120
1
The Pareto diagram is:
Management Systen Cause Category
35
30
25
P er cent
2.146
200
20
15
10
5
0
E ng&D es
c.
P roc&P ract
M gmt&O v er
C ategor y
Trn&C omm
The category with the highest relative frequency of incidents is Engineering and Design. The
category with the lowest relative frequency of incidents is Training and Communication.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.147
a.
81
The relative frequency for each response category is found by dividing the frequency by the total
sample size. The relative frequency for the category “Global Marketing” is 235/2863 = .082. The
rest of the relative frequencies are found in a similar manner and are reported in the table.
Area
Global Marketing
Sales Management
Buyer Behavior
Relationships
Innovation
Marketing Strategy
Channels/Distribution
Marketing Research
Services
TOTAL
Number
235
494
478
498
398
280
213
131
136
2,863
Relative Frequencies
235/2863 = .082
494/2863 = .173
478/2863 = .167
498/2863 = .174
398/2863 = .139
280/2863 = .098
213/2863 = .074
131/2863 = .046
136/2863 = .048
1.00
Relationships and sales management had the most articles published with 17.4% and 17.3%,
respectively. Not far behind was Buyer Behavior with 16.7%. Of the rest of the areas, only
innovation had more than 10%.
b.
Using MINITAB, the pie chart of the data is:
Pie Chart of Number vs Area
Serv ices
Mark eting research 4.8%
4.6%
Global Mark eting
8.2%
C hannells/Distribution
7.4%
Sales Management
17.3%
Mark eting Strategy
9.8%
Inov ation
13.9%
C ategory
Global Mark eting
Sales Management
Buy er Behavior
Relationships
Inovation
Mark eting Strategy
C hannells/Distribution
Mark eting research
Serv ices
Buy er Behav ior
16.7%
Relationships
17.4%
The slice for Marketing Research is smaller than the slice for Sales Management because there were
fewer articles on Marketing Research than for Sales Management.
2.148
a.
The data are time series data because the numbers of bankruptcies were collected over a
period of 10 months.
Copyright © 2014 Pearson Education, Inc.
82
Chapter 2
b.
Using MINITAB, the time series plot is:
Time Series Plot of Bankrupties
120000
100000
Bankrupties
80000
60000
40000
20000
0
0
Jan
c.
2.149
Feb
Mar
Apr
May
Jun
Month
Jul
Aug
Sep
Oct
There is a generally increasing trend in the number of bankruptcies as the months increase.
Using MINITAB, the pie chart is:
Pie Chart of F vs DrivStar
2
4, 4.1%
5
18, 18.4%
3
17, 17.3%
C ategory
2
3
4
5
4
59, 60.2%
60% of cars have 4-star rating and only 4% have 2-star ratings.
2.150
a.
The average driver’s severity of head injury in head-on collisions is 603.7.
b.
Since the mean and median are close in value, the data should be fairly symmetric. Thus, we can use
the Empirical Rule. We know that about 95% of all observations will fall within 2 standard
deviations of the mean. This interval is x 2 s 603.7 2(185.4) 603.7 370.8 (232.9, 974.5)
Most of the head-injury ratings will fall between 232.9 and 974.5.
c.
x x 408 603.7
1.06
185.4
s
Since the absolute value is not very big, this is not an unusual value to observe.
The z-score would be: z
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.151
a.
83
Using MINITAB, a Pareto diagram for the data is:
Chart Defects
70
60
Frequency
50
40
30
20
10
0
Body
Accessories
Electrical
Defect
Transmission
Engine
The most frequently observed defect is a body defect.
b.
Using MINITAB, a Pareto diagram for the Body Defect data is:
Chart of Body Defects
30
Frequency
25
20
15
10
5
0
Paint
Dents
Upolstery
Body Defect
Windshield
Chrome
Most body defects are either paint or dents. These two categories account for
30 25 / 70 55 / 70 .786 of all body defects. Since these two categories account for so much of
the body defects, it would seem appropriate to target these two types of body defects for special
attention.
2.152
a.
The data collection method was a survey.
b.
Since the data were 4 different categories, the variable is qualitative.
Copyright © 2014 Pearson Education, Inc.
84
Chapter 2
c.
Using MINITAB, a pie chart of the data is:
Pie Chart of Made USA
Category
< 50%
100%
50-74%
75-99%
< 50%
4, 3.8%
75-99%
20, 18.9%
50-74%
18, 17.0%
100%
64, 60.4%
About 60% of those surveyed believe that “Made in USA” means 100% US labor and
materials.
2.153
a.
From the information given, we have x 375 and s = 25. From Chebyshev's Rule, we know that at
least three-fourths of the measurements are within the interval: x 2 s , or (325, 425)
Thus, at most one-fourth of the measurements exceed 425. In other words, more than 425 vehicles
used the intersection on at most 25% of the days.
b.
According to the Empirical Rule, approximately 95% of the measurements are within the interval:
x 2 s , or (325, 425)
This leaves approximately 5% of the measurements to lie outside the interval. Because of the
symmetry of a mound-shaped distribution, approximately 2.5% of these will lie below 325, and the
remaining 2.5% will lie above 425. Thus, on approximately 2.5% of the days, more than 425
vehicles used the intersection.
2.154
The percentile ranking of the age of 25 years would be 100% 75% = 25%. Thus, an age of 25 would
correspond to the 25th percentile.
2.155
a.
Using MINITAB, the stem-and-leaf display is:
Stem-and-Leaf of PENALTY
Leaf Unit = 10
(28)
10
5
5
4
3
3
3
3
2
1
b.
N = 38
0 0011111222222223333334444899
1 00239
2
3 0
4 0
5
6
7
8 5
9 3
10 0
See the highlighted leaves in part a.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
c.
2.156
85
Most of the penalties imposed for Clean Air Act violations are relatively small compared to the
penalties imposed for other violations. All but two of the penalties for Clean Air Act violations are
below the median penalty imposed.
Using MINITAB, the pie charts are:
Color
F (82, 26.6%)
E (44, 14.3%)
D (16, 5.2%)
I (40, 13.0%)
G (65, 21.1%)
H (61, 19.8%)
VS1 (81, 26.3%)
IF
(44, 14.3%)
VS2 (53, 17.2%)
VVS2 (78, 25.3%)
VVS1 (52, 16.9%)
Clarity
The F color occurs the most often with 26.6%. The clarity that occurs the most is VS1 with 26.3%. The D
color occurs the least often with 5.2%. The clarity that occurs the least is IF with 14.3%.
a.
Using MINITAB, the relative frequency histogram is:
Histogram of CARAT
.20
Relative frequency
2.157
.15
.10
.05
0
0.30
0.45
0.60
CARAT
0.75
0.90
1.05
Copyright © 2014 Pearson Education, Inc.
Chapter 2
b.
Using MINITAB, the relative frequency histogram for the GIA group is:
Histogram for GIA
.14
Relative frequency
.12
.10
.08
.06
.04
.02
0
0.30
0.45
0.60
CARAT
0.75
0.90
1.05
Using MINITAB, the relative frequency histograms for the HRD and IGI
groups are:
Histogram for HRD
.40
Relative frequency
c.
.30
.20
.10
0
0.5
0.6
0.7
0.8
CARAT
0.9
1.0
1.1
Histogram for IGI
.35
.30
Relative frequency
86
.25
.20
.15
.10
.05
0
0.2
0.4
0.6
CARAT
0.8
1.0
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
d.
The HRD group does not assess any diamonds less than .5 carats and almost 40% of the diamonds
they assess are 1.0 carat or higher. The IGI group does not assess very many diamonds over .5 carats
and more than half are .3 carats or less. More than half of the diamonds assessed by the GIA group
are more than .5 carats, but the sizes are less than those of the HRD group.
x
n
The sample mean is: x i 1
n
e.
i
194.32
.631
308
The average number of carats for the 308 diamonds is .631.
f.
The median is the average of the middle two observations once they have been ordered.
The 154th and 155th observations are .62 and .62. The average of these two observations is .62.
Half of the diamonds weigh less than .62 carats and half weigh more.
g
The mode is 1.0. This observation occurred 32 times.
h.
Since the mean and median are close in value, either could be a good descriptor of central
tendency.
i.
From Chebyshev’s Theorem, we know that at least ¾ or 75% of all observations will fall within 2
standard deviations of the mean. From part e, x .63 .
xi
194.322
xi2 i
146.19
n
308 .0768 square carats
The variance is: s 2 i
308 1
n 1
2
The standard deviation is: s s 2 .0768 .277 carats
This interval is: x 2 s .631 2(.277) .631 .554 (.077, 1.185)
Using MINITAB, the scatterplot is:
Scatterplot of PRICE vs CARAT
18000
16000
14000
12000
PRICE
2.158
87
10000
8000
6000
4000
2000
0
0.2
0.3
0.4
0.5
0.6
0.7
CARAT
0.8
0.9
1.0
1.1
As the number of carats increases the price of the diamond tends to increase. There appears to be an
upward trend.
Copyright © 2014 Pearson Education, Inc.
88
Chapter 2
2.159
a.
Using MINITAB, a bar graph of the data is:
Chart of Cause
12
10
Count
8
6
4
2
0
Collision
Fire
Grounding
Cause
HullFail
Unknown
Fire and grounding are the two most likely causes of puncture.
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Spillage
Variable
Spillage
N
42
Mean
66.19
StDev
56.05
Minimum
25.00
Q1
32.00
Median
43.00
Q3
77.50
Maximum
257.00
The mean spillage amount is 66.19 thousand metric tons, while the median is 43.00. Since the
median is so much smaller than the mean, it indicates that the data are skewed to the right. The
standard deviation is 56.05. Again, since this value is so close to the value of the mean, it indicates
that the data are skewed to the right.
Since the data are skewed to the right, we cannot use the Empirical Rule to describe the data.
Chebyshev’s Rule can be used. Using Chebyshev’s Rule, we know that at least 8/9 of the
observations will fall within 3 standard deviations of the mean.
x 3s 66.19 3(56.05) 66.19 168.15 ( 101.96, 234.34) or (0, 234.34) since we cannot
have negative spillage.
Thus, at least 8/9 of all oil spills will be between 0 and 234.34 thousand metric tons.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.160
Using MINITAB, a pie chart of the data is:
Pie Chart of Defectt
Category
False
True
True
49, 9.8%
False
449, 90.2%
A response of ‘true’ means the software contained defective code. Thus, only 9.8% of the modules
contained defective software code.
2.161
a.
Since no information is given about the distribution of the velocities of the Winchester bullets, we
can only use Chebyshev's Rule to describe the data. We know that at least 3/4 of the velocities will
fall within the interval:
x 2 s 936 2(10) 936 20 (916, 956)
Also, at least 8/9 of the velocities will fall within the interval:
x 3s 936 3(10) 936 30 (906, 966)
b.
Since a velocity of 1,000 is much larger than the largest value in the second interval in part a, it is
very unlikely that the bullet was manufactured by Winchester.
Copyright © 2014 Pearson Education, Inc.
89
90
Chapter 2
2.162
a.
First, we must compute the total processing times by adding the processing times of the three
departments. The total processing times are as follows:
Request
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Total
Processing
Time
13.3
5.7
7.6
20.0*
6.1
1.8
13.5
13.0
15.6
10.9
8.7
14.9
3.4
13.6
14.6
14.4
Request
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
Total
Processing
Time
19.4*
4.7
9.4
30.2
14.9
10.7
36.2*
6.5
10.4
3.3
8.0
6.9
17.2*
10.2
16.0
11.5
Request
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Total
Processing
Time
23.4*
14.2
14.3
24.0*
6.1
7.4
17.7*
15.4
16.4
9.5
8.1
18.2*
15.3
13.9
19.9*
15.4
14.3*
19.0
The stem-and-leaf displays with the appropriate leaves highlighted are as follows:
Stem-and-leaf of Mkt
Leaf Unit = 0.10
6 0
7 1
14 2
16 3
22 4
(10) 5
18 6
8 7
4 8
2 9
2 10
1 11
0112446
3
0024699
25
001577
0344556889
0002224799
0038
07
0
0
Stem-and-leaf of Engr
Leaf Unit = 0.10
7
14
19
23
(5)
22
19
14
9
9
7
6
5
2
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
4466699
3333788
12246
1568
24688
233
01239
22379
66
0
3
023
0
4
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
Stem-and-leaf of Accnt
Leaf Unit = 0.10
19
(8)
23
21
19
15
15
13
11
11
11
11
10
9
9
8
8
0
0
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
HI
111111111112 2333444
55556888
00
79
0023
23
78
8
2
91
Stem-and-leaf of Total
Leaf Unit = 1.00
1
3
5
11
17
21
(5)
24
14
10
6
5
4
0 1
0 33
0 45
0 666677
0 888999
1 0000
1 33333
1 4444445555
1 6677
1 8999
2 0
2 3
2 44
HI 30, 36
0
4
99, 105, 135, 144,
182, 220, 300
Of the 50 requests, 10 were lost. For each of the three departments, the processing times for the lost
requests are scattered throughout the distributions. The processing times for the departments do not
appear to be related to whether the request was lost or not. However, the total processing times for
the lost requests appear to be clustered towards the high side of the distribution. It appears that if the
total processing time could be kept under 17 days, 76% of the data could be maintained, while
reducing the number of lost requests to 1.
b.
For the Marketing department, if the maximum processing time was set at 6.5 days, 78% of the
requests would be processed, while reducing the number of lost requests by 4. For the Engineering
department, if the maximum processing time was set at 7.0 days, 72% of the requests would be
processed, while reducing the number of lost requests by 5. For the Accounting department, if the
maximum processing time was set at 8.5 days, 86% of the requests would be processed, while
reducing the number of lost requests by 5.
c.
Using MINITAB, the summary statistics are:
Descriptive Statistics: REQUEST, MARKET, ENGINEER, ACCOUNT
Variable
MARKET
ENGINEER
ACCOUNT
TOTAL
N
Mean
50 4.766
50 5.044
50 3.652
50 13.462
StDev
2.584
3.835
6.256
6.820
Minimum
0.100
0.400
0.100
1.800
Q1
2.825
1.775
0.200
8.075
Median
Q3
5.400 6.250
4.500 7.225
0.800 3.725
13.750 16.600
Copyright © 2014 Pearson Education, Inc.
Maximum
11.000
14.400
30.000
36.200
92
Chapter 2
d.
The z-scores corresponding to the maximum time guidelines developed for each department and the
total are as follows:
Marketing: z
Engineering: z
x x 7.0 5.04
.51
3.84
s
Accounting: z
x x 8.5 3.65
.77
6.26
s
Total: z
e.
x x 6.5 4.77
.67
2.58
s
x x 17 13.46
.52
6.82
s
To find the maximum processing time corresponding to a z-score of 3, we substitute in the values of
z, x , and s into the z formula and solve for x.
z
xx
x x zs x x zs
s
Marketing:
x 4.77 3(2.58) 4.77 7.74 12.51
None of the orders exceed this time.
Engineering:
x 5.04 3(3.84) 5.04 11.52 16.56
None of the orders exceed this time.
These both agree with both the Empirical Rule and Chebyshev's Rule.
Accounting:
x 3.65 3(6.26) 3.65 18.78 22.43
One of the orders exceeds this time or 1/50 = .02.
Total:
x 13.46 3(6.82) 13.46 20.46 33.92
One of the orders exceeds this time or 1/50 = .02.
These both agree with Chebyshev's Rule but not the Empirical Rule. Both of these last two
distributions are skewed to the right.
f.
Marketing:
x 4.77 2(2.58) 4.77 5.16 9.93
Two of the orders exceed this time or 2/50 = .04.
Engineering:
x 5.04 2(3.84) 5.04 7.68 12.72
Two of the orders exceed this time or 2/50 = .04.
Accounting:
x 3.65 2(6.26) 3.65 12.52 16.17
Three of the orders exceed this time or 3/50 = .06.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
93
x 13.46 2(6.82) 13.46 13.64 27.10
Two of the orders exceed this time or 2/50 = .04.
Total:
All of these agree with Chebyshev's Rule but not the Empirical Rule.
g.
No observations exceed the guideline of 3 standard deviations for both Marketing and Engineering.
One observation exceeds the guideline of 3 standard deviations for both Accounting (#23, time =
30.0 days) and Total (#23, time = 36.2 days). Therefore, only (1/10) 100% of the "lost" quotes
have times exceeding at least one of the 3 standard deviation guidelines.
Two observations exceed the guideline of 2 standard deviations for both Marketing (#31, time = 11.0
days and #48, time = 10.0 days) and Engineering (#4, time = 13.0 days and #49, time = 14.4 days).
Three observations exceed the guideline of 2 standard deviations for Accounting (#20, time = 22.0
days; #23, time = 30.0 days; and #36, time = 18.2 days). Two observations exceed the guideline of 2
standard deviations for Total (#20, time = 30.2 days and #23, time = 36.2 days). Therefore, (7/10)
100% = 70% of the "lost" quotes have times exceeding at least one the 2 standard deviation
guidelines.
We would recommend the 2 standard deviation guideline since it covers 70% of the lost quotes, while
having very few other quotes exceed the guidelines.
One reason the plot may be interpreted differently is that no scale is given on the vertical axis. Also,
since the plot almost reaches the horizontal axis at 3 years, it is obvious that the bottom of the plot
has been cut off. Another important factor omitted is who responded to the survey.
b.
A scale should be added to the vertical axis. Also, that scale should start at 0.
a.
Using MINITAB, the time series plot of the data is:
Time Series Plot of Acquisitions
900
800
700
600
500
400
300
200
100
2000
1999
1998
1997
1996
1995
1994
1993
1992
1991
1990
1989
1988
1987
1986
1985
1984
1983
1982
1981
0
1980
2.164
a.
Acquisitions
2.163
Year
Copyright © 2014 Pearson Education, Inc.
Chapter 2
b.
To find the percentage of the sampled firms with at least one acquisition, we divide number with
acquisitions by the total sampled and then multiply by 100%. For 1980, the percentage of firms with at
least on acquisition is (18/1963)*100% = .92%. The rest of the percentages are found in the same
manner and are listed in the following table:
Year
Number of firms
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
TOTAL
1,963
2,044
2,029
2,187
2,248
2,238
2,277
2,344
2,279
2,231
2,197
2,261
2,363
2,582
2,775
2,890
3,070
3,099
2,913
2,799
2,778
51,567
Number with
Acquisitions
18
115
211
273
317
182
232
258
296
350
350
370
427
532
626
652
751
799
866
750
748
9,123
Percentage with
Acquisitions
.92%
5.63%
10.40%
12.48%
14.10%
8.13%
10.19%
11.01%
12.99%
15.69%
15.93%
16.36%
18.07%
20.60%
22.56%
22.56%
24.46%
25.78%
29.73%
26.80%
26.93%
Using MINITAB, the time series plot is:
Time Series Plot of Percent
30
25
20
Percent
15
10
5
2000
1999
1998
1997
1996
1995
1994
1993
1992
1991
1990
1989
1988
1987
1986
1985
1984
1983
1982
1981
0
1980
94
Year
c.
In this case, both plots are almost the same. In general, the time series plot of the percents would be
more informative. By changing the observations to percents, one can compare time periods with
different sample sizes on the same basis.
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
2.165
a.
b.
95
Since the mean is greater than the median, the distribution of the radiation levels is skewed to the
right.
x s 10 3 (7, 13) ; x 2 s 10 2(3) (4, 16) ; x 3s 10 3(3) (1, 19)
Interval
(7, 13)
(4, 16)
(1, 19)
Chebyshev's
At least 0
At least 75%
At least 88.9%
Empirical
68%
95%
100%
Since the data are skewed to the right, Chebyshev's Rule is probably more appropriate in this case.
c.
The background level is 4. Using Chebyshev's Rule, at least 75% or .75(50) 38 homes are above
the background level. Using the Empirical Rule, 97.5% or .975(50) 49 homes are above the
background level.
d.
z
x x 20 10
3.333
3
s
It is unlikely that this new measurement came from the same distribution as the other 50. Using
either Chebyshev's Rule or the Empirical Rule, it is very unlikely to see any observations more than 3
standard deviations from the mean.
2.167
a.
Since it is given that the distribution is mound-shaped, we can use the Empirical Rule. We know that
1.84% is 2 standard deviations below the mean. The Empirical Rule states that approximately 95%
of the observations will fall within 2 standard deviations of the mean and, consequently,
approximately 5% will lie outside that interval. Since a mound-shaped distribution is symmetric,
then approximately 2.5% of the day's production of batches will fall below 1.84%.
b.
If the data are actually mound-shaped, it would be extremely unusual (less than 2.5%) to observe a
batch with 1.80% zinc phosphide if the true mean is 2.0%. Thus, if we did observe 1.8%, we would
conclude that the mean percent of zinc phosphide in today's production is probably less than 2.0%.
a.
Both the height and width of the bars (peanuts) change. Thus, some readers may tend to equate the
area of the peanuts with the frequency for each year.
b.
Using MINITAB, the frequency bar chart is:
Chart of Peanut
5
4
Peanut
2.166
3
2
1
0
1975
1980
1985
1990
1995
2000
2005
2010
Year
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96
Chapter 2
2.168
a.
Clinic A claims to have a mean weight loss of 15 during the first month and Clinic B claims to have a
median weight loss of 10 pounds in the first month. With no other information, I would choose
Clinic B. It is very likely that the distributions of weight losses will be skewed to the right – most
people lose in the neighborhood of 10 pounds, but a couple might lose much more. If a few people
lost much more than 10 pounds, then the mean will be pulled in that direction.
b.
For Clinic A, the median is 10 and the standard deviation is 20. For Clinic B, the mean is 10 and the
standard deviation is 5.
For Clinic A:
The mean is 15 and the median is 10. This would indicate that the data are skewed to the right.
Thus, we will have to use Chebyshev’s Rule to describe the distribution of weight losses.
x 2 s 15 2(20) 15 40 ( 25, 55)
Using Chebyshev’s Rule, we know that at least 75% of all weight losses will be between -25 and 55
pounds. This means that at least 75% of the people will have weight losses of between a loss of 55
pounds to a gain of 25 pounds. This is a very large range.
For Clinic B:
The mean is 10 and the median is 10. This would indicate that the data are symmetrical. Thus, the
Empirical Rule can be used to describe the distribution of weight losses.
x 2 s 10 2(5) 10 10 (0, 20)
Using the Empirical Rule, we know that approximately 95% of all weight losses will be between 0
and 20 pounds. This is a much smaller range than in Clinic A.
I would still recommend Clinic B. Using Clinic A, a person has the potential to lose a large amount
of weight, but also has the potential to gain a relatively large amount of weight. In Clinic B, a person
would be very confident that he/she would lose weight.
c.
2.169
One would want the clients selected for the samples in each clinic to be representative of all clients in
that clinic. One would hope that the clinic would not choose those clients for the sample who lost the
most weight just to promote their clinic.
First we make some preliminary calculations.
Of the 20 engineers at the time of the layoffs, 14 are 40 or older. Thus, the probability that a randomly
selected engineer will be 40 or older is 14/20 = .70. A very high proportion of the engineers is 40 or over.
In order to determine if the company is vulnerable to a disparate impact claim, we will first find the median
age of all the engineers. Ordering all the ages, we get:
29, 32, 34, 35, 38, 39, 40, 40, 40, 40, 40, 41, 42, 42, 44, 46, 47, 52, 55, 64
The median of all 20 engineers is
40 40 80
40
2
2
Now, we will compute the median age of those engineers who were not laid off. The ages underlined
40 40 80
40 .
above correspond to the engineers who were not laid off. The median of these is
2
2
Copyright © 2014 Pearson Education, Inc.
Methods for Describing Sets of Data
97
The median age of all engineers is the same as the median age of those who were not laid off. The median
40 41 81
40.5 , which is not that much different from the median age of those
age of those laid off is
2
2
not laid off. In addition, 70% of all the engineers are 40 or older. Thus, it appears that the company would
not be vulnerable to a disparate impact claim.
2.170
Answers will vary. The graph is made to look like the amount of money spent on education has risen
dramatically from 1980 to 2000, but the 4th grade reading scores have not increased at all. The graph does
not take into account that the number of school children has also increased dramatically in the last 20 years.
A better portrayal would be to look at the per capita spending rather than total spending.
2.171
There is evidence to support this claim. The graph peaks at the interval above 1.002. The heights of the
bars decrease in order as the intervals get further and further from the peak interval. This is true for all bars
except the one above 1.000. This bar is greater than the bar to its right. This would indicate that there are
more observations in this interval than one would expect, suggesting that some inspectors might be passing
rods with diameters that were barely below the lower specification limit.
Copyright © 2014 Pearson Education, Inc.
Chapter 3
Probability
3.1
a.
Since the probabilities must sum to 1,
P ( E3 ) 1 P ( E1 ) P ( E 2 ) P ( E 4 ) P ( E5 ) 1 .1 .2 .1 .1 .5
b.
3.2
P ( E3 ) 1 P ( E1 ) P ( E2 ) P ( E4 ) P ( E5 ) 1 P ( E3 ) P ( E2 ) P ( E4 ) P ( E5 )
2 P ( E3 ) 1 .1 .2 .1 2 P ( E3 ) .6 P ( E3 ) .3
c.
P ( E3 ) 1 P ( E1 ) P ( E 2 ) P ( E 4 ) P ( E5 ) 1 .1 .1 .1 .1 .6
a.
This is a Venn Diagram.
b.
If the sample points are equally likely, then
P (1) P (2) P (3) P (10)
1
10
Therefore,
1
1
1
3
.3
10 10 10 10
1
1
2
P ( B ) P (6) P (7)
.2
10 10 10
P ( A) P (4) P (5) P (6)
c.
3.3
1
1
3
5
.25
20 20 20 20
3
3
6
P ( B ) P (6) P (7)
.3
20 20 20
P ( A) P (4) P (5) P (6)
P( A) P(1) P(2) P(3) .05 .20 .30 .55
P( B ) P (1) P (3) P (5) .05 .30 .15 .50
P(C ) P(1) P(2) P(3) P(5) .05 .20 .30 .15 .70
3.4
a.
b.
c.
d.
9
9!
9 8 7 6 5 4 3 2 1
126
4 4!(9 4)! 4 3 2 1 5 4 3 2 1
7
7!
7 6 5 4 3 2 1
21
2
2!(7
2)!
2 1 5 4 3 2 1
4
4!
4 3 2 1
1
4
4!(4
4)!
4
3 2 1 1
5
5!
5 4 3 2 1
1
0 0!(5 0)! 1 5 4 3 2 1
98
Copyright © 2014 Pearson Education, Inc.
Probability
3.5
3.6
99
e.
6
6!
6 5 4 3 2 1
6
5
5!(6 5)! 5 4 3 2 1 1
a.
N 5
5!
5 4 3 2 1 120
10
n 2 2!(5 2)! 2 1 3 2 1 12
b.
N 6
6!
6 5 4 3 2 1 720
20
n
3
3!(6
3)!
3 2 1 3 2 1 36
c.
N 20
20!
20 19 18 3 2 1
2.432902008 1018
15,504
14
n 5 5!(20 5)! 5 4 3 2 1 15 14 13 3 2 1 1.569209242 10
a.
The tree diagram of the sample points is:
b.
If the dice are fair, then each of the sample points is equally likely. Each would have a probability of
1/36 of occurring.
Copyright © 2014 Pearson Education, Inc.
100
Chapter 3
c.
There is one sample point in A: (3,3). Thus, P ( A)
1
.
36
There are 6 sample points in B: (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1). P ( B)
6 1
.
36 6
There are 18 sample points in C: (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4)
18 1
.
(4,6) (5,1) (5,3) (5,5) (6,2) (6,4) and (6,6). Thus, P(C )
36 2
3.7
a.
If we denote the marbles as B1, B2, R1, R2, and R3, then the ten sample points are:
(B1, B2) (B1, R1) (B1, R2) (B1, R3) (B2, R1) (B2, R2) (B2, R3) (R1, R2) (R1, R3) (R2, R3)
b.
1
.
10
There are 6 sample points in B: (B1, R1) (B1, R2) (B1, R3) (B2, R1) (B2, R2) (B2, R3).
1 6 3
Thus, P ( B ) 6
.
10 10 5
1 3
.
There are 3 sample points in C: (R1, R2) (R1, R3) (R2, R3). Thus, P (C ) 3
10 10
Each student will obtain slightly different proportions. However, the proportions should be close to
c.
3.8
Each of the sample points would be equally likely. Thus, each would have a probability of 1/10 of
occurring.
There is one sample point in A: (B1, B2). Thus, P( A)
P ( A) 1 / 10, P ( B ) 6 / 10, and P (C ) 3 / 10.
3.9
a.
The sample points of this experiment correspond to each of the 6 possible colors of the M&M’s. Let
B r = brown, Y = yellow, R = red, Bl = blue, O = orange, G = green. The six sample points are: Br,
Y, R, Bl, O, and G
b.
From the problem, the probabilities of selecting each color are:
P(Br) = 0.13, P(Y) = 0.14, P(R) = 0.13, P(Bl) = 0.24, P(O) = 0.2, P(G) = 0.16
c.
The probability that the selected M&M is brown is P(Br) = 0.13
d.
The probability that the selected M&M is red, green or yellow is:
P ( R or G or Y ) P ( R ) P (G ) P (Y ) 0.13 0.16 0.14 0.43
e.
3.10
P (not Bl ) P ( R ) P (G ) P (Y ) P ( Br ) P (O ) 0.13 0.16 0.14 0.13 0.20 0.76
Define the following events:
I:
{personal illness}
F: {family issues}
N: {personal needs}
E: {entitlement mentality}
S:
{stress}
Copyright © 2014 Pearson Education, Inc.
Probability
a.
The 5 sample points are: I, F, N, E, S
b.
The probability of each sample points are:
P ( I ) 0.34, P ( F ) 0.22, P ( N ) 0.18, P ( E ) 0.13, P ( S ) 0.13
c.
The probability that the absence is due to something other than “personal illness” (I) is:
P (not I ) P ( F ) P ( N ) P ( E ) P ( S ) 0.22 0.18 0.13 0.13 0.66
3.11
Define the following events:
M: {Nanny who was placed in a job last year is a male}
P(M )
3.12
a.
24
.0057
4,176
Define the following events:
H5: {Hurricane develops from 5th tropical storm}
H12: {Hurricane develops before the 12th or higher tropical storm}
3.13
P(H 5 )
11
.164
67
b.
P ( H 12 )
67 5
.925
67
a.
The 5 sample points are the possible responses of a randomly selected person who participated in
Harris Poll:
None, 1-2, 3-5, 6-9, 10 or more
b.
The probabilities are:
P (none) 0.19, P (1 2) 0.31, P (3 5) 0.26, P (6 9) 0.05, P (10 or more) 0.19
c.
Define the following event:
A: {Respondent looks for healthcare information online more than two times per month}
P ( A) P (3 5) P (6 9) P (10 or more) 0.25 0.05 0.19 0.50
3.14
a.
Define the following events:
A: {Respondent works during summer vacation}
B: {Respondent does not work during summer vacation}
C: {Respondent unemployed}
The sample points are A, B, and C.
b.
Reasonable probabilities are: P ( A) .46, P ( B ) .35, and P (C ) .19
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101
102
3.15
Chapter 3
c.
P ( B or C ) P ( B ) P (C ) .35 .19 .54
a.
The international consumer is most likely to use the Certification mark on a label to identify a green
product.
b.
Define the following events:
A: {Certification mark on label}
B: {Packaging}
C: {Reading information about the product}
D: {Advertisement}
E: {Brand website}
F: {Other}
P ( A or B ) P ( A) P ( B ) .45 .15 .60
3.16
c.
P (C or E ) P (C ) P ( E ) .12 .04 .16
d.
P (not D ) P ( A) P ( B ) P (C ) P ( E ) P ( F ) .45 .15 .12 .04 .18 .94
a.
Define the following events:
A: {Total visitors}
B: {Paying visitors}
C: {Big shows}
D: {Funds raised}
E: {Members}
P( A or D) P( A) P( D)
b & c.
8
7 15
.5
30 30 30
A tree diagram with the corresponding probabilities for this problem follows. To compute the
probabilities, we have to assume that this sample is representative of all such museums. In
addition, we have to assume that each selection of a museum is independent of the second
selection. The probability of selecting a particular type of museum is estimated by the number of
museums in that category divided by 30. Each sample point consists of two museums. The
probabilities of each type of museum in the pair are then multiplied together to find the probability
of the sample point. The probabilities are shown in the tree.
Copyright © 2014 Pearson Education, Inc.
Probability
3.17
d.
P ( AA or DD or AD or DA) P ( AA) P ( DD ) P ( AD ) P ( DA) .071 .054 .062 .062 .249
a.
Define the following event:
C: {Slaughtered chicken passes inspection with fecal contamination}
P (C )
1
.01
100
Copyright © 2014 Pearson Education, Inc.
103
104
Chapter 3
b.
3.18
306
.0095 .01
32, 075
Yes. The probability of a slaughtered chicken passing inspection with fecal contamination rounded
off to 2 decimal places is .01.
Based on the data, P (C )
Define the following events:
B: {Bitel}
C: {Cybernet}
F: {Fujian Landi}
G: {Glint (Pava Rede)}
I: {Intelligent}
K: {Kwang Woo}
O: {Omron}
PT: {Pax Tech}
PC: {Provenco Cadmus}
S: {SZZT Electronics}
T: {Toshiba TEC}
U: {Urmet}
To compute the probability of each event, we first must sum the number of units shipped by all the
manufacturers. The sum is 334,039.
P ( B ) 13, 500 / 344, 039 .040; P (C ) 16, 200 / 344, 039 .048; P ( F ) 119, 000 / 344, 039 .356;
P (G ) 5, 990 / 344, 039 .018; P ( I ) 4, 562 / 344, 039 .014; P ( K ) 42, 000 / 344, 039 .126;
P (O ) 20, 000 / 344, 039 .060; P ( PT ) 10, 072 / 344, 039 .030; P ( PC ) 20, 000 / 344, 039 .060;
P ( S ) 67, 300 / 344, 039 .201; P (T ) 12, 415 / 344, 039 .037; P (U ) 3, 000 / 344, 039 .009
a.
P ( F or S ) P ( F ) P ( S ) 0.356 0.201 0.557
b.
Define the event:
D: {PIN pad is defective}
P ( D ) 1000 / 334, 039 .003
3.19
a.
The probability that any network is selected on a particular day is 1/8. Therefore,
P( ESPN selected on July 11) = 1/8.
b.
The number of ways to select four networks for the weekend days is a combination of 8 networks
8
8!
8 7 6 5 4 3 2 1
taken 4 at a time. The number of ways to do this is
70 .
4 4!(8 4)! 4 3 2 1 4 3 2 1
c.
First, we need to find the number of ways one can choose the 4 networks where ESPN is one of the 4.
If ESPN has to be chosen, then the number of ways of doing this is a combination of one thing taken
1
1!
1
one at a time or
1 . The number of ways to select the remaining 3 networks is a
1 1!(1 1)! 1 1
Copyright © 2014 Pearson Education, Inc.
Probability
105
7
7!
7 6 5 4 3 2 1
combination of 7 things taken 3 at a time or
35 . Thus, the total
3 3!(7 3)! 3 2 1 4 3 2 1
number of ways of selecting 4 networks of which one has to be ESPN is 1(35) = 35.
Finally, the probability of selecting ESPN as one of the 4 networks for the weekend analysis is
35 / 70 .5 .
3.20
a.
Since order does not matter, the number of different bets would be a combination of 8 things taken 2 at
8
a time. The number of ways would be
2
b.
3.21
8!
8 7 6 5 4 3 2 1 40,320
28 .
2!(8 2)! 2 1 6 5 4 3 2 1
1440
If all players are of equal ability, then each of the 28 sample points would be equally likely. Each
would have a probability of occurring of 1/28. There is only one sample point with values 2 and 7.
Thus, the probability of winning with a bet of 2-7 would be 1/28 or .0357.
Since one would be selecting 3 stocks from 15 without replacement, the total number of ways to select the
3 stocks would be a combination of 15 things taken 3 at a time. The number of ways would be
15
15!
15 14 13 3 2 1
1.307674368 1012
455
2874009600
3 3!(15 3)! 3 2 1 12 11 10 3 2 1
3.22
Denote Pu = public, Pr = private, B = bedrocks, U = unconsolidated, BL = below limit, D = detect.
a.
The 8 sample points for this experiment in which the well class (public or private), aquifer (bedrocks
or unconsolidated) and detectible (below limit or detect) MTBE level of a well are observed are as
follows:
(Pu, B, BL)
(Pu, U, BL)
b.
c.
(Pr, B, BL) (Pu, B, D)
(Pr, U, BL) (Pu, U, D)
(Pr, B, D)
(Pr, U, D)
P ( Pu , B , BL ) 57 / 223 0.256
P ( Pr , B , BL ) 81 / 223 0.363
P ( Pu , B , D ) 41 / 223 0.184
P ( Pr , B , D ) 22 / 223 0.099
P ( Pu , U , BL ) 15 / 223 0.067
P ( Pr , U , BL ) 0 / 223 0.000
P ( Pu , U , D ) 7 / 223 0.031
P ( Pr , U , D ) 0 / 223 0.000
Define the following event:
D = {Well has a detectible level or MTBE}
P ( D ) P ( Pu , B , D ) P ( Pu , U , D ) P ( Pr , B , D ) P ( Pr , U , D ) 0.184 0.031 0.099 0 0.314
This means that if one well is chosen at random, the probability that it has a detectible level of MTBE
is .314.
3.23
a.
Since we want to maximize the purchase of grill #2, grill #2 must be one of the 3 grills in the display.
Thus, we have to pick 2 more grills from the 4 remaining grills. Since order does not matter, the
number of different ways to select 2 grill displays from 4 would be a combination of 4 things taken 2
at a time. The number of ways is:
4
4!
4 3 2 1 24
6
2 2!(4 2)! 2 1 2 1 4
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106
Chapter 3
Let Gi represent Grill i. The possibilities are:
G1G2G3, G1G2G4, G1G2G5, G2G3G4, G2G3 G5, G2G4G5
b.
3.24
To find reasonable probabilities for the 6 possibilities, we divide the frequencies by the total sample
size of 124. The probabilities would be:
P (G1G 2 G3 ) 35 / 124 .282
P (G1G 2 G 4 ) 8 / 124 .065
P (G1G 2 G5 ) 42 / 124 .339
P (G 2 G3 G 4 ) 4 / 124 .032
P (G 2 G3 G5 ) 1 / 124 .008
P (G2 G4 G5 ) 34 / 124 .274
c.
P( display contained Grill #1) P (G1G 2 G3 ) P (G1G2 G 4 ) P (G1G2 G5 ) .282 .065 .339 .686
a.
Let H = Hyundai Elantra, T = Toyota Prius, and S = Subaru Forrester. All possible rankings are as
follows, where the first car listed is ranked first, the second car listed is ranked second, and the third
car listed is ranked third:
H,T,S
b.
H,S,T
S,H,T
S,T,H
T,H,S
T,S,H
If each set of rankings is equally likely, then each has a probability of 1/6.
The probability that the Toyota Prius is ranked first
P (T , H , S ) P (T , S , H ) 1 / 6 1 / 6 2 / 6 1 / 3
The probability that the Hyundai Elantra is ranked third
P ( S , T , H ) P (T , S , H ) 1 / 6 1 / 6 2 / 6 1 / 3 .
The probability that the Toyota Prius is ranked first and the Subaru Forrester is ranked second
P (T , S , H ) 1 / 6 .
3.25
1
3
1 2
or 1 to 2.
3 3
a.
The odds in favor of an Oxford Shoes win are to 1
b.
If the odds in favor of Oxford Shoes are 1 to 1, then the probability that Oxford Shoes wins is
1
1
.
11 2
c.
If the odds against Oxford Shoes are 3 to 2, then the odds in favor of Oxford Shoes are
2 to 3. Therefore, the probability that Oxford Shoes wins is
3.26
2
2
.
23 5
First, we need to compute the total number of ways we can select 2 bullets (pair) from 1,837 bullets. This
is a combination of 1,837 things taken 2 at a time.
1,837
1,837!
1837 1836 1
1837 1836
1,686,366
2
2!(1,837
2)!
2
1
1835
1834
1
2
The number of pairs is:
The probability of a false positive is the number of false positives divided by the number of pairs and is:
P(false positive) = # false positives / # pairs 693 / 1,686,366 .0004
This probability is very small. There would be only about 4 false positives out of every 10,000. I would
have confidence in the FBI’s forensic evidence.
Copyright © 2014 Pearson Education, Inc.
Probability
3.27
a.
The number of ways the 5 commissioners can vote is 2(2)(2)(2)(2) = 25 = 32 (Each of the 5
commissioners has 2 choices for his/her vote – For or Against.)
b.
Let F denote a vote ‘For’ and A denote a vote ‘Against’. The 32 sample points would be:
107
FFFFF FFFFA FFFAF FFAFF FAFFF AFFFF FFFAA FFAFA FAFFA AFFFA
FFAAF FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA FAAAF
AFFAA AFAFA AFAAF AAFFA AAFAF AAAFF FAAAA AFAAA AAFAA AAAFA
AAAAF AAAAA
Each of the sample points should be equally likely. Thus, each would have a probability of 1/32.
c.
The sample points that result in a 2-2 split for the other 4 commissioners are:
FFAAF FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA
AFFAA AFAFA AAFFA
There are 12 sample points.
d.
Let V = event that your vote counts. P (V ) 12 / 32 0.375 .
e.
If there are now only 3 commissioners in the bloc, then the total number of ways the bloc can vote is
2(2)(2) 23 8 . The sample points would be:
FFF
FFA
FAF
AFF
FAA
AFA
AAF
AAA
The number of sample points where your vote would count is 4: FAF, AFF, FAA, AFA
Let W = event that your vote counts in the bloc. P (W ) 4 / 8 0.5 .
3.28
3.29
a.
P ( B c ) 1 P ( B ) 1 .7 .3
b.
P ( Ac ) 1 P ( A) 1 .4 .6
c.
P ( A B ) P ( A) P ( B ) P ( A B ) .4 .7 .3 .8
a.
A: {HHH, HHT, HTH, THH, TTH, THT, HTT}
B: {HHH, TTH, THT, HTT}
A B : {HHH, HHT, HTH, THH, TTH, THT, HTT}
Ac: {TTT}
A B : {HHH, TTH, THT, HTT}
b.
P ( A)
c.
P ( A B ) P ( A) P ( B ) P ( A B )
d.
No. P ( A B )
7
8
P(B)
4 1
8 2
P( A B)
7
8
P ( Ac )
1
8
7 1 1 7
8 2 2 8
1
which is not 0.
2
Copyright © 2014 Pearson Education, Inc.
P( A B)
4 1
8 2
108
3.30
Chapter 3
The experiment consists of rolling a pair of fair dice. The sample points are:
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2, 1
2, 2
2, 3
2, 4
2, 5
2, 6
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4, 1
4, 2
4, 3
4, 4
4, 5
4, 6
5, 1
5, 2
5, 3
5, 4
5, 5
5, 6
6, 1
6, 2
6, 3
6, 4
6, 5
6, 6
Since each die is fair, each sample point is equally likely. The probability of each sample point is 1/36.
a.
A: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B: {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}
A B : {(3, 4), (4, 3)}
A B : {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5),
(4, 6), (1, 6), (2, 5), (5, 2), (6, 1)}
Ac: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5),
(3, 6), (4, 1), (4, 2), (4, 4), (4, 5), (4, 6), (5, 1), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4),
(6, 5), (6, 6)}
b.
1 6 1
P ( A) 6
36 36 6
1 11
P ( B ) 11
36 36
1 15 5
P ( A B ) 15
36 36 12
1 30 5
P ( Ac ) 30
36 36 6
1 11 1 6 11 2 15 5
6 36 18
36
36 12
c.
P ( A B ) P ( A) P ( B ) P ( A B )
d.
A and B are not mutually exclusive. To be mutually exclusive, P ( A B ) must be 0. Here,
P( A B)
3.31
1
1 2
P( A B) 2
36 36 18
1
.
18
1 1 1 1
1 15 3
5 5 5 20 10 20 4
a.
P ( A) P ( E1 ) P ( E2 ) P ( E3 ) P ( E5 ) P ( E6 )
b.
P ( B ) P ( E2 ) P ( E3 ) P ( E4 ) P ( E7 )
c.
P ( A B ) P ( E1 ) P ( E 2 ) P ( E3 ) P ( E 4 ) P ( E5 ) P ( E 6 ) P ( E 7 )
1 1 1 1 13
5 5 20 5 20
1 1 1 1
1
1 1
1
5 5 5 20 20 10 5
d.
P ( A B ) P ( E2 ) P ( E3 )
e.
P ( Ac ) 1 P ( A) 1
1 1 2
5 5 5
3 1
4 4
Copyright © 2014 Pearson Education, Inc.
Probability
3.32
3.33
13 7
20 20
f.
P( B c ) 1 P( B) 1
g.
P ( A Ac ) P ( E1 ) P ( E2 ) P ( E3 ) P ( E4 ) P ( E5 ) P ( E6 ) P ( E7 )
109
1 1 1 1
1
1 1
1
5 5 5 20 20 10 5
1 1 5 1
20 5 20 4
h.
P ( Ac B ) P ( E4 ) P ( E7 )
a.
P ( Ac ) P ( E3 ) P ( E6 ) .2 .3 .5
b.
P ( B c ) P ( E1 ) P ( E7 ) .10 .06 .16
c.
P ( Ac B ) P ( E3 ) P ( E6 ) .2 .3 .5
d.
P ( A B ) P ( E1 ) P ( E 2 ) P ( E3 ) P ( E 4 ) P ( E5 ) P ( E 6 ) P ( E 7 )
.10 .05 .20 .20 .06 .30 .06 .97
e.
P ( A B ) P ( E 2 ) P ( E 4 ) P ( E5 ) .05 .20 .06 .31
f.
P ( Ac B c ) P ( E8 ) .03
g.
No. A and B are mutually exclusive if P ( A B ) 0 . Here, P ( A B ) .31 .
a.
P ( A) .50 .10 .05 .65
b.
P ( B ) .10 .07 .50 .05 .72
c.
P (C ) .25
d.
P ( D ) .05 .03 .08
e.
P ( Ac ) .25 .07 .03 .35 (Note: P ( Ac ) 1 P ( A) 1 .65 .35 )
f.
P ( A B ) P ( B ) .10 .07 .50 .05 .72
g.
P( A C ) 0
h.
Two events are mutually exclusive if they have no sample points in common or if the probability of
their intersection is 0.
P ( A B ) P ( A) .50 .10 .05 .65 . Since this is not 0, A and B are not mutually exclusive.
P ( A C ) 0 . Since this is 0, A and C are mutually exclusive.
P ( A D ) .05 . Since this is not 0, A and D are not mutually exclusive.
P ( B C ) 0 . Since this is 0, B and C are mutually exclusive.
Copyright © 2014 Pearson Education, Inc.
110
Chapter 3
P ( B D ) .05 . Since this is not 0, B and D are not mutually exclusive.
P (C D ) 0 . Since this is 0, C and D are mutually exclusive.
3.34
3.35
3.36
a.
The outcome "On" and "High" is A D .
b.
The outcome "Low" or "Medium" is Dc.
a.
The analyst makes an early forecast and is only concerned with accuracy is the event ( A B ) .
b.
The analyst is not only concerned with accuracy is the event Ac.
c.
The analyst is from a small brokerage firm or makes an early forecast is the event C B .
d.
The analyst makes a late forecast and is not only concerned with accuracy is the event B c Ac .
Define the following events:
A: {problems with absenteeism}
T: {problems with turnover}
From the problem, P ( A) .55, P (T ) .41 , and P ( A T ) .22
P(problems with either absenteeism or turnover)
P ( A T ) P ( A) P (T ) P ( A T ) .55 .41 .22 .74
3.37
a.
Define the following events:
L: {Legs only}
W: {Wheels only}
B: {Both legs and wheels}
N: {Neither legs nor wheels}
The sample points are: L, W, B, and N
b.
From the given data:
P( L)
3.38
63
.594
106
P (W )
20
.189
106
P(B)
8
.075
106
c.
P (Wheels) P (W or B ) P (W ) P ( B ) .189 .075 .264
d.
P (Legs) P ( L or B ) P ( L ) P ( B ) .594 .075 .669
e.
P (Either legs or wheels) 1 P ( N ) 1 .142 .858
P( N )
Define the following event:
A: {Store violates the NIST scanner accuracy standard}
Then P ( Ac ) 1 P ( A) 1 52 / 60 8 / 60 .133
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15
.142
106
Probability
3.39
Define the following events:
A: {oil structure is active}
I: {oil structure is inactive}
C: {oil structure is caisson}
W: {oil structure is well protector}
F: {oil structure is fixed platform}
a.
The simple events are all combinations of structure type and activity type. The simple events are:
AC, AW, AF, IC, IW, IF
b.
3.40
Reasonable probabilities would be the frequency divided by the sample size of 3,400. The
probabilities are:
P ( AC ) 503 / 3, 400 .148
P ( AW ) 225 / 3, 400 .066
P ( AF ) 1, 447 / 3, 400 .426
P ( IC ) 598 / 3, 400 .176
P ( IW ) 177 / 3, 400 .052
P ( IF ) 450 / 3, 400 .132
c.
P ( A) P ( AC ) P ( AW ) P ( AF ) .148 .066 .426 .640
d.
P (W ) P ( AW ) P ( IW ) .066 .052 .118
e.
P ( IC ) .176
f.
P ( I F ) P ( IC ) P ( IW ) P ( IF ) P ( AF ) .176 .052 .132 .426 .786
g.
P(C c ) 1 P(C ) 1 P ( AC ) P( IC ) 1 .148 .176 1 .324 .676
Define the following events:
M: {UK citizen visits MySpace}
B: {UK citizen visits Bebo}
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111
112
a.
3.41
Chapter 3
The Venn Diagram that illustrates the use of social networking sites in UK is:
M
M∩B
B
4%
1%
3%
b.
P ( M B ) P ( M ) P ( B ) P ( M B ) 0.04 0.03 0.01 0.06
c.
P ( M c B c ) 1 P ( M B ) 1 0.06 0.94
First, define the following events:
F: {Fully compensated}
P: {Partially compensated}
N: {Non-compensated}
R: {Left because of retirement}
From the text, we know
127
45
72
7 11 10 28
, P( P)
, P( N )
, and P( R)
P( F )
244
244
244
244
244
3.42
127
244
a.
P( F )
b.
P( F R)
c.
P( F c ) 1 P( F ) 1
d.
P( F R) P( F ) P( R) P( F R)
7
244
127 117
244 244
127 28
7
148
244 244 244 244
Define the following events:
I: {Invests in Market}
N: {No investment}
a.
P(I )
44, 651
.283
158, 044
Copyright © 2014 Pearson Education, Inc.
Probability
3.43
31, 943 17, 958 12,145 9, 531 71, 577
.453
158, 044
158, 044
b.
P (IQ 6)
c.
P ( I {IQ 6})
d.
P ( I {IQ 6}) P ( I ) P (IQ 6) P ( I {IQ 6}) .283 .453 .168 .568
e.
P ( I c ) 1 P ( I ) 1 .283 .717
f.
Two events are mutually exclusive if the probability of their intersection is 0.
893
P ( I {IQ 1})
.006 . Since this value is not 0, these two events are not mutually
158, 044
exclusive.
a.
P S A . Products 6 and 7 are contained in this intersection.
b.
P(possess all the desired characteristics) P( P S A) P(6) P(7)
c.
10, 270 6, 698 5,135 4, 464 26, 567
.168
158, 044
158, 044
A S
P ( A S ) P (2) P (3) P (5) P (6) P (7) P (8) P (9) P (10)
d.
1
1
1
1
1
1
1
1
8 4
10 10 10 10 10 10 10 10 10 5
PS
P ( P S ) P (2) P (6) P (7)
3.44
a.
1
1
1
3
10 10 10 10
Define the following events:
G: {Student is assigned to the guilty state}
C: {Student chooses the stated option}
Then P (G ) 57 / 171 .333 .
3.45
b.
P (C ) 60 / 171 .351
c.
P (G C ) 45 / 171 .263
d.
P (G C ) P (G ) P (C ) P (G C ) .333 .351 .263 .421
Define the following events:
M1: {Model 1}
M2: {Model 2}
a.
P (5)
85
.531
160
Copyright © 2014 Pearson Education, Inc.
1 1 1
10 10 5
113
114
3.46
Chapter 3
b.
P (5 0) P (5) P (0) P (5 0) .531
c.
P ( M 2 0)
35
0 .531 .219 .75
160
15
.094
160
Define the following events:
A: {Individual tax return is audited by the IRS}
B: {Corporation tax return is audited by the IRS}
3.47
1, 581, 394
.0111
142,823,105
a.
P ( A)
b.
P ( Ac ) 1 P ( A) 1 .0111 .9889
c.
P( B)
d.
P ( B c ) 1 P ( B ) 1 .0139 .9861
29,803
.0139
2,143,808
Define the following events:
A: {Air pressure is over-reported by 4 psi or more}
B: {Air pressure is over-reported by 6 psi or more}
C: {Air pressure is over-reported by 8 psi or more
3.48
a.
For gas station air pressure gauges that read 35 psi, P ( B ) .09 .
b.
For gas station air pressure gauges that read 55 psi, P (C ) .09 .
c.
For gas station air pressure gauges that read 25 psi, P ( Ac ) 1 P ( A) 1 .16 .84 .
d.
No. If air pressure is over-reported by 6 psi or more, then it is also over-reported by 4 psi or more.
Thus, these 2 events are not mutually exclusive.
e.
The columns in the table are not mutually exclusive. All events in the last column
(% Over-reported by 8 psi or more) are also part of the events in the first and second columns. All
events in the second column are also part of the events in the first column. In addition, there is no
column for the event ‘Over-reported by less than 4 psi or not over-reported’.
There are a total of 6 6 6 216 possible outcomes from throwing 3 fair dice. To help demonstrate this,
suppose the three dice are different colors – red, blue and green. When we roll these dice, we will record
the outcome of the red die first, the blue die second, and the green die third. Thus, there are 6 possible
outcomes for the first position, 6 for the second, and 6 for the third. This leads to the 216 possible
outcomes.
Copyright © 2014 Pearson Education, Inc.
Probability
115
The Grand Duke argued that the chance of getting a sum of 9 and the chance of getting a sum of 10 should
be the same since the number of partitions for 9 and 10 are the same. These partitions are:
9
126
135
144
225
234
333
10
136
145
226
235
244
334
In each case, there are 6 partitions. However, if we take into account the three colors of the dice, then there
are various ways to get each partition. For instance, to get a partition of 126, we could get 126, 162, 216,
261, 612, and 621 (again, think of the red die first, the blue die second, and the green die third). However,
to get a partition of 333, there is only 1 way. To get a partition of 144, there are 3 ways: 144, 414, and
441. The numbers of ways to get each of the above partitions are:
9
126
135
144
225
234
333
# ways
6
6
3
3
6
_ 1
25
10
136
145
226
235
244
334
# ways
6
6
3
6
3
_3
27
Thus, there are a total of 25 ways to get a sum of 9 and 27 ways to get a sum of 10. The chance of
throwing a sum of 9 (25 chances out of 216 possibilities) is less than the chance of throwing a 10 (27
chances out of 216 possibilities).
3.49
3.50
3.51
3.52
a.
P( A | B)
P ( A B ) .1
.5
.2
P( B)
b.
P ( B | A)
P ( A B ) .1
.25
.4
P ( A)
c.
Events A and B are said to be independent if P ( A | B ) P ( A) . In this case, P ( A | B ) .5 and
P ( A) .4 . Thus, A and B are not independent.
a.
P ( A B ) P ( A | B ) P ( B ) .6(.2) .12
b.
P ( B | A)
a.
If two events are independent, then P ( A B ) P ( A) P ( B ) .4(.2) .08 .
b.
If two events are independent, then P ( A | B ) P ( A) .4 .
c.
P ( A B ) P ( A) P ( B ) P ( A B ) .4 .2 .08 .52
a.
Since A and B are mutually exclusive events, P ( A B ) P ( A) P ( B ) .30 .55 .85
P ( A B ) .12
.3
P ( A)
.4
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116
3.53
Chapter 3
b.
Since A and C are mutually exclusive events, P ( A C ) 0
c.
P( A | B)
d.
Since B and C are mutually exclusive events, P ( B C ) P ( B ) P (C ) .55 .15 .70
e.
No, B and C cannot be independent events because they are mutually exclusive events.
a.
P ( A) P ( E1 ) P ( E 2 ) P ( E3 ) .2 .3 .3 .8
P( A B) 0
0
.55
P( B)
P ( B ) P ( E 2 ) P ( E3 ) P ( E5 ) .3 .3 .1 .7
P ( A B ) P ( E 2 ) P ( E3 ) .3 .3 .6
b.
P ( E1 | A)
P ( E 1 A) P ( E 1) .2
.25
P ( A)
P ( A) .8
P ( E2 | A)
P ( E 2 A) P ( E 2) .3
.375
P ( A)
P ( A) .8
P ( E3 | A)
P ( E 3 A) P ( E 3) .3
.375
P ( A)
P ( A) .8
The original sample point probabilities are in the proportion .2 to .3 to .3 or 2 to 3 to 3.
The conditional probabilities for these sample points are in the proportion .25 to .375 to .375 or 2 to 3
to 3.
c.
(1)
P ( B | A) P ( E 2 | A) P ( E3 | A) .375 .375 .75 (from part b)
(2)
P ( B | A)
P ( A B ) .6
.75 (from part a)
.8
P ( A)
The two methods do yield the same result.
3.54
d.
If A and B are independent events, P ( B | A) P ( B ) . From part c, P ( B | A) .75 . From part a,
P ( B ) .7 . Since .75 .7 , A and B are not independent events.
a.
If two fair coins are tossed, there are 4 possible outcomes or simple events. They are:
E1 = HH
E2 = HT E3 = TH E4 = TT
Event A contains the simple events E1, E2, and E3. Event B contains the simple events E2 and E3.
Copyright © 2014 Pearson Education, Inc.
Probability
117
A Venn diagram of this would be:
A
B
E2
E3
E1
E4
Since the coins are fair, each of the sample points is equally likely. Each would have probabilities
of ¼.
b.
1 3
P ( A) 3 .75
4 4
P ( A B ) P ( E2 )P ( E3 )
3.55
1 2 1
P ( B ) 2 .5
4 4 2
1 1 2 1
.5
4 4 4 2
P ( A B ) .5
1
.5
P( B)
P ( B | A)
P ( A B ) .5
.667
.75
P ( A)
c.
P( A | B)
a.
P ( A) P ( E1 ) P ( E3 ) .22 .15 .37
b.
P ( B ) P ( E 2 ) P ( E3 ) P ( E 4 ) .31 .15 .22 .68
c.
P ( A B ) P ( E3 ) .15
d.
P( A | B)
e.
P(B C ) 0
f.
P (C | B )
g.
For pair A and B: A and B are not independent because P ( A | B ) P ( A) or .2206 .37 .
P ( A B ) .15
.2206
P( B)
.68
P (C B ) 0
0
.68
P( B)
For pair A and C: P ( A C ) P ( E1 ) .22
P( A | C )
P (C ) P ( E1 ) P ( E5 ) .22 .10 .32
P ( A C ) .22
.6875
.32
P (C )
Copyright © 2014 Pearson Education, Inc.
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Chapter 3
A and C are not independent because P ( A | C ) P ( A) or .6875 .37 .
For pair B and C: B and C are not independent because P (C | B ) P (C ) or 0 .32 .
3.56
The 36 possible outcomes obtained when tossing two dice are listed below:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
A: {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3),
(4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
B: {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5), (6, 6)}
A B : {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5)}
If A and B are independent, then P ( A) P ( B ) P ( A B ) .
P ( A)
18 1
36 2
P ( A) P ( B )
3.57
a.
P( B)
7
36
P( A B)
6 1
36 6
1 7
7 1
P ( A B ) . Thus, A and B are not independent.
2 36 72 6
P ( A C ) 0 A and C are mutually exclusive.
P ( B C ) 0 B and C are mutually exclusive.
b.
P ( A) P (1) P (2) P (3) .20 .05 .30 .55
P ( B ) P (3) P (4) .30 .10 .40
P (C ) P (5) P (6) .10 .25 .35
P ( A B ) P (3) .30
P( A | B)
P ( A B ) .30
.75
.40
P( B)
A and B are independent if P ( A | B ) P ( A) . Since P ( A | B ) .75 and P ( A) .55 ,
A and B are not independent.
Since A and C are mutually exclusive, they are not independent. Similarly, since B and C are
mutually exclusive, they are not independent.
c.
Using the probabilities of sample points,
P ( A B ) P (1) P (2) P (3) P (4) .20 .05 .30 .10 .65
Using the additive rule,
P ( A B ) P ( A) P ( B ) P ( A B ) .55 .40 .30 .65
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Probability
Using the probabilities of sample points,
P ( A C ) P (1) P (2) P (3) P (5) P (6) .20 .05 .30 .10 .25 .90
Using the additive rule,
P ( A C ) P ( A) P (C ) P ( A C ) .55 .35 0 .90
3.58
3.59
From the Exercise, P ( A) .15 , P ( B ) .10 , and P ( A B ) .05 .
a.
If events A and B are mutually exclusive then P ( A B ) 0 . For this problem, P ( A B ) .05 .
Therefore, events A and B are not mutually exclusive.
b.
P ( B | A)
c.
Events A and B are independent if P ( B | A) P ( B ) . For this exercise, P ( B | A) .333 and
P ( B ) .10 . Since these are not equal, events A and B are not independent.
P ( A B ) .05
.333
.15
P ( A)
Define the following events:
A: {Company is a banking/investment company}
B: {Company is based in United States}
From the problem, we know that P ( A B )
P( A | B)
3.60
4
9
.20 and P ( B )
.45
20
20
P ( A B ) .20
.444 .
P( B)
.45
Define the following events:
G: {The respondent is assigned to the guilt state}
A: {The respondent is assigned to the anger state}
C: {The respondent chooses the stated option to repair car}
a.
From Exercise 3.44, we know P (G ) 57 / 171 .333 and P (G C ) 45 / 171 .263
P (C | G )
b.
P (G C ) .263
.790
.333
P (G )
From Exercise 3.44, we know P (C ) 60 / 171 .351 .
Thus, P (C c ) 1 .351 .649
P( A | C c )
P ( A C c ) 50 / 171 .292
P ( A C c ) .292
.450
.649
P (C c )
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119
120
Chapter 3
c.
Two events C and G are independent if P (C G ) P (C ) P (G ) . From Exercise 3.44, P (G ) .333 ,
P (C ) .351 , and P (G C ) .263 .
P (G ) P (C ) .333(.351) .117 .263 P (G C ) .
Thus C and G are not independent.
3.61
Define the following events:
A: {Internet user has wireless connection via mobile device}
B: {Internet user uses Twitter}
From the exercise, P ( A ) .54 and P ( B | A) .25 .
P ( A B ) P ( B | A) P ( A) .25(.54) .135
3.62
Define the following events:
A: {Person is victim of identity theft}
B: {Theft occurred from unauthorized use of credit card}
From the exercise, P ( A) .05 and P ( B | A) .53
3.63
a.
P ( A) .05
b.
P ( A B ) P ( B | A) P ( A) .05(.53) .0265
Define the following events:
F: {Worker is fully compensated}
P: {Worker is partially compensated}
N: {Worker is non-compensated}
R: {Worker retired}
From the exercise, P ( F ) 127 / 244 .520 , P ( P ) 45 / 244 .184 , P ( R | F ) 7 / 127 .055 ,
P ( R | P ) 11 / 45 .244 , and P ( R | N ) 10 / 72 .139 .
a.
P ( R | F ) 7 / 127 .055
b.
P ( R | N ) 10 / 72 .139
c.
The two events are independent if P ( R | F ) P ( R ) .
7 11 10 28
.115 and P ( R | F ) 10 / 72 .055 . Since these are not equal, events R
244
244
and F are not independent.
P( R)
Copyright © 2014 Pearson Education, Inc.
Probability
3.64.
121
Define the following events:
A: {Respondent works during summer vacation}
B: {Respondent does not work during summer vacation}
C: {Respondent unemployed}
D: {Respondent monitors business emails}
From Exercise 3.14: P ( A ) .46 , P ( B ) .35 , P (C ) .19 . From this exercise, P ( D | A) .35 .
3.65
a.
P ( D | A) .35
b.
P ( A D ) P ( D | A) P ( A) .35(.46) .161
c.
P ( B D ) 0 (If an employee is not working, then he/she will not monitor business emails.)
Define the following events:
I: {Invests in Market}
N: {No investment}
a.
10, 270 6, 698 5,135 4, 464
P ( I {IQ 6})
26,567
158, 044
.371
P ( I | IQ 6)
31,943 17,958 12,145 9,531 71,577
P (IQ 6)
158, 044
b.
44, 651 26,567
P( I {IQ 5})
18, 084
158, 044
.209 .
P( I | IQ 5)
158, 044 71,577 86, 467
P(IQ 5)
158, 044
c.
3.66
Yes, it appears that investing in the stock market is dependent on IQ. If investing in the stock market
and IQ were independent, then P ( I | IQ 5) P ( I | IQ 6) P ( I ) . Since
P ( I | IQ 5) P ( I | IQ 6) , then investing in the stock market and IQ are dependent.
Define the following events:
th
Ai : {i CEO has bachelor’s degree}
13
.325
40
a.
P ( A1 )
b.
If the first 4 CEO’s have just bachelor’s degree, then on the next pick there are only 9 left to choose
from. Similarly, after picking 4 CEO’s, there are only 36 observations left to choose from.
P ( A5 | A1 A2 A3 A4 )
9
.25
36
Copyright © 2014 Pearson Education, Inc.
122
3.67
Chapter 3
Define the following events:
A: {Ambulance can travel to location A under 8 minutes}
B: {Ambulance can travel to location B under 8 minutes}
C: {Ambulance is busy}
We are given P ( A) .58 , P ( B ) .42 , and P (C ) .3 .
3.68
a.
P ( A C c ) P ( A | C c ) P (C c ) .58(1 .3) .406
b.
P ( B | C c ) P (C c ) .42(1 .3) .294
If A and B are independent, then P ( A B ) P ( A) P ( B ) . For this Exercise,
P ( A)
1174 416 1590
1174 89 1263
.883 , P ( B )
.702 , and
1800
1800
1800
1800
P( A B)
1174
.652 .
1800
P ( A) P ( B ) .883(.702) .620 .652 P ( A B ) . Thus, A and B are not independent.
3.69
Define the following events:
A: {Alarm A sounds alarm}
B: {Alarm B sounds alarm}
I: {Intruder}
a.
From the problem P A | I .9, P B | I .95, P( A | I c ) .2 and P( B | I c ) .1 .
b.
Since the two systems are operating independently of each other,
P ( A B | I ) P ( A | I ) P ( B | I ) .9(.95) .855
3.70
c.
P ( A B | I c ) P ( A | I c ) P ( B | I c ) .2(.1) .02
d.
P ( A B | I ) P ( A | I ) P ( B | I ) P ( A B | I ) .9 .95 .855 .995
a.
Since there are 2 vineyards and 3 years, there are a total of 2(3) = 6 combinations.
b.
Of the 6 combinations, 3 of them are from the Llarga vineyard. Thus, P (Llarga) 3 / 6 .5 .
c.
Of the 6 combinations, 2 of them are Year 3. Thus, P (Year 3) 2 / 6 .333
d.
If the tasters are independent, then the probability that each selects Llarga is
P (Llarga) P (Llarga) P (Llarga) P (Llarga) .5(.5)(.5)(.5) .0625 .
Copyright © 2014 Pearson Education, Inc.
Probability
3.71
123
Define the following event:
A: {The specimen labeled “red snapper” was really red snapper}
a.
The probability that you are actually served red snapper the next time you order it at a restaurant is
P ( A) 1 .77 .23
b.
P(at least one customer is actually served red snapper)
= 1 – P(no customer is actually served red snapper)
1 P ( A c A c A c Ac Ac ) 1 P ( A c ) P ( Ac ) P ( A c ) P ( A c ) P ( Ac )
1 .775 1 .271 .729
Note: In order to compute the above probability, we had to assume that the trials or events are
independent. This assumption is likely to not be valid. If a restaurant served one customer a look-alike variety, then it probably served the next one a look-a-like variety.
3.72
First, define the following event:
A: {CVSA correctly determines the veracity of a suspect} P(A) = .98 (from claim)
a.
The event that the CVSA is correct for all four suspects is the event A A A A .
P ( A A A A) .98(.98)(.98)(.98)(.98) .9224
b.
The event that the CVSA is incorrect for at least one of the four suspects is the event
( A A A A) c . P ( A A A A) c 1 P ( A A A A) 1 .9224 .0776
c.
If the CVSA had an accuracy of .98, then the probability of observing 2 incorrect results is less than
.0776. Since 2 incorrect results were observed, it was either a rare event or the accuracy of the CVSA is
not .98 but something less than .98.
3.73
Define the following events:
A: {Patient receives PMI sheet}
B: {Patient was hospitalized}
P ( A ) .20 ,
3.74
P ( A B ) .12 ,
P ( B | A)
P ( A B ) .12
.60
.20
P ( A)
Define the following events:
I: {Leak ignites immediately (jet fire)}
D: {Leak has delayed ignition (flash fire)}
From the problem, P ( I ) .01 and P ( D | I c ) .01
The probability of a jet fire or a flash fire P( I D) P( I ) P( D) P( I D)
P ( I ) P ( D | I c ) P ( I c ) P ( I D ) .01 .01(1 .01) 0 .01 .0099 .0199
Copyright © 2014 Pearson Education, Inc.
124
Chapter 3
A tree diagram of this problem is:
I
I
.01
D(.01)
IcD
.99(.01)=.0099
.01
.99
Ic
Dc
(.99)
3.75
a.
IcDc .99(.99)=.9801
If the coin is balanced, then P ( H ) .5 and P (T ) .5 on any trial. Also, we can assume
that the results of any coin toss is independent of any other. Thus,
P( H H H H H H H H H H )
P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H ) P( H )
.5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5) .510 .0009766
P( H H T T H T T H H H )
P( H ) P( H ) P(T ) P(T ) P( H ) P(T ) P(T ) P( H ) P( H ) P( H )
.5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5) .510 .0009766
P(T T T T T T T T T T )
P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T ) P(T )
.5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5) .510 .0009766
b.
Define the following events:
A: {10 coin tosses result in all heads or all tails}
B: {10 coin tosses result in mix of heads and tails}
P( A) P( H H H H H H H H H H )
P(T T T T T T T T T T )
.0009766 .0009766 .0019532
c.
d.
3.76
P ( B ) 1 P ( A) 1 .0019532 .9980468
From the above probabilities, the chances that either all heads or all tails occurred is extremely rare.
Thus, if one of these sequences really occurred, it is most likely sequence #2.
Define the following events:
A: {Algorithm predicts defects}
B: {Module has defects}
C: {Algorithm is correct}
Copyright © 2014 Pearson Education, Inc.
Probability
a.
Accuracy P (C ) P ( A B ) P ( Ac B c )
b.
Detection rate P ( A | B )
d
bd
c.
False alarm P ( A | B c )
c
ac
d.
Precision P ( B | A)
e.
From the SWDEFECTS file the table is:
d
a
ad
abcd abcd abcd
d
cd
Module has Defects
Algorithm
Predicts
Defects
False
True
No
400
29
Yes
49
20
Accuracy P (C ) P ( A B ) P ( Ac B c )
20 400
420
d
a
d a
.843
a b c d a b c d a b c d 400 29 49 20 498
The probability that the algorithm is correct is .843.
Detection rate P ( A | B )
20
20
d
.408
b d 29 20 49
The probability that the algorithm predicts a defect given the module is actually defective is .408.
False alarm P ( A | B c )
49
49
c
.109
a c 400 49 449
The probability that the algorithm predicts a defect given the module is not defective is .109.
Precision P ( B | A)
20
20
d
.290
c d 49 20 69
The probability that the module is defective given the algorithm predicted a defect is .290.
3.77
a.
P ( B1 A) P ( A | B1 ) P ( B1 ) .3(.75) .225
b.
P ( B2 A) P ( A | B2 ) P ( B2 ) .5(.25) .125
c.
P ( A) P ( B1 A) P ( B2 A) .225 .125 .35
d.
P ( B1 | A)
P ( B1 A) .225
.643
.35
P ( A)
e.
P ( B2 | A)
P ( B2 A) .125
.357
.35
P ( A)
Copyright © 2014 Pearson Education, Inc.
125
126
3.78
Chapter 3
First, we find the following probabilities:
P ( A B1 ) P ( A | B1 ) P ( B1 ) .4(.2) .08
P ( A B2 ) P ( A | B2 ) P ( B2 ) .25(.15) .0375
P ( A B3 ) P ( A | B3 ) P ( B3 ) .6(.65) .39
P ( A) P ( A B1 ) P ( A B2 ) P ( A B3 ) .08 .0375 .39 .5075
3.79
a.
P ( B1 | A)
P ( A B1 )
.08
.158
.5075
P ( A)
b.
P ( B2 | A)
P ( A B2 ) .0375
.074
.5075
P ( A)
c.
P ( B3 | A)
P ( A B3 )
.39
.768
P ( A)
.5075
If A is independent of B1, B2, and B3, then P ( A | B1 ) P ( A) .4 .
Then P ( B1 | A)
3.80
P ( A | B1 ) P ( B1 ) .4(.2)
.2
.4
P ( A)
From the information given, P( D) 1 / 80 , P ( D c ) 79 / 80 , P ( N | D ) 1 / 2 , P ( N c | D ) 1 / 2 ,
P ( N | D c ) 1 , and P ( N c | D c ) 0 . Using Bayes’ Rule
P( D N )
P( N | D) P ( D)
P( N )
P ( N | D) P ( D) P ( N | D c ) P( D c )
1 1
1
1
1
2
80
160
160
.0063
1 1
79
1
79
1 158 159
1
2 80
80 160 80 160 160
P( D | N )
3.81
Define the following events:
E: {Expert makes the correct decision}
N: {Novice makes the correct decision}
M: {Matched condition}
E: {Similar distracter condition}
E: {Non-similar distracter condition}
a.
P ( E c | M ) 1 .9212 .0788
b.
P ( N c | M ) 1 .7455 .2545
c.
Since P ( N c | M ) .2545 P ( E c | M ) .0788 , it is more likely that the participant is a Novice.
Copyright © 2014 Pearson Education, Inc.
Probability
3.82
P ( E1 error )
P (error )
P (error | E1 ) P ( E1 )
P (error | E1 ) P ( E1 ) P (error | E2 ) P ( E2 ) P (error | E3 ) P ( E3 )
P ( E1 | error )
a.
.01(.30)
.003
.003
.158
.01(.30) .03(.20) .02(.50) .003 .006 .01 .019
P ( E2 error )
P (error )
P (error | E2 ) P ( E2 )
P (error | E1 ) P ( E1 ) P(error | E2 ) P ( E2 ) P (error | E3 ) P ( E3 )
P ( E2 | error )
b.
.03(.20)
.006
.006
.316
.01(.30) .03(.20) .02(.50) .003 .006 .01 .019
P ( E3 error )
P (error )
P (error | E3 ) P ( E3 )
P (error | E1 ) P ( E1 ) P (error | E2 ) P ( E2 ) P(error | E3 ) P ( E3 )
P ( E3 | error )
c.
3.83
.02(.50)
.01
.01
.526
.01(.30) .03(.20) .02(.50) .003 .006 .01 .019
d.
If there was a serious error, the probability that the error was made by engineer 3 is .526.
This
probability is higher than for any of the other engineers. Thus engineer #3 is most likely responsible
for the error.
a.
Converting the percentages to probabilities,
P (275 300) .52 , P (305 325) .39 , and P (330 350) .09 .
b.
Using Bayes Theorem,
P(275 300 CC )
P(CC )
P(CC | 275 300) P(275 300)
P(CC | 275 300) P(275 300) P(CC | 305 325) P(305 325) P(CC | 330 350) P(330 350)
P(275 300 | CC )
3.84
127
.775(.52)
.403
.403
.516
.775(.52) .77(.39) .86(.09) .403 .3003 .0774 .7807
Define the following events:
U: {Athlete uses testosterone}
P: {Test is positive}
a.
Sensitivity is P ( P | U )
50
.5
100
Copyright © 2014 Pearson Education, Inc.
128
Chapter 3
9
1 .01 .99
900
b.
Specificity is P ( P c | U c ) 1
c.
First, we need to find the probability that an athlete is a user: P (U ) 100 / 1000 .1 .
Next, we need to find the probability of a positive test:
P ( P ) P ( P | U ) P (U ) P ( P | U c ) P (U c ) .5(.1) .01(.9) .05 .009 .059
Positive predictive value is P (U | P )
3.85
P (U P ) P ( P | U ) P (U ) .5(.1)
.847
P( P)
P( P)
.059
Define the following events:
S: {Shale}
D: {Dolomite }
G: {Gamma ray reading > 60 }
From the exercise: P ( D )
476
295
34
280
.617 , P ( S )
.383 , P (G | D )
.071 , and P (G | S )
.949 .
771
771
476
295
P ( D G ) P (G | D ) P ( D ) .071(.617) .0438 and
P (G ) P (G | D ) P ( D ) P (G | S ) P ( S ) .071(.617) .949(.383) .0438 .3635 .4073 .
P ( D G ) .0438
.1075 . Since this probability is so small, we would suggest that the
.4073
P (G )
area should not be mined.
Thus, P ( D | G )
3.86
Define the following events:
D: {Defect in steel casting}
H: {NDE detects ‘Hit” or defect in steel casting}
From the problem, P( H | D) .97 , P ( H | D c ) .005 , and P ( D ) .01 .
P ( H ) P ( H | D ) P ( D ) P ( H | D c ) P ( D c ) .97(.01) .005(.99) .0097 .00495 .01465
P( D | H )
3.87
P ( D H ) P ( H | D ) P ( D ) .97(.01) .0097
.6621
.01465 .01465
P( H )
P( H )
Define the following event:
D: {Chip is defective}
From the Exercise, P ( S1 ) .15 , P ( S 2 ) .05 , P ( S 3 ) .10 , P ( S 4 ) .20 , P ( S 5 ) .12 , P ( S 6 ) .20 , and
P ( S 7 ) .18 . Also, P ( D | S1 ) .001 , P ( D | S 2 ) .0003 , P ( D | S 3 ) .0007 , P ( D | S 4 ) .006 ,
P ( D | S 5 ) .0002 , P ( D | S 6 ) .0002 , and P ( D | S 7 ) .001 .
Copyright © 2014 Pearson Education, Inc.
Probability
a.
P ( S1 | D )
129
We must find the probability of each supplier given a defective chip.
P ( S1 D )
P( D)
P ( D | S1 ) P ( S1 )
P ( D | S1 ) P ( S1 ) P ( D | S 2 ) P ( S 2 ) P ( D | S3 ) P ( S3 ) P ( D | S 4 ) P ( S 4 ) P ( D | S5 ) P ( S5 ) P ( D | S6 ) P( S6 ) P( D | S7 ) P( S7 )
.001(.15)
.001(.15) .0003(.05) .0007(.10) .006(.20) .0002(.12) .0002(.02) .001(.18)
.00015
.00015
.0893
.00015 .000015 .00007 .0012 .000024 .00004 .00018 .001679
P( S2 | D)
P ( S 2 D ) P ( D | S 2 ) P ( S 2 ) .0003(.05) .000015
.0089
.001679
.001679
P( D)
P( D)
P ( S3 | D )
P ( S3 D ) P ( D | S3 ) P ( S3 ) .0007(.10) .00007
.0417
P( D)
P( D)
.001679
.001679
P(S4 | D)
P ( S 4 D ) P ( D | S 4 ) P ( S 4 ) .006(.20)
.0012
.7147
.001679 .001679
P( D)
P( D)
P ( S5 | D )
P ( S5 D ) P ( D | S5 ) P ( S5 ) .0002(.12) .000024
.0143
P( D)
P( D)
.001679
.001679
P ( S6 | D )
P ( S6 D ) P ( D | S6 ) P ( S6 ) .0002(.20) .00004
.0238
P( D)
P( D)
.001679
.001679
P ( S7 | D )
P ( S7 D ) P ( D | S7 ) P ( S7 ) .001(.18) .00018
.1072
.001679 .001679
P( D)
P( D)
Of these probabilities, .7147 is the largest. This implies that if a failure is observed, supplier number
4 was most likely responsible.
b. If the seven suppliers all produce defective chips at the same rate of .0005, then P ( D | S i ) .0005 for
all i = 1, 2, 3, … 7 and P ( D ) .0005 .
For any supplier i, P ( S i D ) P ( D | S i ) P ( S i ) .0005 P ( S i ) and
P ( Si | D )
P ( Si D ) P ( D | Si ) P ( Si ) .0005 P ( Si )
P ( Si )
.0005
.0005
P( D)
Thus, if a defective is observed, then it most likely came from the supplier with the largest proportion
of sales (probability). In this case, the most likely supplier would be either supplier 4 or supplier 6.
Both of these have probabilities of .20.
Copyright © 2014 Pearson Education, Inc.
130
3.88
Chapter 3
Define the following events:
A: {Alarm A sounds alarm}
B: {Alarm B sounds alarm}
I: {Intruder}
From the problem:
P ( A | I ) .9 , P ( B | I ) .95 , P ( A | I c ) .2 , P ( B | I c ) .1 , and P ( I ) .4
Since the two systems are operating independently of each other,
P ( A B | I ) P ( A | I ) P ( B | I ) .9(.95) .855
P ( A B I ) P ( A B | I ) P ( I ) .855(.4) .342
P ( A B | I c ) P ( A | I c ) P ( B | I c ) .2(.1) .02
P ( A B I c ) P ( A B | I c ) P ( I c ) .02(.6) .012
Thus, P ( A B ) P ( A B I ) P ( A B I c ) .342 .012 .354
Finally, P ( I | A B )
3.89
a.
b.
P ( A B I ) .342
.966
P( A B)
.354
P (T | E )
1 , then P (T | E ) P (T c | E ) . Thus, the probability of more than two bullets given the
c
P (T | E )
evidence is greater than the probability of two bullets given the evidence. This supports the theory of
more than two bullets were used in the assassination of JFK.
If
Using Bayes Theorem,
P (T | E )
P (T ) P ( E | T )
P (T c ) P ( E | T c )
.
and P (T c | E )
c
c
P (T ) P ( E | T ) P (T ) P ( E | T )
P (T ) P ( E | T ) P (T c ) P ( E | T c )
P(T ) P( E | T )
P(T | E )
P (T ) P( E | T )
P(T ) P( E | T ) P(T c ) P( E | T c )
.
Thus,
c
P(T | E )
P(T c ) P( E | T c )
P (T c ) P( E | T c )
P(T ) P( E | T ) P(T c ) P( E | T c )
3.90
a.
If the Dow Jones Industrial Average increases, a large New York bank would tend to decrease the
prime interest rate. Therefore, the two events are not mutually exclusive since they could occur
simultaneously.
b.
The next sale by a PC retailer could not be both a notebook and a desktop computer. Since the two
events cannot occur simultaneously, the events are mutually exclusive.
c.
Since both events cannot occur simultaneously, the events are mutually exclusive.
Copyright © 2014 Pearson Education, Inc.
Probability
3.91
a.
131
The two probability rules for a sample space are that the probability for any sample point is between
0 and 1 and that the sum of the probabilities of all the sample points is 1.
For this Exercise, all the probabilities of the sample points are between 0 and 1 and
P ( S ) P ( S ) P ( S ) P ( S ) P ( S ) .2 .1 .3 .4 1.0
4
i 1
b.
1
i
2
3
4
P ( A) P ( S1 ) P ( S 4 ) .2 .4 .6
3.92
P ( A B ) P ( A) P ( B ) P ( A B ) .7 .5 .4 .8
3.93
a.
If events A and B are mutually exclusive, then P ( A B ) 0 .
P( A | B)
3.94
P( A B) 0
0
.3
P( B)
b.
No. If events A and B are independent, then P ( A | B ) P ( A) . However, from the Exercise we
know P ( A) .2 and from part a, we know P ( A | B ) 0 . Thus, events A and B are not independent.
a.
Because events A and B are independent, we have:
P ( A B ) P ( A) P ( B ) .3(.1) .03
Thus, P ( A B ) 0 , and the two events cannot be mutually exclusive.
3.95
P ( A B ) .03
.3
.1
P( B)
P ( B | A)
P ( A B ) .03
.1
.3
P ( A)
b.
P( A | B)
c.
P ( A B ) P ( A) P ( B ) P ( A B ) .3 .1 .03 .37
P ( A B ) .4 , P ( A | B ) .8
Since P( A | B )
.8
P( A B)
, substitute the given probabilities into the formula and solve for P(B).
P( B)
.4
.4
P ( B ) .5
P( B)
.8
3.96
The number of ways to select 5 things from 50 is a combination of 50 things taken 5 at a time or
50
50!
50!
50 49 48 47 46 45!
2,118, 760 .
5
5!(50
5)!
5!45!
5 4 3 2 1 45!
3.97
a.
P( A B) 0
P ( B C ) P (2) .2
P ( A C ) P (1) P (2) P (3) P (5) P (6) .3 .2 .1 .1 .2 .9
P ( A B C ) P (1) P (2) P (3) P (4) P (5) P (6) .3 .2 .1 .1 .1 .2 1
Copyright © 2014 Pearson Education, Inc.
132
Chapter 3
P ( B c ) P (1) P (3) P (5) P (6) .3 .1 .1 .2 .7
P ( Ac B ) P (2) P (4) .2 .1 .3
P( B | C )
.2
.2
P( B C )
P (2)
.4
P (C )
P (2) P (5) P (6) .2 .1 .2 .5
P ( B | A)
0
P ( B A)
0
P ( A)
P ( A)
b.
Since P ( A B ) 0 , and P ( A) P ( B ) 0 , these two would not be equal, implying A and B are not
independent. However, A and B are mutually exclusive, since P ( A B ) 0 .
c.
P ( B ) P (2) P (4) .2 .1 .3 . But P ( B | C ) , calculated above, is .4. Since these are not equal, B
and C are not independent. Since P ( B C ) .2 , B and C are not mutually exclusive.
3.98
3.99
a.
6! 6 5 4 3 2 1 720
b.
10
10!
10 9 8 1
10
9
8 7 1 1
9!(10
9)!
9
c.
10
10!
10 9 8 1
10
1 1!(10 1)! 1 9 8 1
d.
6
6!
6 5 4 3 2 1
20
3
3!(6
3)!
3 2 1 3 2 1
e.
0! 1
Define the following events:
E: {Industrial accident caused by faulty Engineering & Design}
P: {Industrial accident caused by faulty Procedures & Practices}
M: {Industrial accident caused by faulty Management & Oversight}
T: {Industrial accident caused by faulty Training & Communication}
a.
The sample points for this problem are: E, P, M, and T. Reasonable probabilities are:
P ( E ) 27 / 83 .3253 , P ( P ) 24 / 83 .2892 , P ( M ) 22 / 83 .2651 , and P (T ) 10 / 83 .1205 .
b.
P ( E ) 27 / 83 .3253 . Approximately 32.53% of all industrial accidents are caused by faulty
Engineering and Design.
c.
P(Industrial accident caused by something other than procedures & practices)
1 P ( P c ) 1 .2892 .7108 . Approximately 71.08% of all industrial accidents are caused by
something other than faulty procedures & practices.
Copyright © 2014 Pearson Education, Inc.
Probability
3.100
a.
133
Define the following events:
J: {Raise based on job performance}
C: {Raise based on cost of living}
U: {Unsure.}
The 3 sample points are: J, C, and U
3.101
b.
We will base the probabilities on the proportions of the 10,000 U.S. workers surveyed who responded
in each category. Thus, P ( J ) .35 , P (C ) .50 , and P (U ) .15
c.
P(Raise based on either job performance or cost of living) P ( J ) P (C ) .35 .50 .85
Define the event:
B: {Small business owned by non-Hispanic white female}
From the problem, P ( B ) .27
The probability that a small business owned by a non-Hispanic white is male-owned is
P ( B c ) 1 P ( B ) 1 .27 .73 .
3.102
Define the following events:
C: {Public school building has inadequate plumbing}
D: {Public school has plans for repairing building}
From the problem, we know P (C ) .25 and P ( D | C ) .38 .
P (C D ) P ( D | C ) P (C ) .38(.25) .095
3.103
a.
This statement is false. All probabilities are between 0 and 1 inclusive. One cannot have a
probability of 4.
b.
If we assume that the probabilities are the same as the percents (changed to proportions), then this is a
true statement.
P (4 or 5) P (4) P (5) .6020 .1837 .7857
3.104
c.
This statement is true. There were no observations with one star. Thus, P (1) 0 .
d.
This statement is false. P (2) .0408 and P (5) .1837 . P (5) P (2) .
Define the following events:
S: {cause of fatal crash is speeding}
C: {cause of fatal crash is missing a curve}
From the problem, we know P(S) = .3 and P ( S C ) .12 .
P (C | S )
P (C S ) .12
.4
.3
P(S )
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134
3.105
3.106
Chapter 3
a.
B C
b.
Ac
c.
CB
d.
A Cc
a.
The 5 sample points are:
Total population, Agricultural change, Presence of industry, Growth, and Population concentration.
b.
The probabilities are best estimated with the sample proportions. Thus,
P(Total population) = .18
P(Agricultural change) = .05
P(Presence of industry) = .27
P(Growth) = .05
P(Population concentration) = .45
c.
Define the following event:
A: {Factor specified is population-related}
P(A) = P(Total population) + P(Growth) + P(Population concentration) .18 .05 .45 .68 .
3.107
Define the following events:
G: {regularly use the golf course}
T: {regularly use the tennis courts}
Given: P (G ) .7 and P (T ) .5
The event "uses neither facility" can be written as G c T c or (G T ) c . We are given
P (G c T c ) P[(G T ) c ] .05 . The complement of the event "uses neither facility" is the event "uses at
least one of the two facilities" which can be written as G T .
P (G T ) 1 P (G T ) c 1 .05 .95
From the additive rule, P (G T ) P (G ) P (T ) P (G T ) .95 .7 .5 P (G T ) P (G T ) .25
Copyright © 2014 Pearson Education, Inc.
Probability
a.
The Venn Diagram is:
G
.45
T
.25
.25
.05
3.108
b.
P (G T ) .95 from above.
c.
P (G T ) .25 from above.
d.
P (G | T )
S
P (G T ) .25
.5
P (T )
.5
Define the following events:
A: {electrical switch monitors quality of power}
B: {electrical switch not wired properly}
From the problem, P ( A ) .90 and P ( B | A) .90 .
P ( A B c ) P ( B c | A) P ( A) (1 .90)(.90) .09 .
3.109
a.
P ( A)
1, 465
.684
2,143
b.
P( B)
265
.124
2,143
c.
No. There is one sample point that they have in common: Plaintiff trial win – reversed, Jury
d.
P ( Ac ) 1 P ( A) 1 .684 .316
e.
P( A B)
194 71 429 111 731 1, 536
.717
2,143
2,143
f.
P( A B)
194
.091
2,143
Copyright © 2014 Pearson Education, Inc.
135
136
3.110
Chapter 3
Since there are 11 individuals who are willing to serve on the panel, the number of different panels of 5
experts is a combination of 11 things taken 5 at a time or
11 11! 11 10 9 8 7 6 5 4 3 2 1
462
5 5!6! (5 4 3 2 1)(6 5 4 3 2 1)
3.111
Define the following events:
A: {The watch is accurate}
N: {The watch is not accurate}
Assuming the manufacturer's claim is correct,
P ( N ) .05 and P ( A) 1 P ( N ) 1 .05 .95
The sample space for the purchase of four of the manufacturer's watches is listed below.
(A, A, A, A) (N, A, A, A) (A, N, N, A) (N, A, N, N)
(A, A, A, N) (A, A, N, N) (N, A, N, A) (N, N, A, N)
(A, A, N, A) (A, N, A, N) (N, N, A, A) (N, N, N, A)
(A, N, A, A) (N, A, A, N) (A, N, N, N) (N, N, N, N)
a.
All four watches not being accurate as claimed is the sample point (N, N, N, N).
Assuming the watches purchased operate independently and the manufacturer's claim is correct,
P N , N , N , N P N P N P N P N .054 .00000625
b.
The sample points in the sample space that consist of exactly two watches failing to meet the claim
are listed below.
(A, A, N, N) (N, A, A, N)
(A, N, A, N) (N, A, N, A)
(A, N, N, A) (N, N, A, A)
The probability that exactly two of the four watches fail to meet the claim is the sum of the
probabilities of these six sample points.
Assuming the watches purchased operate independently and the manufacturer's claim is correct,
P ( A, A, N , N ) P ( A) P ( A) P ( N ) P ( N ) .95(.95)(.05)(.05) .00225625
All six of the sample points will have the same probability. Therefore, the probability that exactly
two of the four watches fail to meet the claim when the manufacturer's claim is correct is
6(.00225625) .0135
Copyright © 2014 Pearson Education, Inc.
Probability
c.
137
The sample points in the sample space that consist of three of the four watches failing to meet the
claim are listed below.
(A, N, N, N) (N, N, A, N)
(N, A, N, N) (N, N, N, A)
The probability that three of the four watches fail to meet the claim is the sum of the probabilities of
the four sample points.
Assuming the watches purchased operate independently and the manufacturer's claim is correct,
P ( A, N , N , N ) P ( A) P ( N ) P ( N ) P ( N ) .95(.05)(.05)(.05) .00011875
All four of the sample points will have the same probability. Therefore, the probability that three of
the four watches fail to meet the claim when the manufacturer's claim is correct is
4(.00011875) .000475
If this event occurred, we would tend to doubt the validity of the manufacturer's claim since its
probability of occurring is so small.
d.
All four watches tested failing to meet the claim is the sample point (N, N, N, N).
Assuming the watches purchased operate independently and the manufacturer's claim is correct,
P ( N , N , N , N ) P ( N ) P ( N ) P ( N ) P ( N ) .05(.05)(.05)(.05) .00000625
Since the probability of observing this event is so small if the claim is true, we have strong evidence
against the validity of the claim. However, we do not have conclusive proof that the claim is false.
There is still a chance the event can occur (with probability .00000625) although it is extremely
small.
3.112
The possible ways of ranking the blades are:
GSW
SGW
WGS
GWS
SWG
WSG
If the consumer had no preference but still ranked the blades, then the 6 possibilities are equally likely.
Therefore, each of the 6 possibilities has a probability of 1/6 of occurring.
1 1 2 1
6 6 6 3
a.
P(Ranks G first) P (GSW ) P (GWS )
b.
P(Ranks G last) P ( SWG ) P (WSG )
c.
P(ranks G last and W second) P ( SWG )
d.
P (WGS )
1 1 2 1
6 6 6 3
1
6
1
6
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3.113
Chapter 3
Define the following events:
A: {Never smoked cigars}
B: {Former cigar smoker}
C: {Current cigar smoker}
D: {Died from cancer}
E: {Did not die from cancer}
3.114
3.115
a.
P ( D | A)
782
782
P ( D A)
137, 243
.006
121,529
121,529
P ( A)
137, 243
b.
P( D | B)
91
91
P( D B)
137, 243
.012
7,848
7,848
P( B)
137, 243
c.
P( D | C )
141
141
P( D C )
137, 243
.018
7,866
7,866
P (C )
137, 243
a.
Consecutive tosses of a coin are independent events since what occurs one time would not affect the
next outcome.
b.
If the individuals are randomly selected, then what one individual says should not affect what the
next person says. They are independent events.
c.
The results in two consecutive at-bats are probably not independent. The player may have faced the
same pitcher both times which may affect the outcome.
d.
The amount of gain and loss for two different stocks bought and sold on the same day are probably
not independent. The market might be way up or down on a certain day so that all stocks are
affected.
e.
The amount of gain or loss for two different stocks that are bought and sold in different time periods
are independent. What happens to one stock should not affect what happens to the other.
f.
The prices bid by two different development firms in response to the same building construction
proposal would probably not be independent. The same variables would be present for both firms to
consider in their bids (materials, labor, etc.).
Define the following events:
A: {Wheelchair user had an injurious fall}
B: {Wheelchair user had all five features installed in the home}
C: {Wheelchair user had no falls}
D: {Wheelchair user had none of the features installed in the home}
a.
P ( A)
48
.157
306
b.
P(B)
9
.029
306
Copyright © 2014 Pearson Education, Inc.
Probability
139
89
.291
306
c.
P (C D )
d.
P( A | B)
2
P( A B )
2
306 .222
9
P( B)
9
306
e.
P( A | D)
20
P ( A D)
306 20 .183
109
P ( D)
109
306
3.116 Define the following events:
A1: {Paraguay is assigned to Group A}
A2: {Ecuador is assigned to Group A}
B1: {Paraguay is assigned to Group B}
B2: {Sweden or top team in pot 3 is assigned to Group B}
D1: {Paraguay is assigned to Group D}
D2: {Ecuador is assigned to Group D}
If the teams are drawn at random from each pot, the probability that any team is assigned to a group is 1/8.
a.
P ( A1 ) 1 / 8 .125
b.
P ( A1 A2 ) P ( A1 ) P ( A2 ) P ( A1 A2 ) 1 / 8 1 / 8 0 2 / 8 .25
c.
P ( B1 B2 ) P ( B1 ) P ( B2 ) (1 / 8)(2 / 8) 2 / 64 .03125
d.
We can look at this probability by looking at how the slots can be filled. We will just look at how the
teams from pot 2 can be put into Groups A, B, C, and D. The order of filling these really does not
matter, so we will look at the ways to fill Group C, then Group D, then Group A, then Group B.
First, we will find the total number of ways we can fill these 4 slots or Groups where Group C cannot
have Paraguay or Ecuador. Since Group C cannot have Paraguay or Ecuador, then there are only 6
ways to fill Group C. There would then be 7 ways to fill Group D, 6 ways to fill Group A and 5 ways
to fill Group B. The total ways to fill these 4 Groups without having Paraguay or Ecuador in Group C
is 6(7)(6)(5) = 1,260.
Now, we will find the number of ways we can fill these 4 Groups where Group C cannot have
Paraguay or Ecuador and Group D does have either Paraguay or Ecuador. There will be 6 ways to fill
Group C, 2 ways to fill Group D, 6 ways to fill Group A, and 5 ways to fill Group B. The total ways to
fill these 4 Groups without having Paraguay or Ecuador in Group C and having either Paraguay or
Ecuador in Group D is 6(2)(6)(5) = 360.
Thus, the probability that Group C does not have either Paraguay or Ecuador and Group D does have
either Paraguay or Ecuador is 360 / 1, 260 2 / 7 .286 .
Finally, the probability that Group D does not have either Paraguay or Ecuador is 1 .286 .714 .
Copyright © 2014 Pearson Education, Inc.
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3.117
Chapter 3
Define the following events:
S1: {Salesman makes sale on the first visit}
S2: {Salesman makes a sale on the second visit}
P ( S1 ) .4
P ( S 2 | S1c ) .65
The sample points of the experiment are:
S1 S 2c , S1c S 2 , S1c S2c
The probability the salesman will make a sale is:
P ( S1 S 2c ) P ( S1c S 2 ) P ( S1 ) P ( S 2 | S1c ) P ( S1c ) .4 .65(1 .4) .4 .39 .79
3.118
Define the following events:
S: {System shuts down}
F1: {Hardware failure}
F2: {Software failure}
F3: {Power failure}
From the Exercise, we know:
P ( F1 ) .01 , P ( F2 ) .05 , and P ( F3 ) .02 . Also, P ( S | F1 ) .73 , P ( S | F2 ) .12 , and P ( S | F3 ) .88 .
The probability that the current shutdown is due to a hardware failure is:
P( F1 | S )
P( F1 S )
P( S | F1 ) P( F1 )
P( S )
P( S | F1 ) P( F1 ) P( S | F2 ) P( F2 ) P( S | F3 ) P( F3 )
.73(.01)
.0073
.0073
.2362
.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309
The probability that the current shutdown is due to a software failure is:
P( F2 | S )
P( F2 S )
P( S | F2 ) P( F2 )
P( S )
P( S | F1 ) P( F1 ) P( S | F2 ) P( F2 ) P( S | F3 ) P( F3 )
.12(.05)
.006
.006
.1942
.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309
The probability that the current shutdown is due to a power failure is:
P( F3 | S )
P( F3 S )
P( S | F3 ) P( F3 )
P( S )
P( S | F1 ) P( F1 ) P( S | F2 ) P( F2 ) P( S | F3 ) P( F3 )
.88(.02)
.0176
.0176
.5696
.73(.01) .12(.05) .88(.02) .0073 .006 .0176 .0309
Copyright © 2014 Pearson Education, Inc.
Probability
3.119
a.
Suppose we let the four positions in a sample point represent in order (1) Raise a broad mix of crops,
(2) Raise livestock, (3) Use chemicals sparingly, and (4) Use techniques for regenerating the soil,
such as crop rotation. A farmer is either likely (L) to engage in an activity or unlikely (U). The
possible classifications are:
LLLL LLLU LLUL LULL ULLL LLUU LULU LUUL ULLU ULUL UULL
LUUU ULUU UULU UUUL UUUU
b.
Since there are 16 classifications or sample points and all are equally likely, then each has a
probability of 1/16.
P (UUUU )
c.
1
16
The probability that a farmer will be classified as likely on at least three criteria is
1 5
.
P ( LLLL ) P ( LLLU ) P ( LLUL ) P ( LULL ) P (ULLL ) 5
16 16
3.120
Define the following events:
C: {Committee judges joint acceptable}
I: {Inspector judges joint acceptable}
The sample points of this experiment are:
C I , C I c , C c I , Cc I c
a.
The probability the inspector judges the joint to be acceptable is:
P ( I ) P (C I ) P (C c I )
101 23 124
.810
153 153 153
The probability the committee judges the joint to be acceptable is:
P (C ) P (C I ) P (C I c )
b.
101 10 111
.725
153 153 153
The probability that both the committee and the inspector judge the joint to be acceptable is:
P (C I )
101
.660
153
The probability that neither judge the joint to be acceptable is: P (C c I c )
c.
141
The probability the inspector and committee disagree is:
P (C I c ) P (C c I )
10
23 33
.216
153 153 153
Copyright © 2014 Pearson Education, Inc.
19
.124
153
142
Chapter 3
The probability the inspector and committee agree is:
P (C I ) P (C c I c )
3.121
101 19 120
.784
153 153 153
Define the following events:
O1:
O2:
O3:
A:
{Component #1 in System A operates properly}
{Component #2 in System A operates properly}
{Component #3 in System A operates properly}
{System A works properly}
P (O1 ) 1 P O1c 1 .12 .88
a.
P (O2 ) 1 P O2c 1 .09 .91
P (O3 ) 1 P O3c 1 .11 .89
P ( A) P (O1 O2 O3 ) P (O1 ) P (O2 ) P (O3 ) .88(.91)(.89) .7127
(since the three components operate independently)
b.
P ( Ac ) 1 P ( A) 1 .7127 .2873
(see part a)
c.
Define the following events:
C1: {Component 1 in System B works properly}
C2: {Component 2 in System B works properly}
D3: {Component 3 in System B works properly}
D4: {Component 4 in System B works properly}
C: {Subsystem C works properly}
D: {Subsystem D works properly}
The probability a component fails is .1, so the probability a component works properly is1 .1 .9 .
Subsystem C works properly if both components 1 and 2 work properly.
P (C ) P (C1 C 2 ) P (C1 ) P (C 2 ) .9(.9) .81
(since the components operate independently)
Similarly, P ( D ) P ( D1 D2 ) P ( D1 ) P ( D2 ) .9(.9) .81
The system operates properly if either subsystem C or D operates properly.
The probability that System B operates properly is:
P (C D ) P (C ) P ( D ) P (C D ) P (C ) P ( D ) P (C ) P ( D ) .81 .81 .81(.81) .9639
d.
The probability exactly one subsystem fails in System B is:
P (C D c ) P (C c D ) P (C ) P ( D c ) P (C c ) P ( D )
.81(1 .81) (1 .81)(.81) .1539 .1539 .3078
Copyright © 2014 Pearson Education, Inc.
Probability
e.
143
The probability that System B fails is the probability that both subsystems fail:
P (C c D c ) P (C c ) P ( D c ) (1 .81)(1 .81) .0361
f.
The system operates correctly 99% of the time means it fails 1% of the time. The probability one
subsystem fails is .19. The probability n subsystems fail is .19n. Thus, we must find n such that
(.19) n .01 n 3
3.122 Define the following events:
R: {Successful regime change is achieved}
M: {Mission is extended to support a weak Iraq government)
a.
The probability that a successful regime change is not achieved is
P ( R c ) 1 P ( R ) 1 .7 .3
b.
P ( R | M ) .26
c.
Given that P ( M ) .55 , the probability that the mission is extended and results in a successful regime
change is P ( M R ) P ( R | M ) P ( M ) .26(.55) .143
3.123
The probability of a false positive is P ( A | B ) .
3.124
Define the following events:
A1: {Fuse made by line 1}
A2: {Fuse made by line 2}
D: {Fuse is defective}
From the Exercise, we know P ( D | A1 ) .06 and P ( D | A2 ) .025 . Also, P ( A1 ) P ( A2 ) .5 .
Two fuses are going to be selected and we need to find the probability that one of the two is defective. We
can get one defective fuse out of two by getting a defective on the first and non-defective on the second
( D D c ) or non-defective on the first and defective on the second ( D c D ) . The probability of getting
one defective out of two fuses given line 1 is:
P ( D D c | A1 ) P ( D c D | A1 ) P ( D | A1 ) P( D c | A1 ) P( D c | A1 ) P( D | A1 )
.06(1 .06) (1 .06)(.06) .06(.94) .94(.06) .1128 P(1 D | A1 )
The probability of getting one defective out of two fuses given line 2 is:
P ( D D c | A2 ) P( D c D | A2 ) P( D | A2 ) P( D c | A2 ) P( D c | A2 ) P( D | A2 )
.025(1 .025) (1 .025)(.025) .025(.975) .975(.025) .04875 P(1 D | A2 )
The probability of getting one defective out of two fuses is:
P (1 D ) P (1 D A1 ) P (1 D A2 ) P (1 D | A1 ) P ( A1 ) P (1 D | A2 ) P ( A2 )
.1128(.5) .04875(.5) .0564 .024375 .080775
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Chapter 3
Finally, we want to find:
P ( A1 | 1 D )
3.125
P (1 D A1 )
.0564
.6982
P (1 D)
.080775
Define the following events:
A: {Press is correctly adjusted}
B: {Press is incorrectly adjusted}
D: {part is defective}
From the exercise, P ( A ) .90 , P ( D | A) .05 , and. We also know that event B is the complement of
event A. Thus, P ( B ) 1 P ( A) 1 .90 .10 .
P( B | D)
3.126
P( B D)
P( D | B) P( B)
.50(.10)
.05
.05
.526
P( D)
P ( D | B ) P ( B ) P ( D | A) P( A) .50(.10) .05(.90) .05 .045 .095
There are a total of 6 6 = 36 outcomes when rolling 2 dice. If we let the first number in the pair represent
the outcome of die number 1 and the second number in the pair represent the outcome of die number 2,
then the possible outcomes are:
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
3,1
3,2
3,3
3,4
3,5
3,6
4,1
4,2
4,3
4,4
4,5
4,6
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
If both dice are fair, then each of these outcomes are equally like and have a probability of 1/36.
a.
To win on the first roll, a player must roll a 7 or 11. There are 6 ways to roll a 7 and 2 ways to roll an
11. Thus the probability of winning on the first roll is:
P (7 or 11)
b.
To lose on the first roll, a player must roll a 2 or 3. There is 1 way to roll a 2 and 2 ways to roll a 3.
Thus the probability of losing on the first roll is:
P (2 or 3)
c.
8
.2222
36
3
.0833
36
If a player rolls a 4 on the first roll, the game will end on the next roll if the player rolls 4 (player
wins) or if the player rolls a 7 (player loses). There are 3 ways to roll a 4 and 6 ways to roll a 7.
Thus, P (4or 7 on 2 nd roll)
36 9
.25 .
36
36
Copyright © 2014 Pearson Education, Inc.
Probability
3.127
145
Define the flowing events:
A: {Dealer draws a blackjack}
B: {Player draws a blackjack}
a.
For the dealer to draw a blackjack, he needs to draw an ace and a face card. There are
4
4!
4 3 2 1
4 ways to draw an ace and
1 1!(4 1)! 1 3 2 1
12
12!
12 11 10 1
12 ways to draw a face card (there are 12 face
1 1!(12 1)! 1 11 10 9 1
cards in the deck).
The total number of ways a dealer can draw a blackjack is 4 12 = 48.
The total number of ways a dealer can draw 2 cards is
52
52!
52 51 50 1
1326
2 2!(52 2)! 2 1 50 49 48 1
Thus, the probability that the dealer draws a blackjack is P ( A)
b.
48
.0362
1326
In order for the player to win with a blackjack, the player must draw a blackjack and the dealer does
not. Using our notation, this is the event B AC . We need to find the probability that the player
draws a blackjack P( B) and the probability that the dealer does not draw a blackjack given the
player does P ( Ac | B ) . Then, the probability that the player wins with a blackjack is P ( Ac | B ) P ( B ) .
The probability that the player draws a blackjack is the same as the probability that the dealer draws a
blackjack, which is P ( B ) .0362 .
There are 5 scenarios where the dealer will not draw a blackjack given the player does. First, the
dealer could draw an ace and not a face card. Next, the dealer could draw a face card and not an ace.
Third, the dealer could draw two cards that are not aces or face cards. Fourth, the dealer could draw
two aces, and finally, the dealer could draw two face cards.
The number of ways the dealer could draw an ace and not a face card given the player draws a
blackjack is
3 36
3!
36!
3 2 1 36 35 34 1
3(36) 108
1
1
2 1 1 35 34 33 1
1!(3
1)!
1!(36
1)!
1
(Note: Given the player has drawn blackjack, there are only 3 aces left and 36 non-face cards.)
The number of ways the dealer could draw a face card and not an ace given the player draws a
blackjack is
11 36
11!
36!
11 10 9 1 36 35 34 1
11(36) 396
1
1
10 9 8 1 1 35 34 33 1
1!(11
1)!
1!(36
1)!
1
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Chapter 3
The number of ways the dealer could draw neither a face card nor an ace given the player draws a
blackjack is
36
36!
36 35 34 1
630
2
2!(36
2)!
2
1 34 33 32 1
The number of ways the dealer could draw two aces given the player draws a blackjack is
3
3!
3 2 1
3
2
2!(3
2)!
2 1 1
The number of ways the dealer could draw two face cards given the player draws a blackjack is
11
11!
11 10 9 1
55
2
9 8 7 1
2!(11
2)!
2
The total number of ways the dealer can draw two cards given the player draws a blackjack is
50
50!
50 49 48 1
1225
2
48 47 46 1
2!(50
2)!
2
1
The probability that the dealer does not draw a blackjack given the player draws a blackjack is
P ( Ac | B )
108 396 630 3 55 1192
.9731
1225
1225
Finally, the probability that the player wins with a blackjack is
P ( B Ac ) P ( Ac | B ) P ( B ) .9731(.0362) .0352
3.128
a.
Define the following events:
W:
F:
{Player wins the game Go}
{Player plays first (black stones)}
P (W F ) 319 / 577 .553
b.
P (W F | CA) 34 / 34 1
P (W F | CB ) 69 / 79 .873
P (W F | CC ) 66 / 118 .559
P (W F | BA) 40 / 54 .741
P (W F | BB ) 52 / 95 .547
P (W F | BC ) 27 / 79 .342
P (W F | AA) 15 / 28 .536
P (W F | AB ) 11 / 51 .216
P (W F | AC ) 5 / 39 .128
Copyright © 2014 Pearson Education, Inc.
Probability
c.
147
There are three combinations where the player with the black stones (first) is ranked higher than the
player with the white stones: CA, CB, and BA.
P (W F | CA CB BA) (34 69 40) / (34 79 54) 143 / 167 .856
d.
There are three combinations where the players are of the same level: CC, BB, and AA.
P (W F | CC BB AA) (66 52 15) / (118 95 28) 133 / 241 .552
3.129
First, we will list all possible sample points for placing a car (C) and 2 goats (G) behind doors #1, #2, and
#3. If the first position corresponds to door #1, the second position corresponds to door #2, and the third
position corresponds to door #3, the sample space is:
(C G G) (G C G) (G G C)
Now, suppose you pick door #1. Initially, the probability that you will win the car is 1/3 – only one of the
sample points has a car behind door #1.
The host will now open a door behind which is a goat. If you pick door #1 in the first sample point
(C G G), the host will open either door #2 or door #3. Suppose he opens door #3 (it really does not matter).
If you pick door #1 in the second sample point (G C G), the host will open door #3. If you pick door #1 in
the third sample point (G G C), the host will open door #2. Now, the new sample space will be:
(C G) (G C) (G C)
where the first position corresponds to door #1 (the one you chose) and the second position
corresponds to the door that was not opened by the host.
Now, if you keep door #1, the probability that you win the car is 1/3. However, if you switch to the
remaining door, the probability that you win the car is now 2/3. Based on these probabilities, it is to your
advantage to switch doors.
The above could be repeated by selecting door #2 initially or door #3 initially. In either of these cases,
again, the probability of winning the car is 1/3 if you do not switch and 2/3 if you switch. Thus, Marilyn
was correct.
3.130
Suppose we define the following event:
E: {Error produced when dividing}
From the problem, we know that P ( E ) 1 / 9, 000, 000, 000
The probability of no error produced when dividing is
P ( E c ) 1 P ( E ) 1 1 / 9,000,000,000 8,999,999,999 / 9,000,000,000 .999999999 1.0000
Suppose we want to find the probability of no errors in 2 divisions (assuming each division is
independent):
P ( E c E c ) .999999999(.999999999) .999999999 1.0000
Thus, in general, the probability of no errors in k divisions would be:
c
c k
k
P (
Ec Ec
Ec E
) P ( E ) [8, 999, 999, 999 / 9, 000, 000, 000]
k
Copyright © 2014 Pearson Education, Inc.
148
Chapter 3
Suppose a user ran a program that performed 1 billion divisions. The probability of no errors in these 1
billion divisions would be:
P ( E c )1,000,000,000 [8,999,999,999 / 9,000,000,000]1,000,000,000 .8948
Thus, the probability of at least 1 error in 1 billion divisions would be
1 P ( E c )1,000,000,000 1 [8,999,999,999 / 9,000,000,000]1,000,000,000 1 .8948 .1052
Copyright © 2014 Pearson Education, Inc.
Chapter 4
Random Variables
and Probability Distributions
4.1
4.2
a.
The number of newspapers sold by New York Times each month can take on a countable number of
values. Thus, this is a discrete random variable.
b.
The amount of ink used in printing the Sunday edition of the New York Times can take on an infinite
number of different values. Thus, this is a continuous random variable.
c.
The actual number of ounces in a one gallon bottle of laundry detergent can take on an infinite number
of different values. Thus, this is a continuous random variable.
d.
The number of defective parts in a shipment of nuts and bolts can take on a countable number of values.
Thus, this is a discrete random variable.
e.
The number of people collecting unemployment insurance each month can take on a countable number
of values. Thus, this is a discrete random variable.
a.
The closing price of a particular stock on the New York Stock Exchange is discrete. It can take on only
a countable number of values.
b.
The number of shares of a particular stock that are traded on a particular day is discrete. It can take on
only a countable number of values.
c.
The quarterly earnings of a particular firm is discrete. It can take on only a countable number of values.
d.
The percentage change in yearly earnings between 2011 and 2012 for a particular firm is continuous. It
can take on any value in an interval.
e.
The number of new products introduced per year by a firm is discrete. It can take on only a countable
number of values.
f.
The time until a pharmaceutical company gains approval from the U.S. Food and Drug Administration
to market a new drug is continuous. It can take on any value in an interval of time.
4.3
Since there are only a fixed number of outcomes to the experiment, the random variable, x, the number of
stars in the rating, is discrete.
4.4
The number of customers, x, waiting in line can take on values 0, 1, 2, 3, … . Even though the list is never
ending, we call this list countable. Thus, the random variable is discrete.
4.5
The variable x, total compensation in 2011 (in $ millions), is reported in whole number dollars. Since there
are a countable number of possible outcomes, this variable is discrete.
4.6
A banker might be interested in the number of new accounts opened in a month, or the number of mortgages
it currently has, both of which are discrete random variables.
149
Copyright © 2014 Pearson Education, Inc.
150
Chapter 4
4.7
An economist might be interested in the percentage of the work force that is unemployed, or the current
inflation rate, both of which are continuous random variables.
4.8
The manager of a hotel might be concerned with the number of employees on duty at a specific time, or the
number of vacancies there are on a certain night.
4.9
The manager of a clothing store might be concerned with the number of employees on duty at a specific time
of day, or the number of articles of a particular type of clothing that are on hand.
4.10
A stockbroker might be interested in the length of time until the stock market is closed for the day.
4.11
a.
p (22) .25
b.
P( x 20 or x 24) P( x 20) P( x 24) .15 .20 .35
c.
P( x 23) P( x 20) P( x 21) P( x 22) P( x 23) .15 .10 .25 .30 .80
a.
The variable x can take on values 1, 3, 5, 7, and 9.
b.
The value of x that has the highest probability associated with it is 5. It has a probability of .4.
c.
Using MINITAB, the probability distribution of x as a graph is:
4.12
.4
p(x)
.3
.2
.1
0
1
3
4
5
x
6
7
8
9
d.
P( x 7) .2
e.
P( x 5) p(5) p(7) p(9) .4 .2 .1 .7
f.
P( x 2) p(3) p(5) p(7) p(9) .2 .4 .2 .1 .9
g.
4.13
2
E( x) xp( x) 1(.1) 3(.2) 5(.4) 7(.2) 9(.1) .1 .6 2.0 1.4 .9 5.0
p(x) 1 . Thus, p(2) p(3) p(5) p(8) p(10) 1
a.
We know
b.
P( x 2 or x 10) P( x 2) P( x 10) .15 .25 .40
c.
P( x 8) P( x 2) P( x 3) P( x 5) P( x 8) .15 .10 .25 .25 .75
p(5) 1 p(2) p(3) p(8) p(10) 1 .15 .10 .25 .25 .25
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.14
4.15
p( x) .1 .3 .3 .2 .9 1 .
a.
This is not a valid distribution because
b.
This is a valid distribution because 0 p( x) 1 for all values of x and
c.
This is not a valid distribution because p(4) .3 0 .
d.
The sum of the probabilities over all possible values of the random variable is
p( x) .15 .15 .45 .35 1.1 1 , so this is not a valid probability distribution.
a.
When a die is tossed, the number of spots observed on the upturned face can be 1, 2, 3, 4, 5, or 6. Since
the six sample points are equally likely, each one has a probability of 1/6.
p(x) .25 .5 .25 1 .
The probability distribution of x may be summarized in tabular form:
x
1
2
3
4
5
6
p(x)
1
6
1
6
1
6
1
6
1
6
1
6
The probability distribution of x may also be presented in graphical form:
p(x)
b.
1/6
0
1
2
3
4
5
6
x
4.16
151
a.
The sample points are (where H = head, T = tail):
x = # heads
b.
HHH HHT HTH THH HTT THT TTH TTT
3
2
2
2
1
1
1
0
If each event is equally likely, then P(sample point)
1 1
k 8
1
1 1 1 3
1 1 1 3
1
p(3) , p(2) , p (1) , and p(0)
8
8 8 8 8
8 8 8 8
8
Copyright © 2014 Pearson Education, Inc.
152
Chapter 4
c.
Using Minitab, the graph of p(x) is:
.500
p(x)
.375
.250
.125
0
0
1
2
3
x
d.
a.
3 1 4 1
8 8 8 2
E( x) xp( x) 4(.02) (3)(.07) (2)(.10) (1)(.15) 0(.3)
1(.18) 2(.10) 3(.06) 4(.02)
.08 .21 .2 .15 0 .18 .2 .18 .08 0
2 E[( x )2 ] ( x ) 2 p( x)
( 4 0) 2 (.02) ( 3 0) 2 (.07) ( 2 0) 2 (.10)
( 1 0) 2 (.15) (0 0) 2 (.30) (1 0) 2 (.18)
(2 0) 2 (.10) (3 0) 2 (.06) (4 0) 2 (.02)
.32 .63 .4 .15 0 .18 .4 .54 .32 2.94
2.94 1.715
b.
Using MINITAB, the graph is:
Histogram of x
.30
.25
.20
p(x)
4.17
P( x 2 or x 3) p (2) p(3)
.15
.10
.05
0
-4
-3
2
-2
-1
0
1
2
0
3
4
2
2 0 2(1.715) 0 3.430 (3.430, 3.430)
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
c.
4.18
a.
153
P(3.430 x 3.430) p (3) p (2) p (1) p (0) p(1) p (2) p (3)
.07 .10 .15 .30 .18 .10 .06 .96
E( x) xp( x)
10(.05) 20(.20) 30(.30) 40(.25) 50(.10) 60(.10) .5 4 9 10 5 6 34.5
2 E( x )2 ( x )2 p( x)
(10 34.5)2 (.05) (20 34.5) 2 (.20) (30 34.5) 2 (.30)
(40 34.5)2 (.25) (50 34.5) 2 (.10) (60 34.5) 2 (.10)
30.0125 42.05 6.075 7.5625 24.025 65.025 174.75
174.75 13.219
b.
Using MINITAB, the graph is:
Histogram of x
.30
.25
p(x)
.20
.15
.10
.05
0
10
20
30
40
50
60
x
2
c.
34.5
2
2 34.5 2(13.219) 34.5 26.438 (8.062, 60.938)
P(8.062 x 60.938) p(10) p(20) p(30) p(40) p(50) p(60)
.05 .20 .30 .25 .10 .10 1.00
4.19
a.
It would seem that the mean of both would be 1 since they both are symmetric distributions centered at
1.
b.
P(x) seems more variable since there appears to be greater probability for the two extreme values of 0
and 2 than there is in the distribution of y.
c.
For x: E( x)
xp( x) 0(.3) 1(.4) 2(.3) 0 .4 .6 1
2 E[( x ) 2 ] ( x )2 p( x)
(0 1)2 (.3) (1 1)2 (.4) (2 1)2 (.3) .3 0 .3 .6
Copyright © 2014 Pearson Education, Inc.
154
Chapter 4
For y: E( y)
yp( y) 0(.1) 1(.8) 2(.1) 0 .8 .2 1
2 E[( y )2 ] ( y ) 2 p( y )
(0 1) 2 (.1) (1 1) 2 (.8) (2 1) 2 (.1) .1 0 .1 .2
The variance for x is larger than that for y.
4.20
4.21
a.
The possible values of x are 1, 2, 3, and 4 or more.
b.
P( x 1) .26
c.
P( x 4) .25
d.
We cannot compute E(x) because the last value of x has more than one value (4 or more). To find the
E(x), each possible value of x can have only one value.
a.
The probability distribution for x is found by converting the Percent column to a probability column by
dividing the percents by 100. The probability distribution of x is:
x
2
3
4
5
b.
P( x 5) p(5) .1837 .
c.
P( x 2) p(2) .0408 .
p(x)
.0408
.1735
.6020
.1837
E ( x) xi p ( xi ) 2(.0408) 3(.1735) 4(.6020) 5(.1837)
4
d.
i 1
.0816 .5205 2.4080 .9185 3.9286 3.93
The average star rating for a car’s drivers-side star rating is 3.93.
4.22
a.
Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly
used to estimate unknown probabilities. In addition, relative frequencies have the same properties as
the probabilities in a probability distribution, namely
1. all relative frequencies are greater than or equal to zero
2. the sum of all the relative frequencies is 1
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
155
Using MINITAB, the graph of the probability distribution is:
.16
.14
.12
p(age)
.10
.08
.06
.04
.02
0
20
c.
22
24
26
age
28
30
32
Let x = age of employee. Then P( x 30) .13 .15 .12 .40 .
P( x 40) 0
P( x 30) .02 .04 .05 .07 .04 .02 .07 .02 .11 .07 .51
4.23
d.
P( x 25or x 26) .02 .07 .09
a.
In order for this to be a valid probability distribution, all probabilities must be between 0 and 1 and the
sum of all the probabilities must be 1. For this data, all the probabilities are between 0 and 1. If you
sum all of the probabilities, the sum is 1.
b.
P( x 10) P( x 10) P( x 11) P( x 20)
.02 .02 .02 .02 .01 .01 .01 .01 .01 .005 .005 .14
c.
The mean of x is
E ( x) xp( x) 0(.17) 1(.10) 2(.11) 20(.005)
0 .1 .22 .33 .1 4.655
The variance of x is
2 E ( x ) 2 ( x )2 p( x) (0 4.655) 2 (.17) (1 4.655) 2 (.1) (2 4.655) 2 (.11)
(20 4.655)2 (.005)
3.6837 1.3359 .7754 1.1773 19.8560
d.
From Chebyshev’s Rule, we know that at least .75 of the observations will fall within 2 standard
deviations of the mean. The standard deviation is 19.8560 4.456 .
The interval is: 2 4.655 2(4.456) 4.655 8.912 (4.257, 13.567) .
Copyright © 2014 Pearson Education, Inc.
156
4.24
Chapter 4
a.
The probability distribution for x is:
Grill Display
Combination
1-2-3
1-2-4
1-2-5
2-3-4
2-3-5
2-4-5
4.25
x
6
7
8
9
10
11
p(x)
35 /124 .282
8 /124 .065
42 /124 .339
4 /124 .032
1/124 .008
34 /124 .274
b.
P( x 10) p(11) .274
a.
The possible values of x are 0, 2, 3, and 4.
b.
To find the probability distribution of x, we first find the frequency distribution of x. We then divide
the frequencies by n 106 to get the probabilities. The probability distribution of x is:
0
35
.3302
x
f(x)
p(x)
c.
2
58
.5472
3
5
.0472
4
8
.0755
E( x) xp( x) 0(.3302) 2(.5472) 3(.0472) 4(.0755) 1.538 . For all social robots, the
average number of legs on the robot is 1.538.
4.26
a.
For this problem, x = sequence number of a Florida tropical storm within a season that develops into a
hurricane. Thus, x can take on values 1, 2, 3, … . Since this is a countable number of outcomes, x is a
discrete random variable.
b.
The probability distribution of x is found by dividing the number of storms by the total number of
storms, n = 67. The probability distribution of x is:
x
1
2
3
4
5
6
7
8
f(x)
4
10
5
6
11
5
5
5
p(x)
.0597
.1493
.0746
.0896
.1642
.0746
.0746
.0746
x
9
10
11
12
13
14
15
22
f(x)
4
2
5
1
1
1
1
1
p(x)
.0597
.0299
.0746
.0149
.0149
.0149
.0149
.0149
c.
P( x 5) .1642
d.
P( x 5) P( x 1) P( x 2) P( x 3) P( x 4) .0597 .1493 .0746 .0896 .3732
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
e.
157
The expected value is
E ( x) xp( x) 1(.0597) 2(.1493) 3(.0746) 22(.0149)
.0597 .2986 .2238 .3278 6.1174
The average sequence number of a Florida tropical storm within a season that develops into a hurricane
is 6.1174.
4.27
f.
No, it is not likely. The probability is only .0149.
a.
The random variable x is a discrete random variable because it can take on only values 0, 1, 2, 3, 4, or 5
in this example.
b.
p (0)
5!(.35)0 (.65)5 0 5 4 3 2 1(1)(.65)5
.655 .1160
0!(5 0)!
1 5 4 3 2 1
p(1)
5!(.35)1 (.65)51 5 4 3 2 1(.35)1 (.65) 4
5(.35)(.65) 4 .3124
1!(5 1)!
1 4 3 2 1
p(2)
5!(.35) 2 (.65)5 2 5 4 3 2 1(.35) 2 (.65)3
10(.35) 2 (.65)3 .3364
2!(5 2)!
2 1 3 2 1
p(3)
5!(.35)3 (.65)53 5 4 3 2 1(.35)3 (.65) 2
10(.35)3 (.65) 2 .1811
3!(5 3)!
3 2 1 2 1
p(4)
5!(.35) 4 (.65)5 4 5 4 3 2 1(.35) 4 (.65)1
5(.35) 4 (.65)1 .0488
4!(5 4)!
4 3 2 1 1
p(5)
5!(.35)5 (.65)55 5 4 3 2 1(.35)5 (.65)0
(.35)5 .0053
5!(5 5)!
5 4 3 2 1 1
c.
The two properties of discrete random variables are that 0 p( x) 1 for all x and
above, all probabilities are between 0 and 1 and
p(x) 1 . From
p( x) .1160 .3124 .3364 .1811 .0488 .0053 1
4.28
d.
P( x 4) p(4) p(5) .0488 .0053 .0541
a.
First, we must find the probability distribution of x. Define the following events:
C: {Chicken is contaminated}
N: {Chicken is not contaminated}
If 3 slaughtered chickens are randomly selected, then the possible outcomes are:
CCC, CCN, CNC, NCC, CNN, NCN, NNC, and NNN
Each of these outcomes are NOT equally likely since P(C ) 1/100 .01 .
Copyright © 2014 Pearson Education, Inc.
158
Chapter 4
P( N ) 1 P(C ) 1 .01 .99 .
P(CCC ) P(C C C ) P(C ) P(C ) P(C ) .01(.01)(.01) .000001
P(CCN ) P(CNC ) P( NCC ) P(C C N ) P(C ) P(C ) P( N ) .01(.01)(.99) .000099
P(CNN ) P( NCN ) P( NNC ) P(C N N ) P(C ) P( N ) P( N ) .01(.99)(.99) .009801
P( NNN ) P( N N N ) P( N ) P( N ) P( N ) .99(.99)(.99) .970299
The variable x is defined as the number of contaminated chickens in the sample. The value of x for each
of the outcomes is:
x
3
2
2
2
1
1
1
0
Event
CCC
CCN
CNC
NCC
CNN
NCN
NNC
NNN
p(x)
.000001
.000099
.000099
.000099
.009801
.009801
.009801
.970299
The probability distribution of x is:
x
0
1
2
3
b.
p(x)
.970299
.029403
.000297
.000001
Using MINITAB, the probability graph for x is:
1
.8
p(x)
.6
.4
.2
0
0
1
2
3
x
4.29
c.
P( x 1) P( x 0) P( x 1) .970299 .029403 .999702
a.
p(1) .23(.77)11 .23(.77)0 .23 . The probability that one would encounter a
contaminated cartridge on the first trial is .23.
b.
p(5) .23(.77)51 .23(.77)4 .0809 . The probability that one would encounter a
the first contaminated cartridge on the fifth trial is .0809.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.30
c.
P( x 2) 1 P( x 1) 1 P( x 1) 1 .23 .77 . The probability that the first contaminated
cartridge is found on the second trial or later is .77.
a.
If the first letters of consumers’ last names are all equally likely, then P( x i) 1/ 26 for
i = 1, 2, …, 26.
b.
The expected value is
159
E ( x) xp( x) 1
1 1 1
1
2 3 26 13.5
26 26 26
26
The average number given to a consumer based on his last name is 13.5.
c.
4.31
4.32
This probability distribution is probably not realistic. Very few consumers have last names that begin
with Q or U. However, many consumers have last names that begin with S and T. One could estimate
the true probability distribution of x by taking a random sample of names from a phone book and
looking at the relative frequency distribution of the values of x assigned to the sampled names.
a.
20 100 20
20!
80!
20! 80!
0 3 - 0 0!(20 0)! 3!(80 3)! 0!20! 3!77! 82,160
p (0)
.508
100!
100!
161, 700
100
3!(100 3)!
3!97!
3
b.
20 100 20
20!
80!
20! 80!
1 3 - 1 1!(20 1)! 2!(80 2)! 1!19! 2!78! 63, 200
p (1)
.391
100!
100!
161, 700
100
3!(100 3)!
3!97!
3
c.
20 100 20
20!
80!
20! 80!
2
3
2
2!(20 2)!1!(80 1)! 2!18!1!79! 15, 200
p (2)
.094
100!
100!
161, 700
100
3!(100 3)!
3!97!
3
d.
20 100 20
20!
80!
20!
1
3
3
0
1,140
3!(20 3)! 0!(80 0)! 3!17!
p (3)
.007
100!
100!
161, 700
100
3!(100 3)!
3!97!
3
a.
E ( x ) xp ( x )
All x
Firm A:
E ( x ) 0(.01) 500(.01) 1000(.01) 1500(.02) 2000(.35) 2500(.30)
3000(.25) 3500(.02) 4000(.01) 4500(.01) 5000(.01)
0 5 10 30 700 750 750 70 40 45 50 2450
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Chapter 4
Firm B:
E ( x ) 0(.00) 200(.01) 700(.02) 1200(.02) 1700(.15) 2200(.30)
2700(.30) 3200(.15) 3700(.02) 4200(.02) 4700(.01)
0 2 14 24 255 660 810 480 74 84 47 2450
b.
2
2 ( x ) 2 p ( x)
All x
Firm A:
2 (0 2450) 2 (.01) (500 2450) 2 (.01) (5000 2450) 2 (.01)
60, 025 38, 025 21, 025 18, 050 70,875 750 75, 625
22, 050 24, 025 42, 025 65, 025
437,500
437,500 661.44
Firm B:
2 (0 2450) 2 (.00) (200 2450) 2 (.01) (4700 2450) 2 (.01)
0 50, 625 61, 250 31, 250 84,375 18, 750 84, 375
31, 250 61, 250 50, 625
492, 500
492,500 701.78
Firm B faces greater risk of physical damage because it has a higher variance and standard deviation.
4.33
To find the probability distribution of x, we sum the probabilities associated with the same value of x. The
probability distribution is:
x
p(x)
4.34
8.5
.462189
9
.288764
9.5
.141671
10
.069967
10.5
.025236
11
.011657
12
.000518
To determine which group of Finnish citizens has the highest average IQ score, we must find the expected
value for each group. To do this, we first find the probability distribution for each group by dividing the
frequency for each IQ level in each group by the group total. The probability distributions are:
IQ
1
2
3
4
5
6
7
8
9
Invest
.020
.030
.045
.120
.190
.230
.150
.115
.100
No Invest
.041
.083
.088
.174
.217
.191
.099
.062
.045
For Investors, E( x) xp( x) 1(.020) 2(.030) 3(.045) 9(.100) 5.895
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
For Non-investors, E( x)
xp(x) 1(.041) 2(.083) 3(.088) 9(.045) 4.992
161
Thus, the investors had a higher average IQ than the non-investors.
4.35
a.
Let x = the potential flood damages. Since we are assuming if it rains the business will incur damages
and if it does not rain the business will not incur any damages, the probability distribution of x is:
0
.7
x
p(x)
b.
300,000
.3
The expected loss due to flood damage is
E ( x) xp( x) 0(.7) 300, 000(.3) 0 90, 000 $90, 000
All x
4.36
Let x = winnings in the Florida lottery. The probability distribution for x is:
x
$1
$6,999,999
p(x)
22,999,999/23,000,000
1/23,000,000
The expected net winnings would be:
E ( x ) xp ( x) ( 1)
All x
1
22, 999, 999
6,999,999
$.70
23,
000,
000
23,
000,
000
The average winnings of all those who play the lottery is $.70.
4.37
a.
Since there are 20 possible outcomes that are all equally likely, the probability of any of the 20 numbers
is 1/20. The probability distribution of x is:
P( x 5) 1/ 20 .05 ; P( x 10) 1/ 20 .05 ; etc.
x
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
p(x) .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05
b.
E( x) xp( x) 5(.05) 10(.05) 15(.05) 20(.05) 25(.05) 30(.05) 35(.05)
40(.05) 45(.05) 50(.05) 55(.05) 60(.05) 65(.05) 70(.05) 75(.05)
80(.05) 85(.05) 90(.05) 95(.05) 100(.05) 52.5
c.
2 E ( x ) 2 ( x ) 2 p( x) (5 52.5) 2 (.05) (10 52.5) 2 (.05)
(15 52.5) 2 (.05) (20 52.5) 2 (.05) (25 52.5) 2 (.05) (30 52.5) 2 (.05)
(35 52.5) 2 (.05) (40 52.5) 2 (.05) (45 52.5) 2 (.05) (50 52.5) 2 (.05)
(55 52.5) 2 (.05) (60 52.5) 2 (.05) (65 52.5) 2 (.05) (70 52.5) 2 (.05)
(75 52.5) 2 (.05) (80 52.5) 2 (.05) (85 52.5) 2 (.05) (90 52.5) 2 (.05)
(95 52.5) 2 (.05) (100 52.5) 2 (.05)
831.25
Copyright © 2014 Pearson Education, Inc.
162
Chapter 4
831.25 28.83
Since the uniform distribution is not mound-shaped, we will use Chebyshev's theorem to describe the
data. We know that at least 8/9 of the observations will fall with 3 standard deviations of the mean and
at least 3/4 of the observations will fall within 2 standard deviations of the mean. For this problem,
2 52.5 2(28.83) 52.5 57.66 (5.16, 110.16) . Thus, at least 3/4 of the data will fall
between 5.16 and 110.16. For our problem, all of the observations will fall within 2 standard
deviations of the mean. Thus, x is just as likely to fall within any interval of equal length.
d.
If a player spins the wheel twice, the total number of outcomes will be 20(20) = 400. The sample space
is:
5, 5
10, 5
15, 5
20, 5
25, 5...
100, 5
5,10 10,10
15,10
20,10
25,10... 100,10
5,15 10,15
15,15
20,15
25,15... 100,15
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5,100 10,100
15,100
20,100 25,100... 100,100
Each of these outcomes are equally likely, so each has a probability of 1/400 = .0025.
Now, let x equal the sum of the two numbers in each sample. There is one sample with a sum of 10, two
samples with a sum of 15, three samples with a sum of 20, etc. If the sum of the two numbers exceeds
100, then x is zero. The probability distribution of x is:
x
0
10
15
20
25
30
35
40
45
50
e.
f.
p(x)
.5250
.0025
.0050
.0075
.0100
.0125
.0150
.0175
.0200
.0225
x
55
60
65
70
75
80
85
90
95
100
p(x)
.0250
.0275
.0300
.0325
.0350
.0375
.0400
.0425
.0450
.0475
We assumed that the wheel is fair, or that all outcomes are equally likely.
E( x) xp( x) 0(.5250) 10(.0025) 15(.0050) 20(.0075) 100(.0475) 33.25
2 E ( x ) 2 ( x - ) 2 p ( x) (0 33.25) 2 (.5250) (10 33.25) 2 (.0025)
(15 33.25) 2 (.0050) (20 33.25) 2 (.0075) (100 33.25) 2 (.0475) 1, 471.3125
1,471.3125 38.3577
g.
P ( x 0) .525
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
h.
Given that the player obtains a 20 on the first spin, the possible values for x (sum of the two spins) are
0 (player spins 85, 90, 95, or 100 on the second spin), 25, 30, ..., 100. To get an x of 25, the player
would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second spin
order to get an x of 30, etc. Since all of the outcomes are equally likely on the second spin, the
distribution of x is:
x
0
25
30
35
40
45
50
55
60
p(x)
.20
.05
.05
.05
.05
.05
.05
.05
.05
x
65
70
75
80
85
90
95
100
p(x)
.05
.05
.05
.05
.05
.05
.05
.05
i.
The probability that the players total score will exceed one dollar is the probability that x is zero.
P( x 0) .20
j.
Given that the player obtains a 65 on the first spin, the possible values for x (sum of the two spins) are
0 (player spins 40, 45, 50, up to 100 on second spin), 70, 75, 80,..., 100. In order to get an x of 70, the
player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second
spin in order to get an x of 75, etc. Since all of the outcomes are equally likely on the second spin, the
distribution of x is:
x
0
70
75
80
85
90
95
100
p(x)
.65
.05
.05
.05
.05
.05
.05
.05
The probability that the players total score will exceed one dollar is the probability that x is zero.
P( x 0) .65 .
4.38
163
a.
Each point in the system can have one of 2 status levels, “free” or “obstacle”. Define the following
events:
AF: {Point A is free}
BF: {Point B is free}
CF: {Point C is free}
AO: {Point A is obstacle}
BO: {Point B is obstacle}
CO: {Point C is obstacle}
Thus, the sample points for the space are:
AFBFCF, AFBFCO, AFBOCF, AFBOCO, AOBFCF, AOBFCO, AOBOCF, AOBOCO
Copyright © 2014 Pearson Education, Inc.
164
Chapter 4
b.
Since it is stated that the probability of any point in the system having a “free” status is .5, the
probability of any point having an “obstacle” status is also .5. Thus, the probability of each of the
sample points above is P( Ai Bi Ci ) .5(.5)(.5) .125 .
The values of Y, the number of free links in the system, for each sample point are listed below. A link
is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A link from
B to C is free if B is free and C is free.
Sample point
Y
Probability
AFBFCF
AFBFCO
AFBOCF
AFBOCO
AOBFCF
AOBFCO
AOBOCF
AOBOCO
2
1
0
0
1
0
0
0
.125
.125
.125
.125
.125
.125
.125
.125
The probability distribution for Y is:
Y
0
1
2
4.39
Probability
.625
.250
.125
Let x = bookie's earnings per dollar wagered. Then x can take on values $1 (you lose) and $-5 (you win). The
only way you win is if you pick 3 winners in 3 games. If the probability of picking 1 winner in 1 game is .5,
then P( www) p( w) p( w) p( w) .5(.5)(.5) .125 (assuming games are independent).
Thus, the probability distribution for x is:
x
p(x)
$1 .875
$-5 .125
E( x) xp( x)1(.875) 5(.125) .875 .625 $.25
4.40
a.
6!
6!
6 5 4 3 2 1
15
2!(6 2)! 2!4! (2 1)(4 3 2 1)
b.
5
5!
5!
5 4 3 2 1
10
2
1)(3 2 1)
2!(5
2)!
2!3!
(2
c.
7
7!
7!
7 6 5 4 3 2 1
1
0 0!(7 0)! 0!7! (1)(7 6 5 4 3 2 1)
(Note: 0! = 1)
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.41
d.
6
6!
6!
6 5 4 3 2 1
1
6 6!(6 6)! 6!0! (6 5 4 3 2 1)(1)
e.
4
4!
4!
4 3 2 1
4
3
3!(4
3)!
3!1!
(3
2 1)(1)
a.
x is discrete. It can take on only six values.
b.
This is a binomial distribution.
c.
5
5!
5 4 3 2 1
(.7) 0 (.3)5
(1)(.00243) .00243
p (0) (.7) 0 (.3)5 0
0!5!
1 5 4 3 2 1
0
165
5
5!
(.7)1 (.3) 4 .02835
p (1) (.7)1 (.3)5 1
1
1!4!
5
5!
(.7) 2 (.3)3 .1323
p (2) (.7) 2 (.3)5 2
2
2!3!
5
5!
(.7)3 (.3) 2 .3087
p (3) (.7)3 (.3)5 3
3
3!2!
5
5!
(.7) 4 (.3)1 .36015
p (4) (.7) 4 (.3)5 4
4
4!1!
5
5!
(.7)5 (.3) 0 .16807
p (5) (.7)5 (.3)5 5
5!0!
5
Histogram of x
.4
p(x)
.3
.2
.1
0
0
1
2
2
4.42
3
4
3.5
5
2
npq 5(.7)(.3) 1.0247
d.
np 5(.7) 3.5
e.
2 3.5 2(1.0247) 3.5 2.0494 (1.4506, 5.5494)
a.
3
3!
3 2 1
p (0) (.3) 0 (.7)3 0
(.3)0 (.7)3
(1)(.7)3 .343
0!3!
1 3 2 1
0
3
3!
(.3)1 (.7) 2 .441
p (1) (.3)1 (.7)31
1
1!2!
3
3!
(.3) 2 (.7)1 .189
p (2) (.3) 2 (.7)3 2
2
2!1!
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166
Chapter 4
3
3!
(.3)3 (.7) 0 .027
p (3) (.3)3 (.7)3 3
3!0!
3
b.
4.43
4.44
The probability distribution in tabular form is:
x
p(x)
0
1
2
3
.343
.441
.189
.027
a.
P( x 1)
5!
5 4 3 2 1
(.2)1(.8) 4
(.2)1(.8) 4 5(.2)1 (.8)4 .4096
1!4!
(1)(4 3 2 1)
b.
P( x 2)
4!
4 3 2 1
(.6) 2(.4) 2
(.6) 2(.4) 2 6(.6) 2 (.4) 2 .3456
2!2!
(2 1)(2 1)
c.
P( x 0)
3!
3 2 1
(.7) 0(.3) 3
(.7) 0(.3) 3 1(.7)0 (.3)3 .027
0!3!
(1)(3 2 1)
d.
P( x 3)
5!
5 4 3 2 1
(.1) 3(.9) 2
(.1) 3(.9) 2 10(.1)3 (.9) 2 .0081
3!2!
(3 2 1)(2 1)
e.
P( x 2)
4!
4 3 2 1
(.4) 2(.6) 2
(.4) 2(.6) 2 6(.4)2 (.6) 2 .3456
2!2!
(2 1)(2 1)
f.
P( x 1)
a.
P x 2 P( x 2) P( x 1) .167 .046 .121 (from Table I, Appendix D with n = 10 and p = .4)
b.
P ( x 5) .034
c.
P( x 1) 1- P( x 1) 1 .919 .081
d.
P( x 10) P( x 9) 0
e.
P( x 10) 1 P( x 9) 1 .002 .998
f.
P( x 2) P( x 2) P( x 1) .206 .069 .137
3!
3 2 1
(.9)1(.1) 2
(.9)1(.1) 2 3(.9)1 (.1)2 .027
1!2!
(1)(2 1)
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.45
a.
167
np 25(.5) 12.5
2 np(1 p) 25(.5)(.5) 6.25 and 2 6.25 2.5
b.
np 80(.2) 16
2 np(1 p) 80(.2)(.8) 12.8 and 2 12.8 3.578
c.
np 100(.6) 60
2 np(1 p) 100(.6)(.4) 24 and 2 24 4.899
d.
np 70(.9) 63
2 np(1 p) 70(.9)(.1) 6.3 and 2 6.3 2.510
e.
np 60(.8) 48
2 np(1 p) 60(.8)(.2) 9.6 and 2 9.6 3.098
f.
np 1,000(.04) 40
2 np(1 p) 1,000(.04)(.96) 38.4 and 2 38.4 6.197
4.46
x is a binomial random variable with n = 4.
a.
If the probability distribution of x is symmetric, p(0) = p(4) and p(1) = p(3).
n
We know p ( x ) p x q n x x = 0, 1, ... , n,
x
When n = 4,
4
4
4! 0 4
4! 4 0
4
4
p (0) p (4) p 0 q 4 p 4 q 0
p q
p q q p pq
0!4!
4!0!
0
4
Since p q 1 , p = .5
Therefore, the probability distribution of x is symmetric when p = .5.
b.
If the probability distribution of x is skewed to the right, then the mean is greater than the median.
Therefore, there are more small values in the distribution (0, 1) than large values (3, 4). For this to
happen, p must be smaller than .5. If we pick p .2 , the probability distribution of x will be skewed to
the right.
c.
If the probability distribution of x is skewed to the left, then the mean is smaller than the median.
Therefore, there are more large values in the distribution (3, 4) than small values (0, 1). For this to
happen, p must be larger than .5. If we pick p .8 , the probability distribution of x will be skewed to
the left.
Copyright © 2014 Pearson Education, Inc.
Chapter 4
d.
In part a, x is a binomial random variable with n = 4 and p = .5.
4
p ( x ) .5 x.54 x
x
x = 0, 1, 2, 3, 4
4
4! 4
.5 1(.5) 4 .0625
p (0) .5 0.5 4
0
0!4!
4
4! 4
.5 4(.5) 4 .25
p (1) .51.5 3
1
1!3!
4
4! 4
.5 6(.5) 4 .375
p (2) .5 2.5 2
2!2!
2
p(3) p(1) .25 (since the distribution is symmetric)
p(4) p(0) .0625
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.0625
.25
.375
.25
.0625
np 4(.5) 2
Using MINITAB, the graph of the probability distribution of x when n 4 and p .5 is as follows.
Histogram of x
.4
.3
p(x)
168
.2
.1
0
0
1
2
x
3
4
2
In part b, x is a binomial random variable with n 4 and p .2 .
4
p ( x ) .2 x.8 4 x
x
x = 0, 1, 2, 3, 4
4
p (0) .20.84 1(1).84 .4096
0
4
p (1) .21.83 4(.2).83 .4096
1
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4
p (3) .23.81 4(.2)3 (.8) .0256
3
4
p (2) .2 2.82 6(.2) 2 .82 .1536
2
4
p(4) .24.80 1(.2) 4 (1) .0016
4
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.4096
.4096
.1536
.0256
.0016
np 4(.2) .8 .
Using MINITAB, the graph of the probability distribution of x when n 4 and p .2 is as follows:
Histogram of x
.4
p(x)
.3
.2
.1
0
0
1
2
x
3
4
.8
In part c, x is a binomial random variable with n 4 and p .8 .
4
p ( x ) .8 x.2 4- x
x
x = 0, 1, 2, 3, 4
4
p (0) .80.2 4 1(1).2 4 .0016
0
4
p (1) .81.23 4(.8).23 .0256
1
4
p (2) .82.2 2 6(.8) 2 .2 2 .1536
2
4
p (3) .83.21 4(.8)3 .2 .4096
3
4
p(4) .84.20 1(.8) 4 (1) .4096
4
Copyright © 2014 Pearson Education, Inc.
169
170
Chapter 4
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.0016
.0256
.1536
.4096
.4096
Note: The distribution of x when n 4 and p .8 is the reverse of the distribution of
x when n 4 and p .8 .
np 4(.8) 3.2
Using MINITAB, the graph of the probability distribution of x when n 4 and p .8 is as follows:
Histogram of x
.4
p(x)
.3
.2
.1
0
0
1
2
x
3
4
3.2
4.47
e.
In general, when p .5 , a binomial distribution will be symmetric regardless of the value of n. When
p is less than .5, the binomial distribution will be skewed to the right; and when p is greater than .5, it
will be skewed to the left. (Refer to parts a, b, and c.)
a.
Let S = adult who does not work while on summer vacation.
b.
To see if x is approximately a binomial random variable we check the characteristics:
1.
n identical trials. Although the trials are not exactly identical, they are close. Taking a sample of
reasonable size n from a very large population will result in trials being essentially identical.
2.
Two possible outcomes. The adults can either not work on their summer vacation or they can
work on their summer vacation. S = adult does not work on summer vacation and F = adult does
work on summer vacation.
3.
P(S) remains the same from trial to trial. If we sample without replacement, then P(S) will change
slightly from trial to trial. However, the differences are extremely small and will essentially be 0.
4.
Trials are independent. Again, although the trials are not exactly independent, they are very close.
5.
The random variable x = number of adults who work on their summer vacation in n = 10 trials.
Thus, x is very close to being a binomial. We will assume that it is a binomial random variable.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
c.
For this problem, p .35 .
d.
Using MINITAB n 10 and p .35 , the probability is:
171
Probability Density Function
Binomial with n = 10 and p = 0.35
x
3
P( X = x )
0.252220
Thus, P( x 3) .2522 .
e.
Using MINITAB n 10 and p .35 , the probability is:
Cumulative Distribution Function
Binomial with n = 10 and p = 0.35
x
2
P( X <= x )
0.261607
Thus, P( x 2) P( x 0) P( x 1) P( x 2) .2616 .
4.48
a.
To see if x is approximately a binomial random variable we check the characteristics:
1.
n identical trials. Although the trials are not exactly identical, they are close. Taking a sample of
size n 15 from a very large population will result in trials being essentially identical.
2.
Two possible outcomes. The hotel guests are either aware of and participate in the conservation
efforts or they do not. S = hotel guest is aware of and participates in conservation efforts and F =
hotel guest is not aware of and/or does not participate in conservation efforts.
3.
P(S) remains the same from trial to trial. If we sample without replacement, then P(S) will change
slightly from trial to trial. However, the differences are extremely small and will essentially be 0.
4.
Trials are independent. Again, although the trials are not exactly independent, they are very close.
5.
The random variable x = number of hotel guests who are aware of and participate in conservation
efforts in n 15 trials.
Thus, x is very close to being a binomial. We will assume that it is a binomial random variable.
b.
Define the following events:
P: {hotel guest is aware of conservation program}
A: {hotel guest participates in conservation efforts}
Then, p P( P | A) P( A) .72(.66) .4752 .
Copyright © 2014 Pearson Education, Inc.
172
c.
Chapter 4
Using MINITAB with n 15 and p .45 , the probability is:
Cumulative Distribution Function
Binomial with n = 15 and p = 0.45
x
9
P( X <= x )
0.923071
Thus, P( x 10) 1 P( x 9) 1 .9231 .0769 .
4.49
a.
To see if x is approximately a binomial random variable we check the characteristics:
1.
n identical trials. Although the trials are not exactly identical, they are close. Taking a sample of
size n 250 from a very large population will result in trials being essentially identical.
2.
Two possible outcomes. A U.S. adult has either used the internet and paid to download music or
he/she has not. S = U.S. adult has used the internet and paid to download music and F = U.S. adult
has not used the internet and/or has not paid to download music.
3.
P(S) remains the same from trial to trial. If we sample without replacement, then P(S) will change
slightly from trial to trial. However, the differences are extremely small and will essentially be 0.
4.
Trials are independent. Again, although the trials are not exactly independent, they are very close.
5.
The random variable x = number of U.S. adults who have used the internet and paid to download
music in n 250 trials.
Thus, x is very close to being a binomial. We will assume that it is a binomial random variable.
4.50
b.
For this example, p .5 . Half of the adults in the U.S. have used the internet and have paid to
download music.
c.
E ( x) np 250(.5) 125
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
3.
The experiment consists of n identical trials.
There are only 2 possible outcomes for each trial. Let S = general practice physician in
the United States does not recommend medicine as a career and F = general practice physician in
the United States does recommend medicine as a career.
The probability of success (S) is the same from trial to trial. For each trial p P ( S ) .60 and
4.
q 1 p 1 .60 .40 .
The trials are independent.
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Random Variables and Probability Distributions
5.
173
The binomial random variable x is the number of general practice physicians in the United States
in n trials who do not recommend medicine as a career.
Thus, x is a binomial random variable.
b.
From the information given, p .60 .
c.
E ( x) np 25(.60) 15
npq 25(.60)(.40) 6 2.4495
d.
4.51
From Table I, Appendix D, with n 25 and p .60 , P( x 1) 1 P( x 0) 1 .000 1.000 .
For this problem, let x = number of law librarians who are unsatisfied with their job. Then x is a binomial
random variable with n 20 and p 1 .90 .10 . Using a MINITAB with n 20 and p .10 , the
probability is
Cumulative Distribution Function
Binomial with n = 20 and p = 0.1
x
2
P( X <= x )
0.676927
Thus, P( x 2) .6769 .
4.52
a.
Let x = number of students who initially answer the question correctly in 20 students. Then x is a
binomial random variable with n 20 and p .5 . Using a MINITAB with n 20 and p .5 , the
probability is:
Cumulative Distribution Function
Binomial with n = 20 and p = 0.5
x
10
P( X <= x )
0.588099
Thus, P( x 10) 1 P( x 10) 1 .5881 .4119 .
b.
Let y = number of students who answer the question correctly after immediate feedback in 20 students.
Then y is a binomial random variable with n 20 and p .7 . Using a MINITAB with n 20 and
p .7 , the probability is:
Cumulative Distribution Function
Binomial with n = 20 and p = 0.7
x
10
P( X <= x )
0.0479619
Thus, P( x 10) 1 P( x 10) 1 .0480 .9520 .
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174
4.53
Chapter 4
a.
Let x = number of pairs correctly identified by an expert in 5 trials. Then x is a binomial random
variable with n 5 and p .92 . Using a MINITAB with n 5 and p .92 , the probability is:
Probability Density Function
Binomial with n = 5 and p = 0.92
x
5
P( X = x )
0.659082
Thus, P( x 5) .6591 .
b.
Let y = number of pairs correctly identified by a novice in 5 trials. Then y is a binomial random
variable with n 5 and p .75 . Using a MINITAB with n 5 and p .75 , the probability is:
Probability Density Function
Binomial with n = 5 and p = 0.75
x
5
P( X = x )
0.237305
Thus, P( x 5) .2373 .
4.54
a.
Let x = number of commissioners out of 4 who vote in favor of an issue. Then x is a binomial random
variable with n 4 and p .5 (since they are equally likely to vote for or against an issue). The
probability that your vote counts is equal to P( x 2) .
P( x 2)
b.
Let x = number of commissioners out of 2 who vote in favor of an issue. Then x is a binomial random
variable with n 2 and p .5 (since they are equally likely to vote for or against an issue). The
probability that your vote counts is equal to P ( x 1) .
P( x 1)
4.55
4!
4 3 2 1 2
.52 (.5) 4 2
.5 (.5)2 .375
2!(4 2)!
2 1 2 1
2!
2 1 1 1
.51 (.5)2 1
.5 (.5) .5
1!(2 1)!
1 1
Let x = number of major bridges in Denver that will have a rating of 4 or below in 2020 in 10 trials. Then x
has an approximate binomial distribution with n 10 and p .09 .
a.
P ( x 3) 1 P ( x 2) 1 P ( x 0) P( x 1) P( x 2)
b.
Since the probability of seeing at least 3 bridges out of 10 with ratings of 4 or less is so small, we can
conclude that the forecast of 9% of all major Denver bridges will have ratings of 4 or less in 2020 is too
small. There would probably be more than 9%.
10
10
10
1 .09 0 (.91)10 0 .091 (.91)10 1 .092 (.91)10 2
0
1
2
10!
10! 1 9 10!
1
.090.9110
.09 .91
.092.918 1 .389 .385 .171 .055
0!10!
1!9!
2!8!
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.56
175
Define the following events:
A: {Taxpayer is audited}
B: {Taxpayer has income less than $1 million)
C: {Taxpayer has income of $1 million or higher}
a.
From the information given in the problem,
P( A|B) 1/100 .01
b.
P( A | C ) 9 /100 .09
Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial
random variable with n 5 and p .01 .
5
5!
.011 (.99) 4 .0480
P ( x 1) .011 (.99)5 1
1!4!
1
5
P( x 1) 1 [ P( x 0) P( x 1)] 1 .010 (.99)5 0 .0480
0
5!
.010 (.99)5 .0480 1 [.9510 .0480] 1 .9990 .0010
1
0!5!
c.
Let x = number of taxpayers with incomes of $1 million or more who are audited. Then x is a binomial
random variable with n 5 and p .09 .
5
5!
.091 (.91) 4 .3086
P ( x 1) .091 (.91)5 1
1! 4!
1
5
P( x 1) 1 [ P( x 0) P( x 1)] 1 .090 (.91)5 0 .3086
0
5!
.090 (.91)5 .3086 1 [.6240 .3086] 1 .9326 .0674
1
0!5!
d.
Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial
random variable with n 2 and p .01 .
Let y = number of taxpayers with incomes $1 million or more who are audited. Then y is a binomial
random variable with n 2 and p .09 .
2
2!
.010 (.91) 2 .9801
P ( x 0) .010 (.91) 2 0
0! 2!
0
2
2!
.090 (.91) 2 .8281
P ( y 0) .090 (.91) 2 0
0!2!
0
P( x 0) P( y 0) .9801(.8281) .8116
e.
We must assume that the variables defined as x and y are binomial random variables. We must assume
that the trials are identical, the probability of success is the same from trial to trial, and that the trials are
independent.
Copyright © 2014 Pearson Education, Inc.
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4.57
Chapter 4
a.
b.
E ( x) np 800(.65) 520
npq 800(.65)(.35) 182 13.49
Half of the 800 food items would be 400. A value of x 400 would have a z-score of:
z
x
400 520
8.90
13.49
Since the z-score associated with 400 items is so small (8.90), it would be virtually impossible to
observe less than half with any pesticides if the 65% value was correct.
4.58
Assuming the supplier's claim is true,
np 500(.001) .5 and npq 500(.001)(.999) .4995 .707
If the supplier's claim is true, we would only expect to find .5 defective switches in a sample of size 500.
Therefore, it is not likely we would find 4. Based on the sample, the guarantee is probably inaccurate.
Note: z
4.59
x
4 .5
4.95 . This is an unusually large z-score.
.707
a.
We must assume that the probability that a specific type of ball meets the requirements is always the
same from trial to trial and the trials are independent. To use the binomial probability distribution, we
need to know the probability that a specific type of golf ball meets the requirements.
b.
For a binomial distribution, np and npq .
In this example, n two dozen 2 12 24 , p .10 , and q 1 .10 .90 .
(Success here means the golf ball does not meet standards.)
np 24(.10) 2.4 and npq 24(.10)(.90) 1.47
c.
In this situation, n 24 , p = Probability of success = Probability golf ball does meet standards = .90,
and q 1 .90 .10 .
E ( y ) np 24(.90) 21.6 and npq 24(.10)(.90) 1.47
(Note that is the same as in part b.)
4.60
a.
For this test, n 20 and p .10 . Then x is a binomial random variable with n 20 and p .10 .
Using Table I, Appendix D, with n 20 and p .10 ,
P ( x 1) .392
b.
For the experiment in part a, the level of confidence is 1 P( x 1) 1 .392 .608 . Since this value is
not close to 1, this would not be an acceptable level.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
c.
177
Suppose we increased n from 20 to 25. Using Table I, Appendix D, with n 25 and p .10 ,
P ( x 1) .271 . This value is smaller than the value found in part a. The level of confidence is
1 P( x 1) 1 .271 .729 .
Now, suppose we keep n 20 , but change K to 0 instead of 1. Using Table I, Appendix D, with
n 20 and p .10 ,
P ( x 0) .122 . This value is again, smaller than the value found in part a. The level of confidence is
1 P( x 1) 1 .122 .878 .
d.
Suppose we let K 0 . Now, we need to find n such that the level of confidence .95, which means
that P( x 0) .05 .
n
P( x 0) .10 (.9) n 0 .05
0
n! n
.9 .05
0!n!
.9n .05
ln(.9n ) ln(.05)
nln(.9) ln(.05)
n
ln(.05) 2.99573
28.4
ln(.9)
.10536
Thus, if K 0 , then we need a sample size of 29 or larger to get a level of confidence of at least .95.
Now, suppose K = 1. Now, we need to find n such that the level of confidence is at least .95, which
means that P( x 1) .05 .
n
n
P ( x 1) P ( x 0) P ( x 1) .10 (.9) n 0 .11 (.9) n 1 .05
0
1
n! n
n!
.9
(.1)1 .9n 1 .05
0!n!
1!(n 1)!
.9 n n(.1)1 .9n 1 .05
.9 n 1 .9 (.1) n ln(.05)
From here, we will use trial and error.
Copyright © 2014 Pearson Education, Inc.
178
Chapter 4
30 1
For n 30 , .9
.9 .1(30) .1837
n
.9n1 .9 .1n
30
.9301 .9 .1 30 .1837
.9401 .9 .1 40 .0805
40
.9451 .9 .1 45 .0524
45
.9461 .9 .1 46 .0480
46
Thus, for K 1, we would need a sample size of 46 to get a level of confidence of at least .95.
a.
The random variable x is discrete since it can assume a countable number of values (0, 1, 2, ...).
b.
This is a Poisson probability distribution with 3 .
c.
In order to graph the probability distribution, we need to know the probabilities for the possible values
of x. Using MINITAB with 3 :
Probability Density Function
Poisson with mean = 3
x
0
1
2
3
4
5
6
7
8
9
10
P( X = x )
0.049787
0.149361
0.224042
0.224042
0.168031
0.100819
0.050409
0.021604
0.008102
0.002701
0.000810
Using MINITAB, the probability distribution of x in graphical form is:
Histogram of x
.25
.20
.15
f(x)
4.61
.10
.05
0
0
1
2
3
4
5
6
7
8
9
10
x
d.
3
2 3 and 3 1.7321
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.62
1.5 . Using MINITAB with the Poisson distribution and 1.5 , the probabilities are:
Cumulative Distribution Function
Poisson with mean = 1.5
x
3
2
0
6
4.63
P( X <= x )
0.934358
0.808847
0.223130
0.999074
a.
P( x 3) .934358 .
b.
P( x 3) 1 P( x 2) 1 .808847 .191153 .
c.
P( x 3) P( x 3) P( x 2) .934358 .808847 .125511 .
d.
P( x 0) .22313 .
e.
P( x 0) 1 P( x 0) 1 .22313 .77687 .
f.
P( x 6) 1 P( x 6) 1 .999074 .000926 .
a.
r N r 3 5 3
3! 2!
x n x 1 3 1 1!2! 2!0! 3(1)
.3
P ( x 1)
5!
10
N
5
3!2!
n
3
b.
r N r 3 9 3
3! 6!
x n x 3 5 3 3!0! 2!4! 1(15)
P ( x 3)
.119
9!
126
N
9
5!4!
n
5
c.
r N r 2 4 2
2! 2!
x n x 2 2 2 2!0! 0!2! 1(1)
.167
P ( x 2)
4!
6
N
4
2!2!
n
2
d.
r N r 2 4 2
2! 2!
x n x 0 2 0 0!2! 2!0! 1(1)
.167
P ( x 0)
4!
6
N
4
2!2!
n
2
Copyright © 2014 Pearson Education, Inc.
179
180
4.64
Chapter 4
For N 8, n 3, and r 5 ,
a.
r N r 58 5
5! 3!
x n x 1 3 1 1!4! 2!1! 5(3)
.268
P ( x 1)
8!
56
N
8
3!5!
n
3
b.
r N r 58 5
5! 3!
x n x 0 3 0 0!5! 3!0! 1(1)
.018
P ( x 0)
8!
56
N
8
3!5!
n
3
c.
r N r 58 5
5! 3!
x n x 3 3 3 3!2! 0!3! 10(1)
P ( x 3)
.179
8!
56
N
8
3!5!
n
3
d.
P( x 4) P( x 4) P( x 5) 0
Since the sample size is only 3, there is no way to get 4 or more successes in only 3 trials.
4.65
a.
Using MINITAB with 1 ,and the Poisson distribution, the probability is:
Cumulative Distribution Function
Poisson with mean = 1
x
2
P( X <= x )
0.919699
P( x 2) .919699
b.
Using MINITAB with 2 ,and the Poisson distribution, the probability is:
Cumulative Distribution Function
Poisson with mean = 2
x
2
P( X <= x )
0.676676
P( x 2) .676676
c.
Using MINITAB with 3 ,and the Poisson distribution, the probability is:
Cumulative Distribution Function
Poisson with mean = 3
x
2
P( X <= x )
0.423190
P( x 2) .42319
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
d.
The probability decreases as increases. This is reasonable because is equal to the mean. As the
mean increases, the probability that x is less than a particular value will decrease.
a.
To graph the Poisson probability distribution with 5 , we need to calculate p(x) for x = 0 to 15.
Using MINITAB with 5 , the results are:
Probability Density Function
Poisson with mean = 5
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
P( X = x )
0.006738
0.033690
0.084224
0.140374
0.175467
0.175467
0.146223
0.104445
0.065278
0.036266
0.018133
0.008242
0.003434
0.001321
0.000472
0.000157
Using MINITAB, the graph is:
Histogram of x
.20
.15
f(x)
4.66
181
.10
.05
0
0
2
4
6
8
10
12
14
x
5
2
b.
5 and = 5 2.2361
2 5 2(2.2361) 5 4.4722 (.5278, 9.4722)
Copyright © 2014 Pearson Education, Inc.
182
Chapter 4
c.
Using MINITAB with 5
Cumulative Distribution Function
Poisson with mean = 5
x
9
0
P( X <= x )
0.968172
0.006738
P(.5278 x 9.4722) P(1 x 9) P( x 9) P( x 0) .968172 .006738 .961434
4.67
4.68
For this problem, N 100, n 10, and x 4 .
a.
If the sample is drawn without replacement, the hypergeometric distribution should be used. The
hypergeometric distribution requires that sampling be done without replacement.
b.
If the sample is drawn with replacement, the binomial distribution should be used. The binomial
distribution requires that sampling be done with replacement.
With N 10, n 5, and r 7 , x can take on values 2, 3, 4, or 5.
a.
r N r 7 10 7
7! 3!
x n x 2 5 2 2!5! 3!0! 21(1)
P ( x 2)
.083
10!
252
N
10
5!5!
n
5
r N r 7 10 7
7! 3!
x n x 3 5 3 3!4! 2!1! 35(3)
.417
P ( x 3)
10!
252
N
10
5!5!
n
5
r N r 7 10 7
7! 3!
4
5
4
x
n
x
4!3!1!2! 35(3) .417
P ( x 4)
10!
252
N
10
5!5!
5
n
r N r 7 10 7
7! 3!
x n x 5 5 5 5!2! 0!3! 21(1)
P ( x 5)
.083
10!
252
N
10
5!5!
n
5
The probability distribution of x in tabular form is:
x
2
3
4
5
p(x)
.083
.417
.417
.083
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
nr 5(7)
3.5
10
N
2
c.
183
r ( N r )n( N n) 7(10 7)5(10 5) 525
.583
900
N 2 ( N 1)
102 (10 1)
.5833 .764
2 3.5 2(.764) 3.5 1.528 (1.972, 5.028)
The graph of the distribution is:
0.4
p(x)
0.3
0.2
0.1
0.0
2
1.972
4.69
3
4
x
3.5
d.
P(1.972 x 5.028) P(2 x 5) 1.000
a.
The characteristics of a binomial random variable are:
5
5.028
1.
n identical trials. We are selecting 10 robots from 106. On the first trial, we are selecting 1 robot
out of 106. On the next trial, we are selecting 1 robot out of 105. On the 10th trial, we are selecting
1 robot out of 97. These trials are not identical.
2.
Two possible outcomes. A selected robot either has no legs or wheels or it has some legs or
wheels. S = robot has no legs or wheels and F = robot has either legs and/or wheels. This
condition is met
3.
P(S) remains the same from trial to trial. For this example the probability of success does not stay
constant. On the first trial, there are 106 robots of which 15 have neither legs nor wheels. Thus,
P(S) on the first trial is 15/106. If a robot with neither legs nor wheels is selected on the first trial,
then P(S) on the second trial would be 14/105. If a robot with neither legs nor wheels is not
selected on the first trial, then P(S) on the second trial would be 15/105. The value of P(S) is not
constant from trial to trial. This condition is not met.
4.
Trials are independent. The trials are not independent. The type of robot selected on one trial
affects the type of robot selected on the next trial. This condition is not met.
5.
The random variable x = number of robots selected that do not have legs or wheels in 10 trials.
The necessary conditions for a binomial random variable are not met.
Copyright © 2014 Pearson Education, Inc.
184
Chapter 4
b.
The characteristics of a hypergeometric random variable are:
1. The experiment consists of randomly drawing n elements without replacement from a set of N
elements, r of which are successes and (N – r) of which are failures. For this example there are a
total of N 106 robots, of which r 15 have neither legs nor wheels and N – r 106 –15 95 have
some legs and/or wheels. We are selecting n 10 robots.
2. The hypergeometric random variable x is the number of successes in the draw of n elements. For
this example, x = number of robots selected with no legs or wheels in 20 selections.
c.
4.70
nr 10(15)
1.415 and
106
N
15(106 15)10(106 10)
r ( N r )n( N n)
1.1107 1.0539
2
N ( N 1)
1062 (106 1)
d.
15 106 15
15!
91!
2 10 2 2!(15 2)! 8!(91 8)! 105(8.49869 x1012 )
.2801
P ( x 2)
106!
3.18535 x1013
106
10!(106 10)!
10
a.
E ( x) 45
b.
z
c.
Using MINITAB with 45 , and the Poisson distribution, the probability is:
x
45 6.7082
360 45
46.96
6.7082
Cumulative Distribution Function
Poisson with mean = 45
x
65
P( X <= x )
0.998028
P( x 65) .998028
4.71
a.
With 4.5 , P( x 0)
b.
P ( x 1)
c.
E ( x) 4.5
4.50 e 4.5
0.0111
0!
4.51 e 4.5
0.0500
1!
4.5 2.12
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.72
185
Let x = number of male nannies in 10 trials. For this problem, N = 4,176, r = 24, and n = 10.
r N r
24 4176 24
24! 4152!
0 10 0
x n x
0!24!10!4142!
P ( x 1) 1 P ( x 0) 1
1
1
4176!
N
4176
10!4166!
n
10
1
4.73
1.50613 1036
1 .9439 .0561
1.59559 1036
Let x = number of “clean” cartridges selected in 5 trials. For this problem, N 158, n 5, and r 122 .
r N r 122 36
122! 36!
x n x 5 0 5!117! 0!36!
.2693
P ( x 5)
158!
N
158
5!153!
n
5
4.74
Using MINITAB with 5 , and the Poisson distribution, the probability is:
Cumulative Distribution Function
Poisson with mean = 5
x
10
P( X <= x )
0.986305
P( x 10) 1 P( x 10) 1 .986305 .013695
4.75
Let x = number of times “total visitors” is selected in 5 museums. For this exercise, x has a hypergeometric
distribution with N 30, n 5, r 8 and x 0 .
8 30 8
8!
22!
0
5
0
0!(8
0)!
5!(22
5)!
P ( x 0)
.1848
30!
30
5!(30 5)!
5
4.76
Let x = number of game-day traffic fatalities at the winning team’s location. For this Exercise, x has a
Poisson distribution with .5 .
P ( x 3) 1 P ( x 2) 1 P ( x 0) P( x 1) P( x 2)
1
4.77
.50 e .5 .51 e .5 .52 e .5
1 .6065 .3033 .0758 .0144
0!
1!
2!
Let x = number of times cell phone accesses color code “b” in 7 handoffs. For this problem, x has a
hypergeometric distribution with N 85, n 7, and r 40 .
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186
Chapter 4
40 85 40
40!
45!
2! 40 2 ! 5! 45 5 !
2
72
780(1, 221, 759)
P ( x 2)
.1931
85!
85
4,935,847, 320
7! 85 7 !
7
4.78
4.79
For this exercise, x has a hypergeometric distribution with N 57, n 10, and r 45 .
a.
45 57 45
45!
12!
5! 45 5 ! 5! 12 5 !
5 10 5
1, 221, 759(792)
P ( x 5)
.0224
57!
43,183, 019,880
57
10! 57 10 !
10
b.
45 57 45
45!
12!
8! 45 8 ! 2! 12 2 !
8 10 8
215,553,195(66)
P ( x 8)
.3294
57!
43,183, 019,880
57
10! 57 10 !
10
c.
E ( x)
nr 10(45)
7.895
57
N
Let x = number of flaws in a 4 meter length of wire. For this exercise, x has a Poisson distribution with .8.
The roll will be rejected if there is at least one flaw in the sample of a 4 meter length of wire.
P( x 1) 1 P( x 0) 1
.80 e .8
1 .4493 .5507
0!
We have to assume that the flaws are randomly distributed throughout the roll of wire and that the 4 meter
sample of wire is representative of the entire roll.
4.80
4.81
a.
9 The average of 9 noise events in a unit of time.
b.
9 3
c.
SNR
a.
Using MINITAB with 10 ,
9
3
3
Probability Density Function
Poisson with mean = 10
x
24
P( X = x )
0.0000732
P( x 24) .0000732
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
187
Using MINITAB with 10 ,
Probability Density Function
Poisson with mean = 10
x
23
P( X = x )
0.0001756
P( x 23) .0001756
c.
4.82
Yes, these probabilities are good approximations for the probability of “fire” and “theft”. The
researchers estimated these probabilities to be .0001, indicating that these would be extremely rare
events. Our probabilities of .0001 and .0002 are very close to .0001.
If it takes exactly 5 minutes to wash a car and there are 5 cars in line, it will take 5(5) = 25 minutes to wash
these 5 cars. Thus, for anyone to be in line at closing time, more than 1 car must arrive in the final ½ hour. In
addition, if on average 10 cars arrive per hour, then an average of 5 cars will arrive per ½ hour (30 minutes).
If we let x = number of cars to arrive in ½ hour, then x is a Poisson random variable with 5 .
Using MINITAB with 5 ,
Cumulative Distribution Function
Poisson with mean = 5
x
1
P( X <= x )
0.0404277
P( x 1) 1 P( x 1) 1 .0404277 .9595723
Since this probability is so large, it is very likely that someone will be in line at closing time.
4.83
Let x = number of females promoted in the 72 employees awarded promotion, where x is a
hypergeometric random variable. From the problem, N 302, n 72, and r 73 . We need to find if
observing 5 females who were promoted was fair.
E ( x)
nr 72(73)
17.40
302
N
If 72 employees are promoted, we would expect that about 17 would be females.
V ( x) 2
r ( N r )n( N n) 73(302 73)72(302 72)
10.084
N 2 ( N 1)
3022 (302 1)
10.084 3.176
Using Chebyshev’s Theorem, we know that at least 8/9 of all observations will fall within 3
standard deviations of the mean. The interval from 3 standard deviations below the mean to 3
standard deviations above the mean is:
3 17.40 3(3.176) 17.40 9.528 (7.872, 26.928)
If there is no discrimination in promoting females, then we would expect between 8 and 26
females to be promoted within the group of 72 employees promoted. Since we observed only 5 females
promoted, we would infer that females were not promoted fairly.
Copyright © 2014 Pearson Education, Inc.
188
4.84
4.85
Chapter 4
Table II in the text gives the area between z 0 and z z0 . In this exercise, the answers may thus be read
directly from the table by looking up the appropriate z.
a.
P(0 z 2.0) .4772
b.
P(0 z 3.0) .4987
c.
P(0 z 1.5) .4332
d.
P(0 z .80) .2881
Using Table II, Appendix D:
a.
P( z 1.46) .5 P(0 z 1.46) .5 .4279 .0721
b.
P( z 1.56) .5 P(1.56 z 0) .5 .4406 .0594
c.
P(.67 z 2.41) P(0 z 2.41) P(0 z .67)
.4920 .2486 .2434
d.
P(1.96 z .33) P(1.96 z 0) P(.33 z 0)
.4750 .1293 .3457
e.
f.
P( z 0) .5
P(2.33 z 1.50) P(2.33 z 0) P(0 z 1.50)
.4901 .4332 .9233
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.86
Using Table II, Appendix D,
a.
b.
c.
d.
e.
f.
4.87
P(1 z 1) P(1 z 0) P(0 z 1)
.3413 .3413 .6826
P(2 z 2) P(2 z 0) P(0 z 2)
.4772 .4772 .9544
P(2.16 z 0.55) P(2.16 z 0) P(0 z 0.55)
.4846 .2088 .6934
P(.42 z 1.96) P(.42 z 0) P(0 z 1.96)
.1628 .4750 .6378
P( z 2.33) P(2.33 z 0) P( z 0)
.4901 .5 .9901
P( z 2.33) P( z 0) P(0 z 2.33)
.5 .4901 .9901
Using Table II, Appendix D:
a.
P(1 z 1) P(1 z 0) P(0 z 1)
.3413 .3413 .6826
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190
Chapter 4
b.
c.
d.
4.88
P(1.96 z 1.96) P(1.96 z 0) P(0 z 1.96)
.4750 .4750 .9500
P(1.645 z 1.645) P(1.645 z 0) P(0 z 1.645)
.4500 .4500 .9000
(using interpolation)
P(2 z 2) P(2 z 0) P(0 z 2)
.4772 .4772 .9544
Using Table II, Appendix D:
a.
P ( z z0 ) .05
A1 .5 .05 .4500
Looking up the area .4500 in Table II gives z0 1.645 .
b.
P( z z0 ) .025
A1 .5 .025 .4750
Looking up the area .4750 in Table II gives z 0 1.96 .
c.
P ( z z0 ) .025
A1 .5 .025 .4750
Looking up the area .4750 in Table II gives
z 1.96 . Since z0 is to the left of 0, z 0 1.96 .
d.
P ( z z0 ) .10
A1 .5 .1 .4000
Looking up the area .4000 in Table II gives z 0 1.28 .
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
e.
P ( z z 0 ) .10
A1 .5 .1 .4000
z 0 1.28 (same as in d)
4.89
Using Table II of Appendix D:
a.
P ( z z 0 ) .2090
A .5 .2090 .2910
Looking up the area .2910 in the body of Table II
gives z0 .81 . (z0 is negative since the graph
shows z0 is on the left side of 0.)
b.
P ( z z 0 ) .7090
P( z z0 ) P( z 0) P(0 z z0 )
.5 P(0 z z0 ) .7090
Therefore, P (0 z z 0 ) .7090 .5 .2090 A
Looking up the area .2090 in the body of Table II gives z0 .55 .
c.
P ( z0 z z0 ) .8472
P ( z0 z z0 ) 2 P (0 z z0 ) .8472
Therefore, P (0 z z0 ) .8472 / 2 .4236 .
Looking up the area .4236 in the body of Table II gives z 0 1.43 .
d.
P ( z0 z z0 ) .1664
P ( z0 z z0 ) 2 P (0 z z0 ) .1664
Therefore, P (0 z z0 ) .1664 / 2 .0832 .
Looking up the area .0832 in the body of Table II gives z0 .21 .
e.
P ( z 0 z 0) .4798
P ( z 0 z 0) P (0 z z 0 )
Looking up the area .4798 in the body of Table II gives
z 0 2.05 .
f.
P ( 1 z z 0 ) .5328
P ( 1 z z 0 ) P ( 1 z 0) P (0 z z 0 ) .5328
Copyright © 2014 Pearson Education, Inc.
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192
Chapter 4
P (0 z 1) P (0 z z 0 ) .5328
Thus, P (0 z z 0 ) .5328 .3413 .1915
Looking up the area .1915 in the body of Table II gives z0 .50 .
4.90
4.91
4.92
a.
z 1
b.
z 1
c.
z0
d.
z 2.5
e.
z3
a.
z
b.
z
c.
z
d.
z
e.
z
f.
z
x
x
x
x
x
x
20 30
2.50
4
30 30
0
4
27.5 30
0.625
4
15 30
3.75
4
35 30
1.25
4
25 30
1.25
4
Using Table II of Appendix D:
a.
To find the probability that x assumes a value more
than 2 standard deviations from :
P( x 2 ) P( x 2 ) P( z 2) P( z 2)
2 P( z 2) 2(.5 .4772) 2(.0228) .0456
To find the probability that x assumes a value more than
3 standard deviations from :
P( x 3 ) P( x 3 ) P( z 3) P( z 3)
2 P( z 3) 2(.5 .4987) 2(.0013) .0026
b.
To find the probability that x assumes a value within 1
standard deviation of its mean:
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
P( x ) P(1 z 1)
2 P(0 z 1) 2(.3413) .6826
To find the probability that x assumes a value within 2
standard deviations of :
P( 2 x 2 ) P(2 z 2)
2 P(0 z 2) 2(.4772) .9544
c.
To find the value of x that represents the 80th percentile, we
must first find the value of z that corresponds to the 80th percentile.
P ( z z0 ) .80 . Thus, A1 A2 .80 . Since A1 .50 ,
A2 .80 .50 .30 . Using the body of Table II, z0 .84 .
To find x, we substitute the values into the z-score formula:
z
x
.84
x 1000
x .84(10) 1000 1008.4
10
To find the value of x that represents the 10th percentile, we
must first find the value of z that corresponds to the 10th percentile.
P ( z z0 ) .10 . Thus, A1 .50 .10 .40 . Using the body
of Table II, z 0 1.28 . To find x, we substitute the values
into the z-score formula:
z
4.93
a.
b.
x
1.28
x 1000
x 1.28(10) 1000 987.2
10
12 11
10 11
P(10 x 12) P
z
2
2
P(0.50 z 0.50) A1 A2 .1915 .1915 .3830
10 11
6 11
P(6 x 10) P
z
P(2.50 z 0.50)
2
2
P(2.50 z 0) P(0.50 z 0) .4938 .1915 .3023
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194
Chapter 4
c.
d.
e.
f.
4.94
16 11
13 11
P(13 x 16) P
z
P(1.00 z 2.50)
2
2
P(0 z 2.50) P(0 x 1.00) .4938 .3413 .1525
12.6 11
7.8 11
P(7.8 x 12.6) P
z
2
2
P(1.60 z 0.80) A1 A2 .4452 .2881 .7333
13.24 11
P( x 13.24) P z
2
P( z 1.12) A2 .5 A1 .5000 .3686 .1314
7.62 11
P( x 7.62) P z
2
P( z 1.69) A1 A2 =.4545 .5000 .9545
The random variable x has a normal distribution with 50 and 3 .
a.
P ( x x0 ) .8413
So, A1 A2 .8413
Since A1 .5 , A2 .8413 .5 .3413
Looking up the area .3413 in the body of Table II,
Appendix D gives z0 1.0 .
To find x0, substitute all the values into the z-score formula:
z
b.
x
1.0
x0 50
x0 50 3(1.0) 53
3
P ( x x0 ) .025
So, A .5 .025 .4750
Looking up the area .4750 in the body of Table II,
Appendix D gives z 0 1.96 .
To find x0, substitute all the values into the z-score formula:
z
x
1.96
x0 50
x0 50 3(1.96) 55.88
3
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Random Variables and Probability Distributions
c.
195
P ( x x0 ) .95
So, A1 A2 .95 . Since A2 .5 , A1 .95 .5 .4500 .
Looking up the area .4500 in the body of Table II,
Appendix D gives (since it is exactly between two values, average
the z-scores) z 0 1.645 .
To find x0, substitute into the z-score formula:
x
z
d.
1.645
x0 50
x0 50 3(1.645) 45.065
3
P (41 x x0 ) .8630
x
z
41 50
3
3
A1 P(41 x ) P(3 z 0)
P(0 z 3) .4987
A1 A2 .8630 , since A1 .4987 , A2 .8630 .4987 .3643 . Looking up .3643 in the body of Table
II, Appendix D gives z0 1.1 .
To find x0, substitute into the z-score formula:
z
e.
x
1.1
x0 50
x 50 3(1.1) 53.3
3
P ( x x0 ) .10
So A .5 .10 .4000
Looking up area .4000 in the body of Table II, Appendix D gives z 0 1.28 . Since z0 is to the left of 0,
z 0 1.28 .
To find x0, substitute all the values into the z-score formula:
z
f.
x
1.28
x 50
x 50 3(1.28) 46.16
3
P ( x x0 ) .01
A .5 .01 .4900
Looking up area .4900 in the body of Table II, Appendix D gives
z0 2.33 .
To find x0, substitute all the values into the z-score formula:
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Chapter 4
z
4.95
a.
x
2.33
x 50
x 50 3(2.33) 56.99
3
In order to approximate the binomial distribution with the normal distribution, the interval
3 np 3 npq should lie in the range 0 to n.
When n = 25 and p = .4,
np 3 npq 25 .4 3 25(.4)(1 .4) 10 3 6 10 7.3485 (2.6515, 17.3485)
Since the interval calculated does lie in the range 0 to 25, we can use the normal approximation.
4.96
b.
2
np 25(.4) 10 and npq 25(.4)(.6) 6
c.
P ( x 9) 1 P ( x 8) 1 .274 .726 (Table I, Appendix D)
d.
(9 .5) 10
P( x 9) P z
6
P( z .61) .5000 .2291 .7291
(using Table II, Appendix D)
np 1000(.5) 500 , npq 1000(.5)(.5) 15.811
a.
Using the normal approximation,
(500 .5) 500
P( x 500) P z
P z .03 =.5 .0120 .4880
15.811
(from Table II, Appendix D)
4.97
b.
(500 .5) 500
(490 .5) 500
z
P (490 x 500) P
P ( .66 z .03)
15.811
15.811
.2454 .0120 .2334
(from Table II, Appendix D)
c.
(550 .5) 500
P( x 550) P z
P z 3.19 .5 .49929 .00071
15.811
(from Table II, Appendix D)
a.
Using MINITAB with 105.3 and 8 , the probability is:
Cumulative Distribution Function
Normal with mean = 105.3 and standard deviation = 8
x
120
P( X <= x )
0.966932
P( x 120) 1 P x 120 1 .966932 .033068
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Random Variables and Probability Distributions
b.
197
Using MINITAB with 105.3 and 8 , the probabilities are:
Cumulative Distribution Function
Normal with mean = 105.3 and standard deviation = 8
x
110
100
P( X <= x )
0.721566
0.253825
P (100 x 110) P ( x 110) P ( x 100) .721566 .253825 .467741
c.
Using MINITAB with 105.3 and 8 , the value of a is found:
Inverse Cumulative Distribution Function
Normal with mean = 105.3 and standard deviation = 8
P( X <= x )
0.25
x
99.9041
Thus, a = 99.9041.
4.98
Using MINITAB with 67.755 and 26.871 , the probabilities are:
Cumulative Distribution Function
Normal with mean = 67.755 and standard deviation = 26.871
x
40
120
P( X <= x )
0.150826
0.974070
a.
P ( x 40) .150826
b.
P (40 x 120) P ( x 120) P ( x 40) .974070 .150826 .823244
c.
P ( x 120) 1 P ( x 120) 1 .974070 .02593
d.
We want to find a where P ( x a ) .25 . Using MINITAB with 67.755 and 26.871 , the value
of a is found:
Inverse Cumulative Distribution Function
Normal with mean = 67.755 and standard deviation = 26.871
P( X <= x )
0.25
x
49.6308
Thus, a = 49.6308.
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198
4.99
Chapter 4
a.
Using MINITAB with 59 and 5 , the probability is:
Cumulative Distribution Function
Normal with mean = 59 and standard deviation = 5
x
60
P( X <= x )
0.579260
P ( x 60) 1 P ( x 60) 1 .57926 .42074
b.
Using MINITAB with 43 and 5 , the probability is:
Cumulative Distribution Function
Normal with mean = 43 and standard deviation = 5
x
60
P( X <= x )
0.999663
P ( x 60) 1 P ( x 60) 1 .999663 .000337
4.100
a.
Using Table II, Appendix D,
0 5.26
P ( x 0) P z
P ( z 0.526)
10
.5 P ( 0.53 z 0) .5 .2019 .7019
b.
15 5.26
5 5.26
P(5 x 15) P
z
P(0.026 z 0.974)
10
10
P(.03 z 0) P(0 z .97) .0120 .3340 .3460
c.
1 5.26
P( x 1) P z
P( z 0.426)
10
.5 P(0.43 z 0) .5 .1664 .3336
d.
25 5.26
P( x 25) P z
P( z 3.026)
10
.5 P(3.03 z 0) .5 .4988 .0012
Since the probability of seeing a win percentage of -25% or anything more unusual is so small (p =
.0012), we would conclude that the average casino win percentage is not 5.26%.
4.101
a.
Let x buy-side analyst’s forecast error. Then x has an approximate normal distribution with
.85 and 1.93 . Using Table II, Appendix D,
2.00 .85
P( x 2.00) P z
P( z .60) .5 .2257 .2743
1.93
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Random Variables and Probability Distributions
b.
199
Let y sell-side analyst’s forecast error. Then y has an approximate normal distribution with
.05 and .85 . Using Table II, Appendix D,
2.00 (.05)
P( y 2.00) P z
P( z 2.41) .5 .4920 .0080
.85
4.102
Let x driver’s head injury rating. The random variable x has a normal distribution with 605 and 185 .
Using Table II, Appendix D,
a.
700 605
500 605
z
P(500 x 700) P
P(0.57 z 0.51)
185
185
P ( 0.57 z 0) P (0 z 0.51) .2157 .1950 .4107
b.
500 605
400 605
z
P(400 x 500) P
P(1.11 z 0.57)
185
185
P ( 1.11 z 0) P ( 0.57 z 0) .3665 .2157 .1508
c.
850 605
P( x 850) P z
P( z 1.32) .5 P(0 z 1.32)
185
.5 .4066 .9066
d.
1, 000 605
P( x 1, 000) P z
P( z 2.14) .5 P(0 z 2.14)
185
.5 .4838 .0162
4.103
From Exercise 4.49, we determined that x is a binomial random variable with n = 250 and p = .5
a.
np 250(.5) 125
b.
npq 250(.5)(.5) 62.5 7.9057
c.
z
d.
In order to approximate the binomial distribution with the normal distribution, the interval
x
200 125
9.49
7.9057
3 np 3 npq should lie in the range 0 to n.
When n 250 and p .5 , np 3 npq 125 3(7.9057) 125 23.7171 (101.2829, 148.7171)
Since the interval calculated does lie in the range 0 to 250, we can use the normal approximation.
Copyright © 2014 Pearson Education, Inc.
200
Chapter 4
Using MINITAB with 125 and 7.9057 , the approximate probability is:
Cumulative Distribution Function
Normal with mean = 125 and standard deviation = 7.9057
x
200
P( X <= x )
1
P ( x 200) 1
4.104
Let x = number of patients who undergo laser surgery who have serious post-laser vision problems in 100,000
trials. Then x is a binomial random variable with n 100, 000 and p .01 .
E ( x ) np 100, 000(.01) 1, 000
2 npq 100, 000(.01)(.99) 990 31.464
To see if the normal approximation is appropriate, we use:
3 1, 000 3(31.464) 1, 000 94.392 (905.608, 1, 094.392)
Since the interval lies in the range of 0 to 100,000, the normal approximation is appropriate.
949.5 1000
P( x 950) P z
P( z 1.61) .5 .4463 .0537 (using Table II, Appendix D)
31.464
4.105
If the goal keeper stands in the middle of the goal and can reach any ball within 9 feet, then the only way a
player can score is if he/she shoots the ball within 3 feet of either goal post.
a.
If a player aims at the right goal post, then the player will score if x is between -3 and 0. Using Table II,
00
3 0
z
Appendix D, we get P(3 x 0) P
P(1 z 0) .3413 .
3
3
b.
If a player aims at the center of the goal, then the player will score if x is greater than 9 or less than -9.
Using Table II, Appendix D, we get
9 0
90
P ( x 9) P ( x 9) P z
P z
P ( z 3) P ( z 3)
3
3
(.5 .4987) (.5 .4987) .0026
c.
If a player aims halfway between the right goal post and the outer limit of the goal keeper’s reach, then
the player will score if x is between -1.5 and 1.5. Using Table II, Appendix D, we get
1.5 0
1.5 0
z
P(1.5 x 1.5) P
P(.5 z .5) .1915 .1915 .3830 .
3
3
4.106 Let x = number of defects per million. Then x has an approximate normal distribution with 3 . Using Table
II, Appendix D,
3 1.5 3
3 1.5 3
z
P(3 1.5 x 3 1.5 ) P
P(1.5 z 1.5) .4332 .4332 .8664
It is fairly likely that the goal will be met. Since the probability is .8664, the goal would be met
approximately 86.64% of the time.
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Random Variables and Probability Distributions
4.107
a.
201
Let x = rating. Then x has a normal distribution with 50 and 15 . Using Table II,
Appendix D,
P x xo .10. Find xo.
x 50
P( z zo ) .10
P( x xo ) P z o
15
A1 .5 .10 .4000
Looking up area .4000 in Table II, z o 1.28
zo
b.
xo 50
x 50
1.28 o
xo 50 1.28(15) 69.2
15
15
P x xo .10 .20 .40 .70. Find xo.
x 50
P( z zo ) .70
P( x xo ) P z o
15
A1 .70 .5 .2000
Looking up area .2000 in Table II, zo .52
zo
4.108
a.
xo 50
x 50
.52 o
xo 50 .52(15) 42.2
15
15
Using MINITAB with 7.5 and 2.5 , the probability is:
Cumulative Distribution Function
Normal with mean = 7.5 and standard deviation = 2.5
x
9
P( X <= x )
0.725747
P ( x 9) .725747 . Since this probability is less than .90, the regulations are not being met at EMS
station A.
b.
Using MINITAB with 7.5 and 2.5 , the probability is:
Cumulative Distribution Function
Normal with mean = 7.5 and standard deviation = 2.5
x
2
P( X <= x )
0.0139034
P ( x 2) .0139034 . Since this probability is so small, it would be very unlikely that the call was
serviced by Station A.
4.109
a.
Using Table II, Appendix D, and 75 and 7.5 ,
80 75
P( x 80) P z
P z .67 .5 .2486 .2514
7.5
Thus, 25.14% of the scores exceeded 80.
Copyright © 2014 Pearson Education, Inc.
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Chapter 4
b.
P ( x x0 ) .98. Find x0.
x 75
P( x x0 ) P z 0
P( z z0 ) .98
7.5
A1 .98 .5 .4800
Looking up area .4800 in Table II, zo 2.05 .
z0
4.110
x 0 75
x 75
2.05 0
x 0 90.375
7.5
7.5
Let x = wage rate. The random variable x is normally distributed with 18.50 and 1.25 . Using Table
II, Appendix D,
a.
19.80 18.50
P( x 19.80) P z
P( z 1.04)
1.25
.5 P(0 z 1.04) .5 .3508 .1492
b.
19.80 18.50
P( x 19.80) P z
P( z 1.04)
1.25
.5 P(0 z 1.04) .5 .3508 .1492
c.
P ( x ) P ( x ) .5 . Thus, 18.50 .
(Recall from section 2.4 that in a symmetric distribution, the mean equals the median.)
4.111
Let x = number of additional Electoral College votes a candidate will win if he/she wins California’s 55 votes.
Then x has a normal distribution with 241.5 and 49.8 . In order to be elected, the candidate will have
to win an additional 270 – 55 215 votes or x has to be greater than or equal to 215. Using MINITAB with
241.5 and 49.8 , the probability is:
Cumulative Distribution Function
Normal with mean = 241.5 and standard deviation = 49.8
x
215
P( X <= x )
0.297318
P ( x 215) 1 P ( x 215) 1 .297318 .702682
The probability the candidate becomes the next president if he/she wins California is about .70.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.112
203
Let x = number of guests who participate in the conservation program in 200 trials. Then x is a binomial
random variable with n 200 and p .45 . The mean of the distribution is np 200(.45) 90 and the
standard deviation is npq 200(.45)(.55) 49.5 7.0356 .
In order to approximate the binomial distribution with the normal distribution, the interval
3 np 3 npq should lie in the range 0 to n.
When n 200 and p .45 ,
np 3 npq 90 3(7.0356) 90 21.1068 (68.8932, 111.1068)
Since the interval calculated does lie in the range 0 to 200, we can use the normal approximation.
110 .5 90
P( x 110) P z
P( z 2.91) .5 .4982 .0018
7.0356
Since this probability is so low, it is very unlikely that the claim is true.
4.113
b.
Let v = number of credit card users out of 100 who carry Visa. Then v is a binomial random variable
with n 100 and p .50 .
E (v ) npv 100(.50) 50.
Let d = number of credit card users out of 100 who carry Discover. Then d is a binomial random
variable with n 100 and pd .09 .
E ( d ) np d 100(.09) 9.
c.
To see if the normal approximation is valid, we use:
3 npv 3 npv qv 100(.5) 3 100(.5)(.5) 50 3(5)
50 15 (35, 65)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probability.
(50 .5) 50
P(v 50) P z
P( z .1) .5 .0398 .5398
5
Let a = number of credit card users out of 100 who carry American Express. Then a is a binomial
random variable with n 100 and pa .08 . To see if the normal approximation is valid, we use:
3 npa 3 npa qa 100(.08) 3 100(.08)(.92) 8 3(2.713)
8 8.139 (.139, 16.139)
Since the interval does not lie in the range 0 to 100, using the normal approximation to approximate the
probability is risky.
(50 .5) 8
P(a 50) P z
P( z 15.30) .5 .5 0
2.713
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4.114
Chapter 4
d.
In order for the normal approximation to be valid, 3 must lie in the interval (0, n). This check was
done in part c for both portions of the question. The normal approximation was justified for the first
part but not the second.
a.
Let x = quantity injected per container. The random variable x has a normal distribution with 10
and .2 .
10 10
P x 10 P z
P z 0.0 .5
.2
10 10
P( x 10) P z
P( z 0.0) .5
.2
4.115
b.
Since the container needed to be reprocessed, it cost $10. Upon refilling, it contained 10.60 units with
a cost of 10.60($20) = $212. Thus, the total cost for filling this container is $10 + $212 = $222. Since
the container sells for $230, the profit is $230 $222 = $8.
c.
Let x = quantity injected per container. The random variable x has a normal distribution with
10.10 and .2 . The expected value of x is E ( x ) 10.10 . The cost of a container with 10.10
units is 10.10($20) = $202. Thus, the expected profit would be the selling price minus the cost or
$230 $202 = $28.
We have to find the probability of observing x = .7 or anything more unusual given the two different
values of .
Without receiving executive coaching: Using Table II, Appendix D with .75 and .085 ,
.7 .75
P( x .7) P z
P( z .59) .5 .2224 .2776 .
.085
After receiving executive coaching: Using Table II, Appendix D with .52 and .075 ,
.7 .52
P( x .7) P z
P( z 2.40) .5 .4918 .0082 .
.075
Since the probability of observing x .7 for those not receiving executive coaching is much larger than the
probability of x .7 for those receiving executive coaching, it is more likely that the leader did not receive
executive coaching.
4.116
a.
If z is a standard normal random variable,
QL zL is the value of the standard normal distribution which has 25% of the data to the left and 75% to
the right.
Find zL such that P z zL .25
A1 .50 .25 .25.
Look up the area A1 .25 in the body of Table II of Appendix D; zL .67 (taking the closest value). If
interpolation is used, .675 would be obtained.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
205
QU z U is the value of the standard normal distribution which has 75% of the data to the left and 25%
to the right.
Find zU such that P z zU .75
A1 A2 P( z 0) P(0 z zU )
.5 P(0 z zU ) .75
Therefore, P (0 z z U ) .25.
Look up the area .25 in the body of Table II of Appendix D; z U .67 (taking the closest value).
b.
Recall that the inner fences of a box plot are located 1.5(QU QL ) outside the hinges (QL and QU).
To find the lower inner fence,
QL 1.5(QU QL ) .67 1.5 .67 ( .67) .67 1.5 1.34
2.68 ( 2.70 if zL .675 and z U .675)
The upper inner fence is:
QU 1.5(QU QL ) .67 1.5 .67 ( .67) .67 1.5 1.34
2.68 ( 2.70 if zL .675 and z U .675)
c.
Recall that the outer fences of a box plot are located 3(QU QL ) outside the hinges
(QL and QU).
To find the lower outer fence,
QL -3(QU QL ) .67 3 .67 ( .67) .67 3 1.34
4.69 ( 4.725 if zL .675 and z U .675)
The upper outer fence is:
QU 3(QU QL ) .67 3 .67 ( .67) .67 3 1.34
4.69 (4.725 if z L .675 and z U .675)
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Chapter 4
d.
P ( z 2.68) P z 2.68 2P z 2.68
2(.5000 .4963) 2 .0037 .0074
(Table II, Appendix D)
(or 2(.5000 .4965) .0070 if 2.70 and 2.70 are used)
P( z 4.69) P z 4.69 2P z 4.69
2(.5000 .5000) 0
4.117
4.118
e.
In a normal probability distribution, the probability of an observation being beyond the inner fences is
only .0074 and the probability of an observation being beyond the outer fences is approximately zero.
Since the probability is so small, there should not be any observations beyond the inner and outer
fences. Therefore, they are probably outliers.
a.
The proportion of measurements that one would expect to fall in the interval is about .68.
b.
The proportion of measurements that one would expect to fall in the interval 2 is about .95.
c.
The proportion of measurements that one would expect to fall in the interval 3 is about 1.00.
a.
IQR QU QL = 195 72 123
b.
IQR / s 123 / 95 1.295
c.
Yes. Since IQR is approximately 1.3, this implies that the data are approximately normal.
4.119
If the data are normally distributed, then the normal probability plot should be an approximate straight line.
Of the three plots, only plot c implies that the data are normally distributed. The data points in plot c form an
approximately straight line. In both plots a and b, the plots of the data points do not form a straight line.
4.120
a.
Using MINITAB, the stem-and-leaf display is:
Stem-and-Leaf Display: x
Stem-and-leaf of x
Leaf Unit = 0.10
5
6
8
11
14
14
10
7
2
1
2
3
4
5
6
7
8
9
N
= 28
11266
1
35
035
039
3457
346
24469
47
Since the data do not form a mound-shape, it indicates that the data may not be normally distributed.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
207
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: x
Variable
x
N
28
Mean
5.511
StDev
2.765
Minimum
1.100
Q1
3.350
Median
6.100
Q3
8.050
Maximum
9.700
The standard deviation is 2.765.
c.
Using the printout from MINITAB in part b, QL 3.35 , and QU 8.05 . The
IQR QU QL = 8.05 3.35 4.7 . If the data are normally distributed, then IQR / s 1.3 .
For this data, IQR / s 4.7 / 2.765 1.70 . This is a fair amount larger than 1.3, which indicates that the
data may not be normally distributed.
d.
Using MINITAB, the normal probability plot is:
Probability Plot of x
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
5.511
2.765
28
0.533
0.158
Percent
80
70
60
50
40
30
20
10
5
1
-5
0
5
x
10
15
The data do not form a particularly a straight line. This indicates that the data are not normally
distributed.
a.
Using MINITAB, a histogram of the data is:
Histogram of Support
Normal
80
Mean
StDev
N
70
67.76
26.87
992
60
Frequency
4.121
50
40
30
20
10
0
0
20
40
60
80
Support
100
120
140
The data are fairly mound-shaped. This indicates that the data are probably from a normal distribution.
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Chapter 4
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Support
Variable
Support
N
992
Mean
67.755
StDev
26.871
Minimum
0.000000000
Q1
49.000
Median
68.000
Q3
86.000
Maximum
155.000
If the data are normal, then approximately 68% of the observations should fall within 1 standard
deviation of the mean. For this data, the interval is x s 67.755 26.871 (40.884, 94.626) .
There are 665 out of the 992 observations in this interval which is 665 / 992 .670 or 67%. This is very
close to the 68%.
If the data are normal, then approximately 95% of the observations should fall within 2 standard
deviations of the mean. For this data, the interval is
x 2 s 67.755 2(26.871) 67.755 53.742 (14.013, 121.497) . There are 946 out of the 992
observations in this interval which is 946 / 992 .954 or 95.4%. This is very close to the 95%.
If the data are normal, then approximately 100% of the observations should fall within 3 standard
deviations of the mean. For this data, the interval is
x 3s 67.755 3(26.871) 67.755 80.613 ( 12.858, 148.368) . There are 991 out of the 992
observations in this interval which is 991/ 992 .999 or 99.9%. This is very close to the 100%.
Since these percents are very close to percentages for the normal distribution, this indicates that the data
are probably from a normal distribution.
c.
The IQR QU QL 86 49 37 and the standard deviation is s 26.871 . If the data are normal,
IQR
37
IQR
1.3 . For this data,
1.377 . This is very close to 1.3. This indicates that the
s
s
26.871
data probably come from a normal distribution.
then
d.
Using MINITAB, the normal probability plot is:
Probability Plot of Support
Normal - 95% CI
99.99
Mean
StDev
N
AD
P-Value
99
Percent
95
67.76
26.87
992
0.496
0.214
80
50
20
5
1
0.01
0
50
100
Support
150
200
Except for the several 0’s on the left of the plot, the data are very close to a straight line. This again
indicates that the data probably come from a normal distribution.
4.122
The histogram of the data is very close to a normal distribution. The engineers should use the normal
distribution to model the behavior of shear strength for rack fractures.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.123
a.
209
If the data are normal, then approximately 68% of the observations should fall within 1 standard
deviation of the mean. For this data, the interval is x s 89.2906 3.1834 (86.1072, 92.4740) .
There are 34 out of the 50 observations in this interval which is 34 / 50 .68 or 68%. This is exactly
the 68%.
If the data are normal, then approximately 95% of the observations should fall within 2 standard
deviations of the mean. For this data, the interval is
x 2 s 89.2906 2(3.1834) 89.2906 6.3668 (82.9238, 95.6574) . There are 48 out of the 50
observations in this interval which is 48 / 50 .96 or 96%. This is very close to the 95%.
If the data are normal, then approximately 100% of the observations should fall within 3 standard
deviations of the mean. For this data, the interval is
x 3s 89.2906 3(3.1934) 89.2906 9.5502 (79.7404, 98.8408) . There are 50 out of the 50
observations in this interval which is 50 / 50 1.00 or 100%. This is exactly the 100%.
Since these percents are very close to percentages for the normal distribution, this indicates that the data
are approximately normal.
The IQR QU QL 91.88 87.2725 4.6075 and the standard deviation is s = 3.1834. If the data are
IQR 4.6075
IQR
1.3 . For this data,
1.447 . This is fairly close to 1.3. This
s
s
3.1834
indicates that the data are approximately normal.
normal, then
b.
4.124
4.125
The data on the plot are fairly close to a straight line. This indicates that the data are approximately
normal.
Based on the normal probability plot, it appears that the data are not approximately normal. If the data are
normal, then the probability plot should reflect a straight line. In this graph, the plot of the data is not a
straight line.
The information given in the problem states that x 4.71 , s 6.09 , QL 1 , and QU 6 . To be normal, the
data have to be symmetric. If the data are symmetric, then the mean would equal the median and would be
half way between the lower and upper quartile. Half way between the upper and lower quartiles is 3.5. The
sample mean is 4.71, which is much larger than 3.5. This implies that the data may not be normal. In
addition, the interquartile range divided by the standard deviation will be approximately 1.3 if the data are
normal. For this data,
IQR QU QL 6 1
.82
6.09
s
s
The value of .82 is much smaller than the necessary 1.3 to be normal. Again, this is an indication that the data
are not normal. Finally, the standard deviation is larger than the mean. Since one cannot have values of the
variable in this case less than 0, a standard deviation larger than the mean indicates that the data are skewed to
the right. This implies that the data are not normal.
Copyright © 2014 Pearson Education, Inc.
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Chapter 4
We will look at the 4 methods or determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the failure times of the 50 used panels is:
Histogram of Fail
Normal
12
Mean
StDev
N
1.935
0.9287
50
10
8
Frequency
210
6
4
2
0
0
1
2
Fail
3
4
From the histogram, the data appear to have a somewhat normal distribution.
Next, we look at the intervals x s, x 2 s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the
summary statistics are:
Descriptive Statistics: Fail
Variable
Fail
N
50
Mean
1.935
StDev
0.929
Q1
1.218
Median
1.835
Q3
2.645
x s 1.935 .929 (1.006, 2.864) 33 of the 50 values fall in this interval. The proportion is
33/50 = .66. This is fairly close to the .68 we would expect if the data were normal.
x 2 s 1.935 2(.929) 1.935 1.858 (0.077, 3.793) 49 of the 50 values fall in this interval. The
proportion is 49/50 = .98. This is a fair amount above the .95 we would expect if the data were normal.
x 3s 1.935 3(.929) 1.935 2.787 ( 0.852, 4.722) 50 of the 50 values fall in this interval. The
proportion is 50/50 =1.00. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR = Q U – Q L = 2.645 – 1.218 1.427 .
IQR 1.427
1.54 . This is somewhat larger than the 1.3 we would expect if the data were normal. This
s
.929
method indicates the data may be normal.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
211
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Fail
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
1.935
0.9287
50
0.305
0.557
Percent
80
70
60
50
40
30
20
10
5
1
-1
0
1
2
Fail
3
4
5
Since the data form a fairly straight line, the data may be normal.
From the 4 different methods, all indications are that the failure times are approximately normal.
We will look at the 4 methods for determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the driver’s head injury rating is:
Histogram of DrivHead
Normal
25
Mean
StDev
N
603.7
185.4
98
20
Frequency
4.127
15
10
5
0
200
400
600
800
DrivHead
1000
1200
From the histogram, the data appear to be somewhat skewed to the right, but is fairly mound-shaped. This
indicates that the data are approximately normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each interval
are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary
statistics are:
Descriptive Statistics: DrivHead
Variable
DrivHead
N
98
Mean
603.7
StDev
185.4
Minimum
216.0
Q1
475.0
Median
605.0
Q3
724.3
Maximum
1240.0
x s 603.7 185.4 (418.3, 789.1) 68 of the 98 values fall in this interval. The proportion is .69. This
is very close to the .68 we would expect if the data were normal.
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Chapter 4
x 2 s 603.7 2(185.4) 603.7 370.8 (232.9, 974.5) 96 of the 98 values fall in this interval. The
proportion is .98. This is a fair amount larger than the .95 we would expect if the data were normal.
x 3s 603.7 3(185.4) 603.7 556.2 (47.5, 1,159.9) 97 of the 98 values fall in this interval. The
proportion is .99. This is fairly close to the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR QU QL 724.3 475.0 249.3 .
IQR 249.3
1.3 This is equal to the 1.3 we would expect if the data were normal. This method indicates
s
185.4
the data are approximately normal.
Finally, using MINITAB, the normal probability plot is:
Probability Plot of DrivHead
Normal - 95% CI
99.9
Mean
StDev
N
AD
P-Value
99
95
603.7
185.4
98
0.492
0.214
Percent
90
80
70
60
50
40
30
20
10
5
1
0.1
0
200
400
600
800
DrivHead
1000
1200
1400
Since the data form a fairly straight line, the data are approximately normal.
From the 4 different methods, all indications are that the driver’s head injury rating data are normal.
We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the
data. Using MINITAB, the histogram of the sanitation scores is:
Histogram of Score
60
50
40
Frequency
4.128
30
20
10
0
60.0
67.5
75.0
Score
82.5
90.0
97.5
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
213
From the histogram, the data appear to be skewed to the left. This indicates that the data are not normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each interval
are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary
statistics are:
Descriptive Statistics: DDT
Variable
Score
N
182
Mean
95.044
StDev
5.391
Minimum
56.000
Q1
94.000
Median
96.500
Q3
98.000
Maximum
100.000
x s 95.044 5.391 (89.653, 100.435) 164 of the 182 values fall in this interval. The proportion is
.90. This is much larger than the .68 we would expect if the data were normal.
x 2 s 95.044 2(5.391) 95.044 10.782 (84.262, 105.826) 174 of the 182 values fall in this
interval. The proportion is .96. This is slightly larger than the .95 we would expect if the data were normal.
x 3s 95.044 3(5.391) 95.044 16.173 (78.871, 111.271) 178 of the 182 values fall in this
interval. The proportion is .978. This is somewhat smaller than the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
Next, we look at the ratio of the IQR to s. IQR QU QL 798 94 4 .
IQR
4
.742 This is much smaller than the 1.3 we would expect if the data were normal. This
s
5.391
method indicates the data are not normal.
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Score
Normal - 95% CI
99.9
Mean
StDev
N
AD
P-Value
99
Percent
95
90
95.04
5.391
182
10.307
<0.005
80
70
60
50
40
30
20
10
5
1
0.1
50
60
70
80
90
100
110
120
Score
Since the data do not form a straight line, the data are not normal.
From the 4 different methods, all indications are that the sanitation scores data are not normal.
Copyright © 2014 Pearson Education, Inc.
4.129
Chapter 4
We will look at the 4 methods or determining if the 3 variables are normal.
Distance:
First, we will look at A histogram of the data. Using MINITAB, the histogram of the distance data is:
Histogram of DISTANCE
Normal
18
Mean
StDev
N
16
298.9
7.525
40
14
12
Frequency
214
10
8
6
4
2
0
285
290
295
300
305
DISTANCE
310
315
320
From the histogram, the distance data do not appear to have a normal distribution.
Next, we look at the intervals x s, x 2 s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the
summary statistics are:
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
DISTANCE
N
40
Mean
298.95
StDev
7.53
Minimum
283.20
Q1
294.60
Median
299.05
Q3
302.00
Maximum
318.90
x s 298.95 7.53 (291.42, 306.48) 28 of the 40 values fall in this interval. The proportion is
28 / 40 .70 . This is fairly close to the .68 we would expect if the data were normal.
x 2 s 298.95 2(7.53) 298.95 15.06 (283.89, 314.01) 37 of the 40 values fall in this interval.
The proportion is 37 / 40 .925 . This is a fair amount below the .95 we would expect if the data were
normal.
x 3s 298.95 3(7.53) 298.95 22.59 (276.36, 321.54) 40 of the 40 values fall in this interval.
The proportion is 40 / 40 1.00 . This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the distance data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR Q U – Q L = 302 – 294.6 7.4
IQR 7.4
.983 . This is much smaller than the 1.3 we would expect if the data were normal. This
s
7.53
method indicates the distance data may not be normal.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
215
Finally, using MINITAB, the normal probability plot is:
Probability Plot of DISTANCE
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
298.9
7.525
40
0.521
0.174
Percent
80
70
60
50
40
30
20
10
5
1
280
290
300
DISTANCE
310
320
Since the data do not form a fairly straight line, the distance data may not be normal.
From the 4 different methods, all indications are that the distance data are not normal.
Accuracy:
First, we will look at a histogram of the data. Using MINITAB, the histogram of the accuracy data is:
Histogram of ACCURACY
Normal
14
Mean
StDev
N
12
61.97
5.226
40
Frequency
10
8
6
4
2
0
48
54
60
ACCURACY
66
72
From the histogram, the accuracy data do not appear to have a normal distribution.
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
ACCURACY
N
40
Mean
61.970
StDev
5.226
Minimum
45.400
Q1
59.400
Median
61.950
Q3
64.075
Maximum
73.000
x s 61.97 5.226 (56.744, 67.196) 30 of the 40 values fall in this interval. The proportion is
30 / 40 .75 . This is much greater than the .68 we would expect if the data were normal.
x 2 s 61.97 2(5.226) 61.97 10.452 (51.518, 72.422) 37 of the 40 values fall in this interval.
The proportion is 37 / 40 .925 . This is a fair amount below the .95 we would expect if the data were
normal.
Copyright © 2014 Pearson Education, Inc.
Chapter 4
x 3s 61.97 3(5.226) 61.97 15.678 (46.292, 77.648) 39 of the 40 values fall in this interval.
The proportion is 39 / 40 .975 . This is a fair amount lower than the 1.00 we would expect if the data were
normal.
From this method, it appears that the accuracy data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR Q U – Q L 64.075 – 59.4 4.675 .
IQR 4.675
.895 . This is much smaller than the 1.3 we would expect if the data were normal. This
s
5.226
method indicates the accuracy data may not be normal.
Finally, using MINITAB, the normal probability plot is:
Probability Plot of ACCURACY
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
61.97
5.226
40
0.601
0.111
Percent
80
70
60
50
40
30
20
10
5
1
45
50
55
60
65
ACCURACY
70
75
80
Since the data do not form a fairly straight line, the accuracy data may not be normal.
From the 4 different methods, all indications are that the accuracy data are not normal.
Index:
First, we will look at a histogram of the data. Using MINITAB, the histogram of the index data is:
Histogram of INDEX
Normal
Mean
StDev
N
10
1.927
0.6602
40
8
Frequency
216
6
4
2
0
0.5
1.0
1.5
2.0
INDEX
2.5
3.0
3.5
From the histogram, the index data do not appear to have a normal distribution.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
217
Next, we look at the intervals x s, x 2 s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the
summary statistics are:
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
INDEX
N
40
Mean
1.927
StDev
0.660
Minimum
1.170
Q1
1.400
Median
1.755
Q3
2.218
Maximum
3.580
x s 1.927 .660 (1.267, 2.587) 30 of the 40 values fall in this interval. The proportion is
30 / 40 .75 . This is much greater than the .68 we would expect if the data were normal.
x 2 s 1.927 2(.660) 1.927 1.320 (.607, 3.247) 37 of the 40 values fall in this interval. The
proportion is 37 / 40 .925 . This is a fair amount below the .95 we would expect if the data were normal.
x 3 s 1.927 3(.660) 1.927 1.980 ( .053, 3.907) 40 of the 40 values fall in this interval. The
proportion is 40 / 40 1.000 . This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the index data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR Q U – Q L 2.218 – 1.4 .818 .
IQR .818
1.23 . This is fairly close to the 1.3 we would expect if the data were normal. This method
s
.66
indicates the index data may normal.
Finally, using MINTAB, the normal probability plot is:
Probability Plot of INDEX
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
1.927
0.6602
40
1.758
<0.005
Percent
80
70
60
50
40
30
20
10
5
1
0
1
2
INDEX
3
4
Since the data do not form a fairly straight line, the index data may not be normal.
From 3 of the 4 different methods, the indications are that the index data are not normal.
Copyright © 2014 Pearson Education, Inc.
4.130
Chapter 4
We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the
data. Using MINITAB, the histogram of the tensile strength values is:
Histogram of Strength
3.0
2.5
Fr equency
218
2.0
1.5
1.0
0.5
0.0
330
335
340
345
Str ength
350
355
From the histogram, the data appear to be somewhat skewed to the left. This might indicate that the data are
not normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each interval
are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary
statistics are:
Descriptive Statistics: Strength
Variable
Strength
N
Mean
11 342.13
StDev
7.91
Minimum
328.20
Q1
334.70
Median
343.60
Q3
347.80
Maximum
356.30
x s 342.13 7.91 (334.22, 350.04) 8 of the 11 values fall in this interval. The proportion is .73.
This is somewhat larger than the .68 we would expect if the data were normal.
x 2 s 342.13 2(7.91) 342.13 15.82 (326.31, 357.95) All 11 of the 11 values fall in this interval.
The proportion is 1.00. This is somewhat larger than the .95 we would expect if the data were normal.
x 3 s 342.13 3(7.91) 342.13 23.73 (318.40, 365.86) Again, all 11 of the 11 values fall in this
interval. The proportion is 1.00. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the data are quite normal.
Next, we look at the ratio of the IQR to s. IQR Q U – Q L = 347.80 – 334.70 13.1 .
IQR 13.1
1.656 This is much larger than the 1.3 we would expect if the data were normal. This method
s
7.91
indicates the data are not normal.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
219
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Strength
Normal - 95% CI
99
Mean 342.1
StDev 7.907
N
11
AD
0.154
P-Value 0.937
95
90
80
Percent
70
60
50
40
30
20
10
5
1
310
320
330
340
Strength
350
360
370
Since the data do form a fairly straight line, the data could be normal.
From the 4 different methods, three of the four indicate that the data probably are not from a normal
distribution.
4.131
From Exercise 2.51, it states that the mean number of semester hours for those taking the CPA exam is
141.31 and the median is 140. It also states that most colleges only require 128 semester hours for an
undergraduate degree. Thus, the minimum value for the total semester hours is around 128. The
z-score associated with 128 is:
z
x
128 141.31
.75
17.77
If the data are normal, we know that about .34 of the observations are between the mean and 1 standard
deviation below the mean. Thus, .16 of the observations are more than 1 standard deviation below the mean.
With this distribution, that is impossible. Thus, the data are not normal. The mean is greater than the median,
so we know that the data are skewed to the right.
4.132
a.
f ( x)
1
d c
(c x d )
1
1
1
d c 73 4
1
(3 x 7)
f ( x) 4
0 otherwise
c d 3 7 10
5
2
2
2
b.
c.
5 1.155 3.845, 6.155
d c
12
P ( x ) P (3.845 x 6.155)
73
12
4
12
1.155
b a 6.155 3.845 2.31
.5775
d c
73
4
Copyright © 2014 Pearson Education, Inc.
220
4.133
Chapter 4
a.
f ( x)
1
(c x d )
d c
1
1
1
.04
d c 45 20 25
.04 (20 x 45)
So, f ( x)
0 otherwise
c d 20 45 65
32.5
2
2
2
c.
Using MINITAB, the graph is:
f(x)
b.
d c
12
45 20
12
7.22
1/25
0
20
45
x
2 18.06
32.5
2 46.94
2 32.5 2(7.22) 18.06, 46.94
P 18.06 x 46.94 P 20 x 45 (45 20).04 1
4.134
.04 (20 x 45)
From Exercise 4.133, f ( x)
0 otherwise
a.
P (20 x 30) (30 20)(.04) .4
b.
P 20 x 30 (30 20)(.04) .4
c.
P ( x 30) (45 30)(.04) .6
d.
P ( x 45) (45 45)(.04) 0
e.
P ( x 40) (40 20)(.04) .8
f.
P x 40 (40 20)(.04) .8
g.
P (15 x 35) (35 20)(.04) .6
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
h.
4.135
4.136
4.137
P (21.5 x 31.5) (31.5 21.5)(.04) .4
P( x a) ea / ea /1 . Using a calculator:
a.
P x 1 e1/1 e1 .367879
b.
P x 3 1 P x 3 1 e3/1 1 e3 1 .049787 .950213
c.
P x 1.5 e1.5/1 e1.5 .223130
d.
P x 5 1 P x 5 1 e5/1 1 e5 1 .006738 .993262
a.
P x 4 1 P x 4 1 e4/2.5 1 e1.6 1 .201897 .798103
b.
P x 5 e5/2.5 e2 .135335
c.
P x 2 1 P x 2 1 e2/2.5 1 e.8 1 .449329 .550671
d.
P x 3 e3/2.5 e1.2 .301194
f ( x)
1
1
1
.01
d c 200 100 100
.01 (100 x 200)
f ( x)
0 otherwise
a.
c d 100 200 300
150
2
2
2
d c
12
200 100
12
100
12
28.8675
2 150 2(28.8675) 150 57.735 92.265, 207.735
P x 92.265 P x 207.735 P x 100 P x 200 0 0 0
b.
3 150 3(28.8675) 150 86.6025 63.3975, 236.6025
P 63.3975 x 236.6025 P 100 x 200 (200 100) .01 1
c.
From a, 2 92.265, 207.735 .
P 92.265 x 207.735 P 100 x 200 (200 100) .01 1
Copyright © 2014 Pearson Education, Inc.
221
222
4.138
Chapter 4
With =2 , 2
a.
3 2 3(2) 2 6 ( 4, 8)
Since 3 lies below 0, find the probability that x is more than 3 8 .
P x 8 e8/2 e4 .018316
b.
2 2 2(2) 2 4 ( 2, 6)
Since 2 lies below 0, find the probability that x is between 0 and 6.
P x 6 1 P( x 6) 1 e6/2 1 e3 1 .049787 .950213
c.
.5 2 .5(2) 2 1 (1, 3)
P 1 x 3 P x 1 P x 3 e1/2 e3/2 e.5 e1.5 .606531 .223130 .383401
4.139
For this problem, f ( x )
1
f ( x) 3600
0
1
1
. Thus,
3600 0 3600
0 x 3600
otherwise
The last 15 minutes would represent the last 15(60) = 900 seconds.
P (2700 x 3600) (3600 2700)
4.140
a.
1
900
.25
3600 3600
P(1200 x 1500) P( x 1200) P( x 1500) e1200/1000 e1500/1000
e1.2 e1.5 .3012 .2231 .0781
4.141
b.
P( x 1200) e1200/1000 e1.2 .3012
c.
P ( x 1500 | x 1200)
a.
Let x = temperature with no bolt-on trace elements. Then x has a uniform distribution.
f ( x)
1
d c
P (1200 x 1500) .0781
.2593
.3012
P ( x 1200)
(c x d )
1
1
1
d c 290 260 30
1
Therefore, f ( x ) 30
0
(260 x 290)
otherwise
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
P(280 x 284) (284 280)
1
1
4 .133
30
30
Let y = temperature with bolt-on trace elements. Then y has a uniform distribution.
f ( y)
1
d c
(c y d )
1
1
1
d c 285 278 7
1
Therefore, f ( y ) 7
0
(278 y 285)
otherwise
P(280 y 284) (284 280)
b.
P( x 268) (268 260)
1
1
4 .571
7
7
1
1
8 .267
30
30
P ( y 268) (268 260)(0) 0
4.142 a.
Let x = number of anthrax spores. Then x has an approximate uniform distribution.
f ( x)
1
d c
c x d
1
1
1
.1
d c 10 0 10
.1
Therefore, f ( x)
0
(0 x 10)
otherwise
P x 8 8 – 0.1 .8
4.143
4.144
b.
P 2 x 5 5 – 2.1 .3
a.
P x 2 e2/2.5 e.8 .449329
b.
P x 5 1– P( x 5) 1– e5/2.5 1– e2 1 .135335=.864665 (using a calculator)
a.
Let x = time until the first critical part failure. Then x has an exponential distribution with =.1 .
(using a calculator)
P( x 1) e1/.1 e10 .0000454 (using a calculator)
b.
.5/.1
30 minutes = .5 hours. P( x .5) 1 P( x .5) 1 e
1 e5 1 .0067 .9933
Copyright © 2014 Pearson Education, Inc.
223
4.145
Chapter 4
a.
For this problem, x has a uniform distribution on the interval from 0 to 1. Thus,
b.
1
For this problem, f ( x)
0
c.
4.146
a.
0 x 1
c d 0 1
.5.
2
2
P ( x .7) (1 .7)(1) .3
otherwise
2
2!
1 . Thus, the density can be either 0 or 1.
With n = 2, the total possible connections is
2 2!(2 2)!
Therefore, the uniform model would not be a good approximation for the distribution of network
density.
For layer 2, let x = amount loss. Since the amount of loss is random between .01 and .05 million
dollars, the uniform distribution for x is:
f ( x)
1
d c
(c x d )
1
1
1
25
d c .05 .01 .04
25 (.01 x .05)
Therefore, f ( x)
0 otherwise
A graph of the distribution looks like the following:
f(x)
224
25
0
.01
.05
x
d c .05 .01
c d .01 + .05
2
.0115 , 2 .0115 .00013
=
.03 ,
2
2
12
12
The mean loss for layer 2 is .03 million dollars and the variance of the loss for layer 2 is .00013 million
dollars squared.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
For layer 6, let x = amount loss. Since the amount of loss is random between .50 and 1.00 million dollars, the
uniform distribution for x is:
f ( x)
1
d c
(c x d )
1
1
1
2
d c 1.00 .50 .50
2 (.50 x 1.00)
Therefore, f ( x)
0 otherwise
A graph of the distribution looks like the following:
f(x)
b.
225
2
0
.05
1
x
d c 1.00 .50
c d .50 1.00
.1443 ,
.75 ,
2
2
12
12
2 .1443 .0208
2
The mean loss for layer 6 is .75 million dollars and the variance of the loss for layer 6 is .0208 million
dollars squared.
c.
A loss of $10,000 corresponds to x .01 . P x .01 1
A loss of $25,000 corresponds to x .025 .
P x .025 Base Height (.025 .01)(25) .015 25 .375
d.
A loss of $750,000 corresponds to x .75 . A loss of $1,000,000 corresponds to x 1 .
P .75 x 1 1.00 .75 (2) .25 2 .5
A loss of $900,000 corresponds to x .90 .
P x .9 (1.00 .90)(2) .10 2 .20
P x .9 0
Copyright © 2014 Pearson Education, Inc.
4.147
Chapter 4
a.
The amount dispensed by the beverage machine is a continuous random variable since it can take on
any value between 6.5 and 7.5 ounces.
b.
Since the amount dispensed is random between 6.5 and 7.5 ounces, x is a uniform random variable.
f ( x)
1
d c
(c x d )
1
1
1
1
d c 7.5 6.5 1
1
Therefore, f ( x)
0
(6.5 x 7.5)
otherwise
The graph is as follows:
f(x)
226
1
0
6.5
7.5
x
2 6.422
c.
7
2 7.577
c d 6.5 7.5 14
7
2
2
2
d c
12
7.5 6.5
12
.2887
2 7 2(.2887) 7 .5774 6.422, 7.577
d.
P( x 7) (7.5 7) 1 .5
e.
P x 6 0
f.
P(6.5 x 7.25) (7.25 6.5) 1 .75
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
g.
227
The probability that the next bottle filled will contain more than 7.25 ounces is:
P x 7.25 (7.5 7.25) 1 .25
The probability that the next 6 bottles filled will contain more than 7.25 ounces is:
P[ x 7.25 x 7.25 x 7.25 x 7.25 x 7.25 x 7.25 ]
P x 7.25 .256 .0002
6
4.148
a.
120/95
.2828
Two minutes equals 120 seconds. P( x 120) e
b.
Using MINITAB, a histogram of the data with the exponential distribution displayed on top of it is:
Histogram of INTTIME
Exponential
Mean
N
70
95.52
267
60
Frequency
50
40
30
20
10
0
0
75
150
225
300
INTTIME
375
450
525
The data appear to fit the exponential distribution fairly well.
4.149
a.
Let x = product’s lifetime at the end of its lifetime. Then x has an exponential distribution
with 500, 000 .
P x 700,000 1– P x 700,000 1 – e700000/5000000 1– e1.4 1 .246597 .753403
b.
Let y = product’s lifetime during its normal life. Then y has a uniform distribution.
f ( y)
1
d c
(c y d )
1
1
1
d c 1, 000,000 100, 000 900,000
1
Therefore, f ( y ) 900,000
0
(100,000 y 1,000,000)
otherwise
1
P( y 700, 000) (700, 000 100, 000)
.667
900,
000
Copyright © 2014 Pearson Education, Inc.
228
Chapter 4
c.
P x 830,000 1– P x 830,000 1– e830000/5000000 1– e1.66 1 .190139 .809861
(Using a calculator)
1
P( y 830, 000) (830, 000 100, 000)
.811
900,
000
4.150
Let x = cycle availability, where x has a uniform distribution on the interval from 0 to 1.
d c 1 0
c d 0 1
.289
.5 and
2
2
12
12
The 10th percentile is that value of x such that 10% of all observations are below it.
Let K1 = 10th percentile.
P ( x K1 ) ( K1 0)(1 0) K1 .10
The lower quartile is that value of x such that 25% of all observations are below it.
Let K2 = 25th percentile.
P ( x K 2 ) ( K 2 0)(1 0) K 2 .25
The upper quartile is that value of x such that 75% of all observations are below it.
Let K3 = 75th percentile.
P ( x K 3 ) ( K 3 0)(1 0) K 3 .75
4.151
Let x = number of inches a gouge is from one end of the spindle. Then x has a uniform distribution with f(x)
as follows:
1
1
1
f ( x) d c 18 0 18
0
0 x 18
otherwise
In order to get at least 14 consecutive inches without a gouge, the gouge must be within 4 inches of either end.
Thus, we must find:
P x 4 P x 14 (4 0) 1/18 (18 14) 1/18 4 /18 4 /18 8 /18 .4444
4.152
a.
Let x = life length of CD-ROM. Then x has an exponential distribution with 25, 000 .
R(t ) P( x t ) et /25,000
b.
c.
R(8,760) P( x 8,760) e8,760/25,000 e.3504 .7044
S(t) = probability that at least one of two drives has a length exceeding t hours
= 1 – probability that neither has a length exceeding t hours
1 – P( x1 t ) P( x2 t )
1 – 1 – P x1 t 1 – P x2 t
1 – 1 – e t / 25,000 1 – e t / 25,000
1 – 1 – 2e t / 25,000 e t /12,500 2e t / 25,000 – e t /12,500
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.153
d.
S (8,760) 2e8,760/25,000 e8,760/12,500 2(.7044) .4962 1.4088 .4962 .9126
e.
The probability in part d is greater than that in part b. We would expect this. The probability that at
least one of the systems lasts longer than 8,760 hours would be greater than the probability that only
one system lasts longer than 8,760 hours.
Let x be a random variable with an exponential distribution with mean . Let k = median of the distribution.
Then P ( x k ) .5. We now need to find k.
P( x k ) .5 e k / .5 k / ln(.5)
k ln(.5) .693147
a.
Using MINITAB, a graph is:
f(p)
4.154
1
0
0
1
p
c d 0 1
d c 1 0
.289 ,
.5 ,
2
2
12
12
b.
c.
P p .95 (1 .95) 1 .05
2 .2892 .083
P p .95 (.95 0) 1 .95
d.
The analyst should use a uniform probability distribution with c .90 and d .95 .
1
1
1
30
f ( p ) d c .95 .90 .05
0
4.155
229
a.
For 250 , P x a e
(.90 p .95)
otherwise
a /250
For a 300 and b 200 , show P x a b P x a P x b
P x 300 200 P x 500 e500/250 e2 .1353
P x 300 P x 200 e300/250e200/250 e1.2e.8 .3012 .4493 .1353
Since P x 300 200 P x 300 P x 200 , then P x 300 200 P x 300 P x 200
Copyright © 2014 Pearson Education, Inc.
230
Chapter 4
Also, show P x 300 200 P x 300 P x 200 . Since we already showed that
P x 300 200 P x 300 P x 200 , then P x 300 200 P x 300 P x 200 .
b.
Let a 50 and b 100 . Show P x a b P x a P x b
P x 50 100 P x 150 e150/250 e.6 .5488
P x 50 P x 100 e50/250 e100/250 e.2 e.4 .8187 .6703 .5488
Since P x 50 100 P x 50 P x 100 , then P x 50 100 P x 50 P x 100
Also, show P x 50 100 P x 50 P x 100 . Since we already showed that
P x 50 100 P x 50 P x 100 , then P x 50 100 P x 50 P x 100 .
c.
Show P x a b P x a P x b
P x a b e
a b / 250
4.156
4.157
e a / 250 eb / 250 P x a P x b
a.
This experiment consists of 100 trials. Each trial results in one of two outcomes: chip is defective or
not defective. If the number of chips produced in one hour is much larger than 100, then we can assume
the probability of a defective chip is the same on each trial and that the trials are independent. Thus, x
is a binomial. If, however, the number of chips produced in an hour is not much larger than 100, the
trials would not be independent. Then x would not be a binomial random variable.
b.
This experiment consists of two trials. Each trial results in one of two outcomes: applicant qualified or
not qualified. However, the trials are not independent. The probability of selecting a qualified
applicant on the first trial is 3 out of 5. The probability of selecting a qualified applicant on the second
trial depends on what happened on the first trial. Thus, x is not a binomial random variable. It is a
hypergeometric random variable.
c.
The number of trials is not a specified number in this experiment, thus x is not a binomial random
variable. In this experiment, x is counting the number of calls received.
d.
The number of trials in this experiment is 1000. Each trial can result in one of two outcomes: favor
state income tax or not favor state income tax. Since 1000 is small compared to the number of
registered voters in Florida, the probability of selecting a voter in favor of the state income tax is the
same from trial to trial, and the trials are independent of each other. Thus, x is a binomial random
variable.
n
p ( x) p x q n - x x = 0, 1, 2, ... , n
x
a.
7
7! 3 4
P( x 3) p(3) .53.57 3
.5 .5 35(.125)(.0625) .2734
3
3!4!
b.
4
4! 3 1
P( x 3) p(3) .83.24 3
.8 .2 4 .512 .2 .4096
3
3!1!
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
c.
4.158
a.
15
15! 1 14
P( x 1) p(1) .11.9151
.1 .9 15 .1.228768 .3432
1!14!
1
xp( x) 10 .2 12 .3 18 .1 20 .4 15.4
2 ( x )2 p( x) (10 15.4)2 .2 (12 15.4)2 .3 (18 15.4)2 .1 (20 15.4)2 .4 18.44
18.44 4.294
4.159
b
P x 15 p 10 p 12 .2 .3 .5
c.
2 15.4 2 4.294 6.812, 23.988
d.
P 6.812 x 23.988 .2 .3 .1 .4 1.0
From Table I, Appendix D:
a.
P x 14 P( x 14) P( x 13) .584 .392 .192
b.
P ( x 12) .228
c.
P x 12 1 P( x 12) 1 .228 .772
d.
P (9 x 18) P ( x 18) P ( x 8) .992 .005 .987
e.
P 8 x 18 P( x 17) P( x 8) .965 .005 .960
f.
np 20 .7 14 ,
g.
2 14 2 2.049 14 4.098 9.902, 18.098
2 npq 20 .7.3 4.2 ,
4.2 = 2.049
P 9.902 x 18.098 P(10 x 18) P( x 18) P( x 9) .992 .017 .975
4.160
a.
Using MINITAB with 2 ,
Probability Density Function
Poisson with mean = 2
x
3
P( X = x )
0.180447
p 3 P x 3 .180447
Copyright © 2014 Pearson Education, Inc.
231
232
Chapter 4
b.
Using MINITAB with 1 ,
Probability Density Function
Poisson with mean = 1
x
4
P( X = x )
0.0153283
p 4 P x 4 .0153283
c.
Using MINITAB with .5 ,
Probability Density Function
Poisson with mean = .5
x
2
P( X = x )
0.0758163
p 2 P x 2 .0758163
4.161
4.162
4.163
a.
Poisson
b.
Binomial
c.
Binomial
a.
r N r 38 3
3! 5!
x n x 2 5 2 2!1! 3!2! 3(10)
P ( x 2)
.536
8!
56
N
8
5!3!
n
5
b.
r N r 26 2
2! 4!
x n x 2 2 2 2!0! 0!4! 1(1)
P( x 2)
.067
6!
15
N
6
2!4!
n
2
c.
r N r 4 5 4
4! 1!
x n x 3 4 3 3!1!1!0! 4(1)
P( x 3)
.8
5!
5
N
5
4!1!
n
3
a.
Discrete - The number of damaged inventory items is countable.
b.
Continuous - The average monthly sales can take on any value within an acceptable limit.
c.
Continuous - The number of square feet can take on any positive value.
d.
Continuous - The length of time we must wait can take on any positive value.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.164
a.
1
1
1
,
f ( x ) d c 90 10 80
0
otherwise
b.
c d 10 90
50
2
2
d c
12
90 10
12
23.094011
The interval 2 50 2 23.094 50 46.188 3.812, 96.188 is indicated on the graph.
f(x)
c.
(10 x 90)
1/80
0
10
90
x
2 3.812
4.165
50
2 96.188
1 5
.625
80 8
d.
P ( x 60) (60 10)
e.
P ( x 90) 0
f.
P ( x 80) (80 10)
g.
P ( x ) P (50 23.094 x 50 23.094) P (26.906 x 73.094)
h.
P x 75 (90 75)
a.
P ( z 2.1) A1 A2 .5 .4821 .9821
1 7
.875
80 8
1 46.188
(73.094 26.906)
.577
80
80
1 15
.1875
80 80
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234
4.166
Chapter 4
b.
P ( z 2.1) A2 .5 A1 .5 .4821 .0179
c.
P ( z 1.65) A1 A2 .4505 .5000 .9505
d.
P(2.13 z .41) P(2.13 z 0) P(.41 z 0)
.4834 .1591 .3243
e.
P ( 1.45 z 2.15) A1 A2 .4265 .4842 .9107
f.
P ( z 1.43) A1 .5 A2 .5000 .4236 .0764
a.
P ( z z 0 ) .5080 P (0 z z 0 ) .5080 .5 .0080
Looking up the area .0080 in Table II, gives z0 .02 .
b.
P ( z z 0 ) .5517 P ( z 0 z 0) .5517 .5 .0517
Looking up the area .0517 in Table II, z0 .13 .
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
c.
235
P ( z z 0 ) .1492 P (0 z z 0 ) .5 .1492 .3508
Looking up the area .3508 in Table II, gives z 0 1.04 .
d.
P ( z 0 z .59) .4773 P ( z0 z 0) P (0 z .59) .4773
P (0 z .59) .2224
Thus, P ( z0 z 0) .4773 .2224 .2549
Looking up the area .2549 in Table II, gives z0 .69 .
4.167
x
7
a.
For the probability density function, f ( x)
e
b.
For the probability density function, f ( x )
1
, 5 x 25 , x is a uniform random variable.
20
c.
For the probability function, f ( x)
a.
P( x 1) 1 P x 1 1 e1/3 1 .716531 .283469 (using calculator)
b.
P x 1 e1/3 .716531
c.
P x 1 0
, x 0 , x is an exponential random variable.
7
e.5[( x 10)/5]
2
4.168
5 2
, x is a normal random variable.
(x is a continuous random variable. There is no probability associated with a single
point.)
4.169
d.
P( x 6) 1 P x 6 1 e6/3 1 e2 1 .135335 .864665 (using a calculator)
e.
P(2 x 10) P( x 2) P x 10 e2/3 e10/3 .513417 .035674 .47774 (using calculator)
a.
80 75
P( x 80) P z
P( z .5) .5+.1915 .6915
10
(Table II, Appendix D)
b.
85 75
P( x 85) P z
P( z 1) .5 .3413 .1587
10
(Table II, Appendix D)
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Chapter 4
c.
75 75
70 75
P(70 x 75) P
z
10
10
P ( .5 z 0) P (0 z .5) .1915
(Table II, Appendix D)
d.
P x 80 1 P( x 80) 1 .6915 .3085 (Refer to part
a.)
e.
P x 78 0 , since a single point does not have an area.
f.
110 75
P( x 110) P z
P( z 3.5)
10
.5 .49977 .99977
(Table II, Appendix D)
4.170
np 100 .5 50 , npq 100(.5)(.5) 5
a.
(48 .5) 50
P( x 48) P z
P( z .30)
5
.5 .1179 =.3821
b.
(65 .5) 50
(50 .5) 50
P(50 x 65) P
z
5
5
P ( .10 z 3.10) .0398 .49903 .53883
c.
(70 .5) 50
P( x 70) P z
P( z 3.90)
5
.5 .49995 .00005
d.
(58 .5) 50
(55 .5) 50
P(55 x 58) P
z
5
5
P(.90 z 1.70) P(0 z 1.70) P(0 z .90)
.4554 .3159 .1395
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
e.
(62 .5) 50
(62 .5) 50
P x 62 P
z
5
5
P(2.30 z 2.50) P(0 z 2.50) (0 z 2.30)
.4938 .4893 .0045
f.
P ( x 49 or x 72)
(49 .5) 50
(72 .5) 50
P z
P z
5
5
P ( z .10) P ( z 4.30) (.5 .0398) (.5 .5) .4602
4.171
x is normal random variable with 40 , 2 36 , and 6 .
a.
P ( x x0 ) .10
So, A .5 .10 .4000 . Looking up the area .4000
In the body of Table II, Appendix D gives z 0 1.28 .
To find x0, substitute the values into the z-score formula:
x
x 40
1.28 0
x0 1.28(6) 40 47.68
z0 0
6
b.
P ( x x0 ) .40
Looking up the area .4000 in the body of Table II, Appendix
D gives z 0 1.28 .
To find x0, substitute the values into the z-score formula:
z0
c.
x0
1.28
x0 40
x0 1.28(6) 40 47.68
6
P x x0 .05
So, A .5000 .0500 .4500 .
Looking up the area .4500 in the body of Table II, Appendix
D gives z 0 1.645 . (.45 is halfway between .4495 and
.4505; therefore, we average the z-scores)
1.64 1.65
1.645
2
z0 is negative since the graph shows z0 is on the left side of 0.
Copyright © 2014 Pearson Education, Inc.
237
238
Chapter 4
To find x0, substitute the values into the z-score formula:
z0
d.
x0
1.645
x0 40
x0 1.645(6) 40 30.13
6
P x x0 .40
So, A .5000 .4000 .1000 .
Looking up the area .1000 in the body of Table II,
Appendix D gives z0 .25 .
To find x0, substitute the values into the z-score formula:
z0
e.
x0
.25
x0 40
x0 .25(6) 40 41.5
6
P ( x0 x ) .45
Looking up the area .4500 in the body of Table II,
Appendix D gives z 0 1.645 . (.45 is halfway between
.4495 and .4505; therefore, we average the z-scores)
1.64 1.65
1.645
2
z0 is negative since the graph shows z0 is on the left side of 0.
To find x0, substitute the values into the z-score formula:
z0
4.172
a.
x0
1.645
x0 40
x0 1.645(6) 40 30.13
6
We will check the 5 characteristics of a binomial random variable.
1.
The experiment consists of n = 5 identical trials. We have to assume that the number of bottled
water brands is large.
2.
There are only 2 possible outcomes for each trial. Let S = brand of bottled water used tap water
and F = brand of bottled water did not use tap water.
3.
The probability of success (S) is the same from trial to trial. For each trial, p P S .25 and
q 1 – p 1 .25 .75 .
4.
The trials are independent.
5.
The binomial random variable x is the number of brands in the 5 trials that used tap water.
If the total number of brands of bottled water is large, then the above characteristics will be basically
true. Thus, x is a binomial random variable.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
c.
d.
e.
5
The formula for the probability distribution for x is p( x) .25x (.75)5 x , for x = 0, 1, 2, 3, 4, 5.
x
5
5!
P( x 2) .252 (.75)5 2
.252.753 .2637
2!3!
2
5
5
P( x 1) P( x 0) P ( x 1) .250 (.75)5 0 .251 (.75)5 1
0
1
5!
5!
.250.755
.251.754 .2373 .3955 .6328
0!5!
1!4!
E ( x ) np 65(.25) 16.25
2 npq 65(.25)(.75) 12.1875 3.49
To see if the normal approximation is appropriate, we use:
3 16.25 3 3.49 16.25 10.47 5.78, 26.72
Since this interval lies in the range from 0 to 65, the normal approximation is appropriate.
(20 .5) 16.25
P( x 20) P z
P( z .93) .5 P(0 z .93) .5 .3238 .1762
3.49
(Using Table II, Appendix D)
Since this probability is not small, it is likely that 20 or more brands will contain tap water.
p( x ) p(0) p(1) p(2) p(3) p(4) p(5)
6
4.173
a.
i 1
i
.0102 .0768 .2304 .3456 .2592 .0778 1.0000
b.
P x 4 .2592
c.
P x 2 P x 0 P x 1 .0102 .0768 .0870
d.
P( x 3) P x 3 P x 4 P x 5 .3456 .2592 .0778 .6826
E ( x) xi p ( xi ) 0(.0102) 1(.0768) 2(.2304) 3(.3456) 4(.2592) 5(.0778)
6
e.
i 1
0 .0768 .4608 1.0368 1.0368 .3890 3.0002
On the average, 3 out of every 5 dentists will use nitrous oxide.
4.174
Let x = transmission delay. The random variable x has a normal distribution with 48.5 and 8.5 .
Using Table II, Appendix D,
a.
57 48.5
P( x 57) P z
P( z 1.00)
8.5
.5 P(0 z 1) .5 .3413 .8413
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240
4.175
Chapter 4
b.
60 48.5
40 48.5
P(40 x 60) P
z
P(1 z 1.35)
8.5
8.5
P(1 z 0) P(0 z 1.35) .3413 .4115 .7528
a.
For this problem, c 0 and d 1 .
1
1
1
d c 1 0
1 (0 x 1)
f ( x)
0 otherwise
c d 0 1
.5
2
2
2
( d c ) 2 (1 0) 2 1
.0833
12
12
12
.0833 .289
b.
P .2 x .4 (.4 .2) 1 .2
c.
P x .995 (1 .995) 1 .005 . Since the probability of observing a trajectory greater than .995 is
so small, we would not expect to see a trajectory exceeding .995.
4.176
a.
Using MINITAB with 1.2 ,
Probability Density Function
Poisson with mean = 1.2
x
0
P( X = x )
0.301194
P x 0 .301194
b.
Using MINITAB with 1.2 ,
Cumulative Distribution Function
Poisson with mean = 1.2
x
1
P( X <= x )
0.662627
P ( x 2) 1 – P ( x 1) 1 .662627 .337373
4.177
Let x = interarrival time between patients. Then x is an exponential random variable with a mean of 4
minutes.
a.
P x 1 1 P( x 1) 1 e1/4 1 e.25 1 .778801 .221199
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
Assuming that the interarrival times are independent,
P(next 4 interarrival times are all less than 1 minute)
P x 1 .2211994 .002394
4
c.
4.178
a.
P x 10 e10/4 e2.5 .082085
Let x = number of trees infected with the Dutch elm disease in the two trees purchased. For this
problem, x is a hypergeometric random variable with N 10 , n 2 , and r 3 .
The probability that both trees will be healthy is:
r N r 3 10 3
3! 7!
x n x 0 2 0 0!3! 2!5! 1(21)
.467
P x 0
10!
45
N
10
2!8!
n
2
4.179
b.
The probability that at least one tree will be infected is: P( x 1) 1 P x 0 1 .467 .533 .
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
3.
4.
5.
The experiment consists of n 20 identical trials.
There are only 2 possible outcomes for each trial. Let S = intruding object is detected and
F = intruding object is not detected.
The probability of success (S) is the same from trial to trial. For each trial, p P S .8 and
q 1 – p 1 .8 .2 .
The trials are independent.
The binomial random variable x is the number of intruding objects in the 20 trials that are
detected.
Thus, x is a binomial random variable.
b.
For this experiment, n 20 and p .8 .
c.
Using Table I, Appendix D, with n 20 and p .8 ,
P x 15 P( x 15) P( x 14) .370 .196 .174
d.
Using Table I, Appendix D, with n 20 and p .8 ,
P ( x 15) 1 P ( x 14) 1 .196 .804
e.
E x np 20 .8 16 . For every 20 intruding objects, SBIRS will detect an average of 16.
Copyright © 2014 Pearson Education, Inc.
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242
4.180
Chapter 4
Let x = demand for white bread. Then x is a normal random variable with 7200 and 300 :
a.
P ( x x0 ) .94 . Find x0.
x 7200
P( x x0 ) P z 0
P( z z0 ) .94
300
A1 .94 .50 .4400
Using Table II and area .4400, z0 1.555 .
z0
b.
x 0 7200
x 7200
1.555 0
x0 7666.5 7667
300
300
If the company produces 7,667 loaves, the company will be left with more than 500 loaves if the
demand is less than 7, 667 500 7,167 .
7167 7200
P x 7167 P z
P( z .11) .5 .0438 .4562
300
(from Table II, Appendix D)
Thus, on 45.62% of the days the company will be left with more than 500 loaves.
4.181
a.
Let x1 = repair time for machine 1. Then x1 has an exponential distribution with 1 1 hour.
P x1 1 e1/1 e1 .367879 (using a calculator)
b.
Let x2 = repair time for machine 2. Then x2 has an exponential distribution with 2 2 hours.
P x2 1 e1/2 e.5 .606531 (using a calculator)
c.
Let x3 = repair time for machine 3. Then x3 has an exponential distribution with 3 .5 hours.
P x3 1 e1/.5 e2 .135335 (using a calculator)
Since the mean repair time for machine 4 is the same as for machine 3, P x4 1 P x3 1 .135335.
d.
The only way that the repair time for the entire system will not exceed 1 hour is if all four machines are
repaired in less than 1 hour. Thus, the probability that the repair time for the entire system exceeds 1
hour is:
P(Repair time entire system exceeds 1 hour)
1 P ( x1 1) ( x2 1) ( x3 1) ( x4 1) 1 P ( x1 1) P ( x2 1) P( x3 1) P ( x4 1)
1 (1 .367879)(1 .606531)(1 .135335)(1 .135335)
1 .632121.393469 .864665 .864665 1 .185954 .814046
4.182
a.
In order for the number of deaths to follow a Poisson distribution, we must assume that the probability
of a death is the same for any week. We must also assume that the number of deaths in any week is
independent of any other week.
The first assumption may not be valid. The probability of a death may not be the same for every week.
The number of passengers varies from week to week, so the probability of a death may change. Also,
things such as weather, which varies from week to week may increase or decrease the chance of
derailment.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
E ( x ) 16 and 16 4
c.
The z-score corresponding to x 4 is z
d.
Using MINITAB with 16 , we get the following probability:
243
4 16
3 . Since this z-score is 3 standard deviations from
4
the mean, it would be very unlikely that only 4 or fewer deaths occur next week.
Cumulative Distribution Function
Poisson with mean = 160
x
4
P( X <= x )
0.0004004
P ( x 4) .0004
This probability is consistent with the answer in part c. The probability of 4 or fewer deaths is
essentially zero, which is very unlikely.
4.183
4.184
nr 10(8)
.383
N
209
a.
For N 209 , r 10 , and n 8 , E ( x )
b.
8 209 8
8!
201!
4 10 4 4!(8 4)! 6!(201 6)!
.0002
P ( x 4)
209!
209
10!(209 10)!
10
To construct a relative frequency histogram for the data, we can use 7 measurement classes.
Interval width =
Largest number - smallest number 98.0716 .7434
13.9
Number of classes
7
We will use an interval width of 14 and a starting value of .74335.
The measurement classes, frequencies, and relative frequencies are given in the table.
Class
Measurement Class
Class Frequency
1
2
3
4
5
6
7
.74335 14.74335
14.74335 28.74335
28.74335 42.74335
42.74335 56.74335
56.74335 70.74335
70.74335 84.74335
84.74335 98.74335
6
4
6
6
5
4
9
40
Class Relative
Frequency
6/40 = .15
.10
.15
.15
.125
.10
.225
1.000
The histogram looks like the data could be from a uniform distribution. The last class (84.74335 98.74335)
has a few more observations in it than we would expect. However, we cannot expect a perfect graph from a
sample of only 40 observations.
Copyright © 2014 Pearson Education, Inc.
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Chapter 4
Histogram of Class
Relative frequency
.25
.20
.15
.10
98.74335
84.74335
70.74335
56.74335
42.74335
28.74335
.74335
0
14.74335
.05
Class
Using MINITAB, the histogram with the normal distribution overlaid is:
Histogram of Bankruptcy
Normal
20
Mean
StDev
N
2.549
1.828
49
15
Frequency
4. 185
10
5
0
0
2
4
6
Bankruptcy
8
10
The data are skewed to the right, and do not appear to be normally distributed.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Bankruptcy
Variable
Bankruptcy
N
49
Mean
2.549
StDev
1.828
Minimum
1.000
Q1
1.350
Median
1.700
Q3
3.500
Maximum
10.100
x s 2.549 1.828 0.721, 4.377
x 2s 2.549 2 1.828 2.549 3.656 (1.107, 6.205)
x 3s 2.549 31.828 2.549 5.484 (2.935, 8.033)
Of the 49 measurements, 44 are in the interval (0.721, 4.377). The proportion is 44 / 49 .898 . This is much
larger than the proportion (.68) stated by the Empirical Rule.
Of the 49 measurements, 47 are in the interval (1.107, 6.205). The proportion is 47 / 49 .959 . This is close
to the proportion (.95) stated by the Empirical Rule.
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
245
Of the 49 measurements, 48 are in the interval (2.935, 8.033). The proportion is 48 / 49 .980 . This is
smaller than the proportion (1.00) stated by the Empirical Rule.
This would imply that the data are not normal.
IQR
2.15
1.176 . If the data are normally distributed, this
s
1.828
ratio should be close to 1.3. Since 1.176 is smaller than 1.3, this indicates that the data may not be normal.
IQR QU QL 3.500 1.350 2.15 .
Using MINITAB, the normal probability plot is:
Probability Plot of Bankruptcy
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
2.549
1.828
49
3.160
<0.005
Percent
80
70
60
50
40
30
20
10
5
1
-4
-2
0
2
4
Bankruptcy
6
8
10
Since this plot is not a straight line, the data are not normal.
All four checks indicate that the data are not normal.
Using MINITAB, the histogram with the normal distribution overlaid is:
Histogram of PENALTY
Normal
Mean
StDev
N
25
132309
249632
38
20
Frequency
4.186
15
10
5
0
-400000
-200000
0
200000
400000
600000
800000
1000000
PENALTY
The data are skewed to the right, and do not appear to be normally distributed.
Copyright © 2014 Pearson Education, Inc.
246
Chapter 4
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: PENALTY
Variable
PENALTY
N
38
Mean
132309
StDev
249632
Minimum
2500
Q1
20000
Median
35000
Q3
101250
Maximum
1000000
x s 132,309 249,632 117,323, 381,941
x 2s 132,309 2 249,632 132,309 499,264 (366,955, 631,573)
x 3s 132,309 3 249,632 132,309 748,896 (616,587, 881, 205)
Of the 38 measurements, 34 are in the interval (-117,323, 381,941). The proportion is 34 / 38 .895 . This is
much larger than the proportion (.68) stated by the Empirical Rule.
Of the 38 measurements, 35 are in the interval (-366,955, 631,573). The proportion is 35 / 38 .921 . This is
smaller than the proportion (.95) stated by the Empirical Rule.
Of the 38 measurements, 36 are in the interval (616,587, 881,205). The proportion is 36 / 38 .947 . This is
much smaller than the proportion (1.00) stated by the Empirical Rule.
This would imply that the data are not normal.
IQR 98, 750
.28 . If the data are normally distributed,
s
249, 632
this ratio should be close to 1.3. Since .28 is smaller than 1.3, this indicates that the data are not normal.
IQR QU QL 101, 250 20, 000 98, 750 .
Using MINITAB, the normal probability plot is:
Probability Plot of PENALTY
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
132309
249632
38
7.602
<0.005
Percent
80
70
60
50
40
30
20
10
5
1
-500000 -250000
0
250000
PENALTY
500000
750000 1000000
Since this plot is not a straight line, the data are not normal.
All four checks indicate that the data are not normal.
4.187
Let x equal the difference between the actual weight and recorded weight (the error of measurement). The
random variable x is normally distributed with 592 and 628 .
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
a.
247
We want to find the probability that the weigh-in-motion equipment understates the actual weight of
the truck. This would be true if the error of measurement is positive.
0 592
P x 0 P z
P ( z .94)
628
.5000 .3264 =.8264
b.
P(overstate the weight) = 1 P(understate the weight)
1 .8264 .1736
(Refer to part a.)
For 100 measurements, approximately 100 .1736 =17.36 or 17 times the weight would be overstated.
c.
400 592
P x 400 P z
P( z .31)
628
.5000 .1217 .6217
d.
We want P(understate the weight) = .5
To understate the weight, x 0 . Thus, we want to find so that P x 0 .5
0
P x 0 P z
.5
628
From Table II, Appendix D, z0 0 . To find , substitute
into the z-score formula:
x
0
0
0
z0 0
628
Thus, the mean error should be set at 0.
We want P(understate the weight) = .4
To understate the weight, x 0 . Thus, we want to find so
that P x 0 .4 .
A .5 .40 .1 . Look up the area .1000 in the body of
Table II, Appendix D, z0 .25 .
To find , substitute into the z-score formula:
z0
x0
.25
0
0 .25 628 157
628
Copyright © 2014 Pearson Education, Inc.
248
4.188
Chapter 4
Let x = number of packets observed by a network sensor in 150 trials. Then x has an approximate binomial
distribution with n 150 and p .001 .
The virus will be detected if at least 1 packet is observed.
150
150!
0
150 0
P( x 1) 1 P( x 0) 1
1
.999150 1 .8606 .1394
.001 (.999)
0
0!150!
4. 189
a.
np 25 .05 1.25
npq 25(.05)(.95) 1.09
Since is not an integer, x could not equal its expected value.
b.
The event is ( x 5) . From Table I with n 25 and p .05 :
P ( x 5) 1 P ( x 4) 1 .993 .007
4.190
c.
Since the probability obtained in part b is so small, it is unlikely that 5% applies to this agency. The
percentage is probably greater than 5%.
a.
Let x = crop yield. The random variable x has a normal distribution with 1,500
and 250 .
1, 600 -1,500
P x 1, 600 P z
P z .4 .5 .1554 .6554
250
(Using Table II, Appendix D)
b.
Let x1 = crop yield in first year and x2 = crop yield in second year. If x1 and x2 are independent, then the
probability that the farm will lose money for two straight years is:
1,600 1,500
1,600 1,500
P x1 1, 600 P x2 1, 600 P z1
P z2
250
250
P z1 .4 P z2 .4 .5 .1554 .5 .1554 .6554 .6554 .4295
(Using Table II, Appendix D)
c.
P(1,500 2 x 1,500 + 2) =
[1,500 2 ] 1,500
[1,500 2 ] 1,500
P(1,500 2 x 1,500 2 ) P
z
P(2 z 2) 2P(0 z 2) 2 .4772 .9544
(using Table II, Appendix D)
4.191
Let x = number of grants awarded to the north side in 140 trials. The random variable x has a hypergeometric
distribution with N 743 , n 140 , and r 601 .
a.
E ( x)
2
nr 140(601)
113.24
N
743
r ( N r ) n( N n) 601(743 601)140(743 140)
17.5884
N 2 ( N 1)
7432 (743 1)
17.5884 4.194
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
249
If the grants were awarded at random, we would expect approximately 113 to be awarded to the north
side. We observed 140. The z-score associated with 140 is:
z
x
140 113.24
6.38
4.194
Because this z-score is so large, it would be extremely unlikely to observe all 140 grants to the north
side if they are randomly selected. Thus, we would conclude that the grants were not randomly
selected.
4.192
Let x = length of time a bus is late. Then x is a uniform random variable with probability distribution:
1
f ( x) 20
0
4.193
(0 x 20)
otherwise
0 20
10
2
a.
b.
1 1
P( x 19) (20 19)
.05
20 20
c.
It would be doubtful that the director’s claim is true, since the probability of the being more than 19
minutes late is so small.
a.
The properties of valid probability distributions are:
p(x) 1 and 0 p ( x) 1 for all x.
For ARC a1: 0 p ( x ) 1 for all x and
p( x) .05 .10 .25 .60 1.00
Thus, this is a valid probability distribution.
For ARC a2: 0 p ( x ) 1 for all x and
p(x) .10 .30 .60 0 1.00
Thus, this is a valid probability distribution.
For ARC a3: 0 p ( x ) 1 for all x and
p(x) .05 .25 .70 0 1.00
Thus, this is a valid probability distribution.
For ARC a4: 0 p ( x ) 1 for all x and
p( x) .90 .10 0 0 1.00
Thus, this is a valid probability distribution.
b.
For Arc a1, P ( x 1) P ( x 2) P ( x 3) .25 .6 .85
c.
For Arc a2, P ( x 1) P ( x 2) .60
For Arc a3, P ( x 1) P ( x 2) .70
For Arc a4, P ( x 1) 0
Copyright © 2014 Pearson Education, Inc.
250
Chapter 4
d.
E( x) xp( x) 0(.05) 1(.10) 2(.25) 3(.60) 0 .10 .50 1.80 2.40
For Arc a1,
The average capacity of Arc a1 is 2.40.
E( x) xp( x) 0(.10) 1(.30) 2(.60) 0 .30 1.20 1.50
For Arc a2,
The average capacity of Arc a2 is 1.50.
E( x) xp( x) 0(.05) 1(.25) 2(.70) 0 .25 1.40 1.65
For Arc a3,
The average capacity of Arc a3 is 1.65.
E( x) xp( x) 0(.90) 1(.10) 0 .10 .10
For Arc a4,
The average capacity of Arc a4 is 0.10.
e.
2 E ( x ) ( x ) 2 p( x)
For Arc a1,
2
(0 2.4) 2 (.05) (1 2.4)2 (.10) (2 2.4) 2 (.25) (3 2.4) 2 (.60)
(2.4) 2 (.05) (1.4) 2 (.10) (.4) 2 (.25) (.6) 2 (.60)
.288 .196 .04 .216 .74
.74 .86
2.40 2(.86) 2.40 1.72 (.68, 4.12)
We would expect most observations to fall within 2 standard deviations of the mean or
2 E ( x ) ( x ) 2 p( x )
For Arc a2,
2
(0 1.5)2 (.10) (1 1.5)2 (.30) (2 1.5)2 (.60)
(1.5)2 (.10) (.5)2 (.30) (.5)2 (.60) .225 .075 .15 .45
.45 .67
1.50 2(.67) 1.50 1.34 (.16, 2.84)
We would expect most observations to fall within 2 standard deviations of the mean or
2 E ( x ) ( x ) 2 p ( x)
For Arc a3,
2
(0 1.65)2 (.05) (1 1.65)2 (.25) (2 1.65)2 (.70)
(1.65)2 (.05) (.65)2 (.25) (.35)2 (.70) .136125 .105625 .08575 .3275
.3275 .57
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Random Variables and Probability Distributions
251
1.65 2(.57) 1.65 1.14 (.51, 2.79)
We would expect most observations to fall within 2 standard deviations of the mean or
For Arc a4,
2
2 E ( x ) ( x ) 2 p ( x) (0 .1) 2 (.90) (1 .1) 2 (.10)
( .1) 2 (.90) (.9) 2 (.10) .009 .081 .090
.09 .30
.10 2(.30) .10 .60 ( .50, .70)
We would expect most observations to fall within 2 standard deviations of the mean or
4.194
Let x = number of doctors who refuse ethics consultation in n 10 trials. From Exercise 2.11, we can estimate
p with p .195 . Then x will be a binomial random variable with n 10 and p .195 . Using MINITAB with
n 10 and p .195 , the probability is:
Cumulative Distribution Function
Binomial with n = 10 and p = 0.195
x
1
P( X <= x )
0.391097
P ( x 2) 1 P ( x 1) 1 .391097 .608903
4.195
a.
Using MINITAB with 5 ,
Cumulative Distribution Function
Poisson with mean = 5
x
2
P( X <= x )
0.124652
P x 3 P( x 2) .125
b.
E x 5 . The average number of calls blocked during the peak hour of video conferencing call
time is 5.
4.196
Let x = number of spoiled bottles in the sample of 3. Since the sampling will be done without replacement, x
is a hypergeometric random variable with N 12 , n 3 , and r 1 .
r N r 1 12 1
1! 11!
1
3
1
x
n
x
1!0! 2!9! 55 .25
P( x 1)
12!
220
N
12
3!9!
n
3
4.197
Let x = number of defective CDs in n 1, 600 trials. Then x is a binomial random variable with
n 1, 600 and p .006 .
E x np 1,600 .006 9.6 .
2 npq 1, 600(.006)(.994) 9.5424 3.089
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252
Chapter 4
To see if the normal approximation is appropriate, we use:
3 9.6 3(3.089) 9.6 9.267 (0.333, 18.867)
Since the interval lies in the range of 0 to 1,600, the normal approximation is appropriate.
11.5 9.6
P( x 12) P z
P( z 0.62) .5 .2324 .2676
3.089
(Using Table II, Appendix D)
Since this probability is fairly large, it would not be unusual to see 12 or more defectives in a sample of 1,600
if 99.4% were defect-free. Thus, there would be no evidence to cast doubt on the manufacturer’s claim.
4.198
a.
b.
If a large number of measurements are observed, then the relative frequencies should be very good
estimators of the probabilities.
E( x) xp( x) 1(.01) 2(.04) 3(.04) 4(.08) 5(.10) 6(.15) 7(.25) 8(.20) 9(.08) 10(.05)
.01 .08 .12 .32 .50 .90 1.75 1.60 .72 .50
6.50
The average number of checkout lanes per store is 6.5.
c.
2 ( x )2 p( x) (1 6.5) 2 (.01) (2 6.5) 2 (.04) (3 6.5) 2 (.04) (4 6.5) 2 (.08)
All x
(5 6.5) 2 (.10) (6 6.5) 2 (.15) (7 6.5) 2 (.25)
(8 6.5) 2 (.20) (9 6.5) 2 (.08) (10 6.5) 2 (.05)
.3025 .8100 .4900 .5000 .2250 .0375 .0625 .4500 .5000 .6125
3.99
3.99 1.9975
d.
Chebyshev's Rule says that at least 0 of the observations should fall in the interval .
Chebyshev's Rule says that at least 75% of the observations should fall in the interval
2 .
e.
6.5 1.9975 (4.5025, 8.4975)
P (4.5025 x 8.4975) .10 .15 .25 .20 .70
This is at least 0.
2 6.5 2(1.9975) 6.5 3.995 (2.505, 10.495)
P (2.505 x 10.495) .04 .08 .10 .15 .25 .20 .08 .05 .95
This is at least .75 or 75%.
4.199
a.
The contract will be profitable if total cost, x, is less than $1,000,000.
1, 000, 000 850, 000
P x 1, 000, 000 P z
P z .88 .5 .3106 .8106
170, 000
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
b.
253
The contract will result in a loss if total cost, x, exceeds 1,000,000.
P x 1,000,000 1 P x 1,000,000 1 .8106 .1894
c.
P x R .99 . Find R.
R 850, 000
P x R P z
P z z0 .99
170, 000
A1 .99 .5 .4900
Looking up the area .4900 in Table II, z0 2.33
z0
R 850, 000
R 850, 000
2.33
170, 000
170, 000
R 2.33 170, 000 850, 000 $1, 246,100
a.
For 17 . To graph the distribution, we will pick several values of x and find the value of f(x),
where x = time between arrivals of the smaller craft at the pier.
1
1
f ( x ) e x / e x /17
17
f (1)
1 1/17
e
.0555
17
f (3)
f (7)
1 7 /17
e
.0390
17
1 20/17
e
.0181
17
f (20)
1 3/17
e
.04937
17
f (5)
f (10)
1 10/17
e
.0327
17
f (15)
f (25)
1 25/17
e
.0135
17
Using MINITAB, the graph is:
0.06
0.05
0.04
f(x)
4.200
0.03
0.02
0.01
0.00
0
5
10
15
20
25
x
Copyright © 2014 Pearson Education, Inc.
1 5/17
e
.0438
17
1 15/17
e
.0243
17
254
Chapter 4
b.
We want to find the probability that the time between arrivals is less than 15 minutes.
P x 15 1 P( x 15) 1 e15/17 1 .4138 .5862
4.201
We know from the Empirical Rule that almost all the observations are larger than
2 . ( 95% are between 2 and 2 ). Thus 2 100 .
For the binomial, np n (.4) and npq n(.4)(.6) .24n
2 100 .4n 2 .24n 100 .4n .98 n 100 0
n , we get:
Solving for
n
.98 .98 2 4(.4)(100) .98 12.687
2(.4)
.8
n 17.084 n 17.084 2 291.9 292
4.202
Let x = tensile strength of a particular metal part. Then x is a normal random variable with
25 and 2 . The tolerance limits are 21 and 30.
21 25
P( x 21) P z
P( z 2) .5 .4772 .0228 (Using Table II, Appendix D).
2
30 25
21 25
z
P(21 x 30) P
P(2 z 2.5) .4772 .4938 .9710
2
2
30 25
P( x 30) P z
P( z 2.5) .5 .4938 .0062
2
E Profit $2 .0228 $10 .9710 $1.0062 $.0456 $9.71 $.0062 $9.66
4.203
Let x = load. Then x has a normal distribution with 20, 000 . We are given P 10 x 30 .95 . We
want to find .
P 10,000 x 30,000 .95 z1 z z2 .95 z1 z 0 P 0 z z2 .95 / 2 .4750
Looking up area .4750 in Table II, Appendix D, z 2 1.96 and z1 1.96 .
z2
4.204
a.
x 30
30, 000 20, 000
1.96
10, 000
5,102
1.96
Let x = number of passengers in 1500 who will be detained for luggage inspection. Then x is a binomial
random variable with n 1,500 and p .20 . The expected number of passengers detained will be:
E x np 1,500 .2 300
b.
For n 4,000 , E x np 4,000 .2 800
Copyright © 2014 Pearson Education, Inc.
Random Variables and Probability Distributions
4.205
c.
(600 .5) 800
P x 600 P z
P ( z 7.89) .5 .5 1.0
4000(.2)(.8)
a.
Using Table II, Appendix D.
255
For 1 :
P(1 x 1) P (4 x 6) P(9 x 11)
1 5
65
11 5
1 5
45
95
P
z
z
z
P
P
1
1
1
1
1
1
P(6 z 4) P ( 1 z 1) P(4 z 6) 0 .3413 .3413 0 .6826
For 2 :
P(1 x 1) P(4 x 6) P(9 x 11)
1 5
65
11 5
1 5
45
95
P
z
z
z
P
P
2
2
2
2
2
2
P(3 z 2) P(.5 z .5) P(2 z 3)
(.4987 .4772) (.1915 .1915) (.4987 .4772) .4260
For 4 :
P(1 x 1) P(4 x 6) P(9 x 11)
1 5
65
11 5
1 5
45
95
P
z
z
z
P
P
4
4
4
4
4
4
P(1.5 z 1) P(.25 z .25) P(1 z 1.5)
(.4332 .3413) (.0948 .0948) (.4332 .3413) .3734
b.
For 1 , 764 of the 1100 flechettes hit a target. The proportion is 764/1100 = .6945. This is a little
higher than the probability that was computed in part a.
For 2 , 462 of the 1100 flechettes hit a target. The proportion is 462/1100 = .42. This is very close
to the probability that was computed in part a.
For 4 , 408 of the 1100 flechettes hit a target. The proportion is 408/1100 = .3709. Again, this is
very close to the probability that was computed in part a.
c.
If the Army wants to maximize the chance of hitting the target that the prototype gun us aimed at, then
should be set at 1. The probability of hitting the target is .6826.
If the Army wants to hit multiple targets with a single shot of the weapon, then should be set at 2.
The probability of hitting at least one of the targets is .4260.
4.206
Let x = number of disasters in 25 trials. If NASA’s assessment is correct, then x is a binomial random
variable with n 25 and p 1 / 60, 000 .00001667 . If the Air Force’s assessment is correct, then x is a
binomial random variable with n 25 and p 1 / 35 .02857 .
Copyright © 2014 Pearson Education, Inc.
256
Chapter 4
If NASA’s assessment is correct, then the probability of no disasters in 25 missions would be:
25
P ( x 0) (1/ 60, 000) 0 (59,999 / 60, 000) 25 .9996
0
Thus, the probability of at least one disaster would be
P ( x 1) 1 P ( x 0) 1 .9996 .0004
If the Air Force’s assessment is correct, then the probability of no disasters in 25 missions would be:
25
P ( x 0) (1/ 35) 0 (34 / 35) 25 .4845
0
Thus, the probability of at least one disaster would be
P ( x 1) 1 P ( x 0) 1 .4845 .5155
One disaster actually did occur. If NASA’s assessment was correct, it would be almost
impossible for at least one disaster to occur in 25 trials. If the Air Force’s assessment was correct, one
disaster in 25 trials would not be an unusual event. Thus, the Air Force’s assessment appears to be
appropriate.
Copyright © 2014 Pearson Education, Inc.
Chapter 5
Sampling Distributions
a–b. The different samples of n 2 with replacement and their means are:
x
0
1
2
3
1
2
3
4
Possible Samples
0, 0
0, 2
0, 4
0, 6
2, 0
2, 2
2, 4
2, 6
c.
d.
e.
Possible Samples
4, 0
4, 2
4, 4
4, 6
6, 0
6, 2
6, 4
6, 6
x
2
3
4
5
3
4
5
6
Since each sample is equally likely, the probability of any 1 being selected is
x
0
1
2
3
4
5
6
1
16
1
1
2
P ( x 1)
16 16 16
1
1
1
3
P ( x 2)
16 16 16 16
1
1
1
1
4
P ( x 3)
16 16 16 16 16
1
1
1
3
P ( x 4)
16 16 16 16
1
1
2
P ( x 5)
16 16 16
1
P ( x 6)
16
P ( x 0)
Using MINITAB, the graph is:
Histogram of x-bar
.25
.1875
Probability
5.1
.125
.0625
0
0
1
2
3
x-bar
4
5
6
257
Copyright © 2014 Pearson Education, Inc.
11 1
4 4 16
p( x )
1/16
2/16
3/16
4/16
3/16
2/16
1/16
258
5.2
Chapter 5
Answers will vary. Using a statistical package, 100 samples of size 2 with replacement were generated
from the population containing 0, 2, 4, and 6. The sample mean was computed for each of the 100 samples
of size 2. The relative frequency distribution for these 100 sample means is:
x
Frequency
Relative
frequency
p( x )
0
1
2
3
4
5
6
4
15
17
30
21
10
3
.04
.15
.17
.30
.21
.10
.03
1/16 = .0625
2/16 = .1250
3/16 = .1875
4/16 = .2500
3/16 = .1875
2/16 = .1250
1/16 = .0625
The exact distribution is in the last column headed with p ( x ) . The relative frequencies from this sample
are similar to the probabilities from the exact distribution.
5.3
If the observations are independent of each other, then
P 1, 1 p 1 p 1 .2 .2 .04
P 1, 2 p 1 p 2 .2 .3 .06
P 1, 3 p 1 p 3 .2 .2 .04
etc.
a.
Possible Sample
x
p( x )
Possible Samples
x
p( x )
1, 1
1, 2
1, 3
1, 4
1, 5
2, 1
2, 2
2, 3
2, 4
2, 5
3, 1
3, 2
3, 3
1
1.5
2
2.5
3
1.5
2
2.5
3
3.5
2
2.5
3
.04
.06
.04
.04
.02
.06
.09
.06
.06
.03
.04
.06
.04
3, 4
3, 5
4, 1
4, 2
4, 3
4, 4
4, 5
5, 1
5, 2
5, 3
5, 4
5, 5
3.5
4
2.5
3
3.5
4
4.5
3
3.5
4
4.5
5
.04
.02
.04
.06
.04
.04
.02
.02
.03
.02
.02
.01
Summing the probabilities, the probability distribution of is:
x
1
1.5
2
2.5
3
3.5
4
4.5
5
p( x )
.04
.12
.17
.20
.20
.14
.08
.04
.01
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
b.
259
Using MINITAB, the graph is:
Histogram of x-bar
.20
Probability
.15
.10
.05
0
1
5.4
1.5
2
2.5
3
x-bar
3.5
4
4.5
5
c.
P( x 4.5) .04 .01 .05
d.
No. The probability of observing x 4.5 or larger is small (.05).
E ( x) xp( x) 1.2 2 .3 3 .2 4 .2 5 .1 .2 .6 .6 .8 .5 2.7
E ( x ) xp( x ) 1.0 .04 1.5 .12 2.0 .17 2.5 .20 3.0 .20 3.5 .14 4.0 .08
4.5 .04 5.0 .01 .04 .18 .34 .50 .60 .49 .32 .18 .05 2.7
5.6
a.
For a sample of size n 2 , the sample mean and sample median are exactly the same. Thus, the
sampling distribution of the sample median is the same as that for the sample mean (see Exercise
5.3a).
b.
The probability histogram for the sample median is identical to that for the sample mean (see
Exercise 5.3b).
a.
Answers will vary. A statistical package was used to generate 500 samples of size 15 from a uniform
distribution on the interval from 150 to 200. The sample mean was computed for each sample of size
15. Using MINITAB, a histogram of the sample means is:
Histogram of Mean
120
100
80
Frequency
5.5
60
40
20
0
156
162
168
174
Mean
180
186
192
Copyright © 2014 Pearson Education, Inc.
260
Chapter 5
b.
The sample medians were computed for each of the 500 samples of size 15 used in part a. Using
MINITAB, a histogram of the sample medians is:
Histogram of Median
120
100
Frequency
80
60
40
20
0
156
162
168
174
Median
180
186
192
The sampling distribution of the sample medians is more spread out than the sampling distribution of
the sample means. In addition, there are more observations in the middle of the distribution of the
sample means than the distribution of the sample medians.
a.
Answers will vary. A statistical package was used to generate 500 samples of size 25 from a uniform
distribution on the interval from 00 to 99. The sample mean was computed for each sample of size
25. Using MINITAB, a histogram of the sample means is:
Histogram of Mean
60
50
40
Frequency
5.7
30
20
10
0
30
36
42
48
54
Mean
60
66
72
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
b.
The sample variances were computed for each of the 500 samples of size 25 used in part a. Using
MINITAB, a histogram of the sample variances is:
Histogram of Variance
60
Frequency
50
40
30
20
10
0
400
5.8
a.
600
800
1000
Variance
1200
1400
xp( x) 0 1 4 1.667
3
3
3
3
1
1
1
5
5 1 5 1
5 1 78
2.889
( x ) p ( x) 0 1 4
3
3
3
3
3 3 27
2
2
2
2
2
b.
Sample
0, 0
0, 1
0, 4
1, 0
1, 1
1, 4
4, 0
4, 1
4, 4
x
0
0.5
1
2
2.5
4
c.
x
0
0.5
2
0.5
1
2.5
2
2.5
4
Probability
1/9
1/9
1/9
1/9
1/9
1/9
1/9
1/9
1/9
Probability
1/9
2/9
1/9
2/9
2/9
1/9
1
2 1 2
2 1 15 5
E ( x ) xp( x ) 0 0.5 1 2 2.5 4 1.667
9
9 9 9
9 9 9 3
Since E ( x ) , x is an unbiased estimator for .
Copyright © 2014 Pearson Education, Inc.
261
262
Chapter 5
d.
Recall that s 2
x
x
2
2
n 1
For the first sample, s 2
n
02 02
For the second sample, s 2
0 0
2 1
12 02
2
2
0.
1 0
2 1
2
2
1
1
2
2 1
2 1
2
The rest of the values are shown in the table below.
s2
0
0.5
8
0.5
0
4.5
8
4.5
0
Sample
0, 0
0, 1
0, 4
1, 0
1, 1
1, 4
4, 0
4, 1
4, 4
Probability
1/9
1/9
1/9
1/9
1/9
1/9
1/9
1/9
1/9
The sampling distribution of s 2 is:
s2
0
0.5
4.5
8
e.
Probability
3/9
2/9
2/9
2/9
3
2
2 2 26
2.889
E ( s 2 ) s 2 p( s 2 ) 0 .5 4.5 8
9
9
9 9 9
Since E ( s 2 ) 2 , s 2 is an unbiased estimator for 2 .
5.9
a.
xp( x) 2 4 9 5
3
3
3
3
1
1
1 15
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
b.
The possible samples of size n 3 , the sample means, and the probabilities are:
Possible
Samples
2, 2, 2
2, 2, 4
2, 2, 9
2, 4, 2
2, 4, 4
2, 4, 9
2, 9, 2
2, 9, 4
2, 9, 9
4, 2, 2
4, 2, 4
4, 2, 9
4, 4, 2
p( x )
x
2
8/3
13/3
8/3
10/3
5
13/3
5
20/3
8/3
10/3
5
10/3
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
m
2
2
2
2
4
4
2
4
9
2
4
4
4
Possible
Samples
4, 4, 4
4, 4, 9
4, 9, 2
4, 9, 4
4, 9, 9
9, 2, 2
9, 2, 4
9, 2, 9
9, 4, 2
9, 4, 4
9, 4, 9
9, 9, 2
9, 9, 4
9, 9, 9
x
4
17/3
5
17/3
22/3
13/3
5
20/3
5
17/3
22/3
20/3
22/3
9
p( x )
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
m
4
4
4
4
9
2
4
9
4
4
9
9
9
9
The sampling distribution of x is:
x
2
8/3
10/3
4
13/3
5
17/3
20/3
22/3
9
p( x )
1/27
3/27
3/27
1/27
3/27
6/27
3/27
3/27
3/27
1/27
27/27
1 8 3 10 3
1 13 3
E ( x ) xp ( x ) 2 4
27
3
27
3
27
27 3 27
6 17 3 20 3 22 3 1
5 9
27 3 27 3 27 3 27 27
2
8 10 4 13 30 17 20 22 9 135
5
27 27 27 27 27 27 27 27 27 27 27
c.
Since 5 in part a, and E ( x ) 5 , x is an unbiased estimator of .
The median was calculated for each sample and is shown in the table in part b. The sampling
distribution of m is:
m
2
4
9
p(m)
7/27
13/27
7/27
27/27
Copyright © 2014 Pearson Education, Inc.
263
264
Chapter 5
7 13 7 14 52 63 129
4.778
E (m) mp(m) 2 4 9
27 27 27 27 27 27 27
The E (m) 4.778 5 . Thus, m is a biased estimator of .
d.
5.10
a.
Use the sample mean, x . It is an unbiased estimator.
xp( x) 0 1 2 1
3
3
3
1
1
1
b.
Sample
0, 0, 0
0, 0, 1
0, 0, 2
0, 1, 0
0, 1, 1
0, 1, 2
0, 2, 0
0, 2, 1
0, 2, 2
1, 0, 0
1, 0, 1
1, 0, 2
1, 1, 0
1, 1, 1
x
0
1/3
2/3
1/3
2/3
1
2/3
1
4/3
1/3
2/3
1
2/3
1
Probability
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
Sample
1, 1, 2
1, 2, 0
1, 2, 1
1, 2, 2
2, 0, 0
2, 0, 1
2, 0, 2
2, 1, 0
2, 1, 1
2, 1, 2
2, 2, 0
2, 2, 1
2, 2, 2
x
4/3
1
4/3
5/3
2/3
1
4/3
1
4/3
5/3
4/3
5/3
2
Probability
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
From the above table, the sampling distribution of the sample mean would be:
x
0
1/3
2/3
1
4/3
5/3
2
Probability
1/27
3/27
6/27
7/27
6/27
3/27
1/27
c.
Sample
0, 0, 0
0, 0, 1
0, 0, 2
0, 1, 0
0, 1, 1
0, 1, 2
0, 2, 0
0, 2, 1
0, 2, 2
1, 0, 0
1, 0, 1
1, 0, 2
1, 1, 0
1, 1, 1
m
0
0
0
0
1
1
0
1
2
0
1
1
1
1
Probability
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
Sample
1, 1, 2
1, 2, 0
1, 2, 1
1, 2, 2
2, 0, 0
2, 0, 1
2, 0, 2
2, 1, 0
2, 1, 1
2, 1, 2
2, 2, 0
2, 2, 1
2, 2, 2
m
1
1
1
2
0
1
2
1
1
2
2
2
2
Copyright © 2014 Pearson Education, Inc.
Probability
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
1/27
Sampling Distributions
From the table, the sampling distribution of the sample median would be:
m
0
1
2
d.
Probability
7/27
13/27
7/27
1 1 3 2 6 7 4 6 5 3 1
E ( x ) xp( x ) 0 1 2 1
27 3 27 3 27 27 3 27 3 27 27
Since E ( x ) , x is an unbiased estimator for .
7 13 7
E (m) mp(m) 0 1 2 1
27 27 27
Since E (m) , m is an unbiased estimator for .
e.
x2 ( x )2 p( x ) (0 1) 2
1 1 3 2 6
2 7
1 1 (1 1)
27 3 27 3 27
27
2
2
4 6 5 3
1 2
1 1 (2 1)2 .2222
3 27 3 27
27 9
2
2
m2 (m 1)2 p(m) (0 1)2
14
7
2 13
2 7
.5185
(1 1) (2 1)
27
27
27 27
f.
5.11
Since both the sample mean and median are unbiased estimators and the variance is smaller for the
sample mean, the sample mean would be the preferred estimator of .
Answers will vary. MINITAB was used to generate 500 samples of size n = 25 observations from a
uniform population from 1 to 50. The first 10 samples along with the sample means and medians are
shown in the table below:
Sample
Observations
Mean
Median
1
28 27 11 19 50 30 47 26 9 33 50 15 21 41 31 41 35 32 32 17 6 32 39 34 21 29.08
31
2
8
6 15 47 26 48 28 25.88
26
3
6 20 27 1 50 14 21 37 46 23
8 29 18 28 40 39 49 33 23 28.24
28
4
45 12 26 13 40 17 11 43 8 35 20 8 44 48 13 46 49 17 47 27 5 45
5
40 38 25 37 47 2 17 40 32 6 22 30 23
6
17
8 43 27 21 5 18 45 31 15
7
40
1 22 29
8
25
3 44 34 29 6 33 32 43 6 43 24 49 14 37 8 46 44 1 12 36 18 30 25 4
9
7 33 36 41 30 13 17 19 14 36 20 39 41 20 15 38 12 37 14
10
4 46 49 49 45 49 24
4 32 32
3 45 18
6
9 40 3 42 21 44 50 42 14 24 10 36
1 34 42 47 24 46
2 38 22 18 7
8 22 20 36 18 45 16 29
27.4
26
3 43 47 16 35 35 24.88
23
3 35 23 45 24 39 38 35 37 24.20
23
3 49 34 24 40 27 5 49 11 30 23.16
22
2 18 22 14 6 22
9
6
9
9 21 36
9 19 2 37 15 8
3 25 22 27 28 23 17 14 6 35 5 20 34 4 41
Copyright © 2014 Pearson Education, Inc.
9 15 3
25.84
29
22.88
19
23.88
23
265
266
Chapter 5
Using MINITAB, side-by side histograms of the means and medians of the 500 samples are:
Histogram of Mean, Median
12
18
Mean
24
30
36
42
Median
140
Frequency
120
100
80
60
40
20
0
12
5.12
18
24
30
36
42
a.
Yes, it appears that x and the median are unbiased estimators of the population mean. The centers of
both distributions above appear to be around 25 to 26. In fact, the mean of the sampling distribution
of x is 25.65 and the mean of the sampling distribution of the median is 25.73.
b.
The sampling distribution of the median has greater variation because it is more spread out than the
sampling distribution of x .
a.
The mean of the random variable x is
E ( x) xp( x) 1(.2) 2(.3) 3(.2) 4(.2) 5(.1) 2.7
From Exercise 5.3, the sampling distribution of x is:
x
1
1.5
2
2.5
3
3.5
4
4.5
5
p( x )
.04
.12
.17
.20
.20
.14
.08
.04
.01
E ( x ) xp ( x ) 1(.04) 1.5(.12) 2(.17) 2.5(.20) 3(.20) 3.5(.14)
The mean of the sampling distribution of x is:
4(.08) 4.5(.04) 5(.01) 2.7
Since E ( x ) E ( x) , x is an unbiased estimator of .
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
b.
267
x2 ( x )2 p( x ) (1 2.7) 2 (.04) (1.5 2.7) 2 (.12) (2 2.7) 2 (.17)
The variance of the sampling distribution of x is:
(2.5 2.7) 2 (.20) (3 2.7)2 (.20) (3.5 2.7) 2 (.14)
(4 2.7) 2 (.08) (4.5 2.7)2 (.04) (5 2.7) 2 (.01) .805
c.
2 x 2.7 2 .805 2.7 1.794 (.906, 4.494)
P(.906 x 4.494) .04 .12 .17 .2 .2 .14 .08 .95
5.13
a.
Refer to the solution to Exercise 5.3. The values of s2 and the corresponding probabilities are listed
below:
s2
x
x
2
2
n 1
n
For sample 1, 1; s 2
22
2 0
1
2
For sample 1, 2: s 2
The rest of the values are calculated and shown:
s2
0.0
0.5
2.0
4.5
8.0
0.5
0.0
0.5
2.0
4.5
2.0
0.5
0.0
p(s2)
.04
.06
.04
.04
.02
.06
.09
.06
.06
.03
.04
.06
.04
s2
0.5
2.0
4.5
2.0
0.5
0.0
0.5
8.0
4.5
2.0
0.5
0.0
p(s2)
.04
.02
.04
.06
.04
.04
.02
.02
.03
.02
.02
.01
The sampling distribution of s2 is:
s2
0.0
0.5
2.0
4.5
8.0
b.
c.
p(s2)
.22
.36
.24
.14
.04
2 ( x ) 2 p ( x) (1 2.7) 2 (.2) (2 2.7) 2 (.3) (3 2.7) 2 (.2)
(4 2.7) 2 (.2) (5 2.7) 2 (.1) 1.61
E ( s 2 ) s 2 p( s 2 ) 0(.22) .5(.36) 2(.24) 4.5(.14) 8(.04) 1.61
Copyright © 2014 Pearson Education, Inc.
32
2 .5
1
5
268
Chapter 5
d.
The sampling distribution of s is listed below, where s s 2 :
s
0.000
0.707
1.414
2.121
2.828
e.
p(s)
.22
.36
.24
.14
.04
E ( s) sp( s) 0(.22) .707(.36) 1.41(.24) 2.1212(.14) 2.828(.04) 1.00394
Since E ( s ) 1.00394 is not equal to 2 1.61 1.269 , s is a biased estimator of .
5.14
E ( x) xp( x) 1(.2) 2(.3) 3(.2) 4(.2) 5(.1) 2.7
The mean of the random variable x is:
From Exercise 5.5, the sampling distribution of the sample median is:
m
p(m)
1
.04
1.5
.12
2
.17
2.5
.20
3
.20
3.5
.14
4
.08
4.5
.04
5
.01
E ( m) mp ( m) 1(.04) 1.5(.12) 2(.17) 2.5(.20) 3(.20) 3.5(.14) 4(.08)
The mean of the sampling distribution of the sample median m is:
4.5(.04) 5(.01) 2.7
Since E (m) , m is an unbiased estimator of .
5.15
The sampling distribution is approximately normal only if the sample size is sufficiently large or if the
population being sampled from is normal.
5.16
a.
x 10 , x / n 3 / 25 0.6
b.
x 100 , x / n 25 / 25 5
c.
x 20 , x / n 40 / 25 8
d.
x 10 , x / n 100 / 25 20
a.
x 100, x
b.
x 100, x
c.
x 100, x
5.17
n
n
n
100
100
100
4
25
100
5
2
1
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
5.18
5.19
5.20
100
100
100
269
d.
x 100, x
e.
x 100, x
f.
x 100, x
a.
x 20, x / n 16 / 64 2
b.
By the Central Limit Theorem, the distribution of x is approximately normal. For the Central Limit
Theorem to apply, n must be sufficiently large. For this problem, n 64 is sufficiently large.
c.
z
d.
z
x x
x
x x
x
n
n
n
50
500
1.414
.447
1000
15.5 20
2.25
2
23 20
1.50
2
.316
In Exercise 5.18, it was determined that the mean and standard deviation of the sampling distribution of the
sample mean are 20 and 2 respectively. Using Table II, Appendix D:
a.
16 20
P( x 16) P z
P( z 2) .5 .4772 .0228
2
b.
23 20
P( x 23) P z
P( z 1.50) .5 .4332 .0668
2
c.
25 20
P( x 25) P z
P( z 2.5) .5 .4938 .0062
2
d.
22 20
16 20
z
P(16 x 22) P
P(2 z 1) .4772 .3413 .8185
2
2
e.
14 20
P( x 14) P z
P( z 3) .5 .4987 .0013
2
For this population and sample size,
E ( x ) 100 , x / n 10 / 900 1/ 3
a.
Almost all of the time, the sample mean will be within three standard deviations of the mean, i.e.,
1
3 100 3 100 1 (99, 101) . Thus, the smallest value of x we would expect is 99
3
and the largest value would be 101.
Copyright © 2014 Pearson Education, Inc.
270
Chapter 5
1
No more than three standard deviations, i.e., 3 1
3
No, the previous answer only depended on the standard deviation of the sampling distribution of the
sample mean, not the mean itself.
b.
c.
5.21
By the Central Limit Theorem, the sampling distribution of x is approximately normal with x 30
and x / n 16 / 100 1.6 . Using Table II, Appendix D:
28 30
P( x 28) P z
P( z 1.25) .5 .3944 .8944
1.6
a.
26.8 30
22.1 30
P(22.1 x 26.8) P
z
P(4.94 z 2) .5 .4772 .0228
1.6
1.6
b.
28.2 30
P( x 28.2) P z
P( z 1.13) .5 .3708 .1292
1.6
c.
27.0 30
P( x 27.0) P z
P( z 1.88) .5 .4699 .9699
1.6
d.
Answers will vary. A computer package was used to generate 500 samples of size n 2 . The sample mean
was computed for each of the 500 samples. This was repeated for 500 samples of size n 5 , 500 samples
of size n 10 , 500 samples of size n 30 , and 500 samples of size n 50 . Using MINITAB, the relative
frequency histograms for x for each of the sample sizes are:
Histogram of xbar2, xbar5, xbar10, xbar30, xbar50
15
xbar2
30
45
60
75
90
xbar5
xbar10
.6
.4
Relative frequency
5.22
.2
xbar30
xbar50
0
15
30
45
60
75
90
.6
.4
.2
0
15
30
45
60
75
90
All of the histograms look mound-shaped. As n increases, the spread of the values of x decreases.
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
a.
From Exercise 2.33, the population of interarrival times is skewed to the right.
b.
The population mean and standard deviation are:
x 25, 504.845 95.52
267
N
2
x2
x
N
N
2
4, 665, 241.665
25,504.845
2
267
267
8,348.025727
8,348.025727 91.3675
c.
d.
By the Central Limit Theorem, the sampling distribution of x will be approximately normal.
91.3675
14.4465 .
Theoretically, x 95.52 and x
n
40
90 95.52
P( x 90) P z
P( z .38) .5 .1480 .3520 (Using Table II, Appendix D.)
14.4465
e &f. Answers will vary. A statistical package was used to randomly select 40 interarrival times from the
Phishing data set and x was computed. This was repeated 50 times to simulate 50 students selecting
40 interarrival times and computing x .
Using MINITAB, a histogram of the 50 x values is:
Histogram of Means
16
14
12
Frequency
5.23
271
10
8
6
4
2
0
60
80
100
120
Means
This shape is somewhat normal.
g.
Using MINITAB, the mean and standard deviation of these 50 means is:
Descriptive Statistics: Means
Variable
Means
N
50
Mean
96.09
StDev
14.08
Minimum
52.73
Q1
86.36
Median
95.65
Copyright © 2014 Pearson Education, Inc.
Q3
105.23
Maximum
130.23
272
Chapter 5
The mean of these 50 means is 96.09. This is very close to x 95.52 found in part c. The standard
deviation of these 50 means is 14.08. This is also very close to x
91.54
14.4465 found in
n
40
part c.
5.24
5.25
5.26
5.27
a.
x 96,850
b.
x
c.
By the Central Limit Theorem, the sampling distribution of x is approximately normal.
30, 000
4, 242.6407
n
50
x x
z
e.
P( x 89,500) P( z 1.73) .5 .4582 .9582 (Using Table II, Appendix D)
a.
x 68 . The average value of sample mean level of support is 68.
b.
x
c.
Because the sample size is large (n = 45 > 30), the Central Limit Theorem says that the sampling
distribution of x is approximately normal.
x
89,500 96,850
1.73
4, 242.6407
d.
27
4.0249 The standard deviation of the distribution of the sample means is 4.0249.
n
45
d.
65 68
P( x 65) P z
P( z .75) .5 .2734 .7734 (Using Table II, Appendix D)
4.0249
a.
E ( x ) x .10
b.
Since n 30 , the sampling distribution of x is approximately normal by the Central Limit Theorem.
c.
.13 .10
P( x .13) P z
P( z 2.13) .5 .4834 .0166 (Using Table II, Appendix D)
.0141
x
.10
.0141
n
50
By the Central Limit Theorem, the sampling distribution of x is approximately normal with
8
1.
x 105.3 and x
n
64
103 105.3
P( x 103) P z
P( z 2.3) .5 .4893 .0107 (Using Table II, Appendix D)
1
5.28
a.
c d 0 3, 600
1,800 . The average value of the sample mean number of seconds
2
2
from the start of the hour is 1,800 second.
E ( x ) E ( x)
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
b.
c.
5.29
273
1
1
(d c)2
(3, 600 0) 2
12
12
18, 000
n
60
60
2
x
2
Because the sample size is sufficiently large, by the Central Limit Theorem, the sampling distribution
of x is approximately normal.
d.
1700 1800
1900 1800
P(1700 x 1900) P
z
P(.75 z .75) .2734 .2734 .5468
18, 000
18, 000
(Using Table II, Appendix D)
e.
2000 1800
P( x 2000) P z
P( z 1.49) .5 .4319 .0681 (Using Table II, Appendix D)
18, 000
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with a mean
.193
.0273 .
x .53 and standard deviation x
50
n
b.
.58 .53
P( x .58) P z
P( z 1.83) .5 .4664 .0336
.0273
c.
If Before Tensioning: x .53
.59 .53
P( x .59) P z
P( z 2.20) .5 .4861 .0139
.0273
If After Tensioning: x .58
.59 .58
P( x .59) P z
P( z 0.37) .5 .1443 .3557
.0273
Since the probability of getting a maximum differential of .59 or more Before Tensioning is so small,
it would be very unlikely that the measurements were obtained Before Tensioning. However, since
the probability of getting a maximum differential of .59 or more After Tensioning is not small, it
would not be unusual that the measurements were obtained after tensioning. Thus, most likely, the
measurements were obtained After Tensioning.
5.30
a.
Since the sample size is small, we also have to assume that the distribution from which the sample
.5
.1118
was drawn is normal. x 1.8 , x
20
n
1.85 1.8
P( x 1.85) P z
P( z 0.45) .5 .1736 .3264
.1118
(using Table II, Appendix D)
Copyright © 2014 Pearson Education, Inc.
274
Chapter 5
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Rough
Variable
Rough
N Mean
20 1.881
StDev
0.524
Minimum
1.060
Q1
1.303
Median
2.040
Q3
2.293
Maximum
2.640
From this output, the value of x is 1.881.
c.
For x = 1.881:
1.881 1.8
P( x 1.881) P z
P( z 0.72) .5 .2642 .2358
.1118
Since this probability is so high, observing a sample mean of x 1.881 is not unusual. The
assumptions in part a appear to be valid.
5.31
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with
10
.5538 .
x 6 and x
326
n
7.5 6
P( x 7.5) P z
P( z 2.71) .5 .4966 .0034
.5538
(Using Table II, Appendix D)
b.
We first need to find the probability of observing the current data or anything more unusual
if the true mean is 6.
300 6
P( x 300) P z
P( z 530.88) .5 .5 0
.5538
Since the probability of observing a sample mean of 300 ppb or higher is essentially 0 if the true
mean is 6 ppb, we would infer that the true mean PFOA concentration for the population of people
who live near DuPont’s Teflon facility is not 6 ppb but higher than 6 ppb.
5.32
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with
x and x / n / 100 .
b.
The mean of the x distribution is equal to the mean of the distribution of the fleet or the
fleet mean score.
c.
x 30 and x / n / 100 60 / 100 6 .
45 30
P( x 45) P z
P z 2.5 .5 .4938 .0062 (Using Table II, Appendix D)
6
d.
The sample mean of 45 tends to refute the claim. If the true fleet mean was as high as 30, observing a
sample mean of 45 or higher would be extremely unlikely (probability = .0062). Thus, we would
infer that the true mean is actually not 30 but something higher. Thus, we would refute the
company’s claim that the mean “couldn’t possibly be as large as 30.”
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
5.33
275
By the Central Limit Theorem, the sampling distribution of x is approximately normal with x 40
and x
n
5
100
.5 .
42 40
P( x 42) P z
P( z 4) .5 .5 0 (Using Table II, Appendix D)
.5
Since this probability is so small, it is very unlikely that the sample was selected from the population of
convicted drug dealers.
5.34
For n 36 , x 406 and x / n 10.1/ 36 1.6833 . By the Central Limit Theorem, the
sampling distribution is approximately normal (n is large).
400.8 406
P( x 400.8) P z
(using Table II, Appendix D)
P( z 3.09) .5 .4990 .0010
1.6833
We agree with the first operator. If the true value of is 406, it would be extremely unlikely to observe an
x as small as 400.8 or smaller (probability .0010). Thus, we would infer that the true value of is less
than 406.
5.35
For n 50 , we can use the Central Limit Theorem to decide the shape of the distribution of the sample
mean bacterial counts. For the handrubbing sample, the sampling distribution of x is approximately
59
8.344 . For the handwashing sample,
normal with a mean of x 35 and standard deviation
n
50
the sampling distribution of x is approximately normal with a mean of x 69 and standard deviation
n
106
50
14.991 .
For Handrubbing:
30 35
P( x 30 | 35) P z
P( z .60) .5 .2257 .2743 (using Table II, Appendix D)
8.344
For Handwashing:
30 69
P( x 30 | 69) P z
P( z 2.60) .5 .4953 .0047 (using Table II, Appendix D)
14.991
Since the probability of getting a sample mean of less than 30 for the handrubbing is not small compared
with that for the handwashing, the sample of workers probably came from the handrubbing group.
5.36
a.
pˆ p .2 and pˆ
p (1 p )
.2(1 .2)
.0566
50
n
b.
pˆ p .2 and pˆ
.2(1 .2)
p(1 p)
.0126
1, 000
n
c.
pˆ p .2 and pˆ
p (1 p )
.2(1 .2)
.02
400
n
Copyright © 2014 Pearson Education, Inc.
276
5.37
5.38
5.39
Chapter 5
a.
pˆ p .1 and pˆ
p (1 p )
.1(1 .1)
.0134
500
n
b.
pˆ p .5 and pˆ
p (1 p )
.5(1 .5)
.0224
500
n
c.
pˆ p .7 and pˆ
p (1 p )
.7(1 .7)
.0205
500
n
a.
pˆ p .3 and pˆ
p (1 p )
.3(1 .3)
.0512
80
n
b.
The sampling distribution of p̂ will be approximately normal since the sample size is sufficiently
large.
c.
z
d.
.35 .3
P( z .98) .5 .3365 .1635 (using Table II, Appendix D)
P( pˆ .35) P z
.3(1 .3)
80
a.
E ( pˆ ) pˆ p .85 and pˆ
b.
The sampling distribution of p̂ will be approximately normal since the sample size is sufficiently
large.
c.
5.40.
pˆ p
p (1 p )
n
P( pˆ .9) P z
.35 .3
.3(1 .3)
80
.98
p (1 p )
.85(1 .85)
.0226
250
n
.9 .85
P( z 2.21) .5 .4864 .9864 (using Table II, Appendix D)
.85(1 .85)
250
We would not expect to see any values of p̂ more than 3 standard deviations below or above the mean value
of p̂ .
z
pˆ p
p(1 p)
n
3
pˆ .4
.4(1 .4)
1500
3
.4(1 .4)
pˆ .4 .0379 pˆ .4 pˆ .3621
1500
The smallest value we would expect for p̂ would be .3621.
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
z
pˆ p
p(1 p)
n
3
pˆ .4
.4(1 .4)
1500
3
277
.4(1 .4)
pˆ .4 .0379 pˆ .4 pˆ .4379
1500
The largest value we would expect for p̂ would be .4379.
a.
Answers will vary. Using a statistical package, 500 samples of size 10 were generated from the
population of (0,1). The histogram of the 500 sample proportions is:
Histogram of p-hat10
120
Frequency
100
80
60
40
20
0
0.0
0.2
0.4
0.6
0.8
1.0
p-hat10
b.
Using a statistical package, 500 samples of size 25 were generated from the population of (0,1). The
histogram of the 500 sample proportions is:
Histogram of p-hat25
100
80
Frequency
5.41
60
40
20
0
0.12
0.24
0.36
0.48
p-hat25
0.60
0.72
Copyright © 2014 Pearson Education, Inc.
278
Chapter 5
c.
Using a statistical package, 500 samples of size 100 were generated from the population of (0,1).
The histogram of the 500 sample proportions is:
Histogram of p-hat100
50
Frequency
40
30
20
10
0
0.40
d.
0.44
0.48
0.52
p-hat100
0.56
0.60
0.64
As the sample size increases, the spread of the values of p̂ decreases. In the graph in part a, the
spread of the values of p̂ is from 0 to 1. In the graph in part b, the spread of the values of p̂ is from
.20 to .76. In the graph in part c, the spread of the values of p̂ is from .37 to .64. In all graphs, the
distributions are mound-shaped. As the sample size increases, the distribution becomes more peaked.
5.42
pˆ p .4 and pˆ
b.
The sampling distribution of p̂ will be approximately normal since the sample size is sufficiently
large.
c.
d.
5.43
p (1 p )
.4(1 .4)
.0476
106
n
a.
.59
.4
P( z 3.99) .5 .49997 .00003 (Using Table II, Appendix D)
P( pˆ .59) P z
.4(1 .4)
106
x 63
.59. We found in part c that P( pˆ .59) .00003 . Since this probability is so small,
n 106
it casts doubt on the assumption that 40% of all social robots are designed with legs, but no wheels.
pˆ
a.
pˆ p .67
b.
pˆ
c.
p (1 p )
.67(1 .67)
.0149
1000
n
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large.
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
5.44
d.
.75 .67
P( pˆ .75) P z
P( z 5.38) .5 .5 1 (using Table II, Appendix D)
.67(1 .67)
1000
e.
P( pˆ .5) P z
.5 .67
P( z 11.43) .5 .5 1 (using Table II, Appendix D)
.67(1 .67)
1000
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the sample
size is sufficiently large, with pˆ p .45 and pˆ
5.45
p (1 p )
.45(1 .45)
.0222 .
500
n
a.
.4 .45
.5
.45
P(2.25 z 2.25) .4878 .4878 .9756
z
P(.4 pˆ .5) P
.45(1 .45)
.45(1 .45)
500
500
(using Table II, Appendix D)
b.
.6
.45
P( z 6.74) .5 .5 0 (using Table II, Appendix D)
P( pˆ .6) P z
.45(1 .45)
500
a.
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large, with pˆ p .03 and pˆ
5.46
279
p (1 p )
.03(1 .03)
.0054 .
1000
n
b.
P( pˆ .05) P z
c.
.025
.03
P( z .93) .5 .3238 .8238 (using Table II, Appendix D)
P( pˆ .025) P z
.03(1 .03)
1000
a.
P( z 3.71) .5 .4999 .9999 (using Table II, Appendix D)
.03(1 .03)
1000
.05 .03
Let pˆ H = sample proportion of Finnish citizens with high IQ who invest in the stock market. By the
Central Limit theorem, the sampling distribution of pˆ H will be approximately normal since the sample
size is sufficiently large, with pˆ H pH .44 and pˆ H
pH (1 pH )
.44(1 .44)
.0222 .
500
n
Copyright © 2014 Pearson Education, Inc.
280
Chapter 5
150
.3 .44
P pˆ H
P( z 6.31) .5 .5 1
P pˆ H .3 P z
500
.44(1 .44)
500
(using Table II, Appendix D)
b.
Let pˆ A = sample proportion of Finnish citizens with average IQ who invest in the stock market. By
the Central Limit theorem, the sampling distribution of pˆ A will be approximately normal since the
sample size is sufficiently large, with pˆ A p A .26 and pˆ A
p A (1 p A )
.26(1 .26)
.0196 .
500
n
150
.3 .26
P pˆ A
P( z 2.04) .5 .4793 .0207
P pˆ A .3 P z
500
.26(1 .26)
500
(using Table II, Appendix D)
c.
Let pˆ L = sample proportion of Finnish citizens with low IQ who invest in the stock market. By the
Central Limit theorem, the sampling distribution of pˆ L will be approximately normal since the
sample size is sufficiently large, with pˆ L pL .14 and pˆ L
pL (1 pL )
.14(1 .14)
.0155 .
500
n
150
.3 .14
P pˆ L
P( z 10.31) .5 .5 0
P pˆ L .3 P z
500
.14(1 .14)
500
(using Table II, Appendix D)
5.47
a.
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large, with pˆ p .4 and pˆ
P pˆ .6 P z
b.
p (1 p )
.4(1 .4)
.0693 .
50
n
.6 .4
P( z 2.89) .5 .4981 .0019 (Using Table II, Appendix D)
.4(1 .4)
50
Since the probability of observing a value of p̂ larger than .6 is so small (p = .0019) and we observed
a value of pˆ .62 , we would conclude that the true proportion of adult cell phone owners who
download an “app” is not .4 but something larger than .4.
c.
If the value of pˆ .62 was obtained at a convention for the International Association for the Wireless
Telecommunications Industry, then it is probably not representative of the population of all adult cell
phone owners. Those who attend such a convention would tend to be more “tech” savvy than the
population of all adult cell phone owners. The value of pˆ .62 would be larger than what we would
expect from the general population.
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
5.48
281
From Exercise 4.48, we defined the following events:
P: {hotel guest is aware of conservation program}
A: {hotel guest participates in conservation efforts}
Then, the probability of a hotel guest being aware and participating in the hotel’s conservation efforts is
p P( P | A) P( A) .72(.66) .4752 .
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large, with pˆ p .4752 and pˆ
42
ˆ
.42
P pˆ
P
p
P
z
100
(using Table II, Appendix D)
5.49
p (1 p )
.4752(1 .4752)
.0499 .
100
n
.42 .4752
P( z 1.11) .5 .3665 .1335
.4752(1 .4752)
100
a.
E ( pˆ ) pˆ p .92
b.
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large, with pˆ p .92 and pˆ
p (1 p )
.92(1 .92)
.0086 .
1000
n
900
.9 .92
P pˆ
P( z 2.33) .5 .4901 .0099
P pˆ .9 P z
1000
.92(1 .92)
1000
(using Table II, Appendix D)
5.50
a.
As the sample size increases, the standard error will decrease. This property is important because we
know that the larger the sample size, the less variable our estimator will be. Thus, as n increases, our
estimator will tend to be closer to the parameter we are trying to estimate.
b.
This would indicate that the statistic would not be a very good estimator of the parameter. If the
standard error is not a function of the sample size, then a statistic based on one observation would be
as good an estimator as a statistic based on 1000 observations.
c.
x would be preferred over A as an estimator for the population mean. The standard error of x is
smaller than the standard error of A.
d.
The standard error of x is
n
10
10
1.25 and the standard error of A is 3
2.5 .
64
64
If the sample size is sufficiently large, the Central Limit Theorem says the distribution of x is
approximately normal. Using the Empirical Rule, approximately 68% of all the values of x will fall
between 1.25 and 1.25 . Approximately 95% of all the values of x will fall between 2.50
and 2.50 . Approximately all of the values of x will fall between 3.75 and 3.75 .
Copyright © 2014 Pearson Education, Inc.
282
Chapter 5
Using the Empirical Rule, approximately 68% of all the values of A will fall between 2.50 and
2.50 . Approximately 95% of all the values of A will fall between 5.00 and 5.00 .
Approximately all of the values of A will fall between 7.50 and 7.50 .
5.51
5.52
a.
"The sampling distribution of the sample statistic A" is the probability distribution of the variable A.
b.
"A" is an unbiased estimator of if the mean of the sampling distribution of A is .
c.
If both A and B are unbiased estimators of , then the statistic whose standard deviation is smaller is
a better estimator of .
d.
No. The Central Limit Theorem applies only to the sample mean. If A is the sample mean, x , and n
is sufficiently large, then the Central Limit Theorem will apply. However, both A and B cannot be
sample means. Thus, we cannot apply the Central Limit Theorem to both A and B.
a.
First we must compute and . The probability distribution for x is:
x
1
2
3
4
p(x)
.3
.2
.2
.3
E ( x) xp( x) 1(.3) 2(.2) 3(.2) 4(.3) 2.5
2 E ( x ) 2 ( x )2 p( x) (1 2.5)2 (.3) (2 2.5)2 (.2) (3 2.5) 2 (.2) (4 2.5)2 (.3) 1.45
x 2.5 , x
b.
5.53
n
1.45
40
.1904
By the Central Limit Theorem, the distribution of x is approximately normal. The sample size,
n 40 , is sufficiently large. Yes, the answer depends on the sample size.
By the Central Limit Theorem, the sampling distribution of x is approximately normal.
x 19.6 , x
3.2
68
.388
a.
19.6 19.6
P( x 19.6) P z
P( z 0) .5
.388
b.
19 19.6
P( x 19) P z
P( z 1.55) .5 .4394 .0606
.388
c.
20.1 19.6
P( x 20.1) P z
P( z 1.29) .5 .4015 .0985
.388
d.
(Using Table II, Appendix D)
20.6 19.6
19.2 19.6
z
P(19.2 x 20.6) P
.388
.388
P(1.03 z 2.58) .3485 .4951 .8436
Copyright © 2014 Pearson Education, Inc.
(Using Table II, Appendix D)
(Using Table II, Appendix D)
(Using Table II, Appendix D)
Sampling Distributions
5.54
5.55
p (1 p )
.35(1 .35)
.0213 .
500
n
a.
pˆ p .35 and pˆ
b.
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large.
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large with pˆ p .8 and pˆ
p (1 p )
.8(1 .8)
.0231 .
300
n
a.
.83
.8
P( z 1.30) .5 .4032 .9032 (Using Table II, Appendix D)
P pˆ .83 P z
.8(1 .8)
300
b.
.75
.8
P( z 2.17) .5 .4850 .9850 (Using Table II, Appendix D)
P pˆ .75 P z
.8(1 .8)
300
c.
.79 .8
.81
.8
P(.43 z .43) .1664 .1664 .3328
z
P .79 pˆ .81 P
.8(1 .8)
.8(1 .8)
300
300
(using Table II, Appendix D)
Answers will vary. One hundred samples of size n = 2 were selected from a normal distribution with a
mean of 100 and a standard deviation of 10. The process was repeated for samples of size n = 5, n = 10,
n = 30, and n = 50. For each sample, the value of x was computed. Using MINITAB, the histograms for
each set of 100 x ’s were constructed:
Histogram of xbar2, xbar5, xbar10, xbar30, xbar50
Normal
85
xbar2
90
95
0 5 0 5 0
10 10 11 11 12
xbar5
xbar10
60
45
30
Frequency
5.56
283
15
xbar30
xbar50
60
0
85
90
95 100 105 110 115 120
45
xbar2
Mean 101.1
StDev 6.614
N
100
xbar5
Mean 99.70
StDev 6.278
N
100
xbar10
Mean 99.73
StDev 3.249
N
100
xbar30
Mean 100.2
StDev 2.040
N
100
30
15
0
85
90
95 100 105 110 115 120
Copyright © 2014 Pearson Education, Inc.
xbar50
Mean 100.1
StDev 1.512
N
100
284
Chapter 5
The sampling distribution of x is normal regardless of the sample size because the population we sampled
from was normal. Notice that as the sample size n increases, the variances of the sampling distributions
decrease.
5.57
Answers will vary. One hundred samples of size n = 2 were selected from a uniform distribution on the
interval from 0 to 10. The process was repeated for samples of size n = 5, n = 10, n = 30, and n = 50. For
each sample, the value of x was computed. Using MINITAB, the histograms for each set of 100 x ’s were
constructed:
Histogram of xbar2, xbar5, xbar10, xbar30, xbar50
Normal
0.0 1.5 3.0 4.5 6.0 7.5 9.0
xbar2
xbar5
xbar10
48
36
Frequency
24
12
xbar30
xbar50
0
0.0 1.5 3.0 4.5 6.0 7.5 9.0
48
36
xbar2
Mean 4.935
StDev 2.073
N
100
xbar5
Mean 4.828
StDev 1.610
N
100
xbar10
Mean
5.004
StDev 0.9256
N
100
xbar30
Mean
5.010
StDev 0.5652
N
100
24
12
xbar50
Mean
4.998
StDev 0.4323
N
100
0
0.0 1.5 3.0 4.5 6.0 7.5 9.0
For small sizes of n, the sampling distributions of x are somewhat normal. As n increases, the sampling
distributions of x become more normal.
5.58
a.
Tossing a coin two times can result in:
2 heads (2 ones)
2 tails (2 zeros)
1 head, 1 tail (1 one, 1 zero)
b.
x2 heads
1 0 1
11
00
1 ; x2 tails
0 ; x1H,1T
2
2
2
2
c.
pˆ 2 heads
2
0
1
1 ; pˆ 2 tails 0 ; pˆ 1H ,1T
2
2
2
d.
There are four possible combinations for one coin tossed two times, as shown below:
Coin Tosses
H, H
H, T
T, H
T, T
p̂
1
1/2
1/2
0
p̂
0
1/2
1
Copyright © 2014 Pearson Education, Inc.
p( pˆ )
1/4
1/2
1/4
Sampling Distributions
e.
The sampling distribution of p̂ is given in the histogram shown.
H istogr am of p-hat
0.5
p(p-hat)
0.4
0.3
0.2
0.1
0.0
0.0
5.59
0.5
p-hat
1.0
Given: 100 and 10
1
n
5
10
n
20
30
4.472 3.162 2.236 1.826
The graph of
10
n
40
50
1.581 1.414
against n is given here:
Scatter plot of st er r vs n
10
9
8
st err
7
6
5
4
3
2
1
0
10
20
30
40
50
n
5.60
a.
x 141
b.
x
n
18
100
1.8
Copyright © 2014 Pearson Education, Inc.
285
286
5.61
Chapter 5
c.
By the Central Limit Theorem, the sampling distribution of x is approximately normal.
d.
z
e.
P( x 142) P( z 0.56) .5 .2123 .2877 (Using Table II, Appendix D)
x x
x
142 141
0.56
1.8
By the Central Limit Theorem, the sampling distribution of x is approximately normal with x 19
and x
n
65
100
6.5 .
10 19
P( x 10) P z
P( z 1.38) .5 .4162 .0838 (using Table II, Appendix D)
6.5
5.62
5.63
a.
For x to be a binomial random variable, the n trials must be identical. We can assume that the
process of selecting of a worker is identical from trial to trial. There are two possible outcomes - a
worker missed work due to a back injury or not. The probability of success must be the same from
trial to trial. We can assume that the probability of missing work due to a back injury is constant.
The trials must be independent of each other. We can assume that the outcome of one trial will not
affect the outcome of any other. Thus, x is a binomial random variable.
b.
From the information given in the problem, the estimate of p is .40.
c.
pˆ p .4 and pˆ
d.
.38 .4
P pˆ .38 P z
P( z .41) .5 .1591 .3409 (using Table II, Appendix D)
.4(1 .4)
100
a.
E ( pˆ ) pˆ p .60
b.
pˆ
c.
5.64
p (1 p )
.4(1 .4)
.0490
100
n
p (1 p )
.6(1 .6)
.0566
75
n
By the Central Limit theorem, the sampling distribution of p̂ will be approximately normal since the
sample size is sufficiently large.
d.
.70 .6
P( z 1.77) .5 .4616 .0384 (using Table II, Appendix D)
P( pˆ .70) P z
.6(1 .6)
75
a.
x 89.34 ; x
n
7.74
35
1.3083
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
287
b.
5.65
c.
88 89.34
P( x 88) P z
P( z 1.02) .5 .3461 .8461 (using Table II, Appendix D)
1.3083
d.
87 89.34
P( x 87) P z
P( z 1.79) .5 .4633 .0367 (using Table II, Appendix D)
1.3083
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal
with x and x / n / 50 .
b.
x 40 and x / 50 12 / 50 1.6971 .
44 40
P( x 44) P z
P( z 2.36) .5 .4909 .0091 (using Table II, Appendix D)
1.6971
c.
2 / n 40 2 1.6971 40 3.3942 36.6058, 43.3942
43.3942 40
36.6058 40
z
P(36.6058 x 43.3942) P
1.6971 (using Table II, Appendix D)
1.6971
P(2 z 2) 2 .4772 .9544
5.66
a.
The mean diameter of the bearings, , is unknown with a standard deviation of .001 inch.
Assuming that the distribution of the diameters of the bearings is normal, the sampling distribution of
the sample mean is also normal. The mean and variance of the distribution are:
.001
.0002
x , x
25
n
Having the sample mean fall within .0001 inch of implies
x .0001 or .0001 x .0001
.0001
.0001
z
P(.0001 x .0001) P
P(.50 z .50) .1915 .1915 .3830
.0002
.0002
(using Table II, Appendix D)
b.
The approximation is unlikely to be accurate. In order for the Central Limit Theorem to apply, the
sample size must be sufficiently large. For a very skewed distribution, n 25 is not sufficiently
large, and thus, the Central Limit Theorem will not apply.
Copyright © 2014 Pearson Education, Inc.
288
5.67
Chapter 5
From Exercise 5.66, .001 . We must assume the Central Limit theorem applies
(n is only 25). Thus, the distribution of x is approximately normal with x .501
and x
n
.001
25
.0002 . Using Table II, Appendix D,
.4994 .501
.5006 .501
P( x .4994) P( x .5006) P z
P z
.0002
.0002
P( z 8) P( z 2) (.5 .5) (.5 .4772) .9772
5.68
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with
6
x
.3235.
n
344
b.
19.1 18.5
If 18.5 , P( x 19.1) P z
P( z 1.85) .5 .4678 .0322
.3235
(Using Table II, Appendix D)
c.
19.1 19.5
If 19.5 , P( x 19.1) P z
P( z 1.24) .5 .3925 .8925
.3235
(Using Table II, Appendix D)
d.
19.1
P( x 19.1) P z
.5
.3235
e.
We know that P z 0 .5 . Thus,
19.1
0 19.1
.3235
19.1
P( x 19.1) P z
.2
.3235
Thus, must be less than 19.1. If 19.1 , then P( P( x 19.1) .5. Since P( x 19.1) .5 , then
19.1 .
5.69
.2(1 .2)
p (1 p )
.0253
250
n
a.
E ( pˆ ) pˆ p .2 and pˆ
b.
E ( pˆ ) 2 pˆ .2 2(.0253) .2 .0506 .1494, .2506
c.
By the Central Limit Theorem, the sampling distribution of p̂ will be approximately normal sincef the
sample size is sufficiently large. Thus,
.2506 .2
.1494 .2
z
P(.1494 pˆ .2506) P
P(2 z 2) .4772 .4772 .9544
.0253
.0253
5.70
a.
b.
By the Central Limit Theorem, the sampling distribution of x is approximately normal since n 30
15
2.1213
with x 840 and x
n
50
830 840
P( x 830) P z
P( z 4.71) .5 .5 0
2.1213
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
c.
Since the probability of observing a mean of 830 or less is extremely small ( 0) if the true mean is
840, we would tend to believe that the mean is not 840, but something less.
d.
By the Central Limit Theorem, the sampling distribution of x is approximately normal since n 30
45
6.3640
with x 840 and x
n
50
289
830 840
P( x 830) P z
P( z 1.57) .5 .4418 .0582
6.3640
5.71
a.
Let p1 = probability of an error 1 / 100 .01 and p2 = probability of an error resulting in a significant
problem 1 / 500 .002 .
Let p̂1 = proportion of errors. Then E ( pˆ1 ) pˆ1 p1 .01 .
Let p̂2 = proportion of significant errors. Then E ( pˆ 2 ) pˆ 2 p2 .002 .
b.
Since the distribution of p̂2 will be approximately normal by the Central Limit Theorem, we would
expect the proportion of significant errors to fall within 2 standard deviations of the expected value.
The interval would be:
pˆ 2 2 pˆ 2 .002 2
5.72
Even though the number of flaws per piece of siding has a Poisson distribution, the Central Limit Theorem
implies that the distribution of the sample mean will be approximately normal with x 2.5 and
x
5.73
5.74
.002(1 .002)
.002 .00036 (.00164, .00236)
60, 000
n
2.5
35
2.1 2.5
.2673 . Therefore, P( x 2.1) P z
P( z 1.50) .5 .4332 .9332
2.5/ 35
(using Table II, Appendix D)
a.
If x is an exponential random variable, then E ( x) 60 . The standard deviation of x is
60 .
2 602
36
V ( x ) x2
Then, E ( x ) x 60 ;
n 100
b.
Because the sample size is fairly large, the Central Limit Theorem says that the sampling distribution
of x is approximately normal.
c.
30 60
P( x 30) P z
P( z 5.0) .5 .5 0 (using Table II, Appendix D)
36
a.
By the Central Limit Theorem, the distribution of x is approximately normal, with x 157 and
x
n
3
40
.474 .
The sample mean is 1.3 psi below 157 or x 157 1.3 155.7
Copyright © 2014 Pearson Education, Inc.
290
Chapter 5
155.7 157
P( x 155.7) P z
P( z 2.74) .5 .4969 .0031 (using Table II, Appendix D)
.474
If the claim is true, it is very unlikely (probability = .0031) to observe a sample mean 1.3 psi below
157 psi. Thus, the actual population mean is probably not 157 but something lower.
b.
155.7 156
P( x 155.7) P z
P( z .63) .5 .2357 .2643 (using Table II, Appendix D)
.474
The observed sample is more likely if 156 rather than 157 .
155.7 158
P( x 155.7) P z
P( z 4.85) .5 .5 0
.474
The observed sample is less likely if 158 rather than 157 .
c.
If 2 , x
n
2
40
.316 .
155.7 157
P( x 155.7) P z
P( z 4.11) .5 .5 0 (using Table II, Appendix D)
.316
The observed sample is less likely if 2 than if 3.
If 6 , x
n
6
40
.949 .
155.7 157
P( x 155.7) P z
P( z 1.37) .5 .4147 .0853
.949
(using Table II, Appendix D)
The observed sample is more likely if 6 than if 3.
5.75
Answers will vary. We are to assume that the fecal bacteria concentrations of water specimens follow an
approximate normal distribution. Now, suppose that the distribution of the fecal bacteria concentration at a
beach is normal with a true mean of 360 and with a standard deviation of 40. If only a single sample was
selected, then the probability of getting an observation at the 400 level or higher would be:
400 360
P( x 400) P z
P( z 1) .5 .3413 .1587 (using Table II, Appendix D)
40
Thus, even if the water is safe, the beach would be closed approximately 15.87% of the time.
On the other hand, if the mean was 440 and the standard deviation was still 40, then the probability of
getting a single observation less than the 400 level would be:
400 440
P( x 400) P z
P( z 1) .5 .3413 .1587 (using Table II, Appendix D)
40
Thus, the beach would remain open approximately 15.78% of the time when it should be closed.
Now, suppose we took a random sample of 64 water specimens. The sampling distribution of x is
Copyright © 2014 Pearson Education, Inc.
Sampling Distributions
approximately normal by the Central Limit Theorem with x and x
n
40
64
291
5.
400 360
If 360 , P( x 400) P z
P( z 8) .5 .5 0 . Thus, the beach would never be shut
5
down if the water was actually safe if we took samples of size 64.
400 440
If 440 , P( x 400) P z
P( z 8) .5 .5 0 . Thus, the beach would never be left
5
open if the water was actually unsafe if we took samples of size 64.
The single sample standard can lead to unsafe decisions or inconvenient decisions, but is much easier to
collect than samples of size 64.
Copyright © 2014 Pearson Education, Inc.
Chapter 6
Inferences Based on a Single Sample:
Estimation with Confidence Intervals
6.1
a.
For .10 , / 2 .10 / 2 .05 . z / 2 z.05 is the z-score with .05 of the area to the right of it. The
area between 0 and z.05 is .5 .05 .4500 . Using Table II, Appendix D, z.05 1.645 .
b.
For .01 , / 2 .01/ 2 .005 . z / 2 z.005 is the z-score with .005 of the area to the right of it.
The area between 0 and z.005 is .5 .005 .4950 . Using Table II, Appendix D, z.005 2.575 .
c.
For .05 , / 2 .05 / 2 .025 . z / 2 z.025 is the z-score with .025 of the area to the right of it.
The area between 0 and z.025 is .5 .025 .4750 . Using Table II, Appendix D, z.025 1.96 .
d.
For .20 , / 2 .20 / 2 .10 . z / 2 z.10 is the z-score with .10 of the area to the right of it. The
area between 0 and z.10 is .5 .10 .4000 . Using Table II, Appendix D, z.10 1.28 .
6.2
a.
z / 2 1.96 , using Table II, Appendix D, P(0 z 1.96) .4750 . Thus, / 2 .5 .4750 .025 ,
2(.025) .05 , and 1 1 .05 .95 . The confidence level is 100%(.95) 95% .
b.
z / 2 1.645 , using Table II, Appendix D, P(0 z 1.645) .45 . Thus, / 2 .5 .45 .05 ,
2(.05) .10 , and 1 1 .10 .90 . The confidence level is 100%(.90) 90% .
c.
z / 2 2.575 , using Table II, Appendix D, P(0 z 2.575) .495 . Thus, / 2 .5 .495 .005 ,
2(.005) .01 , and 1 1 .01 .99 . The confidence level is 100%(.99) 99% .
d.
z / 2 1.282 , using Table II, Appendix D, P(0 z 1.282) .4 . Thus, / 2 .5 .4 .1 ,
2(.1) .20 , and 1 1 .20 .80 . The confidence level is 100%(.80) 80% .
e.
z / 2 .99 , using Table II, Appendix D, P(0 z .99) .3389 . Thus, / 2 .5 .3389 .1611 ,
2(.1611) .3222 , and 1 1 .3222 .6778 . The confidence level is 100%(.6778) 67.78% .
6.3
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z.025
b.
c.
x z.025
x z.025
n
n
n
28 1.96
102 1.96
15 1.96
12
75
28 .784 27.216, 28.784
22
200
.3
100
102 .65 101.35, 102.65
15 .0588 14.9412, 15.0588
292
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 293
d.
6.4
x z.025
4.05 1.96
.83
100
4.05 .163 3.887, 4.213
No. Since the sample size in each part was large (n ranged from 75 to 200), the Central Limit
Theorem indicates that the sampling distribution of x is approximately normal.
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
b.
c.
n
25.9 1.96
2.7
90
25.9 .56 25.34, 26.46
n
25.9 1.645
2.7
90
25.9 .47 25.43, 26.37
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z.025
a.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
x z.025
n
25.9 2.58
2.7
90
25.9 .73 25.17, 26.63
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z / 2
s
n
26.2 1.96
4.1
70
26.2 .96 (25.24, 27.16)
b.
The confidence coefficient of .95 means that in repeated sampling, 95% of all confidence intervals
constructed will include .
c.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z / 2
6.6
n
e.
x z.025
6.5
s
n
26.2 2.58
4.1
70
26.2 1.26 (24.94, 27.46)
d.
As the confidence coefficient increases, the width of the confidence interval also increases.
e.
Yes. Since the sample size is 70, the Central Limit Theorem applies. This ensures the distribution of
x is normal, regardless of the original distribution.
If we were to repeatedly draw samples from the population and form the interval x 1.96 x each time,
approximately 95% of the intervals would contain . We have no way of knowing whether our interval
estimate is one of the 95% that contain or one of the 5% that does not.
Copyright © 2014 Pearson Education, Inc.
294
Chapter 6
6.7
A point estimator is a single value used to estimate the parameter, . An interval estimator is two values,
an upper and lower bound, which define an interval with which we attempt to enclose the parameter, .
An interval estimate also has a measure of confidence associated with it.
6.8
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
33.9 1.96
3.3
33.9 1.96
3.3
33.9 .647 33.253, 34.547
x z.025
s
b.
x z.025
s
c.
For part a, the width of the interval is 2(.647) 1.294 . For part b, the width of the interval is
n
n
100
400
33.9 .323 33.577, 34.223
2(.323) .646 . When the sample size is quadrupled, the width of the confidence interval is halved.
6.9
Yes. As long as the sample size is sufficiently large, the Central Limit Theorem says the distribution of x
is approximately normal regardless of the original distribution.
6.10
a.
The confidence coefficient that was used is .95.
b.
Southwest: We are 95% confident that the true mean airfare for Southwest Airlines was between
$412 and $496.
Delta: We are 95% confident that the true mean airfare for Delta Airlines was between
$468 and $500.
USAir: We are 95% confident that the true mean airfare for USAir was between
$247 and $372.
6.11
c.
“95% confident” means that in repeated sampling, 95% of all confidence intervals constructed will
contain the true mean.
d.
To reduce the width of the confidence interval, one would use a smaller confidence coefficient. The
smaller the value of the confidence coefficient, the smaller the z-score associated with the confidence
coefficient. Thus, one would be adding and subtracting a smaller value if the z-score is smaller.
a.
The point estimate of is x 3.11 .
b.
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table II, Appendix D,
z.01 2.33 . The confidence interval is:
x z.01
c.
n
3.11 2.33
.66
307
3.11 .088 (3.022, 3.198)
This statement is incorrect. Once the interval is constructed, there is no probability involved. The
true mean is either in the interval or it is not. A better statement would be: “We are 98% confident
that the true mean GPA will be between 3.022 and 3.198.”
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 295
6.12
d.
Since the sample size is so large ( n 307 ), the Central Limit Theorem applies. Thus, it does not
matter whether the distributions of grades is skewed or not.
a.
The 90% confidence interval is (66.350, 69.160).
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
x z.05
67.755 1.645
n
26.871
992
67.755 1.403 (66.352, 69.158)
This is close to the interval reported in the output.
6.13
c.
We are 90% confident that the true mean level of support for all senior managers is between 66.350
and 69.160.
d.
No. The 90% confidence interval does not contain 75. Therefore, it is not a likely value for the true
mean.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
s
x z.05
n
6,563 1.645
2, 484
1, 751
6,563 97.65 (6, 465.35, 6, 660.65)
We are 90% confident that the true mean expenses per full-time equivalent employee of all U.S. Army
hospitals is between $6,465.35 and $6,660.65.
6.14
6.15
a.
From the printout, the 95% confidence interval is (1.6711, 2.1989).
b.
We are 95% confident that the true mean failure time of used colored display panels is between
1.6711 and 2.1989 years.
c.
If 95% confidence intervals are formed, then approximately .95 of the intervals will contain the true
mean failure time.
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Wheels
Variable
Wheels
N
28
Mean
3.214
StDev
1.371
Minimum
1.000
Q1
2.000
Median
3.000
Q3
4.000
Maximum
8.000
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z.005
n
67.755 2.58
1.371
28
3.214 .668 (2.546, 3.882)
b.
We are 99% confident that the true mean number of wheels used on all social robots is between 2.546
and 3.882.
c.
99% of all similarly constructed confidence intervals will contain the true mean.
Copyright © 2014 Pearson Education, Inc.
296
Chapter 6
6.16
a.
The population of interest is all U.S. women who shop on Black Friday.
b.
The quantitative variable of interest is the number of hours spent shopping.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Hours
Variable
Hours
N
38
Mean
6.079
StDev
2.755
Minimum
3.000
Q1
4.000
Median
5.000
Q3
7.250
Maximum
16.000
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z.025
6.17
n
6.079 1.96
2.755
38
6.079 .876 (5.203, 6.955)
d.
We are 95% confident that the true mean number of hours spent shopping on Black Friday is between
5.203 and 6.995.
e.
No. The confidence interval constructed in part c contains 5.5. Therefore, the 5.5 is not an unusual
value for the mean.
a.
The target parameter is the population mean 20118 salary of these 500 CEOs who participated in the
Forbes’ survey, .
b.
Answers will vary. Using MINITAB, a sample of 50 CEOs was selected. The ranks of the 50
selected are:
9, 10, 14, 18, 19, 22, 25, 32, 38, 45, 49, 50, 55, 60, 66, 69, 77, 96, 104, 106,
115, 147, 152, 167, 192, 197, 209, 213, 229, 241, 245, 261, 268, 292, 305, 309,
325, 337, 342, 358, 364, 370, 376, 384, 405, 417, 423, 433, 470, 482.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Pay ($mil)
Variable
Pay ($mil)
N
50
Mean
12.40
StDev
10.34
Minimum
0.940
Q1
4.20
Median
7.18
Q3
19.47
Maximum
37.90
The sample mean is x 12.40 and the sample standard deviation is s 10.34 .
d.
Using MINITAB, the descriptive statistics for the entire data set is:
Descriptive Statistics: Pay ($mil)
Variable
Pay ($mil)
N
478
Mean
9.247
StDev
9.842
Minimum
0.000000000
Q1
3.413
Median
6.100
From the above, the standard deviation of the population is $9.842 million.
Copyright © 2014 Pearson Education, Inc.
Q3
11.346
Maximum
101.965
Inferences Based on a Single Sample: Estimation with Confidence Intervals 297
e.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z.005
6.18
n
12.40 2.58
9.84
50
12.40 3.59 (8.81, 15.99)
f.
We are 99% confident that the true mean salary of all 500 CEOs in the Forbes’ survey is between
$8.81 million and $15.99 million.
g.
From part d, the true mean salary of all 500 CEOs is $9.247 million. This value does fall within the
99% confidence interval that we found in part e.
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Rate
Variable
Rate
N
30
Mean
79.73
Median
80.00
StDev Minimum
5.96
60.00
Maximum
90.00
Q1
76.75
Q3
84.00
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
x z.05
6.19
s
n
79.73 1.645
5.96
30
79.73 1.79 (77.94, 81.52)
b.
We are 90% confident that the mean participation rate for all companies that have 401(k) plans is
between 77.94% and 81.52%.
c.
We must assume that the sample size ( n 30 ) is sufficiently large so that the Central Limit Theorem
applies.
d.
Yes. Since 71% is not included in the 90% confidence interval, it can be concluded that this
company's participation rate is lower than the population mean.
e.
The center of the confidence interval is x . If 60% is changed to 80%, the value of x will increase,
thus indicating that the center point will be larger. The value of s2 will decrease if 60% is replaced by
80%, thus causing the width of the interval to decrease.
a.
An estimate of the true mean Mach rating score of all purchasing managers is x 99.6 .
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval is:
x z / 2
s
n
99.6 1.96
12.6
122
99.6 2.24 (97.36, 101.84)
c.
We are 95% confident that the true Mach rating score of all purchasing managers is between 97.36
and 101.84.
d.
Yes, there is evidence to dispute this claim. We are 95% confident that the true mean Mach rating
score is between 97.36 and 101.84. It would be very unlikely that the true means Mach scores is as
low as 85.
Copyright © 2014 Pearson Education, Inc.
298
Chapter 6
6.20
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z.025
b.
6.21
n
1.96 1.96
.15
55
1.96 .04 (1.92, 2.00)
No. The value of 2.2 does not fall in the 95% confidence interval. Therefore, it is not a likely value
for the true mean facial WHR.
To answer the question, we will first form 90% confidence intervals for each of the 2 SAT scores.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
The confidence interval for SAT-Mathematics scores is:
x z / 2
s
n
19 1.645
65
265
19 6.57 (12.43, 25.57)
We are 90% confident that the mean change in SAT-Mathematics score is between 12.43 and 25.57 points.
The confidence interval for SAT-Verbal scores is:
x z / 2
s
n
7 1.645
49
265
7 4.95 (2.05, 11.95)
We are 90% confident that the mean change in SAT-Verbal score is between 2.05 and 11.95 points.
The SAT-Mathematics test would be the most likely of the two to have 15 as the mean change in score.
This value of 15 is in the 90% confidence interval for the mean change in SAT-Mathematics score.
However, 15 does not fall in the 90% confidence interval for the mean SAT-Verbal test.
6.22
x
11, 298
2.26
5, 000
For confidence coefficient, .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z.025
s
n
2.26 1.96
1.5
5, 000
2.26 .04 (2.22, 2.30)
We are 95% confident the mean number of roaches produced per roach per week is between 2.22 and 2.30.
6.23
a.
For confidence coefficient .80, .20 and / 2 .20 / 2 .10 . From Table II, Appendix D,
z.05 1.28 . From Table III, with df n 1 5 1 4, t.10 1.533 .
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . From Table III, with df n 1 5 1 4, t.05 2.132 .
c.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . From Table III, with df n 1 5 1 4, t.025 2.776 .
d.
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table II, Appendix D,
z.01 2.33 . From Table III, with df n 1 5 1 4, t.01 3.747 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 299
6.24
6.25
e.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.575 . From Table III, with df n 1 5 1 4, t.005 4.604 .
f.
Both the t- and z-distributions are symmetric around 0
and mound-shaped. The t-distribution is more spread
out than the z-distribution.
a.
If x is normally distributed, the sampling distribution of x is normal, regardless of the sample size.
b.
If nothing is known about the distribution of x, the sampling distribution of x is approximately
normal if n is sufficiently large. If n is not large, the distribution of x is unknown if the distribution
of x is not known.
a.
P(t0 t t0 ) .95 where df = 10
Because of symmetry, the statement can be written
P 0 t t0 .475 where df = 10
P t t0 .025 t0 2.228
6.26
b.
P(t t0 or t t0 ) .05 where df = 10
c.
P (t t0 ) .05 where df = 10
Because of symmetry, the statement can be written
P t t0 .05 t0 1.812
d.
P(t t0 or t t0 ) .10 where df = 20
e.
P(t t0 or t t0 ) .01 where df = 5
a.
P(t t0 ) .025 where df = 11; t0 2.201
b.
P(t t0 ) .01 where df = 9; t0 2.821
c.
P(t t0 ) .005 where df = 6. Because of symmetry, the statement can be rewritten
2 P t t0 .05 P t t0 .025 t0 2.228
2 P t t0 .10 P t t0 .05 t0 1.725
2 P t t0 .01 P t t0 .005 t0 4.032
P(t t0 ) .005 where df = 6; t0 3.707
d.
P(t t0 ) .05 where df = 18; t0 1.734
Copyright © 2014 Pearson Education, Inc.
300
Chapter 6
6.27
First, we must compute x and s.
x
a.
x 30 5 , s
(30) 2
6 26 5.2 , s 5.2 2.2804
6 1
5
176
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n 1 6 1 5, t.05 2.015 . The 90% confidence interval is:
s
n
5 2.015
5 1.88 3.12, 6.88
2.2804
6
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 6 1 5, t.025 2.571 . The 95% confidence interval is:
s
n
5 2.571
2.2804
6
5 2.39 2.61, 7.39
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D, with
df n 1 6 1 5, t.005 4.032 . The 99% confidence interval is:
x t.005
d.
n 1
6
x t.025
c.
2
n
2
n
x t.05
b.
x
x
2
a)
s
n
5 4.032
5 3.75 1.25, 8.75
s
n
5 1.711
2.2804
25
5 .78 4.22, 5.78
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D,
with df n 1 25 1 24, t.025 2.064 . The 95% confidence interval is:
x t.025
c)
6
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D,
with df n 1 25 1 24, t.05 1.711 . The 90% confidence interval is:
x t.05
b)
2.2804
s
n
5 2.064
2.2804
25
5 .94 4.06, 5.94
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D,
with df n 1 25 1 24, t.005 2.797 . The 99% confidence interval is:
x t.005
s
n
5 2.797
2.2804
25
5 1.28 3.72, 6.28
Increasing the sample size decreases the width of the confidence interval.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 301
6.28
For this sample,
x
a.
x 1567 97.9375 , s
2
n
n 1
16
2
n
1567 2
16 159.9292 , s s 2 12.6463
16 1
155,867
For confidence coefficient, .80, .20 and / 2 .20 / 2 .10 . From Table III, Appendix D, with
df n 1 16 1 15, t.10 1.341 . The 80% confidence interval for is:
x t.10
b.
x
x
2
s
n
97.94 1.341
12.6463
16
97.94 4.240 93.700, 102.180
For confidence coefficient, .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 16 1 15, t.025 2.131 . The 95% confidence interval for is:
x t.025
s
n
97.94 2.131
12.6463
16
97.94 6.737 91.203, 104.677
The 95% confidence interval for is wider than the 80% confidence interval for found in part a.
c.
For part a: We are 80% confident that the true population mean lies between 93.700 and 102.180.
For part b: We are 95% confident that the true population mean lies between 91.203 and 104.677.
The 95% confidence interval is wider than the 80% confidence interval because the more confident
you want to be that lies in an interval, the wider the range of possible values.
6.29
a.
The target parameter is = mean trap spacing for the population of red spiny lobster fishermen
fishing in Baja California Sur, Mexico.
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Trap
Variable
Trap
N
7
Mean
89.86
StDev
11.63
Minimum
70.00
Q1
82.00
Median
93.00
Q3
99.00
Maximum
105.00
The point estimate of is x 89.86 .
c.
For this problem, the sample size is n 7 . For a small sample size, the Central Limit Theorem does
not apply. Therefore, we do not know what the sampling distribution of x is.
d.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 7 1 6 , t.025 2.447 . The 95% confidence interval is:
x t.025
e.
s
n
89.86 2.447
11.63
7
89.86 10.756 (79.104, 100.616)
We are 95% confident that the true mean trap spacing for the population of red spiny lobster
fishermen fishing in Baja California Sur, Mexico is between 79.104 and 100.616 meters.
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
f.
6.30
We must assume that the population of trap spacings is normally distributed and that the sample is a
random sample.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 12 1 11 , t.025 2.201 . The 95% confidence interval is:
s
x t.025
n
3, 643 2.201
4, 487
12
3, 643 2,850.92 (792.08, 6,493.92)
We are 95% confident that the true mean level of radon exposure in tombs in the Valley of Kings is
between 792.08 and 6,493.92 Bq/m3.
6.31
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n 1 25 1 24, t.05 1.711 . The 90% confidence interval is:
x t.05
s
n
75.4 1.711
10.9
25
75.4 3.73 (71.67, 79.13)
We are 90% confident that the true mean breaking strength of the white wood is between 71.67 and 79.13.
6.32
We must assume that the distribution of the LOS's for all patients is normal.
a.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n 1 20 1 19 , t.05 1.729 . The 90% confidence interval is:
x t.05
6.33
6.34
s
n
3.8 1.729
1.2
20
3.8 .464 3.336, 4.264
b.
We are 90% confident that the mean LOS is between 3.336 and 4.264 days.
c.
“90% confidence” means that if repeated samples of size n are selected from a population and 90%
confidence intervals are constructed, 90% of all intervals thus constructed will contain the population
mean.
a.
The 95% confidence interval for the mean surface roughness of coated interior pipe is
(1.63580, 2.12620).
b.
No. Since 2.5 does not fall in the 95% confidence interval, it would be very unlikely that the average
surface roughness would be as high as 2.5 micrometers.
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: AAII
Variable
AAII
N
13
Mean
10.82
StDev
7.71
Minimum
-1.60
Q1
5.45
Median
9.80
Copyright © 2014 Pearson Education, Inc.
Q3
16.80
Maximum
24.80
Inferences Based on a Single Sample: Estimation with Confidence Intervals 303
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n 1 13 1 12 , t.05 1.782 . The 90% confidence interval is:
x t.05
s
n
10.82 1.782
7.71
13
10.82 3.81 7.01, 14.63
We are 95% confident that the true average annualized percentage return on investment of all stock
screeners provided by AAII is between 7.01 and 14.63.
b.
Since the confidence interval in part a contains only positive values, then on average, the AAII stock
screeners perform better than the S&P500.
c.
We must assume that the annualized percentage returns on investment for all stock screeners are
normally distributed and that the sample is random. Yes, this assumption seems to be satisfied. A
histogram of the data is:
Histogram of x
5
Frequency
4
3
2
1
0
0
5
10
15
20
25
x
The distribution is fairly mound-shaped.
6.35
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Skid
Variable
Skid
N Mean
20 358.5
StDev
117.8
Minimum
141.0
Q1
276.0
Median
367.5
Q3
438.0
Maximum
574.0
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 20 1 19, t.025 2.093 . The 95% confidence interval is:
x t.05
s
n
358.5 2.093
117.8
20
358.5 55.13 (303.37, 413.63)
b.
We are 95% confident that the mean skidding distance is between 303.37 and 413.63 meters.
c.
In order for the inference to be valid, the skidding distances must be from a normal distribution. We
will use the four methods to check for normality. First, we will look at a histogram of the data.
Copyright © 2014 Pearson Education, Inc.
Chapter 6
Using MINITAB, the histogram of the data is:
Histogram of Skid
4
3
Fr equency
304
2
1
0
200
300
400
500
Skid
From the histogram, the data appear to be fairly mound-shaped. This indicates that the data may be
normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal.
x s 358.5 117.8 (240.7, 476.3) 14 of the 20 values fall in this interval. The proportion is
.70. This is very close to the .68 we would expect if the data were normal.
x 2s 358.5 2(117.8) 358.5 235.6 (122.9, 594.1) 20 of the 20 values fall in this interval.
The proportion is 1.00. This is a larger than the .95 we would expect if the data were normal.
x 3s 358.5 3(117.8) 358.5 353.4 (5.1, 711.9) 20 of the 20 values fall in this interval.
The proportion is 1.00. This is exactly the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR QU – QL = 438 – 276 162 .
IQR 162
1.37 This is fairly close to the 1.3 we would expect if the data were normal. This
s
117.8
method indicates the data may be normal.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 305
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Skid
N ormal - 95% C I
99
95
90
Mean
StDev
358.5
117.8
N
AD
P-Value
20
0.170
0.921
P er cent
80
70
60
50
40
30
20
10
5
1
0
100
200
300
400
Skid
500
600
700
800
Since the data form a fairly straight line, the data may be normal.
From above, all the methods indicate the data may be normal. It appears that the assumption that the
data come from a normal distribution is probably valid.
6.36
d.
No. A distance of 425 meters falls above the 95% confidence interval that was computed in part a. It
would be very unlikely to observe a mean skidding distance of at least 425 meters.
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: MTBE
Variable
MTBE
N
12
Mean
97.17
StDev
113.76
Minimum
8.00
Q1
12.0
Median
50.5
Q3
146.0
Maximum
367.0
A point estimate for the true mean MTBE level for all well sites located near the New Jersey gasoline
service station is x 97.17 .
b.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D, with
df n 1 12 1 11, t.005 3.106 . The 99% confidence interval is:
x t.005
s
n
97.17 3.106
113.76
12
97.17 102.00 (4.83, 199.17)
We are 99% confident that the true mean MTBE level for all well sites located near the New Jersey
gasoline service station is between 4.83 and 199.17.
c.
We must assume that the data were sampled from a normal distribution. We will use the four
methods to check for normality. First, we will look at a histogram of the data. Using MINITAB, the
histogram of the data is:
Copyright © 2014 Pearson Education, Inc.
Chapter 6
Histogram of MTBE
5
4
Fr equency
306
3
2
1
0
0
50
100
150
200
M T BE
250
300
350
From the histogram, the data do not appear to be mound-shaped. This indicates that the data may not
be normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using
MINITAB, the summary statistics are:
x s 97.17 113.76 (16.59, 210.93) 10 of the 12 values fall in this interval. The proportion
is .83. This is not very close to the .68 we would expect if the data were normal.
x 2s 97.17 2(113.76) 97.17 227.52 (130.35, 324.69) 11 of the 12 values fall in this
interval. The proportion is .92. This is a somewhat smaller than the .95 we would expect if the data
were normal.
x 3s 97.17 3(113.76) 97.17 341.28 (244.11, 438.45) 12 of the 12 values fall in this
interval. The proportion is 1.00. This is exactly the 1.00 we would expect if the data were normal.
From this method, it appears that the data may not be normal.
Next, we look at the ratio of the IQR to s. IQR QU – QL = 146.0 –12.0 134.0 .
IQR 134.0
1.18 This is somewhat smaller than the 1.3 we would expect if the data were normal.
s
113.76
This method indicates the data may not be normal.
Finally, using MINITAB, the normal probability plot is:
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 307
Probability Plot of MTBE
N ormal - 95% C I
99
95
90
Mean
StDev
97.17
113.8
N
AD
P-Value
12
0.929
0.012
P er cent
80
70
60
50
40
30
20
10
5
1
-300
-200
-100
0
100
200
M T BE
300
400
500
Since the data do not form a fairly straight line, the data may not be normal.
From above, the all methods indicate the data may not be normal.
are not normal.
6.37
a.
It appears that the data probably
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Diox Amt
Variable
Diox Amt
Crude
No
Yes
N
10
6
Mean
2.590
0.517
StDev
1.542
0.407
Minimum
0.100
0.200
Q1
1.125
0.200
Median
2.850
0.450
Q3
4.000
0.700
Maximum
4.000
1.300
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 6 1 5 , t.025 2.571 . The 90% confidence interval is:
x t.025
s
n
.517 2.571
.407
6
.517 .427 .090, .944
We are 95% confident that the true mean amount of dioxide present in water specimens that contain
oil is between .090 and .944mg/l.
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n 1 10 1 9 , t.025 2.262 . The 90% confidence interval is:
x t.025
s
n
2.590 2.262
1.542
10
2.590 1.103 1.487, 3.693
We are 95% confident that the true mean amount of dioxide present in water specimens that do not
contain oil is between 1.487 and 3.693mg/l.
c.
Since the confidence interval for the mean amount of dioxide present in water specimens that contain
oil is entirely below the confidence interval for the mean amount of dioxide present in water
specimens that do not contain oil, we can conclude that the mean amount of dioxide present in water
containing oil is significantly less than the mean amount of dioxide present in water not containing
oil.
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
6.38
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Comp
Variable
COMP
N
10
Mean
1473
StDev
465
Minimum
825
Q1
1120
Median
1458
Q3
1758
Maximum
2220
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n 1 10 1 9, t.05 1.833 . The 90% confidence interval is:
x t.05
s
n
1473 1.833
465
10
1473 269.54 (1, 203.46, 1, 742.54)
We are 90% confident that the true mean threshold compensation level for all major airlines is between
$1,203.46 and $1,742.54.
6.39
a.
The population from which the sample was drawn is the Forbes 212 Biggest Private companies.
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Revenue
Variable
Revenue
N
15
Mean
5.39
StDev
4.86
Minimum
2.01
Q1
2.33
Median
2.80
Q3
6.99
Maximum
17.77
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table III, Appendix D, with
df n 1 15 1 14 , t.01 2.624 . The 98% confidence interval is:
x t.01
6.40
s
n
5.39 2.624
15
5.39 3.293 (2.097, 8.683)
c.
We are 98% confident that the mean revenue is between $2.097 and $8.683 billion.
d.
The population must be normally distributed in order for the procedure used in part b to be valid.
e.
Yes. The value of $5.0 billion dollars falls in the 98% confidence interval computed in part b.
Therefore, we should believe the claim.
By the Central Limit Theorem, the sampling distribution of p̂ is approximately normal with mean p̂ p and
standard deviation p̂
6.41
4.86
pq
.
n
The sample size is large enough if both npˆ 15 and nqˆ 15 .
a.
When n 400 , pˆ .10 :
npˆ 400(.10) 40 and nqˆ 400 .90 360
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
b.
When n 50 , pˆ .10 :
npˆ 50 .10 5 and nqˆ 50 .90 45
Since npˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 309
c.
When n 20 , pˆ .5 :
npˆ 20 .5 10 and nqˆ 20 .5 10
Since both numbers are less than 15, the sample size is not large enough to conclude the normal
approximation is reasonable.
d.
When n 20 , pˆ .3 :
npˆ 20 .3 6 and nqˆ 20 .7 14
Since both numbers are less than 15, the sample size is not large enough to conclude the normal
approximation is reasonable.
6.42
a.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 121.88 106.48 and nqˆ 121.12 14.52
Since nqˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable. However, 14.52 is very close to 15, so the normal approximation may work fairly well.
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
pˆ z .05
6.43
ˆˆ
.88(.12)
pq
pq
pˆ 1.645
.88 1.645
.88 .049 .831, .929
121
n
n
c.
We must assume that the sample is a random sample from the population of interest and that the
sample size is sufficiently large.
a.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 225 .46 103.5 and nqˆ 225 .54 121.5
b.
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval is:
pˆ z .025
6.44
ˆˆ
.46(.54)
pq
pq
pˆ 1.96
.46 1.96
.46 .065 .395, .525
225
n
n
c.
We are 95% confident the true value of p falls between .395 and .525.
d.
"95% confidence interval" means that if repeated samples of size 225 were selected from the
population and 95% confidence intervals formed, 95% of all confidence intervals will contain the true
value of p.
a.
Of the 50 observations, 15 like the product pˆ
15
.30 .
50
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 50 .3 15 and nqˆ 50 .7 35
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
For the confidence coefficient .80, .20 and / 2 .20 / 2 .10 . From Table II, Appendix D,
z.05 1.28 . The confidence interval is:
ˆˆ
.3(.7)
pq
pq
pˆ 1.28
.3 1.28
.3 .083 .217, .383
50
n
n
pˆ z .10
6.45
b.
We are 80% confident the proportion of all consumers who like the new snack food is between .217
and .383.
a.
pˆ
b.
By the Central Limit Theorem, the sampling distribution of p̂ is approximately normal with p̂ p
x
818
.4
n 2, 045
pq
if n is sufficiently large. The sample size is sufficiently large if npˆ 15 and nqˆ 15 .
n
For this exercise, npˆ 2, 045(.4) 818 and nqˆ 2, 045(.6) 1, 227 . Since both values are greater
than 15, the sample size is sufficiently large.
and p̂
c.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
6.46
d.
We are 95% confident that the true proportion of Arlington Texas homes with market values that are
overestimated by more than 10% by Zillow is between .379 and .421.
e.
No, the claim is not believable. The 95% confidence interval constructed in part c does not contain .3.
Thus, .3 is not a likely value for p.
a.
pˆ
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
x 506
.504
n 1, 003
pˆ z.05
6.47
ˆˆ
pq
.4(.6)
.4 1.96
.4 .021 (.379, .421)
n
2, 045
ˆˆ
pq
.504(.496)
.504 1.645
.504 .026 (.478, .530)
n
1, 003
c.
We are 90% confident that the true proportion of adults living in the U.S. who have paid to download
music from the internet is between .478 and .530.
d.
“90% confident” means that in repeated sampling, 90% of all intervals constructed in a similar
manner will contain the true proportion.
a.
The population of interest is all American adults.
b.
The sample is the 1,000 adults surveyed.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 311
c.
The parameter of interest is the proportion of all American adults who think Starbucks coffee is
overpriced, p.
d.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 1, 000 .73 730 and nqˆ 1, 000 .27 270
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval is:
pˆ z / 2
pq
pˆ z / 2
n
ˆˆ
.73(.27)
pq
.73 1.96
.73 .028 (.702, .758)
1000
n
We are 95% confident that the true proportion of all American adults who say Starbucks coffee is
overpriced is between .702 and .758.
6.48
a.
pˆ
x 63
.594
n 106
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
pˆ z.005
ˆˆ
.594(.406)
pq
.594 2.58
.594 .123 (.471, .717)
106
n
We are 99% confident that the true proportion of all social robots designed with legs but no wheels is
between .471 and .717.
6.49
b.
Since .40 does not fall in the 99% confidence interval, it is very unlikely that the true proportion of all
social robots designed with legs but no wheels is .40.
a.
pˆ
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x 1, 298
.60
n 2,163
pˆ z.025
c.
6.50
p
ˆˆ
pq
.60(.40)
.60 1.96
.60 .02 (.58, .62)
n
2,163
We are 95% confident that the true proportion of all drivers who are using a cell phone while
operating a motor passenger vehicle is between .58 and .62.
22
x 2 20 2
.041
n 4 528 4 532
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Chapter 6
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
The confidence interval is:
p z.025
6.51
pˆ
.041(.959)
pq
.041 1.96
.041 .017 (.024, .058)
528 4
n4
x 144
.41
n 351
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
The confidence interval is:
pˆ z.05
ˆˆ
.41(.59)
pq
.41 1.645
.41 .043 (.367, .453)
351
n
We are 90% confident that the true probability of unauthorized use of computer systems at an organization
is between .367 and .453. If we were to take repeated samples and form similar confidence intervals, 90%
of the confidence intervals would contain the true probability.
6.52
pˆ
x 15
.15
n 100
Suppose we form a 95% confidence interval for the true proportion of minority-owned franchises in
Mississippi. For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
ˆˆ
.15(.85)
pq
.15 1.96
.15 .07 (.08, .22)
100
n
We are 95% confident that the true percentage of minority-owned franchises in Mississippi is between 8%
and 22%. Since 20.5% falls in this interval, we would not conclude that the percentage of minority-owned
franchises in Mississippi is less than the national value.
6.53
Of the 2,778 sampled firms, 748 announced one or more acquisitions during the year 2000. Thus,
pˆ
x
748
.269
n 2, 778
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 2, 778 .269 747 and nqˆ 2,778 .731 2,031
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the
normal approximation is reasonable.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
The 90% confidence interval is:
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 313
pˆ z / 2
ˆˆ
pq
pq
.269(.731)
pˆ 1.645
.269 1.645
.269 .014 (.255, .283)
n
n
2, 778
We are 90% confident that the true proportion of all firms that announced one or more acquisitions during
the year 2000 is between .255 and .283. Changing these to percentages, the results would be 25.5% and
28.3%.
6.54
a.
The population is all senior human resource executives at U.S. companies.
b.
The population parameter of interest is p, the proportion of all senior human resource executives at
U.S. companies who believe that their hiring managers are interviewing too many people to find
qualified candidates for the job.
c.
The point estimate of p is pˆ
x 211
.42 .
n 502
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 502 .42 210.84 and nqˆ 502 .58 291.16
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
d.
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table II, Appendix D,
z.01 2.33 . The confidence interval is:
pˆ z.01
ˆˆ
.42(.58)
pq
.42 2.33
.42 .051 (.369, .471)
502
n
We are 98% confident that the true proportion of all senior human resource executives at U.S.
companies who believe that their hiring managers are interviewing too many people to find qualified
candidates for the job is between .369 and .471.
6.55
e.
A 90% confidence interval would be narrower. If the interval was narrower, it would contain fewer
values, thus, we would be less confident.
a.
In order for the large-sample estimation method to be valid, npˆ 15 and nqˆ 15 . For this exercise,
x
1
.003 , npˆ 333(.003) .999 , and nqˆ 333(.997) 332.001 . Since one of these
n 333
values is less than 15, the large-sample estimation method is not valid.
pˆ
b.
p
1 2
3
x2
.009
n 4 333 4 337
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
p z.025
.009(.991)
pq
.009 1.96
.009 .010 (.001, .019)
333 4
n4
Copyright © 2014 Pearson Education, Inc.
314
Chapter 6
We are 95% confident that the true proportion of all mountain casualties that require a femoral shaft
splint is between 0 and .019. (We know the proportion cannot be negative, so the lower end point
must be 0.)
6.56
a.
x 16
.052 .
n 308
The sample size is large enough if both npˆ 15 and nqˆ 15 .
The point estimate of p is pˆ
npˆ 308 .052 16 and nqˆ 308 .948 292
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
pˆ z.005
b.
ˆˆ
.052(.948)
pq
.052 2.58
.052 .033 (.019, .085)
308
n
We are 99% confident that the true proportion of diamonds for sale on the open market that are
classified as “D” color is between .019 and .085.
x 81
.263 .
The point estimate of p is pˆ
n 308
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 308 .263 81 and nqˆ 308 .737 227
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
pˆ z.005
ˆˆ
.263(.737)
pq
.263 2.58
.263 .065 (.198, .328)
308
n
We are 99% confident that the true proportion of diamonds for sale on the open market that are
classified as “VS1” clarity is between .198 and .328.
6.57
a.
The parameter of interest is p, the proportion of all fillets that are red snapper.
b.
The estimate of p is pˆ
x 22 17
.23
n
22
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 22 .23 5 and nqˆ 22 .77 17
Since npˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 315
c.
We will use Wilson’s adjustment to form the confidence interval.
Using Wilson’s adjustment, the point estimate of the true proportion of all fillets that are not red
snapper is
p
7
x2 52
.269
n 4 22 4 26
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . Wilson’s adjusted 95% confidence interval is:
.269(.731)
pq
.269 1.96
.269 .170 (.099, .439)
22 4
n
We are 95% confident that the true proportion of all fillets that are red snapper is between .099 and
.439.
p z / 2
d.
6.58
a.
For the large-sample estimation method to be valid, npˆ 15 and nqˆ 15 . For this exercise,
x 12
.092 , npˆ 131(.092) 12.05 , and nqˆ 131(.908) 118.95 . Since one of these values
n 131
is less than 15, the large-sample estimation method is not valid. We will use the Wilson’s adjustment.
pˆ
p
14
x 2 12 2
.104
n 4 131 4 135
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
p z.025
.104(.896)
pq
.104 1.96
.104 .051 (.053, .155)
131 4
n4
We are 95% confident that the true proportion of women with cosmetic dermatitis from using eye
shadow who have a nickel allergy is between .053 and .155.
b.
In order for the large-sample estimation method to be valid, npˆ 15 and nqˆ 15 . For this exercise,
x 25
.1 , npˆ 250(.1) 25 , and nqˆ 250(.9) 225 . Since both of these values are greater
n 250
than 15, the large-sample estimation method is valid.
pˆ
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
ˆˆ
.1(.9)
pq
.1 1.96
.1 .037 (.063, .137)
250
n
We are 95% confident that the true proportion of women with cosmetic dermatitis from using
mascara who have a nickel allergy is between .063 and .137.
Copyright © 2014 Pearson Education, Inc.
316
6.59
Chapter 6
c.
No, we cannot determine which group is referenced. The value of .12 falls in both confidence
intervals.
pˆ
x 282, 200
.85
n 332, 000
Suppose we form a 95% confidence interval for the true proportion of first class mail within the same city
that is delivered on time between Dec. 10 and Mar. 3. For confidence coefficient .95, .05 and
/ 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 . The confidence interval is:
pˆ z.025
ˆˆ
pq
.85(.15)
.85 1.96
.85 .001 (.849, .851)
n
332, 000
We are 95% confident that the true proportion of first class mail within the same city that is delivered on
time between Dec. 10 and Mar. 3 is between .849 and .851 or between 84.9% and 85.1%. This interval
does not contain the reported 95% of first class mailed delivered on time. It appears that the performance
of the USPS is below the standard during this time period.
z
To compute the necessary sample size, use n / 2 where .05 and / 2 .05 / 2 .025 . From
ME
Table II, Appendix D, z.025 1.96 .
2
6.60
Thus, n
6.61
a.
1.962 (7.2)
307.328 308 . You would need to take 308 samples.
.32
An estimate of is obtained from: range 4s s
range 34 30
1
4
4
z
To compute the necessary sample size, use n / 2 where .10 and / 2 .10 / 2 .05 . From
ME
Table II, Appendix D, z.05 1.645 .
2
1.645(1)
Thus, n
67.65 68
.2
2
b.
A less conservative estimate of is obtained from range 6s s
range 34 30
.6667
6
6
z 1.645(.6667)
Thus, n / 2
30.07 31
.2
ME
2
2
z / 2 pq
2
6.62
a.
To compute the needed sample size, use n
Table II, Appendix D, z.025 1.96 .
ME 2
where .05 and / 2 .05 / 2 .025 . From
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 317
Thus, n
(1.96)2 (.2)(.8)
.08
2
96.04 97 . You would need to take a sample of size 97.
z / 2 pq
2
b.
To compute the needed sample size, use n
ME 2
(1.96) 2(.5)(.5)
150.0625 151 . You would
.08 2
need to take a sample of size 151.
6.63
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
We know p̂ is in the middle of the interval, so pˆ
The confidence interval is pˆ z.05
.54 .26
.4
2
ˆˆ
pq
.4(.6)
.4 1.645
n
n
.4(.6)
.26
n
.8059
.8059
.8059
.4
.26 .4 .26
n
5.756 n 5.756² 33.1 34
.14
n
n
We know .4 1.645
6.64
a.
For a width of 5 units, ME 5 / 2 2.5 .
z
To compute the needed sample size, use n / 2 where .05 and / 2 .05 / 2 .025 . From
ME
Table II, Appendix D, z.025 1.96 .
2
1.96(14)
Thus, n
120.47 121
2.5
2
You would need to take 121 samples at a cost of 121($10) = $1210. Yes, you do have sufficient
funds.
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 .
1.645(14)
n
84.86 85
2.5
2
You would need to take 85 samples at a cost of 85($10) = $850. You still have sufficient funds but
have an increased risk of error.
6.65
a.
The width of a confidence interval is W 2 ME 2 za /2
n
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 .
Copyright © 2014 Pearson Education, Inc.
318
Chapter 6
For n 16 , W 2za /2
n
For n 25 , W 2za /2
n
For n 49 , W 2za /2
n
For n 100 , W 2za /2
For n 400 , W 2za /2
2 1.96
1
2 1.96
1
2 1.96
1
n
n
16
25
49
2 1.96
2 1.96
0.98
0.784
0.56
1
100
1
400
0.392
0.196
b.
6.66
The sample size will be larger than necessary for any p other than .5.
6.67
From Exercise 6.29, the standard deviation is 11.63. If the width of the interval is 5, then ME 5 / 2 2.5 .
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
z 1.96(11.63)
n /2
83.14 84
2.5
ME
2
6.68
2
From Exercise 6.48, pˆ .594 . For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From
Table II, Appendix D, z.005 2.58 .
n
6.69
z2 / 2 pq 2.582 (.594)(.406)
285.4 286
( ME ) 2
.0752
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
z 1.645(10.9)
n /2
20.09 21 . Thus, we would need a sample of size 21.
4
ME
2
2
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 319
z
To compute the necessary sample size, use n / 2 where .05 and / 2 .05 / 2 .025 . From
ME
Table II, Appendix D, z.025 1.96 .
2
6.70
1.96(12)
Thus, n
245.86 246
1.5
2
6.71
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
Since we have no estimate given for the value of p, we will use .5. The sample size is:
n
6.72
a.
z2 / 2 pq 1.6452 (.5)(.5)
1, 691.3 1, 692
( ME ) 2
.022
Since no level of significance was given, we will use 95%. From Exercise 6.16, s 2.755 . For
confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
z 1.96(2.755)
n /2
116.6 117
.5
ME
2
b.
6.73
Answers will vary. A plan would need to be devised so that the selected shoppers were selected from
a variety of different stores in a variety of locations so that the sample would be representative of the
entire population.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D, z.005 2.575 .
From the previous estimate, we will use pˆ 1 / 3 to estimate p.
n
6.74
2
z2 / 2 pq 2.5752 (1/ 3)(2 / 3)
14, 734.7 14, 735
( ME )2
.012
From Exercise 6.58, the value of p̂ for both groups was close to .1. Since no level of significance was
given, we will use 95%. For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II,
Appendix D, z.025 1.96 .
n
z2 / 2 pq 1.962 (.1)(.9)
384.16 385
( ME )2
.032
z
To compute the needed sample size, use n / 2 where .05 and / 2 .05 / 2 .025 . From Table
ME
II, Appendix D, z.025 1.96 .
2
6.75
1.96(10)
Thus, for s 10 , n
42.68 43
3
2
1.96(20)
For s 20 , n
170.72 171
3
2
Copyright © 2014 Pearson Education, Inc.
320
Chapter 6
1.96(30)
For s 30 , n
384.16 385
3
2
z
To compute the necessary sample size, use n / 2 where .10 and / 2 .10 / 2 .05 . From Table
ME
II, Appendix D, z.05 1.645 .
2
6.76
1.645(10)
Thus, n
270.6 271
1
2
6.77
The bound is ME = .05. For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II,
Appendix D, z.005 2.575 .
z / 2 pq
2
We estimate p with pˆ 11/ 27 .407 . Thus, n
( ME ) 2
2.5752 (.407)(.593)
640.1 641
.052
The necessary sample size would be 641. The sample was not large enough.
z
To compute the needed sample size, use n / 2 where .10 and / 2 .10 / 2 .05 . From
ME
Table II, Appendix D, z.05 1.645 .
2
6.78
a.
1.645(2)
Thus, n
1, 082.41 1, 083
.1
2
b.
As the sample size decreases, the width of the confidence interval increases. Therefore, if we sample
100 parts instead of 1,083, the confidence interval would be wider.
c.
To compute the maximum confidence level that could be attained meeting the management's
specifications,
100(.01)
z
z (2)
n / 2 100 / 2
z2 / 2
.25 z / 2 .5
4
ME
.1
2
2
Using Table II, Appendix D, P(0 z .5) .1915 . Thus, / 2 .5000 .1915 .3085 ,
2 .3085 .617 , and 1 1 .617 .383 .
The maximum confidence level would be 38.3%.
6.79
a.
Percentage sampled =
1000
n
(100%)
(100%) 40%
N
2500
Finite population correction factor:
2500 1000
N n
.6 .7746
2500
N
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 321
b.
Percentage sampled =
1000
n
(100%)
(100%) 20%
N
5000
Finite population correction factor:
c.
Percentage sampled =
1000
n
(100%)
(100%) 10%
N
10, 000
Finite population correction factor:
d.
Percentage sampled =
6.81
x =
n
10, 000 1000
N n
.9 .9487
10, 000
N
1000
n
(100%)
(100%) 1%
N
100, 000
Finite population correction factor:
6.80
5000 1000
N n
.8 .8944
5000
N
100, 000 1000
N n
.99 .995
100, 000
N
N n
N
200
2500 1000
4.90
2500
a.
x
b.
x
c.
x =
200
d.
x
200
a.
ˆ x
s
b.
ˆ x
10, 000 4000
.6124
10, 000
4000
c.
ˆ x
10, 000 10, 000
0
10, 000
10, 000
d.
As n increases, x decreases.
1000
200
1000
5000 1000
5.66
5000
10, 000 1000
6.00
10, 000
1000
100, 000 1000
6.293
100, 000
1000
50
10, 000 2000
N n
1.00
N
10, 000
2000
n
50
50
Copyright © 2014 Pearson Education, Inc.
322
Chapter 6
e.
We are computing the standard error of x . If the entire population is sampled, then x . There is
no sampling error, so x 0 .
6.82
a.
For n 64 , with the finite population correction factor:
s N n 24 5000 64
ˆ x
3 .9872 2.9807
N
5000
n
64
Without the finite population correction factor: ˆ x s / n
24
64
3
ˆ x without the finite population correction factor is slightly larger.
b.
For n 400 , with the finite population correction factor:
24 5000 400
s N n
ˆ x
1.2 .92 1.1510
N
5000
n
400
Without the finite population correction factor: ˆ x s / n
24
400
1.2
ˆ x without the finite population correction factor is larger.
c.
6.83
In part a, n is smaller relative to N than in part b. Therefore, the finite population correction factor
did not make as much of a difference in the answer in part a as in part b.
The approximate 95% confidence interval for p is
pˆ 2ˆ pˆ pˆ 2
6.84
An approximate 95% confidence interval for is:
x 2ˆ x x 2
6.85
.42(.58) 6000 1600
pˆ (1 pˆ ) N n
.42 2
.42 .021 .399, .441
1600
6000
n
N
a.
x
s
n
14 375 40
N n
422 2
422 4.184 417.816, 426.184
375
N
40
x 1081 36.03
n
30
s
2
x
x
2
2
n 1
n
1, 0812
30 96.3782
30 1
41, 747
The approximate 95% confidence interval is:
x 2ˆ x x 2
s
n
96.3782 300 30
N n
36.03 2
36.03 3.40 32.63, 39.43
300
N
30
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 323
b.
pˆ
x 21
.7
n 30
The approximate 95% confidence interval is:
pˆ 2ˆ pˆ pˆ 2
6.86
a.
.7(.3) 300 30
pˆ (1 pˆ ) N n
.7 2
.7 .159 .541, .859
30
300
n
N
For N 2,193 , n 223 , x 116, 754 , and s 39,185 , the 95% confidence interval is:
x 2ˆ x x 2
s
n
39,185 2,193 223
N n
116, 754 2
116,754 4,974.06
2,193
N
223
(111, 779.94, 121, 728.06)
6.87
b.
We are 95% confident that the mean salary of all vice presidents who subscribe to Quality Progress
is between $111,777.94 and $121,728.06.
a.
First, we must estimate p: pˆ
x
759
.560
n 1,355
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . Since n / N 1,355 /1, 696 .799 .05 , we must use the finite population correction
factor. The 95% confidence interval is:
pˆ z.025
6.88
ˆ ˆ N n
pq
.560(.440) 1, 696 1,355
.560 1.96
.560 .012 (.548, .572)
n N
1,355
1, 696
b.
We used the finite correction factor because the sample size was very large compared to the
population size.
c.
We are 95% confident that the true proportion of active NFL players who select a professional coach
as the most influential in their career is between .548 and .572.
a.
The population of interest is the set of all households headed by women that have incomes of $25,000
or more in the database.
b.
Yes. Since n / N 1,333 / 25, 000 .053 exceeds .05, we need to apply the finite population
correction.
c.
The standard error for p̂ should be ˆ pˆ
d.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The approximate 90% confidence interval is:
.708(1 .708) 25,000 1,333
pˆ (1 pˆ ) N n
.012
1333
25,000
n
N
pˆ 1.645ˆ pˆ .708 1.645 .012 .708 .020 .688, .728
Copyright © 2014 Pearson Education, Inc.
324
Chapter 6
6.89
a.
fx
First, we must calculate the sample mean:
15
x i 1
i i
n
3(108) 2(55) 1(500) 19(100) 15, 646
156.46
100
100
The point estimate of the mean value of the parts inventory is x 156.46 .
b.
The sample variance and standard deviation are:
15
s 2 i 1
f x
fx
2
i i
2
i i
n
n 1
6, 776,336
99
3(108) 2 2(55) 2 19(100) 2
100 1
15, 6462
100
2
15, 646
100 43, 720.83677
s s 2 43, 720.83677 209.10
The estimated standard error is ˆ x
c.
s
n
N n 209.10 500 100
18.7025
500
N
100
The approximate 95% confidence interval is:
s N n
156.46 2 18.7025 156.46 37.405 119.055, 193.865
x 2ˆ x x 2
N
n
We are 95% confident that the mean value of the parts inventory is between $119.06 and $193.87.
d.
6.90
Since the interval in part c does not include $300, the value of $300 is not a reasonable value for the
mean value of the parts inventory.
For N 1,500 , n 35 , x 1 , and s 124 , the 95% confidence interval is:
s N n
124 1,500 35
x 2ˆ x x 2
1 2
1 41.43 40.43, 42.43
N
1,500
n
35
We are 95% confident that the mean error of the new system is between -$40.43 and $42.43.
6.91
pˆ
x 15
.086
n 175
The standard error of p̂ is ˆ pˆ
pˆ (1 pˆ ) N n
.086(1 .086) 3000 175
.0206
n
N
175
3000
An approximate 95% confidence interval for p is pˆ 2ˆ pˆ .086 2 .0206 .086 .041 .045, .127
Since .07 falls in the 95% confidence interval, it is not an uncommon value. Thus, there is no evidence that
more than 7% of the corn-related products in this state have to be removed from shelves and warehouses.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 325
6.92
6.93
a.
2
2
16.0128 and .975,7
1.68987
/ 2 .05 / 2 .025 ; .025,7
b.
/ 2 .10 / 2 .05 ;
c.
2
2
39.9968 and .995,20
7.43386
/ 2 .01/ 2 .005 ; .005,20
d.
2
2
34.1696 and .975,20
9.59083
/ 2 .05 / 2 .025 ; .025,20
a.
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using Table IV, Appendix D, with
2
2
67.5048 and .95,49
34.7642 . The 90% confidence interval is:
df n 1 50 1 49 , .05,49
(n 1) s 2
b.
2
.05
2
.05
2
.05
2
.95
(50 1)2.52
(50 1)2.52
2
4.537 2 8.809
67.5048
34.7642
2
(n 1) s 2
2
.95
(15 1).022
(15 1).022
2
.00024 2 .00085
23.6848
6.57063
2
(n 1) s 2
2
.95
(22 1)31.62
(22 1)31.62
2
641.86 2 1,809.09
32.6705
11.5913
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using Table IV, Appendix D, with
2
2
9.48773 and .95,4
.710721 . The 90% confidence interval is:
df n 1 5 1 4 , .05,4
(n 1) s 2
2
.05
6.94
(n 1) s 2
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using Table IV, Appendix D, with
2
2
32.6705 and .95,21
11.5913 . The 90% confidence interval is:
df n 1 22 1 21 , .05,21
(n 1) s 2
d.
2
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using Table IV, Appendix D, with
2
2
23.6848 and .95,14
6.57063 . The 90% confidence interval is:
df n 1 15 1 14 , .05,14
(n 1) s 2
c.
2
2
.05,16
26.2962 and .95,16
7.96164
2
(n 1) s 2
2
.95
(5 1)1.52
(5 1)1.52
2
.94859 2 12.6632
9.48773
.710721
To find the 90% confidence interval for , we need to take the square root of the end points of the 90%
confidence interval for 2 from Exercise 6.93.
a.
The 90% confidence interval for is:
4.537 8.809 2.13 2.97
b.
The 90% confidence interval for is:
.00024 .00085 .016 .029
c.
The 90% confidence interval for is:
641.86 1,809.09 25.34 42.53
d.
The 90% confidence interval for is:
.94859 12.6632 .974 3.559
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
6.95
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: x
Variable N Mean StDev
x
6 6.17
3.31
Minimum
2.00
Q1
2.75
Median
6.50
Q3
8.75
Maximum
11.00
For confidence level .95, .05 and / 2 .05 / 2 .025 . Using Table IV, Appendix D, with
2
2
12.8325 and .975,5
.831211 . The 95% confidence interval is:
df n 1 6 1 5 , .025,5
(n 1) s 2
2
.025
6.96
2
(n 1) s 2
2
.975
(6 1)3.312
(6 1)3.312
2
4.269 2 65.904
12.8325
.831211
a.
The target parameter is 2 , the population variation in WR scores for all convicted drug dealers.
b.
For confidence level .99, .01 and / 2 .01/ 2 .005 . Using Table IV, Appendix D, with
2
2
140.169 and .995,99
67.3276 . The 90% confidence interval is:
df n 1 100 1 99 , .005,99
(n 1) s 2
2
.005
2
(n 1) s 2
2
.995
(100 1)62
(100 1)62
2
25.426 2 52.935
140.169
67.3276
c.
“99% confidence” means that in repeated sampling, 99% of all confidence intervals constructed in a
similar manner will contain the true variance.
d.
We must assume that a random sample was selected and that the population of interest is
approximately normal.
e.
The variance in measured in terms of WR scores-squared. This is difficult to relate to the data. The
standard deviation is measured in WR scores, the same units as the data.
f.
The 99% confidence interval for is:
25.426 52.935 5.042 7.276
We are 99% confident that the true standard deviation of WR scores is between 5.042 and 7.276.
6.97
a.
To find the confidence interval for , we first find the confidence interval for 2 and then take the
square root of the endpoints. For confidence level .95, .05 and / 2 .05 / 2 .025 . Using Table
2
2
71.4202 and .975,54
32.3574 . The 95%
IV, Appendix D, with df n 1 55 1 54 , .025,54
confidence interval is:
(n 1) s 2
2
.025
2
(n 1) s 2
2
.975
(55 1)(.15) 2
(55 1)(.15)2
2
.0170 2 .0375
71.4202
32.3574
The 95% confidence interval for is:
.0170 .0375 0.130 0.194
We are 95% confident that the true standard deviation of the facial WHR values for all CEOs at
publically traded Fortune 500 firms is between .130 and .194.
b.
In order for the interval to be valid, the distribution of WHR values should be approximately normally
distributed. The distribution should look like:
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 327
Normal distribution
0
6.98
a.
The 90% confidence interval for 2 is (672, 779). We are 90% confident that the true population
variance of the level of support for all senior managers at CPA firms is between 672 and 779 points
squared.
b.
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using MINITAB with
2
2
1065.35 and .95,21
918.926 .
df n 1 992 1 991 , .05,991
The 90% confidence interval is:
(992 1)722
(992 1)722
2
671.6 2 778.6
1065.35
918.926
This corresponds to the interval on the printout.
(n 1) s 2
2
.05
2
(n 1) s 2
2
.95
c.
The 90% confidence interval for is (25.9, 27.9).
d.
To form the confidence interval for using the interval in part a, we take the square root of the
endpoints: ( 672, 779) (25.9, 27.9)
This is the same as the interval on the printout.
6.99
e.
We are 90% confident that the true population standard deviation of the level of support for all senior
managers at CPA firms is between 25.9 and 27.9 points.
f.
We must assume that the distribution of the level of support is approximately normally distributed.
From Exercise 4.121, we concluded that the data were approximately normally distributed.
To find the confidence interval for , we first find the confidence interval for 2 and then take the square
root of the endpoints. For confidence level .95, .05 and / 2 .05 / 2 .025 . Using Table IV,
2
2
21.9200 and .975,11
3.81575 . The 95% confidence
Appendix D, with df n 1 12 1 11 , .025,11
interval is:
(n 1) s 2
2
.025
2
(n 1) s 2
2
.975
(12 1)(4, 487) 2
(12 1)(4, 487)2
2
10,103,323.86 2 58, 039, 666.91
21.9200
3.81575
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
The 95% confidence interval for is: 10,103,323.86 58, 039, 666.91 3,178.57 7, 618.38
We are 95% confident that the true standard deviation of radon levels in tombs in the Valley of Kings is
between 3,178.57 and 7,618.38 Bq/m3.
6.100
To find the confidence interval for , we first find the confidence interval for 2 and then take the
square root of the endpoints. For confidence level .95, .05 and / 2 .05 / 2 .025 . Using Table
2
2
30.1910 and .975,17
7.56418 . The 95%
IV, Appendix D, with df n 1 18 1 17 , .025,17
a.
confidence interval is:
(n 1) s 2
2
.025
2
(n 1) s 2
2
.975
(18 1)(6.3)2
(18 1)(6.3) 2
2
22.349 2 89.201
30.1910
7.56418
The 95% confidence interval for is: 22.349 89.201 4.727 9.445
6.101
b.
We are 95% confident that the true standard deviation of conduction times of the prototype system is
between 4.727 and 9.445.
c.
No, the prototype system does not satisfy this requirement. In order to meet the requirement, the
entire confidence interval constructed in part a would have to have values below 7. The interval
constructed in part a contains 7, but also contains values greater than 7.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Drug
Variable
Drug
N
50
Mean
89.291
StDev
3.183
Variance
10.134
Minimum
81.790
Median
89.375
Maximum
94.830
For confidence level .99, .01 and / 2 .01/ 2 .005 . From Table IV, Appendix D, with
2
2
79.4900 and .995,49
27.9907 . The 99% confidence interval is:
df n 1 50 1 49 , .005,49
(50 1)(10.134)
(50 1)(10.134)
2
6.247 2 17.740
79.4900
27.9907
We are 99% confident that the true population variation in drug concentrations for the new method is
between 6.247 and 17.740.
(n 1) s 2
6.102
a.
2
.005
2
(n 1) s 2
2
.995
Answers will vary. Using a statistical package, a random sample of 10 observations is:
148.289, 41.891, 73.051, 29.140, 211.240, 4.777, 49.255, 99.407, 90.823, 84.203
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Sample
Variable
Sample
N
10
Mean
83.2
StDev
60.5
Variance
3665.2
Minimum
4.78
Median
78.6
Copyright © 2014 Pearson Education, Inc.
Maximum
211.2
Inferences Based on a Single Sample: Estimation with Confidence Intervals 329
For confidence level .95, .05 and / 2 .05 / 2 .025 . Using Table IV, Appendix D, with
2
2
19.0228 and .975,9
2.70039 . The 95% confidence interval is:
df n 1 10 1 9 , .025,9
(n 1) s 2
2
.025
2
(n 1) s 2
2
.975
(10 1)(3,665.2)
(10 1)(3, 665.2)
2
1, 734.066 2 12, 215.569
19.0228
2.70039
The measure of reliability for this estimate is 95%.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: INTTIME
Variable
INTTIME
N
267
Mean
95.52
StDev
91.54
Variance
8379.41
Minimum
1.86
Median
70.88
Maximum
513.52
( x ) . The variance reported here has a
2
The true population variance is found by 2
N
8,379.41(266)
8,3478.03 . This
267
value is in the 95% confidence interval constructed in part b. We know that in repeated sampling,
95% of all intervals constructed in a similar manner will contain the true variance and 5% will not.
The interval that we constructed could be one of the 5% that did not contain the true variance.
denominator of 266 instead of 267. The population variance is
6.103
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Spacing
Variable
Spacing
N
7
Mean
89.86
StDev
11.63
Variance
135.14
Minimum
70.00
Median
93.00
Maximum
105.00
For confidence level .99, .01 and / 2 .01/ 2 .005 . Using Table IV, Appendix D, with
2
2
18.5476 and .995,6
.675727 . The 99% confidence interval is:
df n 1 7 1 6 , .005,6
(n 1) s 2
6.104
2
.005
2
(n 1) s 2
2
.995
(7 1)(135.14)
(7 1)(135.14)
2
43.717 2 1,199.952
18.5476
.675727
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Honey, DM
Variable
Honey
DM
a.
N
35
33
Mean
10.714
8.333
StDev
2.855
3.256
Variance
8.151
10.604
Minimum
4.000
3.000
Median
11.000
9.000
Maximum
16.000
15.000
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using MINITAB with
2
2
48.6024 and .95,34
21.6643 . The 90% confidence interval for the
df n 1 35 1 34 , .05,34
variance is:
(n 1) s 2
2
.05
2
(n 1) s 2
2
.95
(35 1)8.151
(35 1)8.151
2
5.702 2 12.792
48.6024
21.6643
The 95% confidence interval for the standard deviations is:
5.702 12.792 2.39 3.58
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
b.
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using MINITAB with
2
2
46.1943 and .95,32
20.0719 . The 90% confidence interval for the
df n 1 33 1 32 , .05,32
variance is:
(n 1) s 2
2
.05
2
(n 1) s 2
2
.95
(33 1)10.604
(33 1)10.604
2
7.346 2 16.906
46.1943
20.0719
The 95% confidence interval for the standard deviations is:
7.346 16.906 2.71 4.11
6.105
6.106
6.107
c.
Since the confidence intervals overlap, the researchers cannot conclude that the variances of the two
groups differ.
a.
P(t t0 ) .05 where df = 20. Thus, t0 1.725 .
b.
P(t t0 ) .005 where df = 9. Thus, t0 3.250 .
c.
P(t t0 or t t0 ) .10 where df = 8 is equivalent to P(t t0 ) .10 / 2 .05 where df = 8. Thus,
t0 1.860 .
d.
P(t t0 or t t0 ) .01 where df = 17 is equivalent to P(t t0 ) .01/ 2 .005 where df = 17. Thus,
t0 2.898 .
a.
For a small sample from a normal distribution with unknown standard deviation, we use the
t-statistic. For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III,
Appendix D, with df n 1 23 1 22, t.025 2.074 .
b.
For a large sample from a distribution with an unknown standard deviation, we can estimate the
population standard deviation with s and use the z-statistic. For confidence coefficient .95, .05
and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
c.
For a small sample from a normal distribution with known standard deviation, we use the z-statistic.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 .
d.
For a large sample from a distribution about which nothing is known, we can estimate the population
standard deviation with s and use the z-statistic. For confidence coefficient .95, .05 and
/ 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
e.
For a small sample from a distribution about which nothing is known, we can use neither z nor t.
a.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.575 . The confidence interval is:
x z.005
s
n
32.5 2.575
30
225
32.5 5.15 27.35, 37.65
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 331
z 2.575(30)
The sample size is n / 2
23,870.25 23,871 .
.5
ME
2
b.
c.
For confidence level .99, .01 and / 2 .01/ 2 .005 . Using MINITAB with
2
2
282.268 and .995,224
173.238 . The 99% confidence interval is:
df n 1 225 1 224 , .005,224
(n 1) s 2
2
.005
d.
6.108
a.
2
2
(n 1) s 2
2
.995
(225 1)(30)2
(225 1)(30)2
2
714.215 2 1,163.717
282.268
173.238
"99% confidence" means that if repeated samples of size 225 were selected from the population and
99% confidence intervals constructed for the population mean, then 99% of all the intervals
constructed will contain the population mean.
Of the 400 observations, 227 had the characteristic pˆ 227 / 400 .5675 .
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 400 .5675 227 and nqˆ 400 .4325 173
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
b.
ˆˆ
pq
pq
.5675(.4325)
pˆ 1.96
.5675 1.96
.5675 .0486 .5189, .6161
n
n
400
For this problem, ME .02 . For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From
Table II, Appendix D, z.025 1.96 . Thus,
n
( z / 2 ) 2 pq (1.96) 2 (.5675)(.4325)
2,357.2 2,358
ME 2
.022
Thus, the sample size was 2,358.
6.109
6.110
a.
2
2
19.0228 and .975,9
2.70039 .
Using Table IV, Appendix D, with df n 1 10 1 9 , .025,9
b.
2
2
32.8523 and .975,19
8.90655 .
Using Table IV, Appendix D, with df n 1 20 1 19 , .025,19
c.
2
2
79.4900 and .995,49
27.9907 .
Using Table IV, Appendix D, with df n 1 50 1 49 , .005,49
a.
The finite population correction factor is:
( N n)
N
b.
The finite population correction factor is:
( N n)
(100 20)
.8944
N
100
(2, 000 50)
.9874
2, 000
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Chapter 6
c.
6.111
6.112
The finite population correction factor is:
(1, 500 300)
.8944
1,500
The parameters of interest for the problems are:
(1)
The question requires a categorical response. One parameter of interest might be the proportion, p, of
all Americans over 18 years of age who think their health is generally very good or excellent.
(2)
A parameter of interest might be the mean number of days, , in the previous 30 days that all
Americans over 18 years of age felt that their physical health was not good because of injury or
illness.
(3)
A parameter of interest might be the mean number of days, , in the previous 30 days that all
Americans over 18 years of age felt that their mental health was not good because of stress,
depression, or problems with emotions.
(4)
A parameter of interest might be the mean number of days, , in the previous 30 days that all
Americans over 18 years of age felt that their physical or mental health prevented them from
performing their usual activities.
a.
A point estimate for the average number of latex gloves used per week by all healthcare workers with
latex allergy is x 19.3 .
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z / 2
s
n
19.3 1.96
11.9
46
19.3 3.44 (15.86, 22.74)
c.
We are 95% confident that the true average number of latex gloves used per week by all healthcare
workers with a latex allergy is between 15.86 and 22.74.
d.
The conditions required for the interval to be valid are:
i.
ii.
6.113
( N n)
N
The sample selected was randomly selected from the target population.
The sample size is sufficiently large, i.e. n 30 .
a.
The point estimate of p is pˆ .11 .
b.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 150 .11 16.5 and nqˆ 150 .89 133.5
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
ˆˆ
pq
.11(.89)
.11 1.96
.11 .05 (.06, .16)
n
150
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 333
c. We are 95% confident that the true proportion of MSDSs that are satisfactorily completed is
between .06 and .16.
6.114
a.
Since all the people surveyed were from Muncie, Indiana, the population of interest is all
consumers in Muncie, Indiana.
b.
The characteristic of interest in the population is the proportion of shoppers who believe that “Made
in the USA” means that 100% of labor and materials are from the USA.
c.
The point estimate of p is pˆ
x 64
.604 .
n 106
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 106(.604) 64.024 and nqˆ 106(.396) 41.976
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
pˆ z.05
d.
ˆˆ
pq
.604(.396)
.604 1.645
.604 .078 (.526, .682)
n
106
e.
We are 90% confident that the true proportion of shoppers who believe that “Made in the USA” means
that 100% of labor and materials are from the USA is between .526 and .682.
“90% confidence” means that if we took repeated samples of size 106 and computed 90% confidence
intervals for the true proportion shoppers who believe that “Made in the USA” means that 100% of
labor and materials are from the USA, 90% of the intervals computed will contain the true proportion.
f.
From above, we will use pˆ .604 . For confidence coefficient .90, .10 and / 2 .10 / 2 .05 .
From Table II, Appendix D, z.05 1.645 .
n
6.115
z2 / 2 pq 1.6452 (.604)(.396)
258.9 259
( ME ) 2
.052
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
For this study,
z 1.96(5)
n /2
96.04 97
ME 1
2
6.116
The sample size needed is 97.
a.
From the printout, the 90% confidence interval for the mean lead level is (0.61, 5.16).
b.
From the printout, the 90% confidence interval for the mean copper level is (0.2637, 0.5529).
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
c.
We are 95% confident that the mean lead level in water specimens from Crystal Lakes Manors is
between .61 and 5.16.
We are 95% confident that the mean copper level in water specimens from Crystal Lakes Manors is
between .2637 and .5529.
d.
6.117
90% confidence means that if repeated samples of size n are selected and 90% confidence intervals
formed, 90% of all confidence intervals will contain the true mean.
First, we must estimate p: pˆ
pˆ 2
x 50
.694 . The 95% confidence interval is:
n 72
ˆˆ N n
.694(.306) 251 72
pq
.694 2
.694 .092 (.602, .786)
72
n N
251
We are 95% confident that the true proportion of all New Jersey Governor’s Council business members
that have employees with substance abuse problems is between .602 and .786.
6.118
a.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
x z.05
n
12.2 1.645
10
100
12.2 1.645 10.555, 13.845
We are 90% confident that the mean number of days of sick leave taken by all the company’s
employees is between 10.555 and 13.845.
b.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 .
z 2.58(10)
The sample size is n / 2
166.4 167
2
ME
You would need to take n = 167 samples.
2
c.
2
To find the confidence interval for , we first find the confidence interval for 2 and then take the
square root of the endpoints. For confidence level .90, .10 and / 2 .10 / 2 .05 . Using
2
2
124.342 and .95,99
77.9295 . The 90% confidence
MINITAB with df n 1 100 1 99 , .05,99
interval is:
(n 1) s 2
2
.05
2
(n 1) s 2
2
.95
(100 1)(10) 2
(100 1)(10) 2
2
79.619 2 127.038
124.342
77.9295
The 90% confidence interval for the standard deviation is:
79.619 127.038 8.923 11.271
We are 90% confident that the true population standard deviation of the number of sick days is
between 8.923 and 11.271.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 335
6.119
a.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The 99% confidence interval is:
x z / 2
6.120
s
n
4.25 2.58
12.02
56
4.25 4.14 (0.11, 8.39)
b.
We are 99% confident that the true mean number of blogs/forums per site of all Fortune 500 firms
that provide blogs and forums for marketing tools is between 0.11 and 8.39.
c.
No. Since our sample size is 56, the sampling distribution of x is approximately normal by the
Central Limit Theorem.
a.
Answers will vary. Using MINITAB, 30 random numbers were generated using the uniform
distribution from 1 to 308. The random numbers were:
9, 15, 19, 36, 46, 47, 63, 73, 90, 92, 108, 112, 117, 127, 144, 145, 150, 151, 172, 178, 218, 229, 230,
241, 242, 246, 252, 267, 274, 282
The 308 observations were numbered in the order that they appear in the file. Using the random
numbers generated above, I selected the 9th, 15th, 19th, etc. observations for the sample. The selected
sample is:
.31, .34, .34, .50, .52, .53, .64, .72, .70, .70, .75, .78, 1.00, 1.00, 1.03, 1.04, 1.07, 1.10, .21, .24, .58,
1.01, .50, .57, .58, .61, .70, .81, .85, 1.00
b.
Using MINITAB, the descriptive statistics for the sample of 30 observations are:
Descriptive Statistics: carats-samp
Variable
carats-s
N
Mean
30 0.6910
Median StDev
0.7000 0.2620
Minimum
0.2100
Maximum
1.1000
Q1
Q3
0.5150 1.0000
From above, x .6910 and s .2620 .
c.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z / 2
6.121
s
n
.691 1.96
.262
30
.691 .094 (.597, .785)
d.
We are 95% confident that the mean number of carats is between .597 and .785.
e.
From Exercise 2.49, we computed the “population” mean to be .631. This mean does fall in the 95%
confidence interval we computed in part d.
There are a total of 96 channel catfish in the sample. The point estimate of p is pˆ
x 96
.667 .
n 144
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 144 .667 96 and nqˆ 144 .333 48
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the
normal approximation is reasonable.
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
The confidence interval is:
pˆ z.05
ˆˆ
pq
.667(.333)
.667 1.645
.667 .065 (.602, .732)
n
144
We are 90% confident that the true proportion of channel catfish in the population is between .602 and
.732.
6.122
a.
For confidence coefficient .99 .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z / 2
s
n
1.13 2.58
2.21
72
1.13 .67 (.46, 1.80)
We are 99% confident that the mean number of pecks at the blue string is between .46 and 1.80.
6.123
b.
Yes. The mean number of pecks at the white string is 7.5. This value does not fall in the 99%
confident interval for the blue string found in part a. Thus, the chickens are more apt to peck at white
string.
a.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D, with
df n 1 3 1 2 , t.005 9.925 . The confidence interval is:
x t.005
b.
c.
s
n
49.3 9.925
1.5
3
49.3 8.60 40.70, 57.90
We are 99% confident that the mean percentage of B(a)p removed from all soil specimens using the
poison is between 40.70% and 57.90%.
We must assume that the distribution of the percentages of B(a)p removed from all soil specimens
using the poison is normal.
d.
Since the 99% confidence interval for the mean percent removed contains 50%, this would be a very
possible value.
e.
For confidence level .90, .10 and / 2 .10 / 2 .05 . Using Table IV, Appendix D, with
2
2
5.99147 and .95,2
.102587 . The 90% confidence interval is:
df n 1 3 1 2 , .05,2
(n 1) s 2
2
.05
2
(n 1) s 2
2
.95
(3 1)(1.5) 2
(3 1)(1.5)2
2
.751 2 43.865
5.99147
.102587
We are 90% confident that the true population variance in the percentages of B(z)p removed is
between .751 and 43.865.
6.124
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
For a width of .06, ME .06 / 2 .03 .
The sample size is n
( z / 2 )2 pq (1.645)2 (.17)(.83)
424.2 425
ME 2
.032
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 337
You would need to take n 425 samples.
6.125
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: IQ25, IQ60
Variable N
IQ25
36
IQ60
36
Mean
66.83
45.31
Median
66.50
45.00
StDev
14.36
12.70
Minimum
41.00
22.00
Maximum
94.00
73.00
Q1
54.25
36.25
Q3
80.00
58.00
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
x z.005
s
n
66.83 2.58
14.36
36
66.83 6.17 (60.66, 73.00)
We are 99% confident that the mean raw IQ score for all 25-year-olds is between 60.66 and 73.00.
b.
We must assume that the sample is random, the observations are independent, and the sample size is
sufficiently large.
c.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
x z / 2
s
n
45.31 1.96
12.7
36
45.31 4.15 (41.16, 49.46)
We are 95% confident that the mean raw IQ score for all 60-year-olds is between 41.16 and 49.46.
6.126
a.
The point estimate of p is pˆ x / n 35 / 55 .636 .
b.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 55 .636 35 and nqˆ 55 .364 20
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
For confidence coefficient, .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The confidence interval is:
pˆ z.005
c.
ˆˆ
pq
.636(.364)
.636 2.58
.636 .167 .469, .803
n
55
We are 99% confident that the true proportion of fatal accidents involving children is between .469
and .803.
Copyright © 2014 Pearson Education, Inc.
338
Chapter 6
d.
The sample proportion of children killed by air bags who were not wearing seat belts or were
improperly restrained is 24 / 35 .686 . This is rather large proportion. Whether a child is killed by
an airbag could be related to whether or not he/she was properly restrained. Thus, the number of
children killed by air bags could possibly be reduced if the child were properly restrained.
e.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.575 . Also, ME .1 .
The sample size is n
6.127
( z / 2 )2 pq (2.575)2 (.636)(.364)
153.5 154
ME 2
.12
x 52
.867 .
n 60
a.
The point estimate of p is pˆ
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
ˆˆ
pq
.867(.133)
.867 1.96
.867 .086 (.781, .953)
n
60
c.
We are 95% confident that the true proportion of Wal-Mart stores in California that have more than 2
inaccurately priced items per 100 scanned is between .781 and .953.
d.
If 99% of the California Wal-Mart stores are in compliance, then only 1% or .01 would not be.
However, we found the 95% confidence interval for the proportion that are not in compliance is
between .781 and .953. The value of .01 is not in this interval. Thus, it is not a likely value. This
claim is not believable.
e.
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 60(.867) 52 and nqˆ 60 .133 8
Since nqˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable. Thus, the confidence interval constructed in part b may not be valid. Any inference based
on this interval is questionable.
f.
From above, the value of p̂ is .867. For confidence coefficient .90, .10 and / 2 .10 / 2 .05 .
From Table II, Appendix D, z.05 1.645 .
n
z2 / 2 pq 1.6452 (.867)(.133)
124.8 125
( ME ) 2
.052
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Estimation with Confidence Intervals 339
6.128
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: r
Variable
r
N
34
Mean
0.4224
Median StDev
0.4300 0.1998
Minimum
-0.0800
Maximum
0.7400
Q1
0.2925
Q3
0.6000
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
. The confidence interval is:
x z / 2
6.129
s
n
.4224 1.96
.1998
34
z.025 1.96
.4224 .0672 (.3552, .4896)
We are 95% confident that the mean value of r is between .3552 and .4896.
a.
Of the 24 observations, 20 were 2 weeks of vacation pˆ 20 / 24 .833 .
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
b.
ˆˆ
pq
.833(.167)
.833 1.96
.833 .149 (.684, .982)
n
24
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 24 .833 20 and nqˆ 24 .167 4
Since nqˆ is less than 15, the sample size is not sufficiently large to conclude the normal
approximation is reasonable. The validity of the confidence interval is in question.
c.
The bound is ME = .02. For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From
Table II, Appendix D, z.025 1.96 . Thus,
z / 2 pq
2
n
( ME ) 2
1.96 2 (.833)(.167)
1, 336.02 1,337
.02 2
Thus, we would need a sample size of 1,337.
6.130
a.
The point estimate for the fraction of the entire market who refuse to purchase bars is:
pˆ
b.
x 23
.094
n 244
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 244 .094 22.9 and nqˆ 244 .906 221.1
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
Copyright © 2014 Pearson Education, Inc.
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Chapter 6
c.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
pˆ z.025
d.
6.131
ˆˆ
pq
(.094)(.906)
.094 1.96
.094 .037 (.057, .131)
n
244
The best estimate of the true fraction of the entire market who refuse to purchase bars six months
after the poisoning is .094. We are 95% confident the true fraction of the entire market who refuse to
purchase bars six months after the poisoning is between .057 and .131.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
From Exercise 6.130, a good approximation for p is .094. Also, ME .02 .
z / 2 pq
2
The sample size is n
( ME ) 2
(1.96) 2(.094)(.906)
817.9 818
.02 2
You would need to take n 818 samples.
6.132
The point estimate of p is pˆ
x 36
.434 .
n 83
The sample size is large enough if both npˆ 15 and nqˆ 15 .
npˆ 83 .434 36 and nqˆ 83 .566 47
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the
normal approximation is reasonable.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
The confidence interval is:
pˆ z.025
ˆˆ
pq
.434(.566)
.434 1.96
.434 .107 (.327, .541)
n
83
We are 95% confident that the true proportion of healthcare workers with latex allergies who suspects that
he/she actually has the allergy is between .327 and .541.
6.133
Sampling error has to do with chance. In a population, there is variation – not all observations are the
same. The sampling error has to do with the variation within a sample. By chance, one might get a sample
that overestimates the mean just because all the observations in the sample happen to be high.
Nonsampling error has to do with errors that have nothing to do with the sampling. These errors could be
due to misunderstanding the question being asked, asking a question that the respondent does not know
how to answer, etc.
6.134
a.
1
of the observations fall within k standard
k2
1
deviations of the mean. We want to find k such that 1 2 .60 .
k
Using Chebyshev’s Theorem, we know that at least 1
1
1
1
1
.60 2 .40 k 2 2.5 k 2.5 1.5811
.4
k2
k
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Inferences Based on a Single Sample: Estimation with Confidence Intervals 341
Thus, s
80th percentile 20th percentile 73, 000 35,100
11,985.3267
2k
2(1.5811)
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table II, Appendix D,
z.025 2.33 .
z 2.33(11,985.3267)
n /2
194.96 195
2, 000
ME
2
2
6.135
b.
See part a.
c.
We have to assume that the estimate of the standard deviation is accurate.
a.
Answers will vary. Using a computer package, the 100 selected invoices are:
3590 1453 3726 2844 1767 1259 1091 1795 4431 4565 4586 1020 2135 1078 2659
4694 2572 4559 4601
965 4553 1052 3448
574 1360 3803 2247 1164 1862 2385
1255 4966
658 4007 4743 3746 3029 3723 3950
4662
217
949 4580 4126 1794 2912
67 2514 3544 1596 2344 1603 3744 1886
151 4258
183 1869 4509 4572 3875
34 3781 4993 1284 2177 4290
13 2717
287 2977 3459 4639 2272 3620 4646 1544
919 3820
1216 2052 4881 2220 3883
346 4744
312 4325
602 3137
121 2373 4684 2025 2254 4018 2304 3503 1634 2470
The observation numbers ending in 0 are highlighted above.
b.
x 10
.10
n 100
pˆ
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
pˆ z.05
c.
6.136
ˆˆ
pq
.10(.90)
.10 1.645
.10 .049 (.051, .149)
n
100
Our sample proportion was pˆ .10 which is equal to the true proportion. The confidence interval
does contain .10.
Since the manufacturer wants to be reasonably certain the process is really out of control before shutting
down the process, we would want to use a high level of confidence for our inference. We will form a 99%
confidence interval for the mean breaking strength.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D, with
df n 1 9 1 8 t.005 3.355 . The 99% confidence interval is:
x t.005
s
n
985.6 3.355
22.9
9
985.6 25.61 (959.99, 1,011.21)
We are 99% confident that the true mean breaking strength is between 959.99 and 1,011.21. Since 1,000 is
contained in this interval, it is not an unusual value for the true mean breaking strength. Thus, we would
recommend that the process is not out of control.
Copyright © 2014 Pearson Education, Inc.
342
Chapter 6
6.137
a.
As long as the sample is random (and thus representative), a reliable estimate of the mean weight of
all the scallops can be obtained.
b.
The government is using only the sample mean to make a decision. Rather than using a point
estimate, they should probably use a confidence interval to estimate the true mean weight of the
scallops so they can include a measure of reliability.
a.
We will form a 95% confidence interval for the mean weight of the scallops. Using MINITAB, the
descriptive statistics are:
Descriptive Statistics: Weight
Variable
Weight
N
Mean StDev
18 0.9317 0.0753
Minimum
0.8400
Q1
0.8800
Median
0.9100
Q3
9800
Maximum
1.1400
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix A, with
df = n – 1 = 18 – 1 = 17, t.025 = 2.110. The 95% confidence interval is:
x t.025
s
n
.932 2.110
.0753
18
.932 .037 (.895, .969)
We are 95% confident that the true mean weight of the scallops is between .8943 and .9691. Recall
that the weights have been scaled so that a mean weight of 1 corresponds to 1/36 of a pound. Since
the above confidence interval does not include 1, we have sufficient evidence to indicate that the
minimum weight restriction was violated.
Copyright © 2014 Pearson Education, Inc.
Chapter 7
Inferences Based on a Single Sample:
Tests of Hypothesis
7.1
The null hypothesis is the "status quo" hypothesis, while the alternative hypothesis is the research hypothesis.
7.2
The test statistic is used to decide whether or not to reject the null hypothesis in favor of the alternative
hypothesis.
7.3
The "level of significance" of a test is . This is the probability that the test statistic will fall in the
rejection region when the null hypothesis is true.
7.4
A Type I error is rejecting the null hypothesis when it is true.
A Type II error is accepting the null hypothesis when it is false.
the probability of committing a Type I error.
the probability of committing a Type II error.
7.5
The four possible results are:
1.
Rejecting the null hypothesis when it is true. This would be a Type I error.
2.
Accepting the null hypothesis when it is true. This would be a correct decision.
3.
Rejecting the null hypothesis when it is false. This would be a correct decision.
4.
Accepting the null hypothesis when it is false. This would be a Type II error.
7.6
We can compute a measure of reliability for rejecting the null hypothesis when it is true. This measure of
reliability is the probability of rejecting the null hypothesis when it is true which is . However, it is
generally not possible to compute a measure of reliability for accepting the null hypothesis when it is false.
We would have to compute the probability of accepting the null hypothesis when it is false, , for every
value of the parameter in the alternative hypothesis.
7.7
When you reject the null hypothesis in favor of the alternative hypothesis, this does not prove the
alternative hypothesis is correct. We are 100(1 )% confident that there is sufficient evidence to conclude
that the alternative hypothesis is correct.
If we were to repeatedly draw samples from the population and perform the test each time, approximately
100(1 )% of the tests performed would yield the correct decision.
7.8
a.
343
Copyright © 2014 Pearson Education, Inc.
344
Chapter 7
b.
c.
d.
e.
f.
g.
P( z 1.96) .025
P( z 1.645) .05
P( z 2.575) .005
P( z 1.28) .1003
P ( z 1.645or z 1.645) .10
P( z 2.575or z 2.575) .01
7.9
a.
Let p = proportion of college presidents who believe that their online education courses are as good as
or superior to courses that utilize traditional face-to-face instruction. The null hypothesis would be:
H 0 : p .68
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Tests of Hypothesis
7.10
b.
The rejection region requires / 2 .01/ 2 .005 in each tail if the z-distribution. From Table II,
Appendix D, z.005 2.575 . The rejection region for a two-tailed test is z 2.575 or z 2.575 .
a.
Let average gain in green fees, lessons, or equipment expenditures for participating golf
facilities. The null and alternative hypotheses would be:
345
H 0 : $2, 400
H a : $2, 400
7.11
b.
The .05 is the Type I error rate. This means that the probability of concluding that the average
gain in fees, lessons, or equipment expenditures for participation golf facilities exceeds $2,400 when
in fact, the average is $2,400 is .05.
c.
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Let p = student loan default rate in this year. To see if the student loan default rate is less than .07, we test:
H 0 : p .07
H a : p .07
7.12
Let p = proportion of U.S. companies that have formal, written travel and entertainment policies for their
employees. The null hypothesis would be:
H 0 : p .80
7.13
Let mean caloric content of Virginia school lunches. To test the claim that after the testing period
ended, the average caloric content dropped, we test:
H 0 : 863
H a : 863
7.14
Let average Libor rate for 1-year loans. Since many Western banks think that the reported average
Libor rate (1.1%) is too high, they want to show that the average is less than 1.1. The appropriate
hypotheses would be:
H 0 : 1.1
H a : 1.1
7.15
a.
Since the company must give proof the drug is safe, the null hypothesis would be the drug is unsafe.
The alternative hypothesis would be the drug is safe.
b.
A Type I error would be concluding the drug is safe when it is not safe. A Type II error would be
concluding the drug is not safe when it is. is the probability of concluding the drug is safe when it
is not. is the probability of concluding the drug is not safe when it is.
c.
In this problem, it would be more important for to be small. We would want the probability of
concluding the drug is safe when it is not to be as small as possible.
Copyright © 2014 Pearson Education, Inc.
346
7.16
Chapter 7
a.
A Type I error would be concluding the proposed user is unauthorized when, in fact, the proposed
user is authorized.
A Type II error would be concluding the proposed user is authorized when, in fact, the proposed user
is unauthorized.
In this case, a more serious error would be a Type II error. One would not want to conclude that the
proposed user is authorized when he/she is not.
b.
The Type I error rate is 1%. This means that the probability of concluding the proposed user is
unauthorized when, in fact, the proposed user is authorized is .01.
The Type II error rate is .00025%. This means that the probability of concluding the proposed user is
authorized when, in fact, the proposed user is unauthorized is .0000025.
c.
The Type I error rate is .01%. This means that the probability of concluding the proposed user is
unauthorized when, in fact, the proposed user is authorized is .0001.
The Type II error rate is .005%. This means that the probability of concluding the proposed user is
authorized when, in fact, the proposed user is unauthorized is .00005.
7.17
a.
A Type I error is rejecting the null hypothesis when it is true. In a murder trial, we would be
concluding that the accused is guilty when, in fact, he/she is innocent.
A Type II error is accepting the null hypothesis when it is false. In this case, we would be concluding
that the accused is innocent when, in fact, he/she is guilty.
7.18
b.
Both errors are serious. However, if an innocent person is found guilty of murder and is put to death,
there is no way to correct the error. On the other hand, if a guilty person is set free, he/she could
murder again.
c.
In a jury trial, is assumed to be smaller than . The only way to convict the accused is for a
unanimous decision of guilt. Thus, the probability of convicting an innocent person is set to be small.
d.
In order to get a unanimous vote to convict, there has to be overwhelming evidence of guilt. The
probability of getting a unanimous vote of guilt if the person is really innocent will be very small.
e.
If a jury is prejudiced against a guilty verdict, the value of will decrease. The probability of
convicting an innocent person will be even smaller if the jury if prejudiced against a guilty verdict.
f.
If a jury is prejudiced against a guilty verdict, the value of will increase. The probability of
declaring a guilty person innocent will be larger if the jury is prejudiced against a guilty verdict.
a.
The null hypothesis is: Ho: There is no intrusion.
b.
The alternative hypothesis is: Ha: There is an intrusion.
c.
P(warning | no intrusion)
1
.001 .
1000
P(no warning | intrusion)
500
.5
1000
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Tests of Hypothesis
.7.19
7.20
7.21
7.22
a.
p P( z 1.20) .5 .3849 .1151
b.
p P( z 1.20) .5 .3849 .1151
c.
The x is p P( z 1.20) P( z 1.20) 2(.1151) .2302
347
We will reject H0 if the p-value .
a.
.06 .05 , do not reject H0.
b.
.10 .05 , do not reject H0.
c.
.01 .05 , reject H0.
d.
.001 .05 , reject H0.
e.
.251 .05 , do not reject H0.
f.
.042 .05 , reject H0.
a.
Since the p-value .10 is greater than .05 , H0 is not rejected.
b.
Since the p-value .05 is less than .10 , H0 is rejected.
c.
Since the p -value .001 is less than .01 , H0 is rejected.
d.
Since the p-value .05 is greater than .025 , H0 is not rejected.
e.
Since the p-value .45 is greater than .10 , H0 is not rejected.
z
x 0
x
49.4 50
4.1/ 100
1.46
p-value p P( z 1.46) .5 .4279 .9279 (using Table II, Appendix D)
Since the p-value is so large, H0 would not be rejected for any reasonable value of . There is no evidence
to indicate the mean is greater than 50.
7.23
p-value p P( z 2.17) .5 P(0 z 2.17) .5 .4850 .0150 (using Table II, Appendix D)
The probability of observing a test statistic of 2.17 or anything more unusual if the true mean is 100 is
.0150. Since this probability is so small, there is evidence that the true mean is greater than 100.
7.24
First, find the value of the test statistic z
x 0
x
10.7 10
3.1/ 50
1.60
p-value p P( z 1.60 or z 1.60) 2 P( z 1.60) 2(.5 .4452) 2(.0548) .1096
(using Table II, Appendix D)
There is no evidence to reject H0 for .10 .
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Chapter 7
7.25
p-value p P ( z 2.17) P ( z 2.17) 2(.5 .4850) 2(.0150) .0300 (using Table II, Appendix D)
7.26
a.
The p-value reported by SPSS is for a two-tailed test. Thus, P ( z 1.63) P ( z 1.63) .1032 . For
this one-tailed test, the p-value p P( z 1.63) .1032 / 2 .0516 .
Since the p-value .0516 .05 , H0 is not rejected. There is insufficient evidence to indicate
75 at .05 .
b.
For this one-tailed test, the p-value P( z 1.63) . Since P( z 1.63) .1032 / 2 .0516 ,
P ( z 1.63) 1 .0516 .9484 .
Since the p-value p .9484 .10 , H0 is not rejected. There is insufficient evidence to indicate
75 at .10 .
c.
For this one-tailed test, the p-value P( z 1.63) .1032 / 2 .0516 .
Since the p-value p .0516 .10 , H0 is rejected. There is sufficient evidence to indicate
75 at .10 .
d.
For this two-tailed test, the p -value .1032 .
Since the p-value .1032 .01 , H0 is not rejected. There is insufficient evidence to indicate
75 at .01 .
7.27
The smallest value of for which the null hypothesis would be rejected is just greater than .06.
7.28
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with x
and x
b.
n
20
400
The test statistic is z
1.
x 0
n
72.5 70
2.5 .
20
400
c.
The p-value is p P ( z 2.5) .5 .4938 .0062 .
d.
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
e.
Since the p-value is less than ( p .0062 .01) , H0 is rejected. There is sufficient evidence to
indicate the true mean is greater than 70 at .01 .
f.
Since the observed value of the test statistics falls in the rejection region ( z 2.5 2.33) , H0 is
rejected. There is sufficient evidence to indicate the true mean is greater than 70 at .01 .
g.
Yes, the conclusions in parts e and f agree.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Tests of Hypothesis
7.29
a.
The decision rule is to reject H0 if x 270 . Recall that z
x 0
x
Therefore, reject H0 if x 270 can be written as reject H0 if z
349
.
x 0
x
270 255
63 / 81
2.14 .
The decision rule in terms of z is to reject H0 if z 2.14 .
b.
7.30
a.
P z 2.14 .5 P 0 z 2.14 .5 .4838 .0162
H 0 : 100
H a : 100
x 0
The test statistic is z
x
x 0
/ n
110 100
60 / 100
1.67
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region, z 1.67 1.645 , H0 is
rejected. There is sufficient evidence to indicate the true population mean is greater than 100 at
.05 .
b.
H 0 : 100
H a : 100
x 0
The test statistic is z
x
110 100
60 / 100
1.67
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region, ( z 1.67 1.96) ,
H0 is not rejected. There is insufficient evidence to indicate does not equal 100 at .05 .
c.
7.31
a.
In part a, we rejected H0 and concluded the mean was greater than 100. In part b, we did not reject
H0. There was insufficient evidence to conclude the mean was different from 100. Because the
alternative hypothesis in part a is more specific than the one in b, it is easier to reject H0.
H 0 : .36
H a : .36
The test statistic is z
x 0
x
.323 .36
.034 / 64
1.61
The rejection region requires .10 in the lower tail of the z-distribution. From Table II,
Appendix D, z.10 1.28 . The rejection region is z 1.28 .
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Chapter 7
Since the observed value of the test statistic falls in the rejection region ( z 1.61 1.28) , H0 is
rejected. There is sufficient evidence to indicate the mean is less than .36 at .10 .
b.
H 0 : .36
H a : .36
The test statistic is z 1.61 (see part a).
The rejection region requires / 2 .10 / 2 .05 in the each tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.61< 1.645) ,
H0 is not rejected. There is insufficient evidence to indicate the mean is different from .36 at .10 .
7.32
a.
Let true mean level of support. To determine if the true mean level of support differs from 75, we
test:
H 0 : 75
H a : 75
b.
For this problem, a Type I error would be concluding the true mean level of support differs from 75
when, in fact, the true mean level of support is 75.
For this problem, a Type II error would be concluding the true mean level of support equals 75
when, in fact, the true mean level of support differs from 75.
7.33
c.
The test statistic is z 8.4923 and the p-value is p .0001 .
d.
Since the p-value is less than ( p .0001 .05) , H0 is rejected. There is sufficient evidence to
indicate the true mean level of support differs from 75 at .05 .
e.
We do not need to make any assumptions about the distribution of support levels. The sample size is
very large (n 992) . Thus, the Central Limit Theorem holds and no assumptions are necessary.
a.
Let true mean willingness to eat the brand of sliced apples. To determine if the true mean
willingness to eat the brand of sliced apples exceeds 3, we test:
H0 : 3
Ha : 3
The test statistic is z
x 0
n
3.69 3
5.71 .
2.44
408
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 5.71 1.645) , H0 is
rejected. There is sufficient evidence to indicate that the true mean willingness to eat the brand of
sliced apples exceeds 3 at .05 .
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Inferences Based on a Single Sample: Tests of Hypothesis
7.34
351
b.
Even though the willingness to eat scores are not normally distributed, the test in part a is valid.
Because the sample size is so large (n 408) , the Central Limit Theorem applies.
a.
Let mean Mach rating score for all purchasing managers. To determine if the mean Mach rating
score is different from 85, we test:
H 0 : 85
H a : 85
b.
The rejection requires / 2 .10 / 2 .05 in each tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
c.
The test statistic is z
d.
Since the observed value of the test statistic falls in the rejection region z 12.80 1.645 , H0 is
x o
x
99.6 85
12.6 / 122
12.80 .
rejected. There is sufficient evidence to indicate that the true mean Mach rating score of all
purchasing managers is not 85 at .10 .
7.35
Let true mean facial width-to-height ratio. To determine if the true mean facial width-to-height ratio
differs from 2.2, we test:
H 0 : 2.2
H a : 2.2
The test statistic is z
x 0
n
1.96 2.2
11.87 .
.15
55
The p-value is p P( z 11.87) P( z 11.87) 0 0 0 .
Since the p-value is so small ( p 0) , H0 will be rejected for any reasonable value of . There is
sufficient evidence to indicate the true mean facial width-to-height ratio differs from 2.2 for .001 .
7.36
a.
Let true mean rate of return of round-trip trades. To determine if the true mean rate of return of
round-trip trades is positive, we test:
H0 : 0
Ha : 0
b.
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
c.
probability of making a Type I error or the probability of rejecting H0 when H0 is true. Thus,
probability of concluding the true mean rate of return of round-trip trades is positive when, in
fact, it is not.
d.
The test statistic is t 4.73 and the p-value is p 0.000 .
Copyright © 2014 Pearson Education, Inc.
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7.37
Chapter 7
e.
Since the p-value is less than ( p 0.000 .05) , H0 is rejected. There is sufficient evidence to
indicate the true mean rate of return of round-trip trades is positive at .05 .
a.
Let true mean weight of golf tees. To determine if the process is not operating satisfactorily, we
test:
H 0 : .250
H a : .250
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tees
Variable N
Mean Median
StDev
Tees
40 0.25248 0.25300 0.00223
Minimum Maximum
Q1
Q3
0.24700 0.25600 0.25100 0.25400
Thus, x .25248 and s .00223 .
x 0
The test statistic is z
d.
The p-value is p P( z 7.03) P( z 7.03) 0 0 0 .
e.
The rejection region requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.575 . The rejection region is z 2.575 or z 2.575 .
f.
Since the observed value of the test statistic falls in the rejection region z 7.03 2.575 , H0 is
x
.25248 .250
c.
.00223 / 40
7.03 .
rejected. There is sufficient evidence to indicate the process is performing in an unsatisfactory
manner at .01 .
g.
is the probability of a Type I error. A Type I error, in this case, is to say the process is
unsatisfactory when, in fact, it is satisfactory. The risk, then, is to the producer since he will be
spending time and money to repair a process that is not in error.
is the probability of a Type II error. A Type II error, in this case, is to say the process is
satisfactory when it, in fact, is not. This is the consumer's risk since he could unknowingly purchase
a defective product.
7.38
Let mean IQ score of all Norway residents who were 6th-born or later. To determine if the mean IQ
score of all Norway residents who were 6th-born or later is lower than the country mean, we test:
H 0 : 5.2
H a : 5.2
The test statistic is z
x 0
x
4.7 5.2
1.8 / 581
6.70 .
The p-value is p P( z 6.70) .5 .5 0 (using Table II, Appendix D)
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Inferences Based on a Single Sample: Tests of Hypothesis
353
Since the p-value is less than ( p 0 .01) , H0 is rejected. There is sufficient evidence to indicate the
true mean IQ score of all Norway residents who were 6th-born or later is lower than the country mean at
.01 .
7.39
a.
Let mean estimated time to read the report. To determine if the students, on average,
overestimate the time it takes to read the report, we test:
H 0 : 48
H a : 48
The test statistic is z
x 0
x
60 48
1.85 .
41/ 40
The p-value is p P( z 1.85) .5 .4678 .0322 (using Table II, Appendix D)
Since the p-value is less than ( p .0322 .10) , H0 is rejected. There is sufficient evidence to
indicate the students, on average, overestimate the time it takes to read the report at .10 .
b.
Let mean estimated number of pages of the report read. To determine if the students, on
average, underestimate the number of report pages read, we test:
H 0 : 32
H a : 32
The test statistic is z
x 0
x
28 32
14 / 42
1.85 .
The p-value is p P( z 1.85) .5 .4678 .0322 (using Table II, Appendix D)
Since the p-value is less than ( p .0322 .10) , H0 is rejected. There is sufficient evidence to
indicate the students, on average, underestimate the number of report pages read at .10 .
c.
7.40
No. In both tests, the sample sizes are greater than 30. Thus, the Central Limit Theorem will apply.
The distribution of x is approximately normal regardless of the population distribution.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: GASTURBINE
Variable
GASTURBINE
N Mean
67 11066
StDev
1595
Minimum
8714
Q1
9918
Median
10656
Q3
11842
Maximum
16243
To determine if the mean heat rate of gas turbines augmented with high pressure inlet fogging exceeds
10,000 kJ/kWh, we test:
H 0 : 10, 000
H a : 10, 000
The test statistic is z
x o
x
11, 066 10, 000
1,595
67
5.47 .
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Chapter 7
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistics falls in the rejection region z 5.47 1.645 , H0 is rejected.
There is sufficient evidence to indicate the true mean heat rate of gas turbines augmented with high
pressure inlet fogging exceeds 10,000 kJ/kWh at .05 .
7.41
To determine if the mean point-spread error is different from 0, we test:
H0 : 0
Ha : 0
The test statistic is z
x 0
x
1.6 0
13.3 / 240
1.86
The rejection region requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II, Appendix
D, z.005 2.575 . The rejection region is z 2.575 or z 2.575 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.86 2.575) , H0 is
not rejected. There is insufficient evidence to indicate that the true mean point-spread error is different
from 0 at .01 .
7.42
a.
Let average full-service fee (in thousands of dollars) of U.S. funeral homes in the current year.
To determine if the average full-service fee exceeds $6,500, we test:
H 0 : 6.50
H a : 6.50
b.
Using MINITAB, the output is:
Descriptive Statistics: FUNERAL
Variable N
Fee
36
Mean
6.819
Median
6.600
StDev
1.265
Minimum
5.200
Maximum
11.600
Q1
6.025
Q3
7.400
H 0 : 6.50
H a : 6.50
The test statistic is z
x 0
x
6.819 6.50
1.265
36
1.51
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
.645) , H0
Since the observed value of the test statistic does not fall in the rejection region ( z 1.51>1
is not rejected. There is insufficient evidence to indicate the true mean full-service fee of U.S. funeral
homes in the current year exceeds $6,500 at .05 .
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Inferences Based on a Single Sample: Tests of Hypothesis
c.
355
No. Since the sample size n 36 is greater than 30, the Central Limit Theorem applies. The
distribution of x is approximately normal regardless of the population distribution.
7.43
a.
To determine if the true mean forecast error for buy-side analysts is positive, we test:
H0 : 0
Ha : 0
The test statistic is z
x o
x
.85 0
1.93 / 3,526
26.15 .
The observed p-value of the test is p P z 26.15 0 (Using Table II, Appendix D)
Since the p-value is less than ( p 0 .01) , H0 is rejected. There is sufficient evidence to indicate
that the true mean forecast error for buy-side analysts is positive at .01 . This means that the buyside analysts are overestimating earnings.
b.
To determine if the true mean forecast error for sell-side analysts is negative; we test:
H0 : 0
Ha : 0
The test statistic is z
x o
x
.05 0
.85 / 58,562
14.24 .
The observed p-value of the test is p P( z 14.24) 0 (using Table II, Appendix D)
Since the p-value is less than ( p 0 .01) , H0 is rejected. There is sufficient evidence to indicate
that the true mean forecast error for sell-side analysts is negative at .01 . This means that the sellside analysts are underestimating earnings.
7.44
a.
To determine if the sample data refute the manufacturer's claim, we test:
H 0 : 10
H a : 10
b.
A Type I error is concluding the mean number of solder joints inspected per second is less than 10
when, in fact, it is 10 or more.
A Type II error is concluding the mean number of solder joints inspected per second is at least 10
when, in fact, it is less than 10.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: PCB
Variable
PCB
N
48
Mean
9.292
Median
9.000
StDev
2.103
Minimum
0.000
Maximum
13.000
H 0 : 10
H a : 10
Copyright © 2014 Pearson Education, Inc.
Q1
9.000
Q3
10.000
356
Chapter 7
x 0
The test statistic is z
9.292 10
2.33
2.103 / 48
x
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 2.33 1.645) , H0 is
rejected. There is sufficient evidence to indicate the mean number of inspections per second is less
than 10 at .05 .
7.45
a.
To determine if CEOs at all California small firms generally agree with the statement, we test:
H 0 : 3.5
H a : 3.5
The test statistic is z
x o
x
3.85 3.5
1.5 137
2.73
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistics falls in the rejection region ( z 2.73 1.645) , H0 is
rejected. There is sufficient evidence to indicate CEOs at all California small firms generally agree
with the statement (true mean scale score exceeds 3.5) at .05 .
b.
Although the sample mean of 3.85 is far enough away from 3.5 to statistically conclude the
population mean score is greater than 3.5, a score of 3.85 may not be practically different from 3.5 to
make any difference.
c.
No. Since the sample size n 137 is greater than 30, the Central Limit Theorem applies. The
distribution of x is approximately normal regardless of the population distribution.
7.46
a.
No. Since the hypothesized value of M 60, 000 falls in the 95% confidence interval, it is a likely
candidate for the true mean. Thus, we would not reject H0. There is no evidence that the mean salary
for males differs from $60,000.
b.
To determine if the true mean salary of males with post-graduate degrees differs from $60,000, we
test:
H 0 : 60, 000
H a : 60, 000
The test statistic is z
x 0
x
61,340 60,000
0.61
2,185
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 0.61 1.96) , H0
is not rejected. There is insufficient evidence to indicate the true mean salary of males with postgraduate degrees differs from $60,000 at .05 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Tests of Hypothesis
357
c.
Parts a and b must agree. In both cases, a two-sided test / confidence interval is used. The z-score
used in both parts is the same, as are x and sx .
d.
No. Since the hypothesized value of F 33, 000 falls in the 95% confidence interval, it is a likely
candidate for the true mean. Thus, we would not reject H0. There is no evidence that the mean salary
for females differs from $33,000.
e.
To determine if the true mean salary of females with post-graduate degrees differs from
$33,000, we test:
H 0 : 33, 000
H a : 33,000
The test statistic is z
x 0
x
32, 227 33, 000
0.83
932
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region
( z 0.83 1.96) , H0 is not rejected. There is insufficient evidence to indicate the true mean
salary of females with post-graduate degrees differs from $33,000 at .05 .
7.47
7.48
f.
Parts d and e must agree. In both cases, a two-sided test / confidence interval is used. The z-score
used in both parts is the same, as are x and sx .
a.
We should use the t-distribution in testing a hypothesis about a population mean if the sample size is
small, the population being sampled from is normal, and the variance of the population is unknown.
b.
Both distributions are mound-shaped and symmetric. The t-distribution is flatter than the zdistribution.
a.
P t 1.440 .10
(Using Table III, Appendix D, with df 6 )
b.
P (t 1.782) .05
(Using Table III, Appendix D, with df 12 )
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358
c.
Chapter 7
P (t 2.060) P t 2.060 .025 .025 .05
(Using Table III, Appendix D, with df 25 )
7.49
7.50
d.
The probability of a Type I error is computed above for each of the parts.
a.
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df n 1 14 1 13 . From Table III, Appendix D, t.025 2.160 . The rejection region is
t 2.160 or t 2.160 .
b.
The rejection region requires .01 in the upper tail of the t-distribution with df n 1 24 1 23 .
From Table III, Appendix D, t.01 2.500 . The rejection region is t 2.500 .
c.
The rejection region requires .10 in the upper tail of the t-distribution with df n 1 9 1 8 .
From Table III, Appendix D, t.10 1.397 . The rejection region is t 1.397 .
d.
The rejection region requires .01 in the lower tail of the t-distribution with df n 1 12 1 11 .
From Table III, Appendix D, t.01 2.718 . The rejection region is t 2.718 .
e.
The rejection region requires / 2 .10 / 2 .05 in each tail of the t-distribution with
df n 1 20 1 19 . From Table III, Appendix D, t.05 1.729 . The rejection region is
t 1.729 or t 1.729 .
f.
The rejection region requires .05 in the lower tail of the t-distribution with df n 1 4 1 3 .
From Table III, Appendix D, t.05 2.353 . The rejection region is t 2.353 .
a.
H0 : 6
Ha : 6
The test statistic is t
x 0
4.8 6
2.064
s / n 1.3 / 5
The necessary assumption is that the population is normal.
The rejection region requires .05 in the lower tail of the t-distribution with df n 1 5 1 4 .
From Table III, Appendix D, t.05 2.132 . The rejection region is t 2.132 .
Since the observed value of the test statistic does not fall in the rejection region (t 2.064 2.132)
, H0 is not rejected. There is insufficient evidence to indicate the mean is less than 6 at .05 .
b.
H0 : 6
Ha : 6
The test statistic is t 2.064 (from a).
The assumption is the same as in a.
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359
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df n 1 5 1 4 . From Table III, Appendix D, t.025 2.776 . The rejection region is
t 2.776 or t 2.776 .
c.
Since the observed value of the test statistic does not fall in the rejection region (t 2.064 2.776)
, H0 is not rejected. There is insufficient evidence to indicate the mean is different from 6 at .05 .
For part a, the p-value P (t 2.064) . Using MINITAB,
Cumulative Distribution Function
Student's t distribution with 4 DF
x
-2.064
P( X <= x )
0.0539809
The p-value is p .05398 .
For part b, the p-value P(t 2.064) P(t 2.064) .
The p-value is p 2(.05398) .10796 .
7.51
a.
We must assume that a random sample was drawn from a normal population.
b.
The hypotheses are:
H 0 : 1, 000
H a : 1, 000
The test statistic is t 1.89 and the p-value is p .038 .
Since the p-value is so small, there is evidence to reject H0. There is evidence to indicate the mean is
greater than 1000 for .038 .
c.
The hypotheses are:
H 0 : 1, 000
H a : 1, 000
The test statistic is t 1.89 and the p-value is 2(.038) .076 .
There is no evidence to reject H0 for .05 . There is insufficient evidence to indicate the mean is
different than 1000 for .05 .
There is evidence to reject H0 for .076 . There is evidence to indicate the mean is different than
1000 for .076 .
7.52
a.
To determine if the mean of the trap spacing measurements differs from 95 meters, we test:
H 0 : 95
H a : 95
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Chapter 7
b.
The value of x varies from sample to sample. The next sample may yield a value of x that is greater
than 95. We must determine how unusual a value of x 89.9 is if the true mean is 95.
c.
The test statistic is t
d.
Using MINITAB, the results are:
x 0 89.9 95
1.16 .
s
11.6
n
7
One-Sample T
Test of mu = 95 vs not = 95
N
7
Mean
89.9000
StDev
11.6000
SE Mean
4.3844
95% CI
(79.1718, 100.6282)
T
-1.16
P
0.289
The p-value is p .289 .
7.53
e.
Suppose we pick .05 . For this problem, probability of concluding the mean trap spacing is
different from 95 when, in fact, the mean trap spacing is equal to 95.
f.
Since the p-value is greater than ( p .289 .05) , H0 is not rejected. There is insufficient
evidence to indicate the mean trap spacing is different from 95 at .05 .
g.
In order for the test to be valid, the population of trap spacing measurements must be normal and the
sample must be random.
h.
From Exercise 6.29, the 95% confidence interval is (79.104, 100.616). Since 95 is contained in this
interval, there is no evidence to indicate the mean trap spacing is different from 95. This agrees with
the test in part f.
To determine if the mean level of radon exposure in the tombs is less than 6,000 Bq/m3, we test:
H 0 : 6, 000
H a : 6, 000
From the printout, the test statistic is t 1.82 .
Since this is a one-tailed test, the p-value is p .096 / 2 .0480 .
Since the p-value is less than ( p .048 .10) , H0 is rejected. There is sufficient evidence to indicate the
mean level of radon exposure is less than 6,000 Bq/m3 at .10 .
7.54
a.
To determine if the true mean breaking strength of the new bonding adhesive is less
than 5.70 Mpa, we test:
H 0 : 5.70
H a : 5.70
b.
The rejection region requires .01 in the lower tail of the t-distribution with df n 1 10 1 9 .
From Table III, Appendix D, t.01 2.821 . The rejection region is t 2.821 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on a Single Sample: Tests of Hypothesis
7.55
x o
5.07 5.70
c.
The test statistic is t
d.
Since the observed value of the test statistic falls in the rejection region (t 4.33 2.821) , H0 is
rejected. There is sufficient evidence to indicate the true mean breaking strength of the new bonding
adhesive is less than 5.70 Mpa at .01 .
e.
We must assume that the sample was random and selected from a normal population.
a.
To determine if the mean surface roughness of coated interior pipe differs from 2 micrometers, we
test:
s
n
361
.46 10
4.33 .
H0 : 2
Ha : 2
7.56
b.
From the printout, the test statistic is t 1.02 .
c.
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df n –1 20 –1 19 . From Table III, Appendix D, t.025 2.093 . The rejection region is
t 2.093 or t 2.093 .
d.
Since the observed value of the test statistic does not fall in the rejection region
(t 1.02 2.093) , H0 is not rejected. There is insufficient evidence to indicate the true mean
surface roughness of coated interior pipe differs from 2 micrometers at .05 .
e.
The p-value is p .322 . Since the p-value is not less than .05 , H0 is not rejected. There is
insufficient evidence to indicate the true mean surface roughness of coated interior pipe differs from 2
micrometers at .05 .
f.
From Exercise 6.33, we found the 95% confidence interval for the mean surface roughness of coated
interior pipe to be (1.636, 2.126). Since the hypothesized value of ( 2) falls in the confidence
interval, it is a likely value. We cannot reject it. The confidence interval and the test of hypothesis
lead to the same conclusion because the critical values for the 2 techniques are the same.
a.
Let mean annualized percentage return on investment. To determine if the mean annualized
percentage return on investment is positive, we test:
H0 : 0
Ha : 0
x 0 10.8231 0
5.06 .
s
7.7115
n
13
b.
From the printout, x 10.8231 and s 7.7115 . The test statistic is t
c.
The p-value is p 0.0001 .
d.
Since the p-value is less than ( p .0001 .05) , H0 is rejected. There is sufficient evidence to
indicate the mean annualized percentage return on investment is positive at .05 .
e.
We must assume that we have selected a random sample from the population and that the population
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Chapter 7
of annualized percentage return on investments for all AAII stock screeners is normally distributed.
7.57
a.
Let mean daily amount of distilled water collected by the new system. To determine if the mean
daily amount of distilled water collected by the new system is greater than 1.4, we test:
H 0 : 1.4
H a : 1.4
b.
For this problem, probability of concluding the mean daily amount of distilled water collected by
the new system is greater than 1.4 when, in fact, the mean daily amount of distilled water collected by
the new system is not greater than 1.4. Since .10 , this means that H0 will be rejected when it is
true about 10% of the time.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Water
Variable
Water
N
3
Mean
5.243
StDev
0.192
Minimum
5.070
Q1
5.070
Median
5.210
Q3
5.450
Maximum
5.450
x 5.243 and s .192 .
d.
The test statistic is t
e.
Using MINITAB:
x 0 5.243 1.4
34.67 .
s
.192
n
3
One-Sample T: Water
Test of mu = 1.4 vs > 1.4
Variable
Water
N
3
Mean
5.24333
StDev
0.19218
SE Mean
0.11096
95%
Lower
Bound
4.91935
T
34.64
P
0.000
The p-value is p 0.000 .
f.
7.58
Since the p-value is less than ( p .000 .10) , H0 is rejected. There is sufficient evidence to
indicate daily amount of distilled water collected by the new system is greater than 1.4at .10 .
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Choice
Variable
Choice
N
11
Mean
58.91
StDev
7.78
Minimum
43.00
Q1
56.00
Median
58.00
Q3
62.00
Maximum
76.00
Let mean choice score for consumers shopping with flexed arms. To determine if the mean choice
score for consumers shopping with flexed arms is higher than 43, we test:
H 0 : 43
H a : 43
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Inferences Based on a Single Sample: Tests of Hypothesis
The test statistic is t
363
x 0 58.91 43
6.78 .
s
7.78
n
11
The rejection region requires .05 in the upper tail of the t-distribution with df n –1 11 –1 10 .
From Table III, Appendix D, t.05 1.812 . The rejection region is t 1.812 .
Since the observed value of the test statistic falls in the rejection region (t 6.78 1.812) , H0 is rejected.
There is sufficient evidence to indicate the mean choice score for consumers shopping with flexed arms is
higher than 43 at .05 .
7.59
Using MINITAB, the descriptive statistics are:
One-Sample T: Skid
Test of mu = 425 vs < 425
Variable
Skid
N
20
Mean
358.450
StDev
117.817
SE Mean
26.345
95%
Upper
Bound
T
404.004 -2.53
P
0.010
To determine if the mean skidding distance is less than 425 meters, we test:
H 0 : 425
H a : 425
The test statistics is t
x o
358.45 425
2.53 .
s n 117.817 20
The rejection region requires .10 in the lower tail of the t-distribution with. From Table III, Appendix
D, t.10 1.328 . The rejection region is t 1.328 .
Since the observed value of the test statistic falls in the rejection region (t 2.53 1.328) , H0 is rejected.
There is sufficient evidence to indicate the true mean skidding distance is less than 425 meters at .10 .
There is sufficient evidence to refute the claim.
7.60
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Dioxide
Variable
Dioxide
a.
Oil
No
Yes
N
10
6
Mean
2.590
0.517
StDev
1.542
0.407
Minimum
0.100
0.200
Q1
1.125
0.200
Median
2.850
0.450
Q3
4.000
0.700
Maximum
4.000
1.300
To determine if the mean amount of dioxide present in water specimens that contain oil is less than
3 mg/l, we test:
H0 : 3
Ha : 3
The test statistic is t
x 0 .517 3
14.94 .
s
.407
n
6
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Chapter 7
The rejection region requires .10 in the lower tail of the t-distribution with df n –1 6 –1 5 .
From Table III, Appendix D, t.10 1.476 . The rejection region is t 1.476 .
Since the observed value of the test statistic falls in the rejection region (t 14.94 1.476) , H0 is
rejected. There is sufficient evidence to indicate the mean amount of dioxide present in water
specimens that contain oil is less than 3 mg/l at .10 .
b.
To determine if the mean amount of dioxide present in water specimens that do not contain oil is less
than 3 mg/l, we test:
H0 : 3
Ha : 3
The test statistic is t
x 0
2.59 3
0.84 .
s
1.542
n
10
The rejection region requires .10 in the lower tail of the t-distribution with df n –1 10 –1 9 .
From Table III, Appendix D, t.10 1.383 . The rejection region is t 1.383 .
Since the observed value of the test statistic does not fall in the rejection region (t 0.83 1.383) ,
H0 is not rejected. There is insufficient evidence to indicate the mean amount of dioxide present in
water specimens that do not contain oil is less than 3 mg/l at .10 .
7.61
To determine if the true mean crack intensity of the Mississippi highway exceeds the AASHTO
recommended maximum, we test:
H 0 : .100
H a : .100
The test statistic is t
x 0
s/ n
.210 .100
.011 / 8
2.97
The rejection region requires .01 in the upper tail of the t-distribution with df n 1 8 1 7 . From
Table III, Appendix D, t.01 2.998 . The rejection region is t 2.998 .
Since the observed value of the test statistic does not fall in the rejection region (t 2.97 2.998) , H0 is
not rejected. There is insufficient evidence to indicate that the true mean crack intensity of the Mississippi
highway exceeds the AASHTO recommended maximum at .01 .
7.62
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Plants
Variable
Plants
N
20
Mean
3.900
StDev
2.770
Minimum
1.000
Q1
1.250
Median
3.500
Q3
5.000
Maximum
11.000
Let mean number of active nuclear power plants operating in all states. To determine if the mean
number of active nuclear power plants operating in all states exceeds 3, we test:
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Inferences Based on a Single Sample: Tests of Hypothesis
365
H0 : 3
Ha : 3
The test statistic is t
x o
s
n
3.9 3
2.77
20
1.45
The rejection region requires .10 in the upper tail of the t-distribution with df n –1 20 –1 19
. From Table III, Appendix D, t.10 1.328 . The rejection region is t 1.328 .
Since the observed value of the test statistic falls in the rejection region t 1.45 1.328 , H0 is
rejected. There is sufficient evidence to indicate the mean number of active nuclear power plants
operating in all states exceeds 3 at .10 .
We will look at the 4 methods for determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the number of power plants is:
Histogram of Plants
Normal
Mean
StDev
N
7
3.9
2.770
20
6
5
Frequency
b.
4
3
2
1
0
-2
0
2
4
6
Plants
8
10
12
From the histogram, the data appear to be skewed to the right. This indicates that the data may not be
normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal.
x s 3.9 2.77 (1.13, 6.67) 12 of the 20 values fall in this interval. The proportion is .60.
This is smaller than the .68 we would expect if the data were normal.
x 2s 3.9 2(2.77) 3.9 5.54 (1.64, 9.44) 19 of the 20 values fall in this interval. The
proportion is .95. This is the same as the .95 we would expect if the data were normal.
x 3s 3.9 3(2.77) 3.9 8.31 (4.41, 12.21) 20 of the 20 values fall in this interval. The
proportion is 1.000. This is equal to the 1.00 we would expect if the data were normal.
From the first interval, it appears that the data might not be normal.
Next, we look at the ratio of the IQR to s. IQR QU – QL 5.00 –1.25 3.75 .
IQR 3.75
1.35 This is pretty close to the 1.3 we would expect if the data were normal. This
s
2.77
method indicates the data may be normal.
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Chapter 7
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Plants
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
3.9
2.770
20
0.664
0.070
80
Percent
366
70
60
50
40
30
20
10
5
1
-5
0
5
Plants
10
15
Since the data do not form a straight line, the data may not be normal.
From 3 of the 4 different methods, the indications are that the number of power plants data are not
normal.
c.
The two largest values are 9 and 11. The two lowest values are 1 and 1. Using MINITAB with the
data deleted yields the descriptive statistics:
Descriptive Statistics: Plants2
Variable
Plants2
N
16
Mean
3.500
StDev
1.826
Minimum
1.000
Q1
2.000
Median
3.500
Q3
5.000
Maximum
7.000
To determine if the mean number of active nuclear power plants operating in all states exceeds 3
(using the reduced data set), we test:
H0 : 3
Ha : 3
The test statistic is t
x o
s
n
3.50 3
1.826 16
1.095
The rejection region requires .10 in the upper tail of the t-distribution with df n – 1 16 –1 15
. From Table III, Appendix D, t.10 1.341 . The rejection region is t 1.341 .
Since the observed value of the test statistic does not fall in the rejection region t 1.095 1.341 , H0
is not rejected. There is insufficient evidence to indicate the mean number of active nuclear power
plants operating in all states exceeds 3 at .10 .
By eliminating the top two and bottom two observations, we have changed the decision about H0.
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Inferences Based on a Single Sample: Tests of Hypothesis
d.
7.63
367
It is very dangerous to eliminate data points to satisfy assumptions. The data may, in fact, not be
normal. By eliminating data points, one has changed the kind of data that come from the parent
population. Thus, incorrect decisions could be made.
Using MINITAB, the descriptive statistics for the 2 plants are:
Descriptive Statistics: AL1, AL2
Variable
AL1
AL2
N
Mean
2 0.00750
2 0.0700
StDev
0.00354
0.0283
Minimum
0.00500
0.0500
Q1
*
*
Median
0.00750
0.0700
Q3
*
*
Maximum
0.01000
0.0900
To determine if plant 1 is violating the OSHA standard, we test:
H 0 : .004
H a : .004
The test statistic is t
x o
s
n
.0075 .004
.00354
2
1.40
Since no level was given, we will use .10 . The rejection region requires .10 in the upper tail of the
t-distribution with df n 1 2 1 1 . From Table III, Appendix D, t.10 3.078 . The rejection region is
t 3.078 .
Since the observed value of the test statistic does not fall in the rejection region
(t 1.40 3.078) , H0 is not rejected. There is insufficient evidence to indicate the OSHA standard is
violated by plant 1 at .10 .
To determine if plant 2 is violating the OSHA standard, we test:
H 0 : .004
H a : .004
The test statistic is t
x o
s
n
.07 .004
.0283
2
3.30
Since no level was given, we will use .10 . The rejection region requires .10 in the upper tail of
the t-distribution with df n 1 2 1 1 . From Table III, Appendix D, t.10 3.078 . The rejection region is
t 3.078 .
Since the observed value of the test statistic falls in the rejection region t 3.30 3.078) , H0 is rejected.
There is sufficient evidence to indicate the OSHA standard is violated by plant 2 at .10 .
7.64
a.
Since the value of p̂ (.63) is much smaller than the hypothesized value of p (.70), it is likely that the
null hypothesis is not correct.
b.
First, check to see if n is large enough.
np0 100(.7) 70 and nq0 100(.3) 30
Since both np0 15 and nq0 15 , the normal approximation will be adequate.
H 0 : p .70
H a : p .70
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Chapter 7
The test statistic is z
pˆ p0
pˆ
pˆ p0
p0 q0
n
.63 .70
.70(.30)
100
1.53
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.53 1.645) ,
H0 is not rejected. There is insufficient evidence to indicate that the proportion is less than .70 at
.05 .
7.65
c.
p-value p P( z 1.53) .5 .4370 .0630 . Since p is not less than .05 , H0 is not rejected.
a.
z
pˆ p 0
p 0q 0
n
.83 .9
.9(.1)
100
2.33
.7(.3)
.9(.1)
.0458 as compared to
.03 in part a. Since the
100
100
denominator in this problem is smaller, the absolute value of z is larger.
b.
The denominator in Exercise 7.64 is
c.
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 2.33 1.645) , H0 is
rejected. There is sufficient evidence to indicate the population proportion is less than .9 at .05 .
7.66
d.
The p-value p P( z 2.33) .5 .4901 .0099 (from Table II, Appendix D). Since the p-value is
less than .05 , H0 is rejected.
a.
No. The p-value is the probability of observing your test statistic or anything more unusual if H0 is
true. For this problem, the p-value .3300 / 2 .1650 .
Given the true value of the population proportion, p, is .5, the probability of observing a test statistic
of z .44 or larger is .1650. Since the p-value is not small ( p .1650) , there is no evidence to reject
H0. There is no evidence to indicate the population proportion is greater than .5.
b.
If the alternative hypothesis were two-tailed, the p-value would be 2 times the p-value for a one-tailed
test. For this problem, the p -value .3300 . The probability of observing your test statistic or
anything more unusual if H0 is true is .3300.
Since the p-value is so large, there is no evidence to reject H0 for .10 . There is no evidence to
indicate that p .5 for .10 .
7.67
From Exercise 6.44, n 50 and since p is the proportion of consumers who do not like the snack food, p̂
will be:
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Inferences Based on a Single Sample: Tests of Hypothesis
369
Number of 0 's in sample 29
.58
n
50
First, check to see if the normal approximation will be adequate:
pˆ
np0 50(.5) 25
nq0 50(.5) 25
Since both np0 15 and nq0 15, the normal distribution will be adequate.
a.
H 0 : p .5
H a : p .5
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
.58 .5
.5(1 .5)
50
1.13 .
The rejection region requires .10 in the upper tail of the z-distribution. From Table II,
Appendix D, z.10 1.28 . The rejection region is z 1.28 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.13 1.28) , H0 is
not rejected. There is insufficient evidence to indicate the proportion of customers who do not like
the snack food is greater than .5 at .10 .
b.
7.68
7.69
p value p P( z 1.13) .5 .3708 .1292 (using Table II, Appendix D)
The sample size is large enough to use the normal approximation if npo 15 and nqo 15 .
a.
npo 900(.975) 877.5 15 and nqo 900(.025) 22.5 15 . Thus, the sample size is large enough.
b.
npo 125(.01) 1.25 15 and nqo 125(.99) 123.75 15 . Thus, the sample size is not large
enough.
c.
npo 40(.75) 30 15 and nqo 40(.25) 10 15 . Thus, the sample size is not large enough.
d.
npo 15(.75) 11.25 15 and nqo 15(.25) 3.75 15 . Thus, the sample size is not large enough.
e.
npo 12(.62) 7.44 15 and nqo 12(.38) 4.56 15 . Thus, the sample size is not large enough.
a.
pˆ
b.
To determine if the true proportion of all internet-using adults who have paid to download music
exceeds .7, we test:
x 506
.67
n 755
H 0 : p .7
H a : p .7
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Chapter 7
c.
7.70
pˆ p 0
The test statistic is z
pˆ
pˆ p 0
p 0q 0
n
.67 .7
.7(1 .7)
755
1.80 .
d.
The rejection region requires .01 in the upper tail of the z-distribution. From Table II,
Appendix D, z.01 2.33 . The rejection region is z 2.33 .
e.
p value p P( z 1.80) .5 .4641 .9641 (using Table II, Appendix D)
f.
Since the observed value of the test statistic does not fall in the rejection region (t 1.80 2.33) , H0
is not rejected. There is insufficient evidence to indicate the true proportion of all internet-using
adults who have paid to download music exceeds .7 at .01 .
g.
Since the p-value is not less than ( p .9641 .01) , H0 is not rejected. There is insufficient
evidence to indicate the true proportion of all internet-using adults who have paid to download music
exceeds .7 at .01 .
a.
p true proportion of all satellite radio subscribers who have a satellite radio receiver in their car.
b.
The null hypothesis is:
H 0 : p .8
c.
To determine if the claim is too high, the alternative hypothesis is:
H a : p .8
d.
pˆ
x 396
.79
n 501
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
.79 .8
.8(1 .8)
501
.56 .
e.
The rejection region requires .10 in the lower tail of the z-distribution. From Table II,
Appendix D, z.10 1.28 . The rejection region is z 1.28 .
f.
p value p P( z .56) .5 .2123 .2877 (using Table II, Appendix D)
g.
Since the observed value of the test statistic does not fall in the rejection region ( z .56 1.28) ,
H0 is not rejected. There is insufficient evidence to indicate the claim is too high at .10 .
Since the p-value is not less than ( p .2877 .10) , H0 is not rejected. There is insufficient
evidence to indicate the claim is too high at .10 .
7.71
To determine if the sample provides sufficient evidence to indicate that the true percentage of all
firms that announced one or more acquisitions during the year 2000 is less than 30%, we test:
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Inferences Based on a Single Sample: Tests of Hypothesis
371
H 0 : p .30
H a : p .30
x
748
.269
n 2, 778
pˆ po
.269 .30
The test statistic is z
3.57
.30(.70)
po qo
2, 778
n
The point estimate is pˆ
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 3.57 1.645) , H0 is
rejected. There is sufficient evidence to indicate that the true percentage of all firms that
announced one or more acquisitions during the year 2000 is less than 30% at .05 .
7.72
a.
If there is no relationship between color and gummy bear flavor, then .5 of the population of students
will correctly identify the color.
b.
To determine if color and flavor are related, we test:
H 0 : p .5
H a : p .5
7.73
c.
From the printout, the p-value is p .000 . Since the p-value is less than ( p .000 .01) , H0 is
rejected. There is sufficient evidence to indicate that color and flavor are related at .01 .
a.
To determine whether the true proportion of toothpaste brands with the ADA seal verifying effective
decay prevention is less than .5, we test:
H 0 : p .5
H a : p .5
7.74
b.
From the printout, the p-value is p .231 .
c.
Since the observed p-value is greater than ( p .231 .10) , H0 is not rejected. There is insufficient
evidence to indicate the true proportion of toothpaste brands with the ADA seal verifying effective
decay prevention is less than .5 at .10 .
a.
pˆ
x 198
.10
n 1982
To determine if the percentage of all residential properties purchased for vacation homes is less than
14%, we test:
H 0 : p .14
H a : p .14
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Chapter 7
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
.10 .14
.14(1 .14)
1982
5.13 .
The rejection region requires .01 in the lower tail of the z-distribution. From Table II,
Appendix D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection region ( z 5.13 2.33) , H0 is
rejected. There is sufficient evidence to indicate the percentage of all residential properties purchased
for vacation homes is less than 14% at .01 .
b.
7.75
pˆ
The return rate is only 1, 982 / 45, 000 .044 . This is a very low return rate. Since this is a selfselected sample (only those who wanted to respond returned their questionnaire), it is very likely that
the sample was not representative. Therefore, the results are very suspect.
x 417 77
.585
n
845
To determine if fewer than 60% of the coffee growers in southern Mexico are either certified or
transitioning to become certified, we test:
H 0 : p .60
H a : p .60
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
.585 .60
.60(1 .60)
845
.89 .
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z .89 1.645) , H0 is
not rejected. There is insufficient evidence to indicate that fewer than 60% of the coffee growers in
southern Mexico are either certified or transitioning to become certified at .05 .
7.76
pˆ
x 53
.106
n 500
To determine if the French unemployment rate dropped after the enactment of the 35-hour work week law,
we test:
H 0 : p .12
H a : p .12
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
.106 .12
.12(1 .12)
500
.96 .
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Inferences Based on a Single Sample: Tests of Hypothesis
373
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z .96 1.645) , H0 is
not rejected. There is insufficient evidence to indicate that the French unemployment rate dropped after the
enactment of the 35-hour work week law at .05 .
7.77
a.
Le p proportion of middle-aged women who exhibit skin improvement after using the cream. For
x 24
.727 .
n 33
First we check to see if the normal approximation is adequate:
np0 33(.6) 19.8 nq0 33(.4) 13.2
this problem, pˆ
Since nq0 13.2 is less than 15, the assumption of normality may not be valid. We will go ahead and
perform the test.
To determine if the cream will improve the skin of more than 60% of middle-aged women, we test:
H 0 : p .60
H a : p .60
The test statistic is z
pˆ p0
p0 q0
n
.727 .60
.60(.40)
33
1.49
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.49 1.645) , H0
is not rejected. There is insufficient evidence to indicate the cream will improve the skin of more
than 60% of middle-aged women at .05 .
b.
The p-value is p P( z 1.49) (.5 .4319) .0681 . (Using Table II, Appendix D.) Since the pvalue is greater than ( p .0681 .05) , H0 is not rejected. There is insufficient evidence to indicate
the cream will improve the skin of more than 60% of middle-aged women at .05 .
7.78
First, check to see if n is large enough.
np0 2,376(.7) 1, 663.2 and nq0 2,376(.3) 712.8
Since both np0 15 and nq0 15 , the normal approximation will be adequate.
pˆ
x 1, 554
.654
n 2, 376
To determine if the true detection rate for pictures of PTW is different from .70, we test:
H 0 : p .70
H a : p .70
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Chapter 7
The test statistic is z
pˆ p0
pˆ
pˆ p0
p0 q0
n
.654 .70
.70(.30)
2,376
4.89
The rejection region requires / 2 .10 / 2 .05 in each tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 4.89 1.645) , H0 is rejected.
There is sufficient evidence to indicate the true detection rate for pictures of PTW is different from .70 at
.10 .
7.79
Answers will vary. The target population will be all households in the United States that have televisions.
The experimental unit will be an individual household in the United States with a television. The variable
to be measured is whether or not the household has a DVR. Let p = proportion of be all households in the
United States that have televisions that also have DVR’s. The hypotheses of interest are:
H 0 : p .41
H a : p .41
The test statistic is z
pˆ p 0
pˆ
pˆ p 0
p 0q 0
n
pˆ .41
.41(1 .41)
n
.
One could take a random number generator to generate 500 random telephone numbers to call to obtain a
random sample.
7.80
Let p = proportion of students choosing the three-grill display so that Grill #2 is a compromise between a
more desirable and a less desirable grill.
pˆ
x 85
.685
n 124
To determine if the proportion of students choosing the three-grill display so that Grill #2 is a compromise
between a more desirable and a less desirable grill is greater than .167, we test:
H 0 : p .167
H a : p .167
The test statistic is z
pˆ po
po qo
n
.685 .167
.167(.833)
124
15.47
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 15.47 1.645) , H0 is rejected.
There is sufficient evidence to indicate that the true proportion of students choosing the three-grill display
so that Grill #2 is a compromise between a more desirable and a less desirable grill is greater than .167 at
.05 .
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Inferences Based on a Single Sample: Tests of Hypothesis
7.81
375
To minimize the probability of a Type I error, we will select .01 .
First, check to see if the normal approximation is adequate:
np0 100(.5) 50
nq0 100(.5) 50
Since both np0 .15 and nq0 .15 , the normal distribution will be adequate
pˆ
x 56
.56
n 100
To determine if more than half of all Diet Coke drinkers prefer Diet Pepsi, we test:
H 0 : p .5
H a : p .5
The test statistic is z
pˆ p 0
p 0q 0
n
.56 .5
.5(.5)
100
1.20
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix D,
z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.20 2.33) , H0 is not
rejected. There is insufficient evidence to indicate that more than half of all Diet Coke drinkers prefer Diet
Pepsi at .01 .
Since H0 was not rejected, there is no evidence that Diet Coke drinkers prefer Diet Pepsi.
7.82
7.83
Using Table IV, Appendix D:
a.
For n 12 , df n 1 12 1 11 , P ( 2 02 ) .10 02 17.2750
b.
For n 9 , df n 1 9 1 8 , P ( 2 02 ) .05 02 15.5073
c.
For n 5 , df n 1 5 1 4 , P ( 2 02 ) .025 02 11.1433
a.
df n 1 16 1 15 ; reject H0 if 2 6.26214 or 2 27.4884
b.
df n 1 23 1 22 ; reject H0 if 2 40.2894
c.
df n 1 15 1 14 ; reject H0 if 2 21.0642
d.
df n 1 13 1 12 ; reject H0 if 2 3.57056
e.
df n 1 7 1 6 ; reject H0 if 2 1.63539 or 2 12.5916
f.
df n 1 25 1 24 ; reject H0 if 2 13.8484
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7.84
Chapter 7
a.
b.
It would be necessary to assume that the population has a normal distribution.
H0 : 2 1
Ha : 2 1
The test statistic is 2
(n 1) s 2
02
(7 1)(4.84)
29.04
1
The rejection region requires .05 in the upper tail of the 2 distribution with df n 1 7 1 6 .
2
12.5916 . The rejection region is 2 12.5916 .
From Table IV, Appendix D, .05
Since the observed value of the test statistic falls in the rejection region ( 2 29.04 12.5916) , H0 is
rejected. There is sufficient evidence to indicate that the variance is greater than 1 at .05 .
c.
H0 : 2 1
Ha : 2 1
The test statistic is 2
(n 1) s 2
2
0
(7 1)(4.84)
29.04
1
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the 2 distribution with
2
2
df n 1 7 1 6 . From Table IV, Appendix D, .975
1.237347 and .025
14.4494 . The
rejection region is 2 1.237347 or 2 14.4494 .
Since the observed value of the test statistic falls in the rejection region ( 2 29.04 14.4494) , H0 is
rejected. There is sufficient evidence to indicate that the variance is not equal to 1 at .05 .
7.85
a.
H0 : 2 1
Ha : 2 1
The test statistic is 2
(n 1) s 2
2
0
(100 1)4.84
479.16
1
The rejection region requires .05 in the upper tail of the 2 distribution with
2
124.342 . The rejection region is
df n 1 100 1 99 . From Table IV, Appendix D, .05
2 124.342 .
Since the observed value of the test statistic falls in the rejection region ( 2 479.16 124.342) , H0
is rejected. There is sufficient evidence to indicate the variance is larger than 1 at .05 .
b.
In part b of Exercise 7.84, the test statistic was 2 29.04 . The conclusion was to reject H0 as it was
in this problem.
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Inferences Based on a Single Sample: Tests of Hypothesis
7.86
377
Some preliminary calculations are:
s2
x
x
2
2
n 1
n
30 2
7 7.9048
7 1
176
To determine if 2 1 , we test:
H0 : 2 1
Ha : 2 1
The test statistic is 2
(n 1) s 2
2
0
(7 1)7.9048
47.43
1
The rejection region requires .05 in the lower tail of the 2 distribution with df n 1 7 1 6 . From
2
1.63539 . The rejection region is 2 1.63539 .
Table IV, Appendix D, .95
Since the observed value of the test statistic does not fall in the rejection region ( 2 47.43 1.63539) , H0
is not rejected. There is insufficient evidence to indicate the variance is less than 1 at .05 .
7.87
a.
To determine whether the population of institutional investors perform consistently, we test:
H 0 : 2 102 100
H a : 2 100
b.
The rejection region requires .05 in the lower tail of the 2 distribution with
df n 1 200 1 199 . Using MINITAB, we get:
Inverse Cumulative Distribution Function
Chi-Square with 199 DF
P( X <= x )
0.05
x
167.361
The rejection region is 2 167.361 .
c.
For this problem, .05 . The probability of concluding the standard deviation is less than 10 when,
in fact, it is equal to 10 is .05. If this test was repeated a large number of times, approximately 5% of
the time we would conclude the standard deviation was less than 10 when it really was 10.
d.
From the printout, 2 154.81 and the p-value is p .009 .
e.
Since the p-value is less than ( p .009 .05) , H0 is rejected. There is sufficient evidence to
indicate the standard deviation is less than 10% at .05 .
f.
We must assume that a random sample was selected from the target population and the population
sampled from is approximately normal.
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7.88
Chapter 7
a.
To determine if the variance of the population of trap spacing measurements is larger than 10, we test:
H 0 : 2 10
H a : 2 10
b.
Using MINITAB, the results are:
Descriptive Statistics: Spacing
Variable N
Spacing 7
Mean
89.86
StDev
11.63
Variance
135.14
Minimum
70.00
Q1 Median
Q3 Maximum
82.00 93.00 99.00 105.00
The sample variance is s 2 135.14 .
7.89
c.
The value of s2 is a variable. The next time a random sample is selected, the value of s2 could be
much greater or much smaller. We need to find out how unusual it is to obtain a value of s2 of 135.14
if 2 10 .
d.
The test statistic is 2
e.
Using MINITAB, the p-value is p P( 2 81.084) 0 .
f.
Since the p-value is so small, H0 is rejected for any reasonable value of . There is sufficient
evidence to indicate the true population variance is greater than 10.
g.
We must assume that a random sample was selected from the target population and the population
sampled from is approximately normal.
a.
Let 2 weight variance of tees. To determine if the weight variance differs from .000004
(injection mold process is out-of-control), we test:
(n 1) s 2
2
0
(7 1)135.14
81.084
10
H 0 : 2 .000004
H a : 2 .000004
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tees
Variable N
Mean
Tees
40 0.25248
The test statistic is 2
Median
0.25300
(n 1) s 2
02
StDev
0.00223
Minimum
0.24700
Maximum
Q1
Q3
0.25600 0.25100 0.25400
(40 1)(.00223)2
48.49
.000004
The rejection region requires / 2 .01/ 2 .005 in each tail of the 2 distribution with
2
2
66.7659 and .995
20.7065 . The
df n 1 40 1 39 . From Table IV, Appendix D, .005
rejection region is 2 66.7659 or 2 20.7065 .
Since the observed value of the test statistic does not fall in the rejection region
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Inferences Based on a Single Sample: Tests of Hypothesis
379
( 2 49.49 66.7659 and 2 49.49 20.7065) , H0 is not rejected. There is insufficient evidence
to indicate the injection mold process is out-of-control at .01 .
c.
We must assume that the distributions of the weights of tees is approximately normal. Using
MINITAB, a histogram of the data is:
Histogram of Tees
Normal
12
Mean
StDev
N
0.2525
0.002230
40
10
Frequency
8
6
4
2
0
0.248
0.250
0.252
Tees
0.254
0.256
The data look fairly mound-shaped, so the assumption of normality seems to be reasonably satisfied.
7.90
a.
To determine if the breaking strength variance of the new adhesive is less than the variance of the
standard composite adhesive, 2 .25 , we test:
H 0 : 2 .25
H a : 2 .25
b.
The rejection region requires .01 in the lower tail of the 2 distribution with df n 1 10 1 9 .
2
2.087912 . The rejection region is 2 2.087912 .
From Table IV, Appendix D, .99
7.91
(n 1) s 2
(10 1).462
7.6176 .
.25
c.
The test statistic is 2
b.
Since the observed value of the test statistic does not fall in the rejection region
( 2 7.6176 2.087912) , H0 is not rejected. There is insufficient evidence to indicate the breaking
strength variance of the new adhesive is less than the variance of the standard composite adhesive,
2 .25 at .01 .
e.
We must assume that the distribution of the breaking strengths is approximately normal and that a
random sample was selected from this population.
o2
To determine whether the true conduction time standard deviation is less than 7 seconds (variance less than
49), we test:
H 0 : 2 72
H a : 2 72
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Chapter 7
The test statistic is 2
(n 1) s 2
02
(18 1)6.32
13.77 .
72
The rejection region requires .01 in the lower tail of the 2 distribution with df n 1 18 1 17 .
2
6.40776 . The rejection region is 2 6.40776 .
From Table IV, Appendix D, .99
Since the observed value of the test statistic does not fall in the rejection region ( 2 13.77 6.40776) , H0
is not rejected. There is insufficient evidence to indicate the true conduction time standard deviation is less
than 7 seconds at .01 . Thus, the prototype system does not satisfy this requirement.
7.92
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Drug
Variable
Drug
N
50
Mean
89.291
StDev
3.183
Variance
10.134
Minimum
81.790
Median
89.375
Maximum
94.830
To determine whether new method of determining drug concentration is less variable than the standard
method, we test:
H0 : 2 9
Ha : 2 9
The test statistic is 2
(n 1) s 2
02
(50 1)10.134
55.174 .
9
The rejection region requires .01 in the lower tail of the 2 distribution with df n 1 50 1 49 .
2
29.7067 . The rejection region is 2 29.7067 .
From Table IV, Appendix D, .99
Since the observed value of the test statistic does not fall in the rejection region ( 2 55.174 29.7067) ,
H0 is not rejected. There is insufficient evidence to indicate the new method of determining drug
concentration is less variable than the standard method at .01 .
7.93
To determine if the diameters of the ball bearings are more variable when produced by the new process,
test:
H 0 : 2 .00156
H a : 2 .00156
The test statistic is 2
(n 1) s 2
02
99(.00211)
133.90
.00156
The rejection region requires use of the upper tail of the 2 distribution with df n 1 100 1 99 .
We will use df 100 99 due to the limitations of the table. From Table IV, Appendix D,
2
2
.025
129.561 133.90 135.807 .010
. The p-value of the test is between .010 and .025. The decision
made depends on the desired . For .010 , there is not enough evidence to show that the variance in
the diameters is greater than .00156. For .025 , there is enough evidence to show that the variance in
the diameters is greater than .00156.
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Inferences Based on a Single Sample: Tests of Hypothesis
7.94
381
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: GASTURBINE
Variable
N Mean
GASTURBINE 67 11066
StDev
1595
Minimum
8714
Q1
9918
Median
10656
Q3
11842
Maximum
16243
To determine if the heat rates of the augmented gas turbine engine are more variable than the heat rates of
the standard gas turbine engine, we test:
H 0 : 2 1,5002
H a : 2 1,5002
The test statistic is 2
(n 1) s 2
o2
(67 1)1,5952
74.625 .
1,5002
The rejection region requires .05 in the upper tail of the 2 distribution with df n 1 67 1 66 .
Using MINITAB,
Inverse Cumulative Distribution Function
Chi-Square with 66 DF
P( X <= x )
0.95
x
85.9649
The rejection region is 2 85.9649 .
Since the observed value of the test statistic does not fall in the rejection region ( 2 74.625 85.9649) ,
H0 is not rejected. There is insufficient evidence to indicate the heat rates of the augmented gas turbine
engine are more variable than the heat rates of the standard gas turbine engine at .05 .
7.95
a.
Since the sample mean of 3.85 is not that far from the value of 3.5, a large standard deviation would
indicate that the value 3.85 is not very many standard deviations from 3.5.
b.
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
The test statistic is z
x o
x
3.85 3.5
137
To reject H0, z 2.33 . Thus, we need to find so z 2.33 .
z
3.85 3.5
137
2.33 3.85 3.5 2.33
137
.35 .199065 1.758
Thus, the largest value of for which we will reject H0 is 1.758.
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Chapter 7
c.
To determine if 1.758 , we test:
H 0 : 2 1.7582
H a : 2 1.7582
The test statistic is 2
(n 1) s 2
2
o
(137 1)1.52
99.011 .
1.7582
The rejection region requires .01 in the lower tail of the 2 distribution with
df n 1 137 1 136 . Since there are no values in the table with df 100 , we will use MINITAB
to compute the p-value of the test statistic.
Cumulative Distribution Function
Chi-Square with 136 DF
x
99.011
P( X <= x )
0.0072496
Since the p-value is less than ( p 0.0072496 .01) , H0 is rejected. There is sufficient
evidence to indicate the standard deviation of the scores is less than 1.758 at .01 .
7.96
a.
The power of a test increases when:
1.
2.
3.
7.97
The distance between the null and alternative values of increases.
The value of increases.
The sample size increases.
b.
The power of a test is equal to 1 . As increases, the power decreases.
a.
By the Central Limit Theorem, the sampling distribution of x is
approximately normal with
100
x 500 and x
20 .
25
n
b.
x0 0 z x 0 z
where z z 1.645 from Table II, Appendix D.
.05
n
Thus, x0 500 1.645(20) 532.9
c.
The sampling distribution of x is approximately normal by the
Central Limit Theorem with x 550 and
x =
n
100
25
20 .
d.
P ( x0 532.9) when 550) P z
e.
Power 1 1 .1949 .8051
532.9 550
P ( z .86) .5 .3051 .1949
100 / 25
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Inferences Based on a Single Sample: Tests of Hypothesis
7.98
383
From Exercise 7.97 we want to test H 0 : 500 against H a : 500 using .05 , 100 , n 25 , and
x 532.9 .
7.99
532.9 575
P( z 2.11) .5 .4826 .0174
100 / 25
(Using Table II, Appendix D)
a.
P( x0 532.9 when 575) P z
b.
Power 1 1 .0174 .9826
c.
In Exercise 7.97, .1949 and power .8051 . The value of has decreased in this exercise since
575 is further from the hypothesized value than 550 . As a result, the power of the test in
this exercise has increased (when decreases, the power of the test increases).
a.
The sampling distribution of x will be approximately normal (by the Central Limit Theorem) with
15
x 75 and x
2.143 .
49
n
b.
The sampling distribution of x will be approximately normal (by the Central Limit Theorem) with
15
x 70 and x
2.143 .
49
n
c.
First, find x0 0 z x 0 z
Thus, x0 75 1.28
15
49
n
where z.10 1.28 from Table II, Appendix D.
72.257
72.257 70
Now, find P( x0 72.257 when 70) P z
P z 1.05 .5 .3531 .1469
15 / 49
7.100
d.
Power 1 1 .1469 .8531
a.
From Exercise 7.99, we want to test H 0 : 75 against H a : 75 using .10 , 15 , n 49 ,
and x 72.257 . Using Table II, Appendix D:
If μ = 74 ,
72.257 74
P z .81 .5 .2910 .7910
15 / 49
72.257 72
P z .12 .5 .0478 .4522
15 / 49
P( x0 72.257 when 74) P z
If μ = 72 ,
P( x0 72.257 when 72) P z
If μ = 70 ,
P ( x0 72.257 when 70) .1469
(Refer to Exercise 7.99, part c.)
Copyright © 2014 Pearson Education, Inc.
Chapter 7
If μ = 68 ,
72.257 68
P z 1.99 .5 .4767 .0233
15 / 49
72.257 66
P z 2.92 .5 .4982 .0018
15 / 49
P( x0 72.257 when 68) P z
If μ = 66 ,
P( x0 72.257 when 66) P z
In summary,
74
.7910
b.
72
.4522
70
.1469
68
.0233
66
.0018
Using MINITAB, the graph is:
Scatterplot of beta vs mu
0.8
0.7
0.6
0.5
beta
384
0.4
0.3
0.2
0.1
0.0
65
66
67
68
69
70
mu
71
72
73
74
c.
Looking at the graph, is approximately .62 when 73 .
d.
Power 1
Therefore,
Power
74
.7910
.2090
72
.4522
.5478
70
.1469
.8531
68
.0233
.9767
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66
.0018
.9982
Inferences Based on a Single Sample: Tests of Hypothesis
385
Scatterplot of Power vs mu
1.1
1.0
0.9
Power
0.8
0.7
0.6
0.5
0.4
0.3
0.2
65
66
67
68
69
70
mu
71
72
73
74
The power curve starts out close to 1 when 66 and decreases as increases, while the curve is
close to 0 when 66 and increases as increases.
7.101
e.
As the distance between the true mean and the null hypothesized mean 0 increases, decreases
and the power increases. We can also see that as increases, the power decreases.
a.
First, the sample is sufficiently large if both np0 15 and nq0 15 .
np0 100(.7) 70 and nq0 100(1 .7) 30 .
Since both np0 15 and nq0 15, the normal distribution will be adequate.
Thus, the sampling distribution of p̂ will be approximately normal with E ( pˆ ) p .7 and
pˆ
b.
The sampling distribution of p̂ will be approximately normal with E ( pˆ ) p .65 and
pˆ
c.
p0 q0
.7(.3)
.0458 .
100
n
p0 q0
.65(.35)
.0477 .
100
n
First, find pˆ 0, L p0 z / 2 pˆ p0 z / 2
Thus, pˆ 0, L .7 1.96
p0 q0
n
where z.05/ 2 z.025 1.96 from Table II, Appendix D.
.7(.3)
.610
100
pˆ 0,U p0 z / 2 pˆ p0 z / 2
p0 q0
.7(.3)
.7 1.96
.790
n
100
Now, find P (.610 pˆ .79 when p .65)
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Chapter 7
.610 .65
.79 .65
P
z
P (.84 z 2.94) .2995 .4984 .7979
.65(.35)
.65(.35)
100
100
7.102
d.
P(.610 pˆ .79 when p .71)
a.
To determine if the mean size of California homes exceeds the national average, we test:
.610 .71
.79
.71
P (2.20 z 1.76) .4861 .4608 .9469
P
z
.71(.29)
.71(.29)
100
100
H 0 : 2,390
H a : 2,390
The test statistic is z
x 0
x
2,507 2,390
257 / 100
4.55
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection region z 4.55 2.33 , H0 is
rejected. There is sufficient evidence to indicate the mean size of California homes exceeds the
national average at .01 .
b.
To compute the power, we must first set up the rejection regions in terms of x .
s
257
x0 0 z x 0 2.33
2,390 2.33
2, 449.88
n
100
We would reject H0 if x 2, 449.88
The power of the test when 2, 490 would be:
x a
2, 449.88 2, 490
Power P( x 2, 449.88 | 2, 490) P z 0
P z
x
257 / 100
P( z 1.56) .5 .4406 .9406
c.
The power of the test when 2, 440 would be:
x a
2, 449.88 2, 440
Power P( x 2, 449.88 | 2, 440) P z 0
P z
x
257 / 100
P( z 0.38) .5 .1480 .3520
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Inferences Based on a Single Sample: Tests of Hypothesis
7.103
a.
We have failed to reject H0 when it is not true. This is a Type II error.
To compute , first find:
x0 0 z x 0 z
Thus, x0 5.0 1.645
where z 1.645 from Table II, Appendix D.
.05
n
.01
100
4.998355
Then find:
P( x0 4.998355 when 4.9975) P z
P z .86 .5 .3051 .1949
7.104.
4.998355 4.9975
.01/ 100
b.
We have rejected H0 when it is true. This is a Type I error. The probability of a Type I error is
.05 .
c.
A departure of .0025 below 5.0 is 4.9975 . Using a, .1949 when 4.9975 . The power of
the test is 1 1 .1949 .8051 .
To compute the power, we must first set up the rejection region in terms of p̂ . The rejection region
requires .10 in the lower tail of the z-distribution. From Table II, Appendix D, z.10 1.28 . The
rejection region is z 1.28 .
Thus, pˆ 0, L p0 z / 2 pˆ p0 z / 2
p0 q0
.8(.2)
.8 1.28
.8 .023 .777 .
501
n
pˆ p0
.777 .82
P z
P( z 2.51) .5 .4940 .0060
Power P( pˆ .777 | p0 .82) P z
pˆ
.82(.18)
501
7.105.
387
To compute the power, we must first set up the rejection region in terms of p̂ . The rejection region
requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II, Appendix D, z.005 2.575 .
The rejection region is z 2.575 or z 2.575 .
Thus, pˆ 0, L p0 z / 2 pˆ p0 z / 2
pˆ 0,U p0 z / 2 pˆ p0 z / 2
p0 q0
.5(.5)
.5 2.575
.5 .117 .383 and
121
n
p0 q0
.5(.5)
.5 2.575
.5 .117 .617 .
121
n
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Chapter 7
pˆ p0
pˆ p0
Power P( pˆ .383 or pˆ .617 | p0 .65) P z
P z
pˆ
pˆ
.383 .65
.617 .65
P z
P z
P( z 6.16) P( z .76) (.5 .5) (.5 .2764) .7764
.65(.35)
.65(.35)
121
121
7.106
a.
To determine if the mean mpg for 2011 Honda Civic autos is greater than 36 mpg, we test:
H 0 : 36
H a : 36
b.
The test statistic is z
x 0
x
38.3 36
6.4 / 50
2.54
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region z 2.54 1.645 , H0 is
rejected. There is sufficient evidence to indicate that the mean mpg for 2011 Honda Civic autos is
greater than 36 mpg at .05 .
We must assume that the sample was a random sample.
c.
First find:
x0 0 z x 0 z
Thus, x0 36 1.645
6.4
50
n
where z 1.645 from Table II, Appendix D.
37.49
For 36.5 :
37.49 36.5
Power P( x 37.49 | 36.5) P z
P z 1.09 .5 .3621 .1379
6.4 / 50
For 37 :
37.49 37
Power P( x 37.49 | 37) P z
P z .54 .5 .2054 .2946
6.4 / 50
For 37.5 :
37.49 37.5
Power P( x 37.49 | 37.5) P z
P z .01 .5 .0040 .5040
6.4 / 50
For 38 :
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Inferences Based on a Single Sample: Tests of Hypothesis
37.49 38
Power P( x 37.49 | 38) P z
P z .56 .5 .2123 .7123
6.4 / 50
For 38.5 :
37.49 38.5
Power P( x 37.49 | 38.5) P z
P z 1.12 .5 .3686 .8686
6.4 / 50
d. Using MINITAB, the plot is:
Scatterplot of Power vs Mu
0.9
0.8
0.7
Power
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
36.5
e.
37.0
37.5
µ
38.0
38.5
From the plot, the power is approximately .60.
For 37.75 :
37.49 37.75
Power P ( x 37.49 | 37.75) P z
P ( z .29) .5 .1141 .6141
6.4 50
f.
From the plot, the power is approximately 1.
For 41:
37.49 41
Power P ( x 37.49 | 41) P z
P ( z 3.88) .5 .5 1.0
6.4 50
If the true value of is 41, the approximate probability that the test will fail to reject H0 is
1 1 0 .
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390
7.107
Chapter 7
First, find x0 such that P( x x0 ) .05 .
x 10
P( x x0 ) P z 0
P z z0 .05 .
1.2 / 48
From Table II, Appendix D, z0 1.645 .
Thus, z0
x0 10
1.2 / 48
x0 1.645(.173) 10 9.715
The probability of a Type II error is:
9.715 9.5
P( x 9.715 | 9.5) P z
P( z 1.24) .5 .3925 .1075
1.2 / 48
7.108
For a large sample test of hypothesis about a population mean, no assumptions are necessary because the
Central Limit Theorem assures that the test statistic will be approximately normally distributed. For a
small sample test of hypothesis about a population mean, we must assume that the population being
sampled from is normal. The test statistic for the large sample test is the z statistic, and the test statistic for
the small sample test is the t statistic.
7.109
The smaller the p-value associated with a test of hypothesis, the stronger the support for the alternative
hypothesis. The p-value is the probability of observing your test statistic or anything more unusual, given
the null hypothesis is true. If this value is small, it would be very unusual to observe this test statistic if the
null hypothesis were true. Thus, it would indicate the alternative hypothesis is true.
7.110
The elements of the test of hypothesis that should be specified prior to analyzing the data are: null
hypothesis, alternative hypothesis, and rejection region based on .
7.111
There is not a direct relationship between and . That is, if is known, it does not mean is known
because depends on the value of the parameter in the alternative hypothesis and the sample size.
However, as decreases, increases for a fixed value of the parameter and a fixed sample size. Thus, if
is very small, will tend to be large.
7.112
P(Type I error) = P(rejecting H0 when it is true). Thus, if rejection of H0 would cause your firm to go
out of business, you would want this probability or to be small.
7.113
a.
H 0 : 80
H a : 80
The test statistic is t
x 0
s/ n
72.6 80
19.4 / 20
7.51
The rejection region requires .05 in the lower tail of the t-distribution with df n 1 20 1 19 .
From Table III, Appendix D, t.05 1.729 . The rejection region is t 1.729 .
Since the observed value of the test statistic falls in the rejection region (t 7.51 1.729) , H0 is
rejected. There is sufficient evidence to indicate that the mean is less than 80 at .05 .
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Inferences Based on a Single Sample: Tests of Hypothesis
b.
391
H 0 : 80
H a : 80
x 0
The test statistic is t
s/ n
72.6 80
19.4 / 20
7.51
The rejection region requires / 2 .01/ 2 .005 in each tail of the t-distribution with
df n 1 20 1 19 . From Table III, Appendix D, t.005 2.861 . The rejection region is
t 2.861 or t 2.861 .
Since the observed value of the test statistic falls in the rejection region (t 7.51 2.861) , H0 is
rejected. There is sufficient evidence to indicate that the mean is different from 80 at .01 .
7.114
a.
H 0 : 8.3
H a : 8.3
The test statistic is z
x 0
x
8.2 8.3
.79 / 175
1.67
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.67 1.96) ,
H0 is not rejected. There is insufficient evidence to indicate that the mean is different from 8.3 at
.05 .
b.
H 0 : 8.4
H a : 8.4
The test statistic is z
x 0
x
8.2 8.4
.79 / 175
3.35
The rejection region is the same as part b, z 1.96 or z 1.96 .
Since the observed value of the test statistic falls in the rejection region ( z 3.35 1.96) , H0 is
rejected. There is sufficient evidence to indicate that the mean is different from 8.4 at .05 .
c.
H0 : 1
Ha : 1
or
H0 : 2 1
Ha : 2 1
The test statistic is 2
(n 1) s 2
02
(175 1)(.79) 2
108.59
1
The rejection region requires / 2 .05 / 2 .025 in each tail of the 2 distribution with
df n 1 175 1 174 . Since there are no values in the table with df 100 , we will use MINITAB
to find the critical values.
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Chapter 7
Inverse Cumulative Distribution Function
Chi-Square with 174 DF
P( X <= x )
0.025
x
139.367
Inverse Cumulative Distribution Function
Chi-Square with 174 DF
P( X <= x )
0.975
x
212.419
2
2
.025
212.419 and .975
139.367 . The rejection region is 2 212.419 or 2 139.367 .
Since the observed value of the test statistic falls in the rejection region ( 2 108.59 139.367) , H0
is rejected. There is sufficient evidence to indicate the standard deviation differs from 1 at .05 .
d.
In part a, the rejection region is z 1.96 or z 1.96 . In terms of x , the rejection region would be:
z
z
x 0
x
x 0
x
1.96
xU 8.3
.79
1.96
175
.117 xU 8.3 xU 8.417
xL 8.3
.79
175
.117 xL 8.3 xL 8.183
Based on x , the rejection region would be: Reject H0 if x 8.183 or x 8.417 .
The power of the test is the probability the test statistic falls in the rejection region, given the
alternative hypothesis is true. In this case, we will let a 8.5 .
Power P ( x 8.183 | a 8.5) P ( x 8.417 | a 8.5)
7.115
a.
8.183 8.5
8.417 8.5
P z
P z
.79 175
.79 175
P ( z 5.31) P ( z 1.39) (.5 .5) (.5 .4177) .9177
(Using Table II, Appendix D)
H 0 : p .35
H a : p .35
The test statistic is z
pˆ p 0
p 0q 0
n
.29 .35
.35(.65)
200
1.78
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 1.78 1.645) , H0 is
rejected. There is sufficient evidence to indicate p .35 at .05 .
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Inferences Based on a Single Sample: Tests of Hypothesis
b.
393
H 0 : p .35
H a : p .35
The test statistic is z 1.78 (from a).
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.78 1.96) ,
H0 is not rejected. There is insufficient evidence to indicate p is different from .35 at .05 .
7.116
a.
The p-value p .1288 P(t 1.174) . Since the p-value is not very small, there is no evidence to
reject H0 for .10 . There is no evidence to indicate the mean is greater than 10.
b.
We must assume that a random sample was selected from a population that is normally distributed.
c.
For the alternative hypothesis H a : 10 , the p-value is 2 times the p-value for the one-tailed test.
The p-value p 2(.1288) .2576 . There is no evidence to reject H0 for .10 . There is no
evidence to indicate the mean is different from 10.
7.117
a.
H 0 : 2 30
H a : 2 30
The test statistic is 2
(n 1) s 2
02
(41 1)(6.9) 2
63.48
30
The rejection region requires .05 in the upper tail of the 2 distribution with
2
df n 1 41 1 40 . From Table IV, Appendix D, .05
55.7585 . The rejection region is
2 55.7585 .
Since the observed value of the test statistic falls in the rejection region ( 2 63.48 55.7585) , H0 is
rejected. There is sufficient evidence to indicate the variance is larger than 30 at .05 .
b.
H 0 : 2 30
H a : 2 30
The test statistic is 2 63.48 (from part a).
The rejection region requires / 2 .05 / 2 .025 in each tail of the 2 distribution with
2
2
df n 1 41 1 40 . From Table IV, Appendix D, .025
59.3417 and .975
24.4331 . The
rejection region is 2 24.4331 or 2 59.3417 .
Since the observed value of the test statistic falls in the rejection region ( 2 63.48 59.3417) , H0 is
rejected. There is sufficient evidence to indicate the variance is not 30 at .05 .
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7.118
7.119
Chapter 7
a.
The rejection region requires .01 in the lower tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
b.
The test statistic is z
c.
Since the observed value of the test statistics does not fall in the rejection region ( z .40 2.33) ,
H0 is not rejected. There is insufficient evidence to indicate the true mean number of latex gloves
used per week by all hospital employees is less than 20 at .01 .
a.
The rejection region requires / 2 .01/ 2 .005 in each tail of the 2 distribution with
x o
x
19.3 20
11.9
46
.40
df n 1 46 1 45 . Using MINITAB,
Inverse Cumulative Distribution Function
Chi-Square with 45 DF
P( X <= x )
0.005
x
24.3110
Inverse Cumulative Distribution Function
Chi-Square with 45 DF
P( X <= x )
0.995
x
73.1661
The rejection region is 2 24.3110 or 2 73.1661 .
7.120
(n 1) s 2
(46 1)11.92
63.7245 .
100
c.
The test statistic is 2
d.
Since the observed value of the test statistic does not fall in the rejection region
( 2 63.7245 24.3110 and 2 63.7245 73.1661) , H0 is not rejected. There is
insufficient evidence to indicate the variance is different from 100 at .01 .
a.
pˆ
b.
o2
x 64
.604
n 106
H 0 : p .70
H a : p .70
pˆ p0
The test statistic is z
d.
The rejection region requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
p0 q0
n
.604 .70
c.
.70(.30)
106
2.16
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Inferences Based on a Single Sample: Tests of Hypothesis
7.121
395
e.
Since the observed value of the test statistic does not fall in the rejection region ( z 2.16 2.58) ,
H0 is not rejected. There is insufficient evidence to indicate the true proportion of consumers who
believe “Made in the USA” means 100% of labor and materials are from the United States is different
from .70 at .01 .
a.
To determine if the average high technology stock is riskier than the market as a whole, we test:
H0 : 1
Ha : 1
b.
The test statistic is t
x 0
s/ n
The rejection region requires .10 in the upper tail of the t-distribution with df n 1 15 1 14 .
From Table III, Appendix D, t.10 1.345 . The rejection region is t 1.345 .
c.
We must assume the population of beta coefficients of technology stocks is normally distributed.
d.
The test statistic is t
x 0
1.23 1
2.41
s / n .37 / 15
Since the observed value of the test statistic falls in the rejection region (t 2.41 1.345) , H0 is
rejected. There is sufficient evidence to indicate the mean high technology stock is riskier than the
market as a whole at .10 .
e.
From Table III, Appendix D, with df n 1 15 1 14 , .01 P (t 2.41) .025 . Thus,
.01 p-value .025 . The probability of observing this test statistic, t 2.41 , or anything more
unusual is between .01 and .025. Since this probability is small, there is evidence to indicate the null
hypothesis is false for .05 .
f.
To determine if the variance of the stock beta values differs from .15, we test:
H 0 : 2 .15
H a : 2 .15
The test statistic is 2
(n 1) s 2
o2
(15 1).372
12.7773 .
.15
The rejection region requires / 2 .05 / 2 .025 in each tail of the 2 distribution with
2
2
5.62872 and .025
26.1190 .
df n 1 15 1 14 . From Table IV, Appendix D, .975
The rejection region is 2 5.62872 or 2 26.1190 .
Since the observed value of the test statistic does not fall in the rejection region
( 2 12.7773 5.62875 and 2 12.7773 26.1190) , H0 is not rejected. There is insufficient
evidence to indicate the variance of the stock beta values differs from .15 at .05 .
7.122
a.
The population parameter of interest is p = proportion of items that had the wrong price scanned at
California Wal-Mart stores.
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Chapter 7
b.
To determine if the true proportion of items scanned at California Wal-Mart stores with the wrong
price exceeds the 2% NIST standard, we test:
H 0 : p .02
H a : p .02
c.
The test statistic is z
pˆ po
po qo
n
.083 .02
.02(.98)
1000
14.23
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
d.
Since the observed value of the test statistic falls in the rejection region ( z 14.23 1.645) , H0 is
rejected. There is sufficient evidence to indicate that the true proportion of items scanned at
California Wal-Mart stores with the wrong price exceeds the 2% NIST standard at .05 . This
means that the proportion of items with wrong prices at California Wal-Mart stores is much higher
than what is allowed.
e.
In order for the inference to be valid, the sampling distribution of p̂ must be approximately normal.
For this assumption to be valid, both np0 15 and nq0 15 .
nq0 1000(.98) 980
np0 1000(.02) 20
Since np0 15 and nq0 15 , we can assume the distribution of p̂ is approximately normal.
7.123
a.
Let p = proportion of time the camera correctly detects liars. The null hypothesis would be:
H 0 : p .75
7.124
b.
A Type I error would be to conclude the camera cannot correctly identify liars 75% of the time when,
in fact, it can. A Type II error would be to conclude the camera can correctly identify liars 75% of the
time when, in fact, it cannot.
a.
A Type I error is rejecting H0 when H0 is true. In this case, we would conclude that the mean number
of carats per diamond is different from .6 when, in fact, it is equal to .6.
A Type II error is accepting H0 when H0 is false. In this case, we would conclude that the mean
number of carats per diamond is equal to .6 when, in fact, it is different from .6.
b.
From Exercise 6.120, the random sample of 30 diamonds yielded x .691 and s .262 . Let
mean number of carats per diamond. To determine if the mean number of carats per diamond is
different from .6, we test:
H 0 : .6
H a : .6
The test statistic is z
x 0
x
.691 .6
.262
30
1.90
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Inferences Based on a Single Sample: Tests of Hypothesis
397
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.90 1.96) , H0
is not rejected. There is insufficient evidence to indicate the mean number of carats per diamond is
different from .6 carats at .05 .
c.
When is changed, H0, Ha, and the test statistic remain the same.
The rejection region requires / 2 .10 / 2 .05 in each tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
Since the observed value of the test statistic falls in the rejection region z 1.90 1.645 H0 is
rejected. There is sufficient evidence to indicate the mean number of carats per diamond is different
from .6 carats at .10 .
d.
7.125
When the value of changes, the decision can also change. Thus, it is very important to include the
level of used in all decisions.
Some preliminary calculations are:
x
x 667.3 95.33
n
s2
7
x
x
2
2
n 1
n
(667.3) 2
7
42.539
7 1
63,867.99
s 42.539 6.5222
a.
To determine if the true mean cost-of-living index for Southeastern cities is different than the mean
national cost-of-living index of 100, we test:
H 0 : 100
H a : 100
b.
Since the sample size is so small, we must assume that the population being sampled is normal. In
addition, we must assume that the sample is random.
c.
The test statistic is t
x 0
s/ n
95.33 100
6.5222 / 7
1.89
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution. From Table III,
Appendix D, with df n 1 7 1 6 , t.025 2.447 . The rejection region is t 2.447 or t 2.447 .
Since the observed value of the test statistic does not fall in the rejection region (t 1.89 2.447) ,
H0 is not rejected. There is insufficient evidence to indicate the true mean cost-of-living index for
Southeastern cities is different than the mean national cost-of-living index of 100 at .05 .
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Chapter 7
d.
The observed significance level is p-value p P(t 1.89) P(t 1.89) . Since we did not reject
H0 in part c, we know that the p-value must be greater than .05. Using MINITAB,
Cumulative Distribution Function
Student's t distribution with 6 DF
x P( X <= x )
-1.89
0.0538261
Thus, the p-value is p 2(.0538261) .1076522 .
7.126
a.
Let p = proportion of shoppers using cents-off coupons. To determine if the proportion of shoppers
using cents-off coupons exceeds .65, we test:
H 0 : p .65
H a : p .65
The test statistic is z
pˆ p0
p0 q0
n
.72 .65
.65(.35)
1, 000
4.64
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 4.64 1.645) , H0 is
rejected. There is sufficient evidence to indicate the proportion of shoppers using cents-off coupons
exceeds .65 at .05 .
b.
The sample size is large enough if the np0 15 and nq0 15 .
np0 1000(.65) 650
nq0 1000(.35) 350
Since both np0 15 and nq0 15, the normal distribution will be adequate.
7.127
c.
The p-value is p P( z 4.64) (.5 .5) .0 . (Using Table II, Appendix D.) Since the p-value is
smaller than .05 , H0 is rejected. There is sufficient evidence to indicate the proportion of
shoppers using cents-off coupons exceeds .65 at .05 .
a.
The hypotheses would be:
H0: Individual does not have the disease
Ha: Individual does have the disease
b.
A Type I error would be: Conclude the individual has the disease when in fact he/she does not. This
would be a false positive test.
A Type II error would be: Conclude the individual does not have the disease when in fact he/she
does. This would be a false negative test.
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Inferences Based on a Single Sample: Tests of Hypothesis
c.
7.128
399
If the disease is serious, either error would be grave. Arguments could be made for either error being
more grave. However, I believe a Type II error would be more grave: Concluding the individual
does not have the disease when he/she does. This person would not receive critical treatment, and
may suffer very serious consequences. Thus, it is more important to minimize .
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tunnel
Variable
Tunnel
N
10
Mean
989.8
Median
970.5
StDev
160.7
Minimum
735.0
Maximum
1260.0
Q1
862.5
Q3
1096.8
To determine whether peak hour pricing succeeded in reducing the average number of vehicles attempting
to use the Lincoln Tunnel during the peak rush hour, we test:
H 0 : 1, 220
H a : 1, 220
The test statistic is t
x 0
s/ n
989.8 1, 220
160.7 / 10
4.53
Since no is given, we will use .05 . The rejection region requires .05 in the lower tail of the
t-distribution with df n 1 10 1 9 . From Table III, Appendix D, t.05 1.833 . The rejection region is
t 1.833.
Since the observed value of the test statistic falls in the rejection region (t 4.53 1.833) , H0 is rejected.
There is sufficient evidence to indicate that peak hour pricing succeeded in reducing the average number of
vehicles attempting to use the Lincoln Tunnel during the peak rush hour at .05 .
7.129
To determine if the true standard deviation of the point-spread errors exceed 15 (variance exceeds 225), we
test:
H 0 : 2 225
H a : 2 225
The test statistic is 2
(n 1) s 2
02
(240 1)13.32
187.896
225
The rejection region requires in the upper tail of the 2 distribution with df n 1 240 1 239 .
The maximum value of df in Table IV is 100. Thus, we cannot find the rejection region using Table IV.
Using a statistical package, the p-value associated with 2 187.896 is p .9938 .
Since the p-value is so large, there is no evidence to reject H0. There is insufficient evidence to indicate
that the true standard deviation of the point-spread errors exceeds 15 for any reasonable value of .
Since the observed standard deviation (13.3) is less than the hypothesized value of the standard deviation
(15) under H0, there is no way H0 will be rejected for any reasonable value of .
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7.130
Chapter 7
a.
To determine if the true mean number of pecks at the blue string is less than 7.5, we test:
H 0 : 7.5
H a : 7.5
The test statistic is z
x 0
x
1.13 7.5
2.21
72
24.46
The rejection region requires .01 in the lower tail of the z-distribution. From Table II,
Appendix D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection region ( z 24.46 2.33) , H0 is
rejected. There is sufficient evidence to indicate the true mean number of pecks at the blue string is
less than 7.5 at .01 .
b.
From Exercise 6.122, the 99% confidence interval is .46, 1.80 . Since the hypothesized value of the
mean ( 7.5) does not fall in the confidence interval, it is not a likely candidate for the true value of
the mean. Thus, you would reject it. This agrees with the conclusion in part a.
7.131
a.
First, check to see if n is large enough:
np0 132(.5) 66
nq0 132(.5) 66
Since both np0 15 and nq0 15 , the normal distribution will be adequate.
To determine if there is evidence to reject the claim that no more than half of all manufacturers are
dissatisfied with their trade promotion spending, we test:
H 0 : p .5
H a : p .5
The test statistic is z
pˆ p 0
p 0q 0
n
.36 .5
.5(.5)
132
3.22
The rejection region requires .02 in the upper tail of the z-distribution. From Table II,
Appendix D, z.02 2.05 . The rejection region is z 2.05 .
Since the observed value of the test statistic does not fall in the rejection region ( z 3.22 2.05) ,
H0 is not rejected. There is insufficient evidence to reject the claim that no more than half of all
manufacturers are dissatisfied with their trade promotion spending at .02 .
b.
The observed significance level is p-value P( z 3.22) .5 .5 1 . Since this p-value is so large,
H0 will not be rejected for any reasonable value of .
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Inferences Based on a Single Sample: Tests of Hypothesis
c.
401
First, we must define the rejection region in terms of p̂ .
pˆ p0 z pˆ .5 2.05
.5(.5)
.589
132
.589 .55
P z .90 .5 .3159 .8159
P ( pˆ .589 | p .55) P z
.55(.45)
132
7.132
a.
pˆ 24 / 40 .6
To determine if the proportion of shoplifters turned over to police is greater than .5, we test:
H 0 : p .5
H a : p .5
The test statistic is z
pˆ p0
p0 q0
n
.6 .5
.5(.5)
40
1.26
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.26 1.645) , H0
is not rejected. There is insufficient evidence to indicate the proportion of shoplifters turned over to
police is greater than .5 at .05 .
b.
To determine if the normal approximation is appropriate, we check:
np0 40(.5) 20
nq0 40(.5) 20
Since both np0 15 and nq0 15 , the normal distribution will be adequate.
c.
The observed significance level of the test is p-value p P( z 1.26) .5 .3962 .1038 .
(Using Table II, Appendix D)
The probability of observing the value of our test statistic or anything more unusual if the true value
of p is .5 is .1038. Since this p-value is so large, there is no evidence to reject H0. There is no
evidence to indicate the true proportion of shoplifters turned over to police is greater than .5.
7.133
d.
Any value of that is greater than the p-value would lead one to reject H0. Thus, for this problem,
we would reject H0 for any value of .1038 .
a.
A Type II error is concluding the percentage of shoplifters turned over to police is 50% when in fact,
the percentage is higher than 50%.
b.
First, calculate the value of p̂ that corresponds to the border between the acceptance region and the
rejection region.
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Chapter 7
c.
P( pˆ po ) P( z zo ) .05.
From Table II, Appendix D, z0 1.645
pˆ 0 po 1.645 pˆ .5 1.645
.5(.5)
.5 .1300 .6300
40
.6300 -.55
P( pˆ .6300 when p .55) P z
P ( z 1.02) .5 .3461 .8461
.55(.45)
40
If n increases, the probability of a Type II error would decrease.
First, calculate the value of p̂0 that corresponds to the border between the acceptance region and the
rejection region.
P( pˆ po ) P( z zo ) .05.
From Table II, Appendix D, z0 1.645
pˆ 0 po 1.645 pˆ .5 1.645
.5(.5)
.5 .082 .582
100
.582
-.55
P ( z 0.64) .5 .2389 .7389
P( pˆ .582 when p .55) P z
.55(.45)
100
7.134
a.
To determine whether the mean profit change for restaurants with frequency programs is greater than
$1,050, we test:
H 0 : 1, 050
H a : 1, 050
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: x
Variable
x
N
12
Mean
2509
The test statistic is t
StDev
2149
x 0
s/ n
Variance
4619332
2509 1050
2149 / 12
Minimum
-2191
Q1
1646
Median
2493
Q3
3426
Maximum
6553
2.35
The rejection region requires .05 in the upper tail of the t-distribution with
df n 1 12 1 11 . From Table III, Appendix D, t.05 1.796 . The rejection region is
t 1.796 .
Since the observed value of the test statistic falls in the rejection region t 2.35 1.796 , H0 is
rejected. There is sufficient evidence to indicate the mean profit change for restaurants with
frequency programs is greater than $1,050 for .05 .
It appears that the frequency program would be profitable for the company if adopted nationwide.
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Inferences Based on a Single Sample: Tests of Hypothesis
7.135
a.
403
To determine if the production process should be halted, we test:
H0 : 3
Ha : 3
where mean amount of vinyl chloride in the air.
The test statistic is z
x 0
x
3.1 3
.5 / 50
1.41
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic does not fall in the rejection region, ( z 1.41 2.33) , H0
is not rejected. There is insufficient evidence to indicate the mean amount of vinyl chloride in the air
is more than 3 parts per million at .01 . Do not halt the manufacturing process.
7.136
b.
As plant manager, I do not want to shut down the plant unnecessarily. Therefore, I want
P(shut down plant when 3 ) to be small.
c.
The p-value is p P( z 1.41) .5 .4207 .0793 . Since the p-value is not less than .01 , H0 is
not rejected.
a.
A Type II error would be concluding the mean amount of vinyl chloride in the air is less than or equal
to 3 parts per million when, in fact, it is more than 3 parts per million.
x
.5
From Exercise 7.135, z 0
x0 z
0 x0 2.33
3 x0 3.165
/ n
n
50
b.
3.165 3.1
P( z .92) .5 .3212 .8212
For 3.1 , P ( x 3.165) P z
.5
50
(from Table II, Appendix D)
c.
Power 1 1 .8212 .1788
d.
3.165 3.2
P ( z .49) .5 .1879 .3121
For 3.2 , P ( x 3.165) P z
.5
50
Power 1 1 .3121 .6879
As the plant's mean vinyl chloride departs further from 3, the power increases.
7.137
a.
No, it increases the risk of falsely rejecting H0, i.e., closing the plant unnecessarily.
b.
First, find x0 such that P( x x0 ) P z z0 .05 .
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Chapter 7
From Table II, Appendix D, z0 1.645
z
x0
/ n
1.645
x0 3
.5 / 50
x0 3.116
Then, compute:
P( x0 3.116 when 3.1) P z
3.116 3.1
P( z .23) .5 .0910 .5910
.5 / 50
Power 1 1 .5910 .4090
7.138
c.
The power of the test increases as increases.
a.
Some preliminary calculations:
x
x 79.93 15.986
n
s2
5
x
x
2
2
n 1
n
1, 277.7627
5 1
79.932
5 .00043
s .00043 .0207
To determine if the mean measurement differs from 16.01, we test:
H 0 : 16.01
H a : 16.01
The test statistic is t
x 0
s/ n
15.986 16.01
.0207 / 5
2.59
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df n 1 5 1 4 . From Table III, Appendix D, t.025 2.776 . The rejection region is
t 2.776 or t 2.776 .
Since the observed value of the test statistic does not fall in the rejection region (t 2.59 2.776) ,
H0 is not rejected. There is insufficient evidence to indicate the true mean measurement differs from
16.01 at .05 .
b.
We must assume that the sample of measurements was randomly selected from a population of
measurements that is normally distributed.
c.
To determine if the standard deviation of the weight measurements is greater than .01, we test:
H 0 : 2 .012
H a : 2 .012
The test statistic is 2
(n 1) s 2
2
o
(5 1).00043
17.2 .
.012
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Inferences Based on a Single Sample: Tests of Hypothesis
405
The rejection region requires .05 in the upper tail of the 2 distribution with
2
9.48773 . The rejection region is
df n 1 5 1 4 . From Table IV, Appendix D, .05
2 9.48773 .
Since the observed value of the test statistic falls in the rejection region ( 2 17.2 9.48773) , H0 is
rejected. There is sufficient evidence to indicate the standard deviation of the weight measurements
is greater than .01 at .05 .
7.139
a.
To determine if the GSR for all scholarship athletes at Division I institutions differs from 60%, we
test:
H 0 : p .60
H a : p .60
x 315
.63
n 500
The point estimate is pˆ
pˆ po
The test statistic is z
po qo
n
.63 .60
.60(.40)
500
1.37
The rejection region requires /2=.01/2=.005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
Since the observed value does not fall in the rejection region ( z 1.37 2.58) , H0 is not rejected.
There is insufficient evidence to conclude that the GSR for all scholarship athletes at Division I
institution differs from 60% at .01 .
b.
To determine if the GSR for all male basketball players at Division I institutions differs from 58%,
we test:
H 0 : p .58
H a : p .58
The point estimate is pˆ
x 84
.42
n 200
pˆ po
The test statistic is z
po qo
n
.42 .58
.58(.42)
200
4.58
The rejection region requires /2=.01/2=.005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
Since the observed value falls in the rejection region ( z 4.58 2.58) , H0 is rejected. There is
sufficient evidence to conclude that the GSR for all male basketball players at Division I institutions
differs from 58% .01 .
7.140
a.
z
x 0
x
10.2 0
31.3 / 50
2.30
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Chapter 7
b.
For this two-sided test, the p-value P( z 2.30) P( z 2.30) (.5 .4893) (.5 .4893) .0214 .
Since this value is so small, there is evidence to reject H0. There is sufficient evidence to indicate the
mean level of feminization is different from 0% for any value of .0214 .
c.
z
x - 0
x
15.0 0
25.1 / 50
4.23
For this two-sided test, the p-value P( z 4.23) P( z 4.23) (.5 .5) (.5 .5) 0 . Since this
value is so small, there is evidence to reject H0. There is sufficient evidence to indicate the mean
level of feminization is different from 0% for any value of 0.0 .
7.141
Let = mean lacunarity measurement for all grassland pixels. To determine if the area sampled is
grassland, we test:
H 0 : 220
H a : 220
The test statistic is z
x o
x
225 220
20 / 100
2.50 .
The rejection region requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
7.142
Since the observed value of the test statistic does not fall in the rejection region ( z 2.50 2.58) ,
H0 is not rejected. There is insufficient evidence to conclude that the area sampled is not grassland at
.01 .
x o 52.3 51
z
1.29
a.
x
7.1 50
The p-value is p P( z 1.29) P( z 1.29) (.5 .4015) (.5 .4015) .1970 .
(Using Table II, Appendix D.)
b.
The p-value is p P( z 1.29) (.5 .4015) .0985 . (Using Table II, Appendix D.)
c.
z
x o
x
52.3 51
10.4
50
0.88
The p-value is p P( z 0.88) P( z 0.88) (.5 .3106) (.5 .3106) .3788 .
(Using Table II, Appendix D.)
d.
In part a, in order to reject H0, would have to be greater than .1970. In part b, in order to reject H0,
would have to be greater than .0985. In part c, in order to reject H0, would have to be greater
than .3788.
e.
For a two-tailed test, / 2 .01/ 2 .005 . From Table II, Appendix D, z.005 2.58 .
z
x o
x
2.58
52.3 51
s
50
2.58
s
50
52.3 51 .3649s 1.3 s 3.56
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Inferences Based on a Single Sample: Tests of Hypothesis
7.143
a.
407
To determine whether the true mean rating for this instructor-related factor exceeds 4, we test:
H0 : 4
Ha : 4
The test statistic is z
x 0
x
4.7 4
1.62 / 40
2.73
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region z 2.73 1.645 , H0 is
rejected. There is sufficient evidence to indicate that the true mean rating for this instructor-related
factor exceeds 4 at .05 .
b.
If the sample size is large enough, one could almost always reject H0. Thus, we might be able to
detect very small differences if the sample size is large enough. This would be statistical
significance. However, even though statistical significance is found, it does not necessarily mean that
there is practical significance. A statistical significance can sometimes be found between the
hypothesized value of a mean and the estimated value of the mean, but, in practice, this difference
would mean nothing. This would be practical significance.
c.
Since the sample size is sufficiently large n 40 , the Central Limit Theorem indicates that the
sampling distribution of x is approximately normal. Also, since the sample size is large, s is a good
estimator of . Thus, the analysis used is appropriate.
7.144
Let p proportion of patients taking the pill who reported an improved condition.
First we check to see if the normal approximation is adequate:
np0 7000(.5) 3500
nq0 7000(.5) 3500
Since both np0 15 and nq0 15 , the normal distribution will be adequate.
To determine if there really is a placebo effect at the clinic, we test:
H 0 : p .5
H a : p .5
The test statistic is z
pˆ p0
p0 q0
n
.7 .5
.5(.5)
7000
33.47
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region z 33.47 1.645 , H0 is rejected.
There is sufficient evidence to indicate that there really is a placebo effect at the clinic at .05 .
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7.145
Chapter 7
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Candy
Variable
Candy
N
5
Mean
22.000
StDev
2.000
Minimum
20.000
Q1
20.500
Median
21.000
Q3
24.000
Maximum
25.000
To give the benefit of the doubt to the students we will use a small value of . (We do not want to reject H0
when it is true to favor the students.) Thus, we will use .001 .
We must also assume that the sample comes from a normal distribution. To determine if the mean number
of candies exceeds 15, we test:
H 0 : 15
H a : 15
The test statistic is z
x o
n
22 15
2
5
7.83
The rejection region requires .001 in the upper tail of the z-distribution. From Table II, Appendix D,
z.001 3.08 . The rejection region is z 3.08 .
Since the observed value of the test statistic falls in the rejection region z 7.83 3.08 , H0 is rejected.
There is sufficient evidence to indicate the mean number of candies exceeds 15 at .001 .
Copyright © 2014 Pearson Education, Inc.
Chapter 8
Inferences Based on Two Samples: Confidence
Intervals and Tests of Hypotheses
8.1
a.
b.
8.2
8.3
1 2 x 1 2
1
2 2 x 2 2
2
1
n1
2
150 2
150 2
n2
900
100
150 6 144, 156
1600
100
150 8 142, 158
12
22
900 1600
100 100
2500
5
100
c.
x x 1 2 150 150 0
d.
( 1 2 ) 2
e.
The variability of the difference between the sample means is greater than the variability of the
individual sample means.
a.
x 1 12
x
b.
x 2 10
x
c.
x x 1 2 12 10 2
x x
d.
Since n1 30 and n2 30 , the sampling distribution of x1 x2 is approximately normal by the
Central Limit Theorem.
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
1
2
12
n1
22
n2
1
n1
2
2
2
12
n1
22
n2
1
2
n1
2
n2
900 1600
0 10 ( 10, 10)
100 100
1
2
( x1 x2 ) z.025
1
(150 150) 2
1
1
x x
n2
4
3
12
n1
64
64
(5, 275 5, 240) 1.96
22
n2
.5
.375
4 2 32
64 64
25
.625
64
150 2 200 2
35 24.5 10.5, 59.5
400
400
We are 95% confident that the difference between the population means is between 10.5 and 59.5.
409
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Chapter 8
b.
The test statistic is z
( x1 x2 ) ( 1 2 )
2
1
n1
2
2
(5, 275 5, 240) 0
2
2
2.8
150 200
400 400
n2
The p-value is p P( z 2.8) P( z 2.8) 2P( z 2.8) 2(.5 .4974) 2 .0026 .0052
Since the p-value is so small, there is evidence to reject H0. There is evidence to indicate the two
population means are different for .0052 .
c.
The p-value would be half of the p-value in part b. The p-value p P( z 2.8) .5 .4974 .0026 .
Since the p-value is so small, there is evidence to reject H0. There is evidence to indicate the mean
for population 1 is larger than the mean for population 2 for .0026 .
d.
The test statistic is z
( x1 x2 ) ( 1 2 )
2
1
n1
2
2
(5, 275 5, 240) 25
n2
2
2
.8
150 200
400 400
The p-value of the test is p P( z .8) P( z .8) 2P( z .8) 2(.5 .2881) 2 .2119 .4238
Since the p-value is so large, there is no evidence to reject H0. There is no evidence to indicate that
the difference in the 2 population means is different from 25 for .10 .
e.
8.4
We must assume that we have two independent random samples.
Assumptions about the two populations:
1.
2.
Both sampled populations have relative frequency distributions that are approximately normal.
The population variances are equal.
Assumptions about the two samples:
The samples are randomly and independently selected from the populations.
8.5
8.6
a.
No. Both populations must be normal.
b.
No. Both population variances must be equal.
c.
No. Both populations must be normal.
d.
Yes.
e.
No. Both populations must be normal.
a.
s 2p
( n1 1) s12 ( n2 1) s22 (25 1)120 (25 1)100 5, 280
110
n1 n2 2
25 25 2
48
b.
s 2p
(20 1)12 (10 1)20 408
14.5714
20 10 2
28
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
c.
s 2p
(6 1).15 (10 1).2 2.55
.1821
6 10 2
14
d.
s 2p
(16 1)3, 000 (17 1)2,500 85, 000
2, 741.9355
16 17 2
31
sp2 falls nearer the variance with the larger sample size.
8.7
Some preliminary calculations are:
x1
x2
x 11.8 2.36
1
5
n1
x 14.4 3.6
2
n2
4
s12
s22
x
x
2
1
2
1
n1
n1 1
x22
x
n2
n2 1
(11.8) 2
5
.733
5 1
30.78
2
2
(14.4) 2
4
.42
4 1
53.1
( n1 1) s12 ( n2 1) s22 (5 1).773 + (4 1).42 4.192
=
=
= .5989
5+42
7
n1 n2 2
a.
sp2
b.
H0 : 1 2 0
Ha : 1 2 0
The test statistic is t
( x1 x2 ) D0
1 1
sp2
n1 n2
(2.36 3.6) 0
1 1
.5989
5 4
1.24
2.39
.5191
The rejection region requires .10 in the lower tail of the t-distribution with
df n1 n2 2 5 4 2 7 . From Table III, Appendix D, t.10 =1.415 . The rejection region is
t 1.415 .
Since the test statistic falls in the rejection region (t 2.39 1.415) , H0 is rejected. There is
sufficient evidence to indicate that 2 1 at .10 .
c.
A small sample confidence interval is needed because n1 5 30 and n2 4 30 .
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n1 n2 2 5 4 2 7 , t.05 =1.895 . The 90% confidence interval for (1 2 ) is:
1 1
1 1
( x1 x2 ) t.05 sp2 (2.36 3.6) 1.895 .5989 1.24 .98 (2.22, 0.26)
5 4
n1 n2
d.
The confidence interval in part c provides more information about ( 1 2 ) than the test of
hypothesis in part b. The test in part b only tells us that 2 is greater than 1 . However, the
confidence interval estimates what the difference is between 1 and 2 .
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Chapter 8
8.8
a.
x x
b.
The sampling distribution of
x1 x2 is approximately normal
by the Central Limit Theorem
since n1 30 and n2 30 .
1
2
12
n1
22
n2
9
16
.25 .5
100 100
x x 1 2 10
1
c.
2
x1 x2 26.6 15.5 11.1
No, it does not appear that x1 x2 11.1 contradicts the null hypothesis H0 : 1 2 10 . The value
11.1 is fairly close to 10.
d.
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
e.
H0 : 1 2 10
Ha : 1 2 10
The test statistic is z
( x1 x2 ) 10
2
1
n1
2
2
(26.6 15.5) 10
2.2
.5
n2
The rejection region is z 1.96 or z 1.96 . (Refer to part d.)
Since the observed value of the test statistic falls in the rejection region ( z 2.2 1.96) , H0 is
rejected. There is sufficient evidence to indicate the difference in the population means is not equal
to 10 at .05 .
f.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The confidence interval is:
9
16
11.1 .98 (10.12, 12.08)
100 100
We are 95% confident that the difference in the two means is between 10.12 and 12.08 .
(26.6 15.5) 1.96
8.9
g.
The confidence interval gives more information.
a.
The p-value p .1150 . Since the p-value is not small, there is no evidence to reject H0 for .10 .
There is insufficient evidence to indicate the two population means differ for .10 .
b.
If the alternative hypothesis had been one-tailed, the p-value would be half of the value for the twotailed test. Here, p-value .1150 / 2 .0575 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
413
There is no evidence to reject H0 for .05 . There is insufficient evidence to indicate the mean for
population 1 is less than the mean for population 2 at .05 .
There is evidence to reject H0 for .0575 . There is sufficient evidence to indicate the mean for
population 1 is less than the mean for population 2 at .0575 .
8.10
Some preliminary calculations:
x1
x2
sp2
a.
x 654 43.6
s12
1
15
n1
x 858 53.625
2
n2
16
s22
x
x
2
1
2
1
n1
n1 1
x22
x
n2
n2 1
6542
15 419.6 29.9714
15 1
14
28934
2
2
8582
16 439.75 29.3167
16 1
15
46450
( n1 1) s12 ( n2 1) s22 (15 1)29.9714 (16 1)29.3167 859.3501
29.6328
15 16 2
29
n1 n2 2
H0 : 2 1 10
Ha : 2 1 10
The test statistic is t
( x2 x1 ) D0
1 1
sp2
n2 n1
(53.625 43.6) 10
1 1
29.6328
16 15
.025
.013
1.9564
The rejection region requires .01 in the upper tail of the t-distribution with
df n1 n2 2 15 16 2 29 . From Table III, Appendix D, t.01 2.462 . The rejection region is
t 2.462 .
Since the test statistic does not fall in the rejection region (t .013 2.462) , H0 is not rejected.
There is insufficient evidence to conclude 2 1 10 at .01 .
b.
For confidence coefficient .98, .02 and / 2 .02 / 2 .01 . From Table III, Appendix D, with
df n1 n2 2 15 16 2 29 , t.01 2.462 . The 98% confidence interval for ( 1 2 ) is:
1 1
1 1
( x2 x1 ) t / 2 sp2 (53.625 43.6) 2.462 29.6328
16 15
n2 n1
10.025 4.817 5.208, 14.842
We are 98% confident that the difference between the mean of population 2 and the mean of
population 1 is between 5.208 and 14.842.
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Chapter 8
8.11
a.
sp
2
( n1 1) s12 ( n2 1) s22 (17 -1)3.4 2 (12 1)4.82
16.237
n1 n2 2
17 12 2
H0 : 1 2 0
Ha : 1 2 0
( x1 x2 ) 0
1 1
sp2
n1 n2
The test statistic is t
=
(5.4 7.9) 0
1
1
16.237 +
17
12
= 1.646
Since no was given, we will use .05 . The rejection region requires / 2 .05 / 2 .025 in each
tail of the t-distribution with df n1 n2 2 17 12 2 27 . From Table III, Appendix D,
t.025 2.052 . The rejection region is t 2.052 or t 2.052 .
Since the observed value of the test statistic does not fall in the rejection region (t 1.646 2.052) ,
H0 is not rejected. There is insufficient evidence to indicate 1 2 is different from 0 at .05 .
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df n1 n2 2 17 12 2 27 , t.025 2.052 . The confidence interval is:
1 1
1 1
( x1 x2 ) t.025 sp2 (5.4 7.9) 2.052 16.237 2.50 3.12 ( 5.62, 0.62)
17 12
n1 n2
8.12
a.
The target parameter is 1 2 difference in mean trap measurements between the Bahia Tortugas
fishing cooperative and the Punta Abreojos fishing cooperative.
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: BT, PA
Variable N
BT
7
PA
8
Mean
89.86
99.63
StDev Variance Minimum
11.63
135.14
70.00
27.38
749.70
66.00
Q1 Median
Q3 Maximum
82.00 93.00 99.00 105.00
76.50 96.00 115.00 153.00
The point estimate is x1 x2 89.86 99.63 9.77 .
c.
Since the sample sizes for both samples are so small, the Central Limit Theorem does not apply. In
addition, the population standard deviations are not known and must be estimated with the sample
standard deviations.
d.
sp2
( n1 1) s12 ( n2 1) s22 (7 1)135.14 (8 1)749.70 6, 058.74
466.0569
782
13
n1 n2 2
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n1 n2 2 7 8 2 13 , t.05 1.771 . The 90% confidence interval for ( 1 2 ) is:
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
415
1 1
1 1
( x1 x2 ) t / 2 sp2 (89.86 99.63) 1.771 466.0569
7 8
n1 n2
9.77 19.787 29.557, 10.017
8.13
e.
Since 0 is not in the 90% confidence interval, there is sufficient evidence to indicate a difference in
the mean trap measurements between the two fishing cooperatives.
f.
We must assume that we have independent random samples from normal populations and that the
population variances are the same.
a.
sp2
( n1 1) s12 ( n2 1) s22 (25 1)10.412 (25 1)7.12 2 3,817.5
79.53125
25 25 2
48
n1 n2 2
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . Using MINITAB with
df n1 n2 2 25 25 2 48 , t.025 2.011 . The 95% confidence interval for ( 1 2 ) is:
1 1
1
1
( x1 x2 ) t / 2 sp2 (25.08 19.38) 2.011 79.53125
25 25
n1 n2
5.7 5.073 .627, 10.773
b.
8.14
Since 0 does not fall in the 95% confidence interval, there is evidence to indicate there is a difference
in the mean response times between the two groups. Since the interval contains only positive
numbers, it indicates that the mean response time for the group of students whose last names begin
with the letters R-Z is shorter than the mean response time for the group of students whose last names
begin with the letters A-I. This supports the researchers’ last name effect theory.
Let 1 the mean test score of students on the SAT reading test in classrooms that used educational
software and 2 the mean test score of students on the SAT reading test in classrooms that did not use the
technology
a.
The parameter of interest is 1 2 .
b.
The null and alternative hypotheses for the test are:
H 0 : 1 2 0
H a : 1 2 0
8.15
c.
Since the p-value of the test is so large ( p 0.62) , we would not reject H0 for any reasonable value
of . There is insufficient evidence to indicate that the mean test score of the students on the SAT
reading test was significantly higher in classrooms using reading software products than in
classrooms that did not use educational software. This agrees with the conclusion of the DOE.
a.
Let 1 mean number of items recalled by those in the video only group and 2 mean number of
items recalled by those in the audio and video group. To determine if the mean number of items
recalled by the two groups is the same, we test:
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
H 0 : 1 2 0
H a : 1 2 0
b.
s 2p
n1 1 s12 n2 1 s22 20 11.982 20 1 2.132
n1 n2 2
The test statistic is t
1 1
s 2p
n1 n2
3.70 3.30 0
4.22865
1
1
4.22865
20
20
0.4
0.62
.65028
c.
The rejection region requires / 2 .10 / 2 .05 in each tail of the t-distribution with
df n1 n2 2 20 20 2 38 . From Table III, Appendix D, t.05 1.684 . The rejection region is
t 1.684 or t 1.684 .
d.
Since the observed value of the test statistic does not fall in the rejection region (t 0.62 1.684) , Ho
is not rejected. There is insufficient evidence to indicate a difference in the mean number of items
recalled by the two groups at .10 .
e.
The p-value is p .542 . This is the probability of observing our test statistic or anything more
unusual if H0 is true. Since the p-value is not less than .10 , there is no evidence to reject H0 .
There is insufficient evidence to indicate a difference in the mean number of items recalled by the
two groups at .10 .
f.
We must assume:
1.
2.
3.
8.16
x1 x2 Do
20 20 2
Both populations are normal
Random and independent samples
12 22
Let 1 mean drug concentration for Site1 and 2 mean drug concentration for Site 2. To determine if
there is a difference in the mean drug concentration between the two Sites, we test:
H 0 : 1 2 0
H a : 1 2 0
From the printout, the test statistic is t .57 and the p-value is p .573 .
Since the p-value is not small, H0 is not rejected. There is insufficient evidence to indicate a difference in
the mean drug concentrations between the two Sites.
8.17
a.
Let 1 mean forecast error of buy-side analysts and 2 mean forecast error of sell-side analysts.
For confidence coefficient 0.95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval is:
( x1 x2 ) z.025
12
n1
22
n2
.85 ( .05) 1.96
1.932
.85 2
.90 .064 (.836, .964)
3, 526 58, 562
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
417
We are 95% confident that the difference in the mean forecast error of buy-side analysts and sell-side
analysts is between .836 and .964.
b.
Based on 95% confidence interval in part a, the buy-side analysts has the greater mean forecast error
because our interval contains positive numbers.
c.
The assumptions about the underlying populations of forecast errors that are necessary for the validity
of the inference are:
1.
2.
8.18
The samples are randomly and independently sampled.
The sample sizes are sufficiently large.
a.
No. Just looking at the sample means, as the students went from no solution to check figures, the
sample mean improvement score increased. However, as the students went from check figures to
complete solutions, the sample mean improvement score dropped to below the no solution group.
b.
The problem with using only the sample means to make inferences about the population mean
knowledge gains for the three groups of students is that we don’t know the variability or the “spread”
of the probability distributions of the populations.
c.
Let 1 mean knowledge gain for students in the “no solutions” group and 2 mean knowledge
gain for students in the “check figures” group. To determine if the test score improvement decreases
as the level of assistance increases, we test:
H 0 : 1 2 0
H a : 1 2 0
d.
Since the observed significance level of the test is not less than .05 ( p .8248 .05) , H0 is not
rejected. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“no solutions” group is greater than the mean knowledge gain of students in the “check figures”
group at .05 .
e.
Let 3 mean knowledge gain of students in the “completed solutions” group. To determine if the
test score improvement decreases as the level of assistance increases, we test:
H 0 : 2 3 0
H a : 2 3 0
f.
Since the observed significance level of the test is not less than .05 ( p .1849 .05) , do not
reject H0. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“check figures” group is greater than the mean knowledge gain of students in the “complete
solutions” group at .05 .
g.
To determine if the test score improvement decreases as the level of assistance increases, we test:
H 0 : 1 3 0
H a : 1 3 0
h.
Since the observed significance level of the test is not less than .05 ( p .2726 .05) , do not
reject H0. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“no solutions” group is greater than the mean knowledge gain of students in the “complete solutions”
group at .05 .
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
8.19
a.
The descriptive statistics are:
Descriptive Statistics: Text-line, Witness-line, Intersection
Variable
N
Text-line
3
Witness-line 6
Intersection 5
Mean
0.3830
0.3042
0.3290
Median
0.3740
0.2955
0.3190
StDev
0.0531
0.1015
0.0443
Minimum
0.3350
0.1880
0.2850
Maximum
Q1
Q3
0.4400 0.3350 0.4400
0.4390 0.2045 0.4075
0.3930 0.2900 0.3730
Let 1 mean zinc measurement for the text-line, 2 mean zinc measurement for the witness-line,
and 3 mean zinc measurement for the intersection.
s 2p
( n1 1) s12 ( n3 1) s32 (3 1).05312 (5 1).04432
.00225
n1 n3 2
352
For .05 , / 2 .05 / 2 .025 . Using Table III, Appendix D, with df n1 n2 2 3 5 2 6 ,
t.025 2.447 . The 95% confidence interval is:
1 1
1 1
( x1 x3 ) t / 2 s 2p (.3830 .3290) 2.447 .00225
3 5
n1 n3
0.0540 .0848 (0.0308, 0.1388)
We are 95% confident that the difference in mean zinc level between text-line and intersection is
between 0.0308 and 0.1388.
To determine if there is a difference in the mean zinc measurement between text-line and intersection,
we test:
H 0 : 1 3 0
H a : 1 3 0
The test statistic is t
( x1 x3 ) Do
1 1
s 2p
n1 n3
(.3830 .3290) 0
1 1
.00225
3 5
1.56
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df n1 n2 2 3 5 2 6 . From Table III, Appendix D, t.025 2.447 . The rejection region is
t 2.447 or t 2.447 .
Since the observed value of the test statistic does not fall in the rejection region (t 1.56 2.447) , H0
is not rejected. There is insufficient evidence to indicate a difference in the mean zinc measurement
between text-line and intersection at .05 .
b.
s 2p
n2 1 s22 n3 1 s32
n2 n3 2
(6 1).10152 (5 1).04432
.006596
652
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
419
For .05 , / 2 .05 / 2 .025 . Using Table III, Appendix D, with df n1 n2 2 6 5 2 9 ,
t.025 2.262 . The 95% confidence interval is:
1 1
1 1
( x2 x3 ) t / 2 s 2p .3042 .3290 2.262 .006596
6 5
n2 n3
.0248 .1112 (.1361, .0864)
We are 95% confident that the difference in mean zinc level between witness-line and intersection is
between 0.1361 and 0.0864.
To determine if the difference in mean zinc measurement between the witness-line and the
intersection, we test:
H 0 : 2 3 0
H a : 2 3 0
The test statistic is t
x2 x3 Do
1 1
s
n2 n3
2
p
.3042 .3290 0
1 1
.006596
6 5
.50
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution. From Table III,
Appendix D, with df n1 n2 2 6 5 2 9 , t.025 2.262 . The rejection region is t 2.262 or
t 2.262 .
Since the observed value of the test statistic does not fall in the rejection region (t .50 2.262) ,
H0 is not rejected. There is insufficient evidence to indicate a difference in mean zinc measurement
between witness-line and intersection at
.05 .
8.20
c.
If we order the sample means, the largest is Text-line, the next largest is intersection and the smallest
is witness-line. In parts a and b, we found that text-line is not different from the intersection and that
the witness-line is not different from the intersection. However, we cannot make any decisions about
the difference between the witness-line and the text-line.
d.
In order for the above inferences to be valid, we must assume:
1.
The three samples are randomly selected in an independent manner from the three target
populations.
2.
All three sampled populations have distributions that are approximately normal.
3.
All three population variances are equal (i.e. 12 22 32 )
Let 1 mean amount of surplus Missouri producers are willing to sell to the biomass market and 2
mean amount of surplus Illinois producers are willing to sell to the biomass market. To determine if there
is a difference in the mean amount of surplus producers are willing to sell to the biomass market between
Missouri and Illinois producers, we test:
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
( x1 x2 ) ( 1 2 )
2
1
n1
2
2
(21.5 22.2) 0
2
2
.31 .
33.4 34.9
431
508
n2
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II, Appendix
D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z .31 1.96) , H0 is
not rejected. There is insufficient evidence to indicate a difference in the mean amount of surplus
producers are willing to sell to the biomass market between Missouri and Illinois producers at .05 .
8.21.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Control, Rude
Variable
Rude
Control
N
45
53
Mean
8.511
11.81
StDev
3.992
7.38
Minimum
0.000
0.00
Q1
5.500
5.50
Median
9.000
12.00
Q3
11.000
17.50
Maximum
18.000
30.00
Let 1 mean performance level of students in the rudeness group and 2 mean performance level of
students in the control group. To determine if the true performance level for students in the rudeness
condition is lower than the true mean performance level for students in the control group, we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
( x1 x2 ) 0
2
1
n1
2
2
n2
(8.511 11.81) 0
3.9222 7.382
45
53
2.81
The rejection region requires .01 in the lower tail of the z-distribution. From Table II, Appendix D,
z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection region ( z 2.81 2.33) , H0 is rejected.
There is sufficient evidence to indicate the true mean performance level for students in the rudeness
condition is lower than the true mean performance level for students in the control group at .01 .
8.22
a.
If the manipulation was successful, then the positive group (requiring a strong display of positive
emotions) should have the higher mean response. The members of this group should disagree with
the statement presented, resulting in higher responses.
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Positive, Neutral
Variable
Positive
Neutral
N
78
67
Mean
4.4872
1.8955
StDev
0.6595
0.4965
Minimum
2.0000
1.0000
Q1
4.0000
2.0000
Median
5.0000
2.0000
Copyright © 2014 Pearson Education, Inc.
Q3
5.0000
2.0000
Maximum
5.0000
3.0000
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
421
Let 1 mean response for the positive group and 2 mean response for the neutral group. To
determine if the manipulation was successful, we test:
H 0 : 1 2 0
H a : 1 2 0
( x1 x2 ) ( 1 2 )
The test statistic is z
2
1
n1
2
2
n2
(4.4872 1.8955) 0
2
.6595 .4965
78
67
2
26.94 .
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 26.94 1.645) , H0 is
rejected. There is sufficient evidence to indicate the mean response for the positive group is greater
than the mean response for the neutral group at .05 . Thus, the manipulation was successful.
c.
8.23
We need to assume that random and independent samples were selected from each of the populations.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Honey, DM
Variable
Honey
DM
N
35
33
Mean
10.714
8.333
StDev
2.855
3.256
Minimum
4.000
3.000
Q1
9.000
6.000
Median
11.000
9.000
Q3
12.000
11.500
Maximum
16.000
15.000
Let 1 mean improvement in total cough symptoms score for children receiving the Honey dosage and
2 mean improvement in total cough symptoms score for children receiving the DM dosage. To test if
honey may be a preferable treatment for the cough and sleep difficulty associated with childhood upper
respiratory tract infection, we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
( x1 x2 ) 0
2
1
n1
2
2
n2
(10.714 8.333) 0
2.8552 3.2562
35
33
3.20
Since no was given, we will use .05 . The rejection region requires .05 in the upper tail of the
z-distribution. From Table II, Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 3.20 1.645) , H0 is rejected.
There is sufficient evidence to indicate that honey may be a preferable treatment for the cough and sleep
difficulty associated with childhood upper respiratory tract infection at .05 .
Copyright © 2014 Pearson Education, Inc.
422
Chapter 8
8.24
a.
Let 1 the mean heat rates of traditional augmented gas turbines and 2 the mean heat rates of
aeroderivative augmented gas turbines.
Some preliminary calculations are:
s 2p
n1 1 s12 n2 1 s22 39 112792 7 1 26522
n1 n2 2
39 7 2
2,371,831.409
To determine if there is a difference in the mean heat rates for traditional augmented gas turbines and
the mean heat rates of aeroderivative augmented gas turbines, we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is t
x1 x2 Do
1 1
s 2p
n1 n2
11,544 12,312 0
1 1
2,371,831.409
39 7
768
1.21
632.1782
The rejection region requires / 2 .05/ 2 .025 in each tail of the t-distribution with
df n1 n2 – 2 39 7 – 2 44 . From Table III, Appendix D, t.025 2.021 . The rejection region
is t 2.021 or t 2.021 .
Since the observed value of the test statistic does not fall in the rejection region (t 1.20 2.021) ,
H0 is not rejected. There is insufficient evidence to indicate that there is a difference in the mean heat
rates for traditional augmented gas turbines and the mean heat rates of aeroderivative augmented gas
turbines at .05 .
b.
Let 3 the mean heat rates of advanced augmented gas turbines and 2 the mean heat rates of
aeroderivative augmented gas turbines.
Some preliminary calculations are:
s 2p
n3 1 s32 n2 1 s22 21 1 6392 7 1 26522
n3 n2 2
21 7 2
1,937,117.077
To determine if there is a difference in the mean heat rates for traditional augmented gas turbines and
the mean heat rates of aeroderivative augmented gas turbines, we test:
H 0 : 3 2 0
H a : 3 2 0
The test statistic is
x x Do
9, 764 12,312 0
2,548
4.19
t 3 2
1 1 607.4329
1
2 1
1,937,117.077
sp
21 7
n3 n2
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
423
The rejection region requires / 2 .05/ 2 .025 in each tail of the t-distribution with
df n1 n2 – 2 21 7 – 2 26 . From Table III, Appendix D, t.025 2.056 . The rejection region
is t 2.056 or t 2.056 .
Since the observed value of the test statistic falls in the rejection region (t 4.19 2.056) , H0 is
rejected. There is sufficient evidence to indicate that there is a difference in the mean heat rates for
advanced augmented gas turbines and the mean heat rates of aeroderivative augmented gas turbines at
.05 .
8.25
a.
We cannot provide a measure of reliability because we have no measure of the variability or variance
of the data.
b.
We would need the variances of the two samples.
c.
Let 1 mean age for self-employed immigrants and 2 mean age for the wage-earning
immigrants. To determine if the mean age for self-employed immigrants is less than the mean age for
wage-earning immigrants, we test:
H 0 : 1 2 0
H a : 1 2 0
The rejection region requires .01 in the lower tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
d.
We use the following to solve for :
z
( x1 x2 ) ( 1 2 )
2
1
n1
1.91
e.
8.26
a.
2
2
(44.88 46.79) 0
2 2
870
n2
2.33
84,875
1
1
(2.33) 24.056
870 84,875
The true value of is likely to be smaller than 24.056. This standard deviation would be too large for
the ages of people.
H 0 : 1 2 0
H a : 1 2 0
The rejection region requires .10 in the lower tail of the z-distribution. From Table II, Appendix
D, z.10 1.28 . The rejection region is z 1.28 .
b.
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
d 0 3.5 0
4.71 .
sd
21
nd
38
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
The rejection region is z 1.28 . (Refer to part a.)
Since the observed value of the test statistic falls in the rejection region ( z 4.71 1.28) , H0 is
rejected. There is sufficient evidence to indicate 1 2 0 at .10 .
c.
Since the sample size of the number of pairs is greater than 30, we do not need to assume that the
population of differences is normal. The sampling distribution of d is approximately normal by the
Central Limit Theorem. We must assume that the differences are randomly selected.
d.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
d z.05
8.27
8.28
3.5 1.645
sd
nd
21
38
3.5 1.223 ( 4.723, 2.277)
e.
The confidence interval provides more information since it gives an interval of possible values for the
difference between the population means.
a.
The rejection region requires .05 in the upper tail of the t-distribution with
df nd 1 12 1 11 . From Table III, Appendix D, t.05 1.796 . The rejection region
is t 1.796 .
b.
From Table III, with df nd 1 24 1 23 , t.10 1.319 . The rejection region is t 1.319 .
c.
From Table III, with df nd 1 4 1 3 , t.025 3.182 . The rejection region is t 3.182 .
d.
Using Minitab, with df nd 1 80 1 79 , t.01 2.374 . The rejection region is t 2.374 .
a.
Pair
Difference
1
2
3
4
5
6
3
2
2
4
0
1
nd
di
nd
(12) 2
i 1
2
34
d
i
6
nd
i 1
2
2
sd
nd 1
6 1
2
d
nd
d i 1
nd
b.
i
12
2
6
d 1 2
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
c.
For confidence coefficient .95, .05 and / 2 .05/ 2 .025 . From Table III, Appendix D, with
df nd 1 6 1 5 , t.025 2.571 . The confidence interval is:
d t / 2
d.
sd
nd
2
2.571
6
2 1.484 .516, 3.484
H 0 : d 0
H a : d 0
d
The test statistic is t
sd
nd
2
2/ 6
3.46
The rejection region requires / 2 .05/ 2 .025 in each tail of the t-distribution with
df nd 1 6 1 5 . From Table III, Appendix D, t.025 2.571 . The rejection region is
t 2.571 or t 2.571 .
Since the observed value of the test statistic falls in the rejection region (t 3.46 2.571) , H0 is
rejected. There is sufficient evidence to indicate that the mean difference is different from 0 at
.05 .
8.29
Let 1 mean of population 1 and 2 mean of population 2.
a.
b.
425
H 0 : d 0
where d 1 2
H a : d 0
Some preliminary calculations are:
d
Pair
Population 1
Population 2
1
2
3
4
5
6
7
8
9
10
19
25
31
52
49
34
59
47
17
51
24
27
36
53
55
34
66
51
20
55
nd
di
nd
2
37
i 1
2
d
i
181
nd
10 4.9
sd2 i 1
nd 1
10 1
2
nd
d i 1
nd
Difference,
d
5
2
5
1
6
0
7
4
3
4
i
37
3.7
10
The test statistic is t
d
sd
nd
3.7
4.9
5.29
10
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426
Chapter 8
The rejection region requires .10 in the lower tail of the t-distribution with df nd 1 10 1 9 .
From Table III, Appendix D, t.10 1.383 . The rejection region is t 1.383 .
Since the observed value of the test statistic falls in the rejection region (t 5.29 1.383) ,
H0 is rejected. There is sufficient evidence to indicate the mean of population 1 is less than the mean
for population 2 at .10 .
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df nd 1 10 1 9 , t.05 1.833 . The 90% confidence interval is:
d t / 2
sd
nd
3.7 1.833
4.9
10
3.7 1.28 ( 4.98, 2.42)
We are 90% confident that the difference in the two population means is between 4.98 and 2.42.
8.30
d.
We must assume that the population of differences is normal, and the sample of differences is
randomly selected.
a.
Some preliminary calculations:
d
nd
di
nd
i 1
2
4682
di
6,880
n
40 36.0103
d
sd2 i 1
nd 1
40 1
2
nd
d i 1
nd
i
468
11.7
40
To determine if 1 2 d is different from 10, we test:
H 0 : d 10
H a : d 10
The test statistic is z
d D0
11.7 10
1.79
sd
36.0103
nd
40
The rejection region requires / 2 .05/ 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.79 1.96) , H0
is not rejected. There is insufficient evidence to indicate 1 2 d is different from 10 at
.05 .
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.31
427
b.
The p-value is p P( z 1.79) P( z 1.79) (.5 .4633) (.5 .4633) .0367 .0367 .0734 .
The probability of observing our test statistic or anything more unusual if H0 is true is .0734. Since
this p-value is not small, there is no evidence to indicate 1 2 d is different from 10 at .05 .
c.
No, we do not need to assume that the population of differences is normally distributed. Because our
sample size is 40, the Central Limit Theorem applies.
a.
Let 1 mean starting BMI and 2 mean ending BMI. To determine if the mean BMI at the end of
the camp is less than the mean BMI at the start of camp, we test:
H 0 : d 0
H a : d 0
where d 1 2
b.
The data should be analyzed as a paired-difference t-test. Each camper had his/her BMI measured at
the start of the camp and at the end. Therefore, these two sets of BMI’s are not independent.
c.
The test statistic is z
( x1 x2 ) ( 1 2 )
n1 n2
2
1
d
2
2
3.3
(34.9 31.6) 0
2
2
3.10 .
6.9 6.2
76
76
d.
The test statistic is z
e.
The test statistic using the paired-difference formula is much larger than the test statistic using the
independent samples formula. The test statistic for the paired-difference provides more evidence to
support the alternative hypothesis.
f.
Since the p-value is less than ( p .0001 .01) , H0 is rejected. There is sufficient evidence to
indicate the mean BMI at the end of camp is less than the mean BMI at the start of camp.
g.
No, the differences in the BMI values do not have to be normally distributed. The sample size is
d
nd
1.5 / 76
19.18 .
n 76 . Thus, the Central Limit Theorem applies and says that the sampling distribution of d will be
approximately normally distributed.
h.
For confidence coefficient .99, .01 and / 2 .01 / 2 .005 . From Table II, Appendix D,
z.005 2.58 . The 99% confidence interval is:
d z / 2
d
nd
3.3 2.58
1.5
76
3.3 .444 2.856, 3.744
We are 99% confident that the true difference in the mean BMI scores between the start of camp and
the end of camp is between 2.857 and 3.743.
8.32
a.
The data should be analyzed using a paired-difference test because that is how the data were
collected. Evaluation scores were collected twice from each agency, once in year 1 and once in year
2. Since the two sets of data are not independent, they cannot be analyzed using independent samples.
Copyright © 2014 Pearson Education, Inc.
428
Chapter 8
b.
The differences between the Year 1 score and the Year 2 score for each agency are:
Agency
GSA
Agriculture
Social Security
USAID
Defense
c.
Score-Yr1
34
33
33
32
17
d
d
Score-Yr2
40
35
33
42
32
Difference
(Yr12-Yr1)
6
2
0
10
15
The mean and standard deviation of the differences are:
d
di
nd
33
6.6
5
sd2
2
i
2
i
nd
nd 1
332
5 147.2 36.8
5 1
4
365
sd 36.8 6.066
d Do 6.6 0
2.43
sd
6.066
5
nd
d.
The test statistic is t
e.
The rejection requires .10 in the upper tail of the t-distribution with df nd 1 5 1 4 . From
Table III, Appendix D, t.10 1.533 . The rejection region is t 1.533 .
8.33
f.
Since the observed value of the test statistic falls in the rejection region (t 2.433 1.533) , H0 is
rejected. There is sufficient evidence to indicate that the true mean evaluation score of government
agencies in Year 2 exceeds the true mean evaluation score in Year 1 at .10 .
a.
Since the data were collected as “twin holes”, it needs to be analyzed as paired differences.
b.
The differences are calculated by finding the difference between the 1st hole and the second hole.
Location
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1st hole
5.5
11.0
5.9
8.2
10.0
7.9
10.1
7.4
7.0
9.2
8.3
8.6
10.5
5.5
10.0
2nd hole
5.7
11.2
6.0
5.6
9.3
7.0
8.4
9.0
6.0
8.1
10.0
8.1
10.4
7.0
11.2
Difference
-0.2
-0.2
-0.1
2.6
0.7
0.9
1.7
-1.6
1.0
1.1
-1.7
0.5
0.1
-1.5
-1.2
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
d
nd
di
nd
2
1
2
d
1 i n 22.65 (2.1)
15 1.597
d
sd2
nd 1
15 1
2
nd
i
d 1
nd
d.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D with
df nd 1 15 1 14 , t.05 1.761 . The 90% confidence interval is:
e.
2.1
0.14
15
c.
d t / 2
429
sd
nd
.14 1.761
1.2637
15
sd 1.597 1.2637 .
.14 .575 .435, .715
We are 90% confident that the true difference in the mean THM measurements between the 1st and
2nd hole is between -.435 and .715.
Yes, the geologists can conclude that there is no evidence of a difference in the true mean THM
measurements between the original holes and their twin holes because 0 falls in the interval at
.10 .
8.34
a.
Let d F C . To determine if the mean score for the fictitious brand is greater than the mean
score for the commercially available brand, we test:
H 0 : d 0
H a : d 0
b.
The data should be analyzed as paired differences. Each child rated both brands, so the samples are
not independent.
c.
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the test statistic falls in the rejection region ( z 5.71 1.645) , H0 is rejected. There is
sufficient evidence to indicate the mean score for the fictitious brand is greater than the mean score
for the commercially available brand at .05 .
8.35
d.
The p-value is p P( z 5.71) .5 .5 0 .
e.
Yes. Since the p-value is less than .01 , the conclusion would still be to reject H0.
a.
The data should be analyzed using a paired-difference experiment because that is how the data were
collected. Response rates were observed twice from each survey using the “not selling” introduction
method and the standard introduction method. Since the two sets of data are not independent, they
cannot be analyzed using independent samples.
b.
Some preliminary calculations are:
s 2p
( n1 1) s12 ( n2 1) s22 (29 1)(.12) 2 (29 1)(.11) 2
.01325
n1 n2 2
29 29 2
Copyright © 2014 Pearson Education, Inc.
430
Chapter 8
Let 1 mean response rate for those using the “not selling” introduction and 2 mean response rate
for those using the standard introduction. Using the independent-samples t-test to determine if the
mean response rate for “not selling” is higher than that for the standard introduction, we test:
H 0 : 1 2 0
H a : 1 2 0
( x1 x2 ) 0
(.262 .246) 0
.53
1
1 1
1
.01325
s
29 29
n1 n2
The rejection region requires .05 in the upper tail of the t-distribution with
df n1 n2 – 2 29 29 – 2 56 . From Table III, Appendix D, t.05 1.671 . The rejection region is
t 1.671 .
The test statistic is t
2
p
Since the observed value of the test statistic does not fall in the rejection region (t .53 1.671) , H0
is not rejected. There is insufficient evidence to indicate the mean response rate for “not selling” is
higher than that for the standard introduction at .05 .
8.36
c.
Since p-value is less than .05 ( p .001 .05) , H0 is rejected. There is sufficient evidence to
indicate the mean response rate for “not selling” is higher than that for the standard introduction at
.05 .
d.
The two inferences in parts b and c have different results because using the independent samples ttest is not appropriate for this study. The paired-difference design is better. There is much variation
in response rates from survey to survey. By using the paired difference design, we can eliminate the
survey to survey differences.
a.
Let 1 mean driver chest injury rating and 2 mean passenger chest injury rating. Because the
data are paired, we are interested in 1 2 d , the difference in mean chest injury ratings between
drivers and passengers.
b.
The data were collected as matched pairs and thus, must be analyzed as matched pairs. Two ratings
are obtained for each car – the driver’s chest injury rating and the passenger’s chest injury rating.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: DrivChst, PassChst, diff
Variable
DrivChst
PassChst
diff
N
98
98
98
Mean
49.663
50.224
-0.561
Median
50.000
50.500
0.000
StDev
6.670
7.107
5.517
Minimum
34.000
35.000
-15.000
Maximum
68.000
69.000
13.000
Q1
45.000
45.000
-4.000
Q3
54.000
55.000
3.000
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 . The 99% confidence interval is:
d z.005
sd
nd
0.561 2.58
5.517
98
0.561 1.438 (1.999, 0.877)
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.37
431
d.
We are 99% confidence that the difference between the mean chest injury ratings of drivers and frontseat passengers is between 1.999 and 0.877. Since 0 is in the confidence interval, there is no
evidence that the true mean driver chest injury rating exceeds the true mean passenger chest injury
rating.
e.
Since the sample size is large, the sampling distribution of d is approximately normal by the Central
Limit Theorem. We must assume that the differences are randomly selected.
Some preliminary calculations are:
Operator
Difference
(Before - After)
5
3
9
7
2
2
1
11
0
5
1
2
3
4
5
6
7
8
9
10
d
d
d
d
a.
To determine if the new napping policy reduced the mean number of customer complaints, we test:
nd
=
39
= 3.9
10
sd2
2
nE
nd 1
2
2
319 39
10 18.5444
10 1
sd 18.5444 4.3063
H 0 : d 0
H a : d 0
The test statistic is t
d 0 3.9 0
2.864
4.3063
sd
10
nd
The rejection region requires .05 in the upper tail of the t-distribution with df nd 1 10 1 9 .
From Table III, Appendix D, t.05 1.833 . The rejection region is t 1.833 .
Since the observed value of the test statistic falls in the rejection region (t 2.864 1.833) , H0 is
rejected. There is sufficient evidence to indicate the new napping policy reduced the mean number of
customer complaints at .05 .
b.
In order for the above test to be valid, we must assume that
1. The population of differences is normal
2. The differences are randomly selected
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
c.
8.38
Variables that were not controlled that could lead to an invalid conclusion include time of day agents
worked, day of the week agents worked, and how much sleep the agents got before working, among
others.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Initial, Final, Diff
Variable
Initial
Final
Diff
N
3
3
3
Mean
5.640
5.453
0.1867
StDev
1.075
1.125
0.1106
Minimum
4.560
4.270
0.0700
Q1
4.560
4.270
0.0700
Median
5.650
5.580
0.2000
Q3
6.710
6.510
0.2900
Maximum
6.710
6.510
0.2900
Let 1 mean initial pH level, 2 mean final pH level, and d 1 2 difference in mean pH levels
between the initial and final time periods. To determine if the mean pH level after 30 days differs from the
initial pH level, we test:
H 0 : d 0
H a : d 0
The test statistic is t
d Do .1867 0
2.924 .
.1106
sd
3
nd
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df nd 1 3 1 2 . From Table III, Appendix D, t.025 4.303 . The rejection region is
t 4.303 and t 4.303 .
Since the observed value of the test statistic does not fall in the rejection region (t 2.924 4.303) , H0 is
not rejected. There is insufficient evidence to indicate the mean pH level after 30 days differs from the
initial pH level at .05 .
8.39
Some preliminary calculations are:
Circuit
Standard Method
1
2
3
4
5
6
7
8
9
10
11
.80
.80
.83
.53
.50
.96
.99
.98
.81
.95
.99
di 1.43
d 1
0.13
11
nd
nd
Huffman-coding
Method
.78
.80
.86
.53
.51
.68
.82
.72
.45
.79
.77
Difference
.02
.00
-.03
.00
-.01
.28
.17
.26
.36
.16
.22
nd
di
nd
2
1
2
1 di n 0.3799 (1.43)
d
11 0.0194
sd2
nd 1
11 1
2
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sd 0.0194 0.1393
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
433
For confidence coefficient .95, .05 and / 2 .05/ 2 .025 . From Table III, Appendix D, with
df nd 1 11 1 10 , t.025 2.228 . The 95% confidence interval is:
d t.025
sd
n
.13 2.228
.1393
11
.13 .094 (0.036, 0.224)
We are 95% confident that the true difference in mean compression ratio between the standard method and
the Huffman-based coding method is between 0.036 and 0.224. Since 0 is not contained in the interval, we
can conclude there is a difference in mean compression ratios between the two methods. Since the values of
the confidence interval are positive, we can conclude that the mean compression ratio for the Huffmanbased method is smaller than the standard method.
8.40
Let 1 mean number of crashes caused by red light running per intersection before camera is installed and
2 mean number of crashes caused by red light running per intersection after camera is installed. The data
are collected as paired data, so we will analyze the data using a paired t-test. Then, let d 1 2 .
Using MINITAB to compute the differences (Di) and the summary statistics, the results are:
Descriptive Statistics: Before, After, Di
Variable
Before
After
Di
N
13
13
13
Mean
2.513
1.506
1.007
StDev
1.976
1.448
1.209
Minimum
0.270
0.000
-0.850
Q1
0.805
0.260
0.265
Median
2.400
1.360
0.560
Q3
3.405
2.380
2.335
Maximum
7.350
4.920
2.780
To determine if the mean number of crashes caused by red light running has been reduced since the
installation of red light cameras, we test:
H 0 : d 0
H a : d 0
The test statistic is t
d Do 1.007 0
3.003
sd
1.209
13
nd
The rejection region requires .05 in the upper tail of the t-distribution with df nd 1 13 1 12 .
From Table III, Appendix D, t.05 1.782 . The rejection region is t 1.782 .
Since the observed value of the test statistic falls in the rejection region (t 3.003 1.782) , H0 is rejected.
There is sufficient evidence to indicate that photo-red enforcement program is effective in reducing redlight-running crash incidents at intersections at .05 .
8.41
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Male, Female, Diff
Variable N
Male
19
Female
19
Diff
19
Mean
5.895
5.526
0.368
Median
6.000
5.000
1.000
StDev
2.378
2.458
3.515
Minimum
3.000
3.000
-5.000
Maximum
Q1
12.000 4.000
12.000 4.000
7.000 -3.000
Copyright © 2014 Pearson Education, Inc.
Q3
8.000
7.000
3.000
434
Chapter 8
Let 1 mean number of swims by male rat pups and 2 mean number of swims by female rat pups.
Then d 1 2 . To determine if there is a difference in the mean number of swims required by male
and female rat pups, we test:
H 0 : d 0
H a : d 0
The test statistic is t
d Do .368 0
0.46
3.515
sd
19
nd
The rejection region requires / 2 .10 / 2 .05 in each tail of the t-distribution with
df nd 1 19 1 18 . From Table III, Appendix D, t.05 1.734 . The rejection region is
t 1.734 or t 1.734 .
Since the observed value of the test statistic does not fall in the rejection region (t .46 1.734) , H0 is not
rejected. There is insufficient evidence to indicate that there is a difference in the mean number of swims
required by male and female rat pups at .10 . (Using Minitab, the p-value .653.)
Since the sample size is not large, we must assume that the population of differences is normally distributed
and that the sample of differences is random. There is no indication that the sample differences are not
from a random sample. However, because the number of swims is discrete, the differences are probably
not normal.
8.42
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: HMETER, HSTATIC, Diff
Variable N
Mean
StDev
Minimum
Q1
Median
HMETER
40
1.0405
0.0403
0.9936
1.0047
1.0232
HSTATIC 40
1.0410
0.0410
0.9930
1.0043
1.0237
Diff
40 -0.000523 0.001291 -0.004480 -0.001078 -0.000165
Q3
1.0883
1.0908
0.000317
Maximum
1.1026
1.1052
0.001580
For confidence coefficient .95, .05 and / 2 .05/ 2 .025 . From Table III, Appendix D, with
df nd 1 40 1 39 , t.025 2.021 . The 95% confidence interval is:
d t.025
sd
n
0.000523 2.021
0.001291
40
0.000523 0.000413 (0.000936, 0.000110)
We are 95% confident that the true difference in mean density measurements between the two methods is
between -0.000936 and -0.000110. Since the absolute value of this interval is completely less than the
desired maximum difference of .002, the winery should choose the alternative method of measuring wine
density.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.43
a.
435
From the exercise, we know that x1 and x2 are binomial random variables with the number of trials
equal to n1 and n2 . From Chapter 7, we know that for large n, the distribution of pˆ1
x1
is
n1
approximately normal. Since x1 is simply p̂1 multiplied by a constant, x1 will also have an
approximate normal distribution. Similarly, the distribution of pˆ 2
x2
is approximately normal, and
n2
thus, the distribution of x2 is approximately normal.
b.
The Central Limit Theorem is necessary to find the sampling distributions of p̂1 and p̂2 when n1 and
n2 are large. Once we have established that both p̂1 and p̂2 have normal distributions, then the
distribution of their difference will also be normal.
8.44
8.45
a.
The rejection region requires .01 in the lower tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
b.
The rejection region requires .025 in the lower tail of the z-distribution. From Table II, Appendix
D, z.025 1.96 . The rejection region is z 1.96 .
c.
The rejection region requires .05 in the lower tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
d.
The rejection region requires .10 in the lower tail of the z-distribution. From Table II, Appendix
D, z.10 1.28 . The rejection region is z 1.28 .
From Section 6.4, it was given that the distribution of p̂ is approximately normal if npˆ 15 and nqˆ 15 .
a.
n1 pˆ1 12(.42) 5.04 15 and n1qˆ1 12(.58) 6.96 15
n2 pˆ 2 14(.57) 7.98 15 and n2 qˆ2 14(.43) 6.02 15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1 pˆ 2 ) is
approximately normal.
b.
n1 pˆ1 12(.92) 11.04 15 and n1qˆ1 12(.08) 0.96 15
n2 pˆ 2 14(.86) 12.04 15 and n2 qˆ2 14(.14) 1.96 15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1 pˆ 2 ) is
approximately normal.
c.
n1 pˆ1 30(.70) 21 15 and n1qˆ1 30(.30) 9 15
n2 pˆ 2 30(.73) 21.9 15 and n2 qˆ2 30(.27) 8.1 15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1 pˆ 2 ) is
approximately normal.
d.
n1 pˆ1 100(.93) 93 15 and n1qˆ1 100(.07) 7 15
n2 pˆ 2 250(.97) 242.5 15 and n2 qˆ2 250(.03) 7.5 15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1 pˆ 2 ) is
approximately normal.
Copyright © 2014 Pearson Education, Inc.
436
Chapter 8
e.
n1 pˆ1 125(.08) 10 15 and n1qˆ1 125(.92) 115 15
n2 pˆ 2 200(.12) 24 15 and n2 qˆ2 200(.88) 176 15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1 pˆ 2 ) is
approximately normal.
8.46
For confidence coefficient .95, .05 and / 2 .05/ 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval for p1 p2 is approximately:
a.
b.
c.
8.47
a.
( pˆ1 pˆ 2 ) z / 2
pˆ1qˆ1 pˆ 2 qˆ2
.65(1 .65) .58(1 .58)
(.65 .58) 1.96
n1
n2
400
400
( pˆ1 pˆ 2 ) z / 2
pˆ1qˆ1 pˆ 2 qˆ2
.31(1 .31) .25(1 .25)
(.31 .25) 1.96
n1
n2
180
250
( pˆ1 pˆ 2 ) z / 2
pˆ1qˆ1 pˆ 2 qˆ2
.46(1 .46) .61(1 .61)
(.46 .61) 1.96
100
120
n1
n2
.07 .067 .003, .137
.06 .086 (.026, .146)
.15 .131 (.281, .019)
H 0 : p1 p2 0
H a : p1 p2 0
Will need to calculate the following:
pˆ 1
320
.40
800
pˆ 2
The test statistic is z
400
.50
800
pˆ1 pˆ 2 0
1 1
ˆ ˆ
pq
n1 n2
pˆ
320 400
.45
800 800
(.40 .50) 0
1
1
(.45)(.55)
800 800
4.02
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 4.02 1.645) ,
H0 is not rejected. There is insufficient evidence to indicate that p1 p2 the proportions are unequal at
.05 .
b.
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z 4.02 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
437
The rejection region requires / 2 .01/ 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
c.
Since the observed value of the test statistic falls in the rejection region ( z 4.02 2.58) , H0 is
rejected. There is sufficient evidence to indicate that the proportions are unequal at .01 .
H 0 : p1 p2 0
H a : p1 p2 0
Test statistic as above z 4.02 .
The rejection region requires .01 in the lower tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection region ( z 4.02 2.33) , H0 is
rejected. There is sufficient evidence to indicate that p1 p2 at .01 .
d.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
( pˆ1 pˆ 2 ) z.05
pˆ1qˆ1 pˆ 2 pˆ 2
(.4)(.6) (.5)(.5)
(.4 .5) 1.645
.10 .04 ( .14, .06)
n1
n2
800
800
We are 90% confident that the difference between p1 and p2 is between .14 and .06.
8.48
pˆ
n1 pˆ1 n2 pˆ 2 55(.7) 65(.6) 77.5
.646
55 65
120
n1 n2
qˆ 1 pˆ 1 .646 .354
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.7 .6) 0
1
1
.646(.354)
55
65
1.14
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix D,
z.05 1.645 . The rejection region is z 1 .645 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.14 1.645) , H0 is not
rejected. There is insufficient evidence to indicate the proportion from population 1 is greater than that for
population 2 at .05 .
8.49
a.
pˆ 1
x1
29
.153
n1 189
b.
pˆ 2
x2
32
.215
n2 149
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Chapter 8
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
( pˆ1 pˆ 2 ) z / 2
pˆ1qˆ1 pˆ 2 qˆ2
.153(.847) .215(.785)
(.153 .215) 1.645
n1
n2
189
149
.062 .070 (.132, .008)
8.50
d.
We are 90% confident that the difference in the proportion of bidders who fall prey to the winner’s
curse between super-experienced bidders and less-experienced bidders is between .132 and .008.
Since this interval contains 0, there is no evidence to indicate that there is a difference in the
proportion of bidders who fall prey to the winner’s curse between super-experienced bidders and lessexperienced bidders.
a.
The point estimate for the proportion of all Democrats who prefer steak as their favorite barbeque
x
662
food is pˆ1 1
.5296 .
n1 1, 250
b.
The point estimate for the proportion of all Republicans who prefer steak as their favorite barbeque
x
586
food is pˆ 2 2
.6301 .
n2 930
The point estimate for the difference between proportions of all Democrats and all Republicans who
prefer steak as their barbeque food is pˆ1 pˆ 2 .5296 .6301 .1005 .
c.
d.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval for the difference between the proportions of all
Democrats and all Republicans who prefer steak as their barbeque food is
pˆ1 pˆ 2 z.025
8.51
pˆ1qˆ1 pˆ 2 qˆ2
.5296(.4704) .6301(.3699)
(.5296 .6301) 1.96
n1
n2
1, 250
930
.1005 .0416 (.1421, .0589)
e.
We are 95 percent confident that the difference in proportions of all Democrats and all Republicans
who prefer steak as their favorite barbeque food is between -.1421 and -.0589. Since this interval does
not contain 0, there is a sufficient evidence to indicate that there is a significant difference between
the proportions of all Democrats and all Republicans who prefer steak as their favorite barbeque food.
f.
“95% confident” means that in repeated sampling, 95% of all confidence intervals constructed in the
same manner will contain the true population difference in proportions and 5% will not.
a.
Let p1 proportion of producers who are willing to offer windrowing services to the biomass market
in Missouri and p2 proportion of producers who are willing to offer windrowing services to the
biomass market in Illinois. The parameter of interest is p1 p2 .
b.
To determine if the proportion of producers who are willing to offer windrowing services differs
between Missouri and Illinois, we test:
H 0 : p1 p2 0
H a : p1 p2 0
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
439
c.
The test statistic is z 2.67 .
d.
The rejection region requires / 2 .01 / 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 and z 2.58 .
e.
The p-value is p .008 .
f.
Since the observed value of the test statistic falls in the rejection region ( z 2.67 2.58) , H0 is
rejected. There is sufficient evidence to indicate that the proportion of producers who are willing to
offer windrowing services differs between Missouri and Illinois at .01 .
Since the p-value is less than ( p .008 .01) , H0 is rejected. This is the same conclusion as above.
8.52
a.
Let p1 proportion of men who prefer to keep track of appointments in their head and
p2 proportion of women who prefer to keep track of appointments in their head. To determine if
the proportion of men who prefer to keep track of appointments in their head is greater than that of
women, we test:
H 0 : p1 p2 0
H a : p1 p2 0
b.
pˆ
n1 pˆ1 n2 pˆ 2 500(.56) 500(.46)
.51 and qˆ 1 pˆ 1 .51 .49
500 500
n1 n2
The test statistic is z
8.53
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.56 .46) 0
1
1
.51(.49)
500 500
3.16
c.
The rejection region requires .01 in the upper tail of the z distribution. From Table II,
Appendix D, z.01 2.33 . The rejection region is z 2.33 .
d.
The p-value is p P( z 3.16) .5 .5 0 .
e.
Since the observed value of the test statistic falls in the rejection region ( z 3.16 2.33) , H0 is
rejected. There is sufficient evidence to indicate the proportion of men who prefer to keep track of
appointments in their head is greater than that of women at .01 .
a.
The first population of interest is all hospital patients admitted in January. The second population of
interest is all hospital patients admitted in May.
b.
pˆ 1
x1
32
.167
n1 192
pˆ 2
x2
34
.084
n2 403
The point estimate for the difference in malaria admission rates in January and May is
pˆ1 pˆ 2 .167 .084 .083 .
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
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Chapter 8
( pˆ1 pˆ 2 ) z.05
pˆ1qˆ1 pˆ 2 qˆ2
.167(.833) .084(.916)
(.167 .084) 1.645
192
403
n1
n2
.083 .050 (.033, .133)
8.54
d.
Since 0 is not contained in the confidence interval, we can conclude that a difference exists in the true
malaria admission rates in January and May.
a.
Let p1 proportion of customers returning the printed survey and p2 proportion of customers
returning the electronic survey. Some preliminary calculations are:
pˆ 1
x1 261
.414
n1 631
pˆ 2
x2 155
.374
n2 414
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The 90% confidence interval is:
( pˆ1 pˆ 2 ) z.05
pˆ1qˆ1 pˆ 2 qˆ2
.414(.586) .374(.626)
(.414 .374) 1.645
n1
n2
631
414
.04 .051 (.011, .091)
We are 90% confidence that the difference in the response rates for the two types of surveys is
between .011 and .091.
b.
8.55
Since the value .05 falls in the 90% confidence interval, it is not an unusual value. Thus, there is no
evidence that the difference in response rates is different from .05. The researchers would be able to
make this inference.
Let p1 proportion of salmonella in the region’s water and p2 proportion of salmonella in the region’s
wildlife.
Some preliminary calculations are:
pˆ 1
x1
18
.071
n1 252
pˆ 2
x2
20
.042
n2 476
pˆ
x1 x2
18 20
38
.052
n1 n2 252 476 728
To determine if the prevalence of salmonella in the region’s water differs from the prevalence of
salmonella in the region’s wildlife, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
.071 .042
1
1
.052(.948)
252
476
1.68
The rejection region requires / 2 .01 / 2 .005 in each tail of the z-distribution. From Table II, Appendix
D, z.005 2.58 . The rejection region is z 2.58 and z 2.58 .
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
441
Since the observed value of the test statistic does not fall in the rejection region ( z 1.68 2.58) , H0 is not
rejected. There is insufficient evidence to indicate the prevalence of salmonella in the region’s water
differs from the prevalence of salmonella in the region’s wildlife at .01 .
8.56
Let p1 proportion of patients in the angioplasty group who had subsequent heart attacks and
p2 proportion of patients in the medication only group who had subsequent attacks.
Some preliminary calculations:
x
211
pˆ1 1
.184
n1 1,145
qˆ1 1 pˆ1 1 .184 .816
x2
202
.177
n2 1,142
qˆ2 1 pˆ 2 1 .177 .823
pˆ 2
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
A 95% confidence interval for the difference in the rate of heart attacks for the two groups is
( pˆ1 pˆ 2 ) z.025
pˆ1qˆ1 pˆ 2 qˆ 2
.184(.816) .177(.823)
(.184 .177) 1.96
.007 .032 ( .025, .039)
n1
n2
1,145
1,142
Since this interval contains 0, there is insufficient evidence to indicate that there is a difference in the rate
of heart attacks between the angioplasty group and the medication only group at = .05. Yes, we agree
with the study’s conclusion.
8.57
Let p1 proportion of African American MBA students who begin their career as entrepreneurs and
p2 proportion of white MBA students who begin their career as entrepreneurs.
Some preliminary calculations:
x
209
pˆ1 1
.1603
n1 1,304
x2
356
.05
n2 7,120
qˆ2 1 pˆ 2 1 .05 .95
x1 x2
209 356
.0671
n1 n2 1,304 7,120
qˆ 1 pˆ 1 .0671 .9329
pˆ 2
pˆ
qˆ1 1 pˆ1 1 .1603 .8397
To determine if African American MBA students are more likely to begin their careers as an entrepreneur
than white MAB students, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
.1603 .05
1
1
.0671(.9329)
1,304 7,120
14.64
Since no was given, we will use .05 . The rejection region requires .05 in the upper tail of the
z-distribution. From Table II, Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
Since the observed value of the test statistic falls in the rejection region ( z 14.64 1.645) , H0 is rejected.
There is sufficient evidence to indicate that the proportion of African American MBA students who begin
their career as entrepreneurs is significantly greater than the proportion of white MBA students who begin
their career as entrepreneurs.
8.58
Let p1 accuracy rate for modules with correct code and p2 accuracy rate for modules with defective
code.
Some preliminary calculations are:
pˆ 1
x1 400
.891
n1 449
pˆ 2
x2 20
.408
n2 49
For confidence coefficient .99, .01 and / 2 .01 / 2 .005 . From Table II, Appendix D, z.005 2.58 .
The 99% confidence interval is:
( pˆ1 pˆ 2 ) z.005
pˆ1qˆ1 pˆ 2 qˆ 2
.891(.109) .408(.592)
(.891 .408) 2.58
.483 .185 (.298, .668)
n1
n2
449
49
We are 99% confident that the difference in accuracy rates between modules with correct code and
modules with defective code is between .298 and .668.
8.59
a.
Let p1 proportion of women who have food cravings and p2 proportion of men who have food
cravings.
We know that pˆ1 .97 and pˆ 2 .67 .
We know that n1 p1 15 and n1q1 15 in order for the test to be valid. Thus,
n1 .97 15 n1 15 / .97 16 and n1 .03 15 n1 15 / .03 500 .
Also, n2 p2 15 and n2 q2 15 . Thus, n2 .67 15 n2 15 / .97 23 and
n2 .33 15 n2 15 / .33 46 .
Thus, n1 500 and n2 46 .
b.
8.60
This study involved 1,000 McMaster University students. It is very dangerous to generalize the
results of this study to the general adult population of North America. The sample of students used
may not be representative of the population of interest.
Let p1 proportion of TV commercials in 1998 that used religious symbolism and p2 proportion of TV
commercials in a recent study that used religious symbolism.
Some preliminary calculations are:
x x2
x1
x
16 51
67
16
51
pˆ 1
.020
pˆ 2 2
.034
.029
n1 n2 797 1, 499 2, 296
n1 797
n2 1, 499
To determine if the percentage of TV commercials that use religious symbolism has changed since the
1998 study, we test:
pˆ 1
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
443
H 0 : p1 p2 0
H a : p1 p2 0
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
The test statistic is z
.020 .034
1
1
.029(.971)
797
1,
499
1.90
Since no was given, we will use .05 . The rejection region requires / 2 .05 / 2 .025 in each tail of
the z-distribution. From Table II, Appendix D, z.025 1.96 . The rejection region is z 1.96 and z 1.96 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.90 1.96) , H0 is
not rejected. There is insufficient evidence to indicate the percentage of TV commercials that use religious
symbolism has changed since the 1998 study at .05 .
8.61
a.
For confidence coefficient .99, .01 and / 2 .01 / 2 .005 . From Table II, Appendix D,
z.005 2.58 .
( z / 2 )2 ( p1q1 p2 q2 ) 2.58 .4(1 .4) .7(1 .7) 2.99538
29,953.8 29,954
.012
.0001
( ME ) 2
2
n1 n 2
b.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . Since we have no prior information about the proportions, we use p1 p2 .5 to get a
conservative estimate. For a width of .05, the margin of error is .025.
( z / 2 )2 ( p1q1 p2 q2 ) (1.645) .5(1 .5) .5(1 .5)
2164.82 2165
( ME ) 2
.0252
2
n1 n 2
c.
From part b, z.05 1.645 .
( z / 2 )2 ( p1q1 p2 q2 ) (1.645) .2(1 .2) .3(1 .3) 1.00123
1112.48 1113
.0009
( ME ) 2
.032
2
n1 n 2
8.62
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 .
z / 2 12 22
2
n1 n2
b.
ME 2
(1.96) 2 (152 17 2 )
192.83 193
3.2 2
If the range of each population is 40, we would estimate by 60 / 4 15
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 .
z / 2 12 22
2
n1 n2
ME 2
(2.58) 2 (152 152 )
46.80 47
82
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Chapter 8
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . For a width of 1, the margin of error is .5.
z / 2 12 22
2
n1 n2
8.63
n1 n2
ME 2
(1.645) 2 (5.82 7.52 )
143.96 144
.52
( z / 2 ) 2 ( 12 22 )
ME 2
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
n1 n2
8.64
1.962 (14 14)
33.2 34
1.82
First, find the sample sizes needed for width 5, or margin of error 2.5.
For confidence coefficient .9, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
z / 2 12 22
2
n1 n2
ME
2
(1.645) 2 (10 2 10 2 )
86.59 87
2.52
Thus, the necessary sample size from each population is 87. Therefore, sufficient funds have been
allocated to meet the specifications since n1 n2 100 are large enough samples.
8.65
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
z / 2 12 22
2
n1 n2
8.66
ME 2
(1.96)2 (92 92 )
155.6 156
22
For confidence coefficient 0.99, .01 and / 2 .01/ 2 .005 . From the Table II, Appendix D,
z.005 2.58 .
z / 2 ( 12 22 )
2
n1 n1
8.67
ME 2
2.582 (12 12 )
53.25 54
.52
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
If we assume that we do not know the return rates, we will use .5 for both.
z.05 ( p1q1 p2 q2 )
2
1.6452 (.5(.5) .5(.5))
13,530.1 13, 531
.012
ME 2
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
n1 n2
8.68
Since no information is given about the values of p1 and p2 , we will be conservative and use .5 for both. A
width of .04 means the margin of error is .04 / 2 .02 .
z / 2 ( p1q1 p2 q2 )
2
n1 n2
ME
2
1.6452 .5(.5) .5(.5)
.02 2
3,382.5 3,383
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.69
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
n1 n 2
8.70
445
a.
( z / 2 ) 2 ( 12 22 ) 1.96 2 (152 152 )
1728.72 1729
ME 2
12
For confidence coefficient .80, .20 and / 2 .20 / 2 .10 . From Table II, Appendix D, z.10 1.28 .
Since we have no prior information about the proportions, we use p1 p2 .5 to get a conservative
estimate. For a width of .06, the margin of error is .03.
z / 2 ( p1q1 p2 q2 )
2
n1 n 2
b.
ME 2
(1.28) 2 .5(1 .5) .5(1 .5)
.032
910.22 911
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . Using the formula for the sample size needed to estimate a proportion,
z / 2 pq
2
n
ME 2
1.6452 .5(1 .5)
.02 2
.6765
1691.27 1692
.0004
No, the sample size from part a is not large enough.
8.71
a.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . From Exercise 8.56, pˆ1 .184 and pˆ 2 .177 .
z ( p1q1 p2 q2 ) 1.962 (.184(.816) .177(.823))
5,050.7 5,051
n1 n2 / 2
2
.0152
ME
2
8.72
b.
The study would involve 5, 051 2 10,102 patients. A study this large would be extremely time
consuming and expensive.
c.
Since a difference of .015 is so small, the practical significance detecting a 0.015 difference may not
be very worthwhile. A difference of .015 is so close to 0, that it might not make any difference.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 .
z / 2 12 22
2
n1 n2
8.73
(ME )2
1.962 (352 802 )
292.9 293
102
a.
With1 9 and 2 6 , F.05 4.10 .
b.
With 1 18 and 2 14 , F.01 3.57 . (Since1 18 is not given, we estimate the value between
those for 1 15 and 1 20 .)
c.
With1 11 and 2 4 , F.025 8.80 . (Since1 11 is not given, we estimate the value by averaging
those given for1 10 and 1 12 .)
d.
With 1 20 and 2 5 , F.10 3.21 .
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Chapter 8
8.74
a.
With 1 2 and 2 30 , P(F 5.39) .01 (Table VIII, Appendix D)
b.
With 1 24 and 2 10 , P( F 2.74) .05 (Table VI, Appendix D)
Thus, P(F 2.74) 1 P( F 2.74) 1 .05=.95 .
c.
With 1 7 and 2 1 , P( F 236.8) .05 (Table VI, Appendix D)
Thus, P(F 236.8) 1 P( F 236.8) 1 .05=.95 .
8.75
8.76
8.77
d.
With1 40 and 2 40 , P(F 2.11) .01 (Table VIII, Appendix D)
a.
Reject H0 if F F.10 1.74 . (From Table V, Appendix D, with1 30 and 2 20 .)
b.
Reject H0 if F F.05 2.04 . (From Table VI, Appendix D, with1 30 and 2 20 .)
c.
Reject H0 if F F.025 2.35 . (From Table VII.)
d.
Reject H0 if F F.01 2.78 . (From Table VIII.)
To test H 0 : 12 22 against H a : 12 22 , the rejection region is F F /2 with 1 10 and 2 12 .
a.
.20 and / 2 .20 / 2 .10 : Reject H0 if F F.10 2.19 (Table V, Appendix D)
b.
.10 and / 2 .10 / 2 .05 : Reject H0 if F F.05 2.75 (Table VI, Appendix D)
c.
.05 and / 2 .05 / 2 .025 : Reject H0 if F F.025 3.37 (Table VII, Appendix D)
d.
.02 and / 2 .02 / 2 .01 : Reject H0 if F F.01 4.30 (Table VIII, Appendix D)
a.
The rejection region requires .05 in the upper tail of the F-distribution with
1 n1 1 25 1 24 and 2 n2 1 20 1 19 . From Table VI, Appendix D, F.05 2.11 . The
rejection region is F 2.11 (if s12 s22 ).
b.
The rejection region requires .05 in the upper tail of the F-distribution with
1 n2 1 15 1 14 and 2 n1 1 10 1 9 . From Table VI, Appendix D, F.05 3.01 . The
rejection region is F 3.01 (if s22 s12 ).
c.
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution. If s12 s22 ,
1 n1 1 21 1 20 and 2 n2 1 31 1 30 . From Table VI, Appendix D, F.05 1.93 . The
rejection region is F 1.93 . If s12 s22 , 1 n2 1 30 and 2 n1 1 20 . From Table VI,
F.05 2.04 . The rejection region is F 2.04 .
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
d.
447
The rejection region requires .01 in the upper tail of the F-distribution with
1 n2 1 41 1 40 and 2 n1 1 31 1 30 . From Table VIII, Appendix D, F.01 2.30 . The
rejection region is F 2.30 (if s22 s12 ).
e.
The rejection region requires .05 and / 2 .05 / 2 .025 in the upper tail of the F-distribution.
If s12 s22 , 1 n1 1 7 1 6 and 2 n2 1 16 1 15 . From Table VII, Appendix D,
F.025 3.41 . The rejection region is F 3.41 . If s12 s22 , 1 n2 1 15 and 2 n1 1 6 . From
Table VII, Appendix D, F.025 5.27 . The rejection region is F 5.27 .
8.78
a.
To determine if a difference exists between the population variances, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
s22 8.75
2.26
s12 3.87
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution with
1 n2 1 27 1 26 and 2 n1 1 12 1 11 . From Table VI, Appendix D, F.05 2.60 . The
rejection region is F 2.60 .
Since the observed value of the test statistic does not fall in the rejection region (F 2.26 2.60) , H0
is not rejected. There is insufficient evidence to indicate a difference between the population
variances at .10 .
b.
The p-value is p 2P( F 2.26) . From Tables V and VI, with1 16and 2 11 ,
2 .05 2P( F 2.26) 2 .10 .10 2P( F 2.26) .20
There is no evidence to reject H0 for .10 .
8.79
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Sample 1, Sample 2
Variable
Sample 1
Sample 2
N
6
5
Mean
2.417
4.36
Median
2.400
3.70
StDev
1.436
2.97
Minimum
0.700
1.40
Maximum
Q1
Q3
4.400 1.075 3.650
8.90 1.84 7.20
To determine if the variance for population 2 is greater than that for population 1, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
s22
2.97 2
4.28
2
s1 1.436 2
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Chapter 8
The rejection region requires .05 in the upper tail of the F-distribution with1 n2 1 5 1 4
and 2 n1 1 6 1 5 . From Table VI, Appendix D, F.05 5.19 . The rejection region is
F 5.19 .
Since the observed value of the test statistic does not fall in the rejection region (F 4.29 5.19) , H0
is not rejected. There is insufficient evidence to indicate the variance for population 2 is greater than
that for population 1 at .05 .
b.
The p-value is p P(F 4.28) . From Tables V and VI, with1 4 and 2 5 ,
.05 p P(F 4.28) .10
There is no evidence to reject H0 for .05 but there is evidence to reject H0 for .10 .
8.80
a.
To determine if a difference exists between the population variances, we test:
2
2
H 0 : BT
PA
2
2
H a : BT
PA
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: BT, PA
Variable
BT
PA
N
7
8
Mean
89.86
99.63
Variance
135.14
749.70
Minimum
70.00
66.00
Q1
82.00
76.50
Median
93.00
96.00
Q3
99.00
115.00
Maximum
105.00
153.00
2
2
sBT
135.14 and sPA
749.70
2
sPA
Larger sample variance 749.70
5.55 .
2
sBT Smaller sample variance 135.14
c.
The test statistic is F
d.
Using MINITAB, the p-value is:
Cumulative Distribution Function
F distribution with 7 DF in numerator and 68 DF in denominator
x
5.55
P( X <= x )
0.973421
The p-value is p 2(1 .973421) 2(.026579) .053158 .
e.
8.81
Since the p-value is not less than ( p .053158 .01) , H0 is not rejected. There is insufficient
evidence to indicate the variances are different at .01 .
Let 12 variance of the number of ads recalled by children in the video only group and 22 variance of
the number of ads recalled by children in the A/V group.
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
a.
449
To determine if the group variances are equal, we test:
H 0 : 12 22
H a : 12 22
s 2 2.132
larger sample variance
22
1.157
smaller sample variance s1 1.982
b.
The test statistic is: F
c.
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution with
1 n2 1 20 1 19 and 2 n1 1 20 1 19 . From the Table VI, Appendix D, F.05 2.16 .
The rejection region is F 2.16 .
d.
Since the observed value of the test statistic does not fall in the rejection region (F 1.157 2.16) ,
H0 is not rejected. There is insufficient evidence to indicate the variances of the number of ads
recalled by the children in the video-only group and the A/V group differ at .10 .
8.82
e.
Since we could not reject H0 that the variances were equal, it indicates that the assumption of equal
variances is probably valid. The inference about the population means is probably valid.
a.
The amount of variability of GHQ scores tells us how similar or different the members of the group
are on GHQ scores. The larger the variability, the larger the differences are among the members on
the GHQ scores. The smaller the variability, the smaller the differences are among the members on
the GHQ scores.
b.
Let 12 variance of the mental health scores of the employed and 22 variance of the mental health
scores of the unemployed. To determine if the variability in mental health scores differs for
employed and unemployed workers, we test:
H 0 : 12 22
H a : 12 22
c.
The test statistic is F
Larger sample variance s22 5.102
2.45
Smaller sample variance s12 3.26 2
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n2 1 49 1 48 and 2 n1 1 142 1 141 . Using MINITAB,
Inverse Cumulative Distribution Function
F distribution with 48 DF in numerator and 141 DF in denominator
P( X <= x )
0.975
x
1.55339
The rejection region is F 1.55 .
Since the observed value of the test statistic falls in the rejection region (F 2.45 1.55) , H0 is
rejected. There is sufficient evidence to indicate that the variability in mental health scores differs for
employed and unemployed workers for .05 .
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Chapter 8
d.
8.83
We must assume that the 2 populations of mental health scores are normally distributed. We must
also assume that we selected 2 independent random samples.
Let 12 variance at site 1 and 22 variance of site 2. To determine if the variances at the two locations
differ, we test:
H 0 : 12 22
H a : 12 22
From the printout, the test statistic is F .844 and the p-value is p .681 .
Since the p-value is not less than ( p .681 .05) , H0 is not rejected. There is insufficient evidence to
indicate the variances at the two locations differ at .05 .
8.84
Let 12 variance zinc measurements from the text-line, 22 variance zinc measurements from the
witness-line, and 32 variance zinc measurements from the intersection. Using MINITAB, the descriptive
statistics are:
Descriptive Statistics: Text-line, Witness-line, Intersection
Variable
Text-lin
WitnessIntersec
a.
N
3
6
5
Mean
0.3830
0.3042
0.3290
Median
0.3740
0.2955
0.3190
StDev
0.0531
0.1015
0.0443
Minimum
0.3350
0.1880
0.2850
Maximum
0.4400
0.4390
0.3930
Q1
0.3350
0.2045
0.2900
Q3
0.4400
0.4075
0.3730
To determine if the variation in the zinc measurements for the text-line and the intersection differ, we
test:
H 0 : 12 32
H a : 12 32
The test statistic is F
Larger sample variance s12 .05312
1.437
Smaller sample variance s32 .04432
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n1 1 3 1 2 and 2 n3 1 5 1 4 . From Table VII, Appendix D, F.025 10.65 . The
rejection region is F 10.65 .
Since the observed value of the test statistic does not fall in the rejection region (F 1.437 10.65) ,
H0 is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements
for the text-line and the intersection differ at .05 .
b.
To determine if the variation in the zinc measurements for the witness-line and the intersection
differ, we test:
H 0 : 22 32
H a : 22 32
The test statistic is F
Larger sample variance s22 .10152
5.250
Smaller sample variance s32 .04432
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
451
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n2 1 6 1 5 and 2 n3 1 5 1 4 . From Table IX, Appendix D, F.025 9.36 . The
rejection region is F 9.36 .
Since the observed value of the test statistic does not fall in the rejection region (F 5.25 9.36) , H0
is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements for
the witness-line and the intersection differ at .05 .
8.85
c.
There is no indication that the variances of the zinc measurements for three locations differ.
d.
With only 3, 6, and 5 measurements, it is very difficult to check the assumptions.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Novice, Experienced
Variable
Novice
Experien
a.
N
12
12
Mean
32.83
20.58
Median
32.00
19.50
StDev
8.64
5.74
Minimum
20.00
10.00
Maximum
48.00
31.00
Q1
26.75
17.25
Q3
39.00
24.75
Let 12 variance in inspection errors for novice inspectors and 22 variance in inspection errors for
experienced inspectors. Since we wish to determine if the data support the belief that the variance is
lower for experienced inspectors than for novice inspectors, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
Larger sample variance s12 8.642
2.27
Smaller sample variance s 22 5.74 2
The rejection region requires .05 in the upper tail of the F-distribution with1 n1 1 12 1 11
and 2 n2 1 12 1 11 . Using MINITAB:
Inverse Cumulative Distribution Function
F distribution with 11 DF in numerator and 11 DF in denominator
P( X <= x )
0.95 2.81793
x
The rejection region is F 2.82 .
Since the observed value of the test statistic does not fall in the rejection region (F 2.27 2.82) , H0
is not rejected. The sample data do not support her belief at .05 .
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
b.
Using MINITAB:
Cumulative Distribution Function
F distribution with 11 DF in numerator and 11 DF in denominator
x
2.27
P( X <= x )
0.905144
The p-value P(F 2.27) 1 P(F 2.27) 1 .905) .095 .
8.86
Let 12 heat rate variance of traditional augmented gas turbines, 22 heat rate variance of aeroderivative
augmented gas turbines, and 32 heat rate variance of advanced augmented gas turbines.
Using MINITAB, some preliminary calculations are:
Descriptive Statistics: HEATRATE
Variable
HEATRATE
a.
ENGINE
N
Advanced
21
Aeroderiv
7
Traditional 39
Mean
9764
12312
11544
StDev Minimum
Q1 Median
Q3
639
9105 9252
9669 10060
2652
8714 9469 12414 14628
1279
10086 10592 11183 11964
Maximum
11588
16243
14796
To determine if the heat rate variances for traditional and aeroderivative augmented gas turbines
differ, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
Larger sample variance s22 2652 2
4.299
Smaller sample variance s32 1279 2
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
numerator df 2 n2 –1 7 –1 6 and denominator df 3 n3 –1 39 –1 38 . From TableVII,
Appendix D, F.025 2.74 . The rejection region is F 2.74 .
Since the observed value of the test statistic falls in the rejection region (F 4.299 2.74) , H0 is
rejected. There is sufficient evidence to indicate the heat rate variances for traditional and
aeroderivative augmented gas turbines differ at .05 .
Since the test in Exercise 8.24 a assumes that the population variances are the same, the validity of
the test is suspect since we just found the variances are different.
b.
To determine if the heat rate variances for advanced and aeroderivative augmented gas turbines
differ, we test:
H 0 : 22 32
H a : 22 32
The test statistic is F
Larger sample variance s22 26522
17.224
Smaller sample variance s32
639 2
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
453
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
numerator df 1 n1 –1 7 –1 6 and denominator df 2 n2 –1 21–1 20 . From Table
VII, Appendix D, F.025 3.13 . The rejection region is F 3.13 .
Since the observed value of the test statistic falls in the rejection region (F 17.224 3.13) , H0 is
rejected. There is sufficient evidence to indicate the heat rate variances for advanced and
aeroderivative augmented gas turbines differ at .05 .
Since the test in Exercise 8.24 b assumes that the population variances are the same, the validity of
the test is suspect since we just found the variances are different.
8.87
a.
Let 12 variance of the order-to-delivery times for the Persian Gulf War and 22 variance of the
order-to-delivery times for Bosnia.
Descriptive Statistics: Gulf, Bosnia
Variable
Gulf
Bosnia
N
9
9
Mean
25.24
7.38
Median
27.50
6.50
StDev
10.5204
3.6537
Minimum
9.10
3.00
Maximum
41.20
15.10
Q1
15.30
5.25
Q3
32.15
9.20
To determine if the variances of the order-to-delivery times for the Persian Gulf and Bosnia
shipments are equal, we test:
H0 :
12
1
22
Ha :
12
1
22
The test statistic is F
Larger sample variance s12 10.52042
8.29
Smaller sample variance s22
3.6537 2
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n1 1 9 1 8 and 2 n2 1 9 1 8 . From Table VII, Appendix D, F.025 4.43 . The
rejection region is F 4.43 .
Since the observed value of the test statistic falls in the rejection region (F 8.29 4.43) , H0 is
rejected. There is sufficient evidence to indicate the variances of the order-to-delivery times for the
Persian Gulf and Bosnia shipments differ at .05 .
b.
No. One assumption necessary for the small sample confidence interval for (1 2 ) is that 12 22 .
For this problem, there is evidence to indicate that 12 22 .
8.88
Let 12 variance of improvement scores in the honey dosage group and 22 variance of improvement
scores in the DM dosage group. From Exercise 8.23, s1 2.855 and s2 3.256 .
To determine if the variability in coughing improvement scores differs for the two groups, we test:
H 0 : 12 22
H a : 12 22
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Chapter 8
The test statistic is F
s 2 3.2562
larger sample variance
22
1.30
smaller sample variance s1 2.8552
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution with numerator
df 2 n2 –1 33 –1 32 and denominator df 1 n1 –1 35 –1 34 . From Table VI, Appendix D,
F.05 1.84 . The rejection region is F 1.84 .
Since the observed value of the test statistic does not fall in the rejection region (F 1.30 1.84) , H0 is not
rejected. There is insufficient evidence to indicate the variability in the coughing improvement scores
differs for the two groups at .10 .
8.89
a.
The 2 samples are randomly selected in an independent manner from the two populations. The
sample sizes, n1 and n2, are large enough so that x1 and x2 each have approximately normal sampling
distributions and so that s12 and s22 provide good approximations to 12 and 22 . This will be true if
n1 30 and n2 30 .
8.90
b.
1.
2.
3.
Both sampled populations have relative frequency distributions that are approximately normal.
The population variances are equal.
The samples are randomly and independently selected from the populations.
c.
1.
2.
The relative frequency distribution of the population of differences is normal.
The sample of differences are randomly selected from the population of differences.
d.
The two samples are independent random samples from binomial distributions. Both samples should
be large enough so that the normal distribution provides an adequate approximation to the sampling
distributions of p̂1 and p̂2 .
e.
The two samples are independent random samples from populations which are normally distributed.
a.
H 0 : 12 22
H a : 12 22
The test statistic is F
Larger sample variance s22 120.1
3.84
Smaller sample variance s12
31.3
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution
1 n2 –1 15 –1 14 and 2 n1 –1 20 –1 19 . From Table VII, Appendix D, F.025 2.66 . The
rejection region is F 2.66 .
Since the observed value of the test statistic falls in the rejection region (F 3.84 2.66) , H0 is
rejected. There is sufficient evidence to conclude 12 22 at .05 .
b.
No, we should not use a small sample t- test to test H0 : (1 2 ) 0 against Ha : (1 2 ) 0
because the assumption of equal variances does not seem to hold since we concluded 12 22 in
part b.
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.91
a.
sp2
455
( n1 1) s12 ( n1 1) s22 11(74.2) 13(60.5)
66.7792
12 14 2
n1 n2 2
H 0 : 1 2 0
H a : 1 2 0
( x1 x2 ) 0
(17.8 15.3) 0
.78
1 1
1 1
66.7792
s
12 14
n1 n2
The rejection region requires .05 in the upper tail of the t-distribution with
df n1 n2 – 2 12 14 – 2 24 . From Table III, Appendix D, t.05 1.711 . The rejection region is
t 1.711 .
The test statistic is t
2
p
Since the observed value of the test statistic does not fall in the rejection region (t 0.78 1.711) , H0
is not rejected. There is insufficient evidence to indicate that 1 2 at .05 .
b.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table III, Appendix D, with
df n1 n2 – 2 12 14 – 2 24 , t.005 2.797 . The confidence interval is:
1 1
1 1
( x1 x2 ) t.005 s 2p (17.8 15.3) 2.797 66.7792
n
n
12 14
2
1
2.50 8.99 (6.49, 11.49)
c.
For confidence coefficient .99, .01 and / 2 .01/ 2 .005 . From Table II, Appendix D,
z.005 2.58 .
n1 n2
8.92
( z / 2 ) 12 22
( ME ) 2
(2.58) 2 (74.2 60.5)
2
2
224.15 225
Some preliminary calculations are:
pˆ1
a.
x1 110
.55
n1 200
pˆ 2
x2 130
.65
n2 200
pˆ
x1 x2 110 130 240
.6
n1 n2 200 200 400
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.55 .65) 0
1
1
.6(1 .6)
200 200
.10
2.04
.049
The rejection region requires .10 in the lower tail of the z-distribution. From Table II,
Appendix D, z.10 1.28 . The rejection region is z 1.28 .
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Chapter 8
Since the observed value of the test statistic falls in the rejection region ( z 2.04 1.28) , H0 is
rejected. There is sufficient evidence to conclude p1 p2 0 at .10 .
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval for ( p1 p2 ) is approximately:
pˆ1qˆ2 pˆ 2 qˆ2
.55(1 .55) .65(1 .65)
(.55 .65) 1.96
n1
n2
200
200
( pˆ1 pˆ 2 ) z / 2
c.
.10 .096 (.196, .004)
From part b, z.025 1.96 . Using the information from our samples, we can use p1 .55 and p2 .65 .
For a width of .01, the margin of error is .005.
z / 2 ( p1q1 p2 q2 )
2
n1 n2
8.93
a.
ME
2
(1.96) 2 .55(1 .55) .65(1 .65)
.0052
1.82476
72,990.4 72, 991
.000025
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . The confidence interval is:
( x1 x2 ) z.05
b.
s12 s22
2.1 3.0
(12.2 8.3) 1.645
n1 n2
135 148
3.90 .31 (3.59, 4.21)
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
( x1 x2 )
2
1
2
2
s
s
n1 n2
(12.2 8.3) 0
2.1 3.0
135 148
20.60
The rejection region requires / 2 .01 / 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
Since the observed value of the test statistic falls in the rejection region ( z 20.60 2.58) , H0 is
rejected. There is sufficient evidence to indicate that 1 2 at .01 .
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 .
n1 n2
( z / 2 ) 12 22
( ME ) 2
(1.645) 2 (2.1 3.0)
345.02 346
.2 2
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.94
a.
457
This is a paired difference experiment. Some preliminary calculations are:
Pair
Difference
(Pop. 1 - Pop. 2)
1
2
3
4
5
6
4
4
3
2
nd
di
nd
2
i 1
2
di
81 19
nd
5 2.2
sd2 i 1
nd 1
5 1
2
d
nd
i
19
d i 1 3.8
nd
5
sd 2.2 1.4832
H 0 : d 0
H a : d 0
The test statistic is t
d 0
sd
nd
3.8 0
1.4832 / 5
5.73
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df nd 1 5 1 4 . From Table III, Appendix D, t.025 2.776 . The rejection region is
t 2.776 or t 2.776 .
Since the observed value of the test statistic falls in the rejection region (t 5.73 2.776) , H0 is
rejected. There is sufficient evidence to indicate that the population means are different at .05 .
b.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . Therefore, we would use the same
t-value as above, t.025 2.776 . The confidence interval is:
xd t / 2
8.95
sd
nd
3.8 3.8 2.776
1.4832
5
3.8 1.84 (1.96, 5.64)
c.
The sample of differences must be randomly selected from a population of differences which has a
normal distribution.
a.
Let 1 average size of the right temporal lobe of the brain for the short-recovery group and
2 average size of the right temporal lobe of the brain for the long-recovery group.
The target parameter is 1 2 . We must assume that the two samples are random and independent,
the two populations being sampled from are approximately normal, and the two population variances
are equal.
b.
Let p1 proportion of athletes who have a good self-image of their body and p2 proportion of nonathletes who have a good self-image of their body.
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Chapter 8
The target parameter for this comparison is p1 p2 . We must assume that the two samples are
random and independent and that the sample sizes are sufficiently large.
c.
Let 1 average weight of eggs produced by a sample of chickens on regular feed and
2 average weight of eggs produced by a sample of chickens fed a diet supplemented by corn oil.
Let d 1 2 average difference in weight between eggs produced by the chickens on regular
feed and then on a diet supplemented with corn oil.
The target parameter is d . We must assume that we have a random sample of differences and that
the population of differences is approximately normal if the sample size is small. If the sample size
of differences is greater than 30, we do not need to assume that the population of differences is
normal.
8.96
a.
Let p1 proportion of Opening Doors students enrolled full time and p2 proportion of traditional
students enrolled full time. The target parameter for this comparison is p1 p2 .
b.
Let 1 mean GPA of Opening Doors students and 2 mean GPA of traditional students. The
target parameter for this comparison is 1 2 .
8.97
a.
Let 1 mean score for males and 2 mean score for females. For confidence coefficient .90,
.10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 . The 90% confidence
interval is:
( x1 x2 ) z / 2
12
n1
22
n2
(39.08 38.79) 1.645
6.732 6.942
127
114
0.29 1.452 (1.162, 1.742)
We are 90% confident that the difference in mean service-rating scores between males and females is
between -1.162 and 1.742.
b.
To determine if the service-rating score variances differ by gender, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
s 2 6.94 2
larger sample variance
22
1.06
smaller sample variance s1 6.732
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution with numerator
df 2 n2 –1 114 –1 113 and denominator df 1 n1 –1 127 –1 126 . Using MINITAB,
we get:
Inverse Cumulative Distribution Function
F distribution with 113 DF in numerator and 126 DF in denominator
P( X <= x )
0.95
x
1.35141
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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
459
F.05 1.35 . The rejection region is F 1.35 .
Since the observed value of the test statistic does not fall in the rejection region (F 1.06 1.35) , H0
is not rejected. There is insufficient evidence to indicate the service-rating score variances differ by
gender at .10 .
c.
8.98
Since we did not reject H0 in part b, the confidence interval in part a is valid. Because 0 falls in the
90% confidence interval, we are 90% confident that there is no difference in the mean service-rating
scores between males and females.
Using MINITAB, some preliminary calculations are:
Descriptive Statistics: Spillage
Variable
Spillage
a.
Cause
Collision
Fire
Grounding
HullFail
Unknown
N
10
11
11
9
1
Mean
76.6
75.0
53.73
63.7
25.000
StDev
70.4
61.9
29.45
63.1
*
Variance
4950.9
3829.6
867.22
3984.5
*
Q1
35.0
33.0
36.00
31.0
*
Median
41.5
50.0
41.00
36.0
25.000
Q3
102.0
82.0
62.00
73.5
*
Let 1 mean spillage for accidents caused by collision and 2 mean spillage for accidents caused
by fire/explosion.
s 2p
n1 1 s12 n2 1 s22 10 1 4,950.9 11 1 3,829.6
n1 n2 2
10 11 2
4,360.7421
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df n1 n2 – 2 10 11– 2 19 , t.05 1.729 . The confidence interval is:
1 1
1 1
( x1 x2 ) t.05 s 2p (76.6 75.0) 1.729 4,360.7421
10 11
n1 n2
1.6 49.89 (48.29, 51.49)
b.
Let 3 mean spillage for accidents caused by grounding and 4 mean spillage for accidents
caused by hull failure.
s 2p
n3 1 s32 n4 1 s42 11 1 867.22 9 1 3,984.5
n3 n4 2
11 9 2
2, 252.6778
To determine if the mean spillage amount for accidents caused by grounding is different from the
mean spillage amount caused by hull failure, we test:
H 0 : 3 4 0
H a : 3 4 0
The test statistic is t
x3 x4 Do
1 1
s 2p
n3 n4
53.73 63.7 0
1 1
2, 252.6778
11 9
6.61
.47
17.1342
Copyright © 2014 Pearson Education, Inc.
Chapter 8
The rejection region requires / 2 .05/ 2 .025 in each tail of the t-distribution with
df n1 n2 – 2 11 9 – 2 18 . From Table III, Appendix D, t.025 2.101 . The
is t 2.101 or t 2.101 .
rejection region
Since the observed value of the test statistic does not fall in the rejection region (t .47 2.101) ,
H0 is not rejected. There is insufficient evidence to indicate the mean spillage amount for accidents
caused by grounding is different from the mean spillage amount caused by hull failure at .05 .
c.
The necessary assumptions are:
We must assume that the distributions from which the samples were selected are approximately
normal, the samples are independent, and the variances of the two populations are equal.
Below are the histograms for each of the samples:
Histogram of Spillage
Normal
-60
Collision
0
60
120
180
240
Fire
6.0
4.5
3.0
Frequency
460
1.5
Grounding
HullFail
0.0
6.0
4.5
3.0
Collision
76.6
Mean
StDev 70.36
N
10
Fire
Mean
75
StDev 61.88
N
11
Grounding
Mean 53.73
StDev 29.45
N
11
HullFail
Mean 63.67
StDev 63.12
N
9
1.5
0.0
-60
0
60
120
180
240
Spillage
Panel variable: Cause
Based on the shapes of the histograms, it does not appear that the data are normally distributed.
Also, we know that if the data are normally distributed, then the Interquartile Range, IQR, divided by
the standard deviation should be approximately 1.3. We will compute IQR/s for each of the samples:
Collision:
Fire:
Grounding:
Hull Failure:
IQR / s 102.0 – 35.0 / 70.4 .95
IQR / s 82 – 33 / 61.9 .79
IQR / s 62.0 – 36 / 29.45 .88
IQR / s 73.5 – 31 / 63.1 .67
Since all of these ratios are quite a bit smaller than 1.3, it indicates that none of the samples come
from normal distributions.
Thus, it appears that the assumption of normal distributions is violated.
The sample standard deviations are:
Collision:
Fire:
s 70.4
s 61.9
Grounding:
Hull Failure:
Copyright © 2014 Pearson Education, Inc.
s 29.45
s 63.1
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
461
Without doing formal tests, it appears that the variances of the groups Collision, Fire, and Hull
Failure are probably not significantly different. However, it appears that the variance for the
Grounding group is smaller than the others.
d.
Let 12 variance of spillage for accidents caused by collision and 32 variance of spillage for
accidents caused by grounding.
To determine if the variances of the amounts of spillage due to collision and grounding differ, we test:
H 0 : 12 32 0
H a : 12 32 0
The test statistic is F
Larger sample variance s12 4, 950.9
5.71
Smaller sample variance s32
867.22
The rejection region requires / 2 .02 / 2 .01 in the upper tail of the F distribution with
1 n1 –1 10 –1 9 and 2 n3 –1 11–1 10 . From Table VIII, Appendix D, F.01 4.94 . The
rejection region is F 4.94 .
Since the observed value of the test statistic falls in the rejection region (F 5.71 4.94) , H0 is
rejected. There is sufficient evidence to indicate the variances of the amounts of spillage due to
collision and grounding differ at .02 .
8.99
a.
Yes. The sample mean of the virtual-reality group is 10.67 points higher than the sample mean of the
simple user interface group.
b.
Let 1 mean improvement score for the virtual-reality group and 2 mean improvement score for
the simple user interface group.
To determine if the mean improvement scores for the virtual-reality group is higher than that for the
simple user interface group, we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is z
x1 x2 Do
n1 n2
2
1
2
2
43.15 32.48 0
12.57
9.26
45
45
2
2
10.67
4.58
2.3274
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 4.58 1.645) , H0 is
rejected. There is sufficient evidence to indicate the mean improvement scores for the virtual-reality
group is higher than that for the simple user interface group .05 .
Copyright © 2014 Pearson Education, Inc.
462
Chapter 8
8.100
a.
Since there is much variability among cars, by using matched pairs, we can block out the variability
among the cars and compare the means of the 2 types of shocks.
b.
Let 1 mean strength of manufacturer’s shock and 2 mean strength of competitor’s shock. Also,
let d 1 2 .
Using MINITAB the descriptive statistics are:
Descriptive Statistics: Manufacturer, Competitor, Di
Variable
Manufacturer
Competitor
Di
N
6
6
6
Mean
10.717
10.300
0.4167
StDev
1.752
1.818
0.1329
Minimum
8.800
8.400
0.2000
Q1
9.400
8.850
0.3500
Median
10.100
9.700
0.4000
Q3
12.675
12.250
0.5250
Maximum
13.200
13.000
0.6000
To determine if there is a difference in the mean strength of the two types of shocks after 20,000
miles, we test:
H 0 : d 0
H a : d 0
The test statistic is t
d 0 .4167 0
7.68
.1329
sd
6
nd
The rejection region requires / 2 .05 / 2 .025 in each tail of the t-distribution with
df nd 1 6 1 5 . From Table III, Appendix D, t.025 2.571 . The rejection region is
t 2.571 or t 2.571 .
Since the observed value of the test statistic falls in the rejection region (t 7.68 2.571) , H0 is
rejected. There is sufficient evidence to indicate a difference in the mean strength of the two types of
shocks after 20,000 miles at .05 .
c.
Using MINITAB:
Cumulative Distribution Function
Student's t distribution with 5 DF
x
7.68
P( X <= x )
0.999702
The observed significance level is
P(t 7.68) P(t 7.68) 2P(t 7.68) 2(1 .999702) .000596
d.
We must assume that the population of differences is normally distributed and that the sample is
random.
e.
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix B, with
df nd –1 6 –1 5 , t.025 2.571 . The 95% confidence interval is:
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
d z.025
sd
nd
.4167 2.571
.1329
6
463
.4167 .1395 (.2772, .5562)
We are 95% confident that the difference in mean strength between the two types of shocks after
20,000 miles is between .2772 and .5562.
f.
Some preliminary calculations are:
s 2p
(n1 1) s12 (n2 1) s22 (6 1)1.7522 (6 1)1.8182
3.1873
n1 n2 2
662
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix B, with
df n1 n2 – 2 6 6 – 2 11 , t.025 2.228 . The 95% confidence interval is:
1 1
1 1
( 1 2 ) t.025 s 2p (10.717 10.300) 2.228 3.1873
n
n
6 6
2
1
.417 2.2965 (1.8795, 2.7135)
We are 95% confident that the difference in mean strength between the two types of shocks after
20,000 miles is between -1.8795 and 2.7135.
8.101
g.
The interval assuming independent sample in part f is (1.8795, 2.7135) while the interval assuming
paired differences in part e is (.2772, .5562). The interval assuming independent samples is much
wider because the interval for the paired-difference eliminated the car to car differences. The interval
from part e gives more information because the interval is narrower.
h.
No. If the data were collected using a paired experiment, then the data must be analyzed as a paired
experiment.
a.
The data should be analyzed as a paired difference experiment because each actor who
won an Academy Award was paired with another actor with similar characteristics who did not win
the award.
b.
Let 1 mean life expectancy of Academy Award winners and 2 mean life expectancy of nonAcademy Award winners. Let d 1 2 . To compare the mean life expectancies of Academy
Award winners and non-winners, we test:
H 0 : d 0
H a : d 0
c.
Since the p-value was so small, there is sufficient evidence to indicate the mean life expectancies of
the Academy Award winners and non-winners are different for any value of .003 . Since the
sample mean life expectancy of Academy Award winners is greater than that for non-winners, we can
conclude that Academy Award winners have a longer mean life expectancy than non-winners.
Copyright © 2014 Pearson Education, Inc.
464
Chapter 8
8.102
a.
Let 1 mean carat size of diamonds certified by GIA and 2 mean carat size of
diamonds certified by HRD. For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From
Table II, Appendix D, z.025 1.96 . The 95% confidence interval is:
12
( x1 x2 ) z / 2
n1
22
n2
(.6723 .8129) 1.96
.24562 .18312
151
79
.1406 .0563 (.1969, .0843)
b.
We are 95% confident that the difference in mean carat size between diamonds certified by GIA and
those certified by HRD is between -.1969 and -.0843. Since both end points are negative, the mean
carat size of diamonds certified by HRD is larger than the mean carat size of diamonds certified by
GIA by anywhere from .0843 and .1969 carats.
c.
Let 3 mean carat size of diamonds certified by IGI.
( x1 x3 ) z / 2
d.
e.
f.
12
n1
32
n3
(.6723 .3665) 1.96
.2456 2 .21632
151
78
.3058 .0620 (.2438, .3678)
We are 95% confident that the difference in mean carat size between diamonds certified by GIA and
those certified by IGI is between .2438 and .3678. Since both end points are positive, the mean carat
size of diamonds certified by GIA is larger than the mean carat size of diamonds certified by IGI by
anywhere from .2438 and .3678 carats.
( x2 x3 ) z / 2
22
n2
32
n3
(.8129 .3665) 1.96
.18312 .21632
79
78
.4464 .0627 (.3837, .5091)
We are 95% confident that the difference in mean carat size between diamonds certified by HRD and
those certified by IGI is between .3837 and .5091. Since both end points are positive, the mean carat
size of diamonds certified by HRD is larger than the mean carat size of diamonds certified by IGI by
anywhere from .3837 and .5091 carats.
Let 12 variance of carat size for diamonds certified by GIA, 22 variance of carat size for diamonds
certified by HRD, and 32 variance of carat size for diamonds certified by IGI.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
g.
465
To determine if the variation in carat size differs for diamonds certified by GIA and diamonds certified
by HRD, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
Larger sample variance s12 .24562
1.799
Smaller sample variance s22 .18312
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n1 1 151 1 150 and 2 n2 1 79 1 78 . Using MINITAB, F.025 1.494 . The rejection
region is F 1.494 .
Since the observed value of the test statistic falls in the rejection region (F 1.799 1.494) , H0 is
rejected. There is sufficient evidence to indicate the variation in carat size differs for diamonds
certified by GIA and those certified by HRD at .05 .
h.
To determine if the variation in carat size differs for diamonds certified by GIA and diamonds certified
by IGI, we test:
H 0 : 12 32
H a : 12 32
The test statistic is F
Larger sample variance s12 .24562
1.289
Smaller sample variance s32 .21632
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n1 1 151 1 150 and 2 n3 1 78 1 77 . Using MINITAB, F.025 1.497 . The rejection
region is F 1.497 .
Since the observed value of the test statistic does not fall in the rejection region (F 1.289 1.497) ,
H0 is not rejected. There is insufficient evidence to indicate the variation in carat size differs for
diamonds certified by GIA and those certified by IGI at .05 .
i.
To determine if the variation in selling price differs for diamonds certified by HRD and
diamonds certified by IGI, we test:
H 0 : 22 32
H a : 22 32
Copyright © 2014 Pearson Education, Inc.
Chapter 8
The test statistic is F
Larger sample variance s22 28982
1.87
Smaller sample variance s32 21212
The rejection region requires / 2 .05 / 2 .025 in the upper tail of the F-distribution with
1 n2 1 79 1 78 and 2 n3 1 78 1 77 . Using MINITAB, F.025 1.567 . The rejection
region is F 1.567 .
Since the observed value of the test statistic falls in the rejection region (F 1.87 1.567) , H0 is
rejected. There is sufficient evidence to indicate the variation in selling price differs for diamonds
certified by HRD and those certified by IGI at .05 .
j.
We will look at the 4 methods for determining if the data are normal. First, we will look at
histograms of the data. Using MINITAB, the histograms of the carat sizes for the 3 certification
bodies are:
Histogram of CARAT
0.30
GIA
0.45
0.60
HRD
0.75
0.90
1.05
30
20
Frequency
466
10
0
IGI
30
20
10
0
0.30
0.45
0.60
0.75
0.90
1.05
CARAT
Panel variable: CERT
From the histograms, none of the data appear to be mound-shaped. It appears that none of the data
sets are normal.
Next, we look at the intervals x s , x 2 s , x 3 s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal.
For GIA:
x s .6723 .2456 (.4267, .9179) 84 of the 151 values fall in this interval. The proportion is
.56. This is much smaller than the .68 we would expect if the data were normal.
x 2s .6723 2(.2456) .6723 .4912 (.1811, 1.1635) 151 of the 151 values fall in this
interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were
normal.
x 3s .6723 3(.2456) .6723 .7368 (.0645, 1.4091) 151 of the 151 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
467
From this method, it appears that the data are not normal.
For IGI:
x s .3665 .2163 (.1502, .5828) 69 of the 78 values fall in this interval. The proportion is
.88. This is much larger than the .68 we would expect if the data were normal.
x 2s .3665 2(.2163) .3665 .4326 (.0661, .7991) 74 of the 78 values fall in this
interval. The proportion is .95. This is the same as the .95 we would expect if the data were normal.
x 3s .3665 3(.2163) .3665 .6489 (.2824, 1.0154) 78 of the 78 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
For HRD:
x s .8129 .1831 (.6298, .9960) 30 of the 79 values fall in this interval. The proportion is
.38. This is much smaller than the .68 we would expect if the data were normal.
x 2s .8129 2(.1831) .8129 .3662 (.4467, 1.1791) 79 of the 79 values fall in this
interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were
normal.
x 3s .8129 3(.1831) .8129 .5493 (.2636, 1.3622) 79 of the 79 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
Next, we look at the ratio of the IQR to s. Using MINITAB, the quartiles are:
Descriptive Statistics: CARAT
Variable
CARAT
CERT
GIA
HRD
IGI
N
151
79
78
Mean
0.6723
0.8129
0.3665
StDev
0.2456
0.1831
0.2163
Q1
0.5000
0.6500
0.2100
Median
0.7000
0.8100
0.2900
Q3
0.9000
1.0000
0.4850
For GIA:
IQR QU – QL .9 .4 .5 .
IQR
.5
2.036 This is much larger than the 1.3 we would expect if the data were normal.
s
.2456
This method indicates the data are not normal.
Copyright © 2014 Pearson Education, Inc.
Chapter 8
For IGI:
IQR QU – QL .485 .210 .275 .
IQR .275
1.27 This is very close to the 1.3 we would expect if the data were normal. This
s
.2163
method indicates the data might be normal.
For HRD:
IQR QU – QL 1.00 .65 .35 .
IQR
.35
1.91 This is much larger than the 1.3 we would expect if the data were normal. This
s
.1831
method indicates the data are not normal.
Finally, using MINITAB, the normal probability plots are:
Probability Plot of CARAT
Normal - 95% CI
-0.5
GIA
0.0
0.5
1.0
1.5
HRD
99.9
99
90
50
10
Percent
468
1
0.1
IGI
99.9
99
90
50
10
1
0.1
-0.5
0.0
0.5
1.0
1.5
CARAT
GIA
0.6723
Mean
StDev
0.2456
N
151
AD
3.268
P-Value <0.005
HRD
Mean
0.8129
StDev
0.1831
N
79
AD
3.405
P-Value <0.005
IGI
Mean
0.3665
StDev
0.2163
N
78
AD
5.561
P-Value <0.005
Panel variable: CERT
Since the data do not form a straight line for GIA, the data are not normal.
Since the data do not form a straight line for IGI, the data are not normal.
Since the data do not form a straight line HRD, the data are not normal.
From the 4 different methods, all indications are that the carat size data are not normal for any of the
certification bodies.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
8.103
a.
469
Let 1 mean response by noontime watchers and 2 mean response by non-noontime watchers.
To determine if the mean response differs for noontime and non-noontime watchers, we test:
H 0 : 1 2 0
H a : 1 2 0
b.
Since the p-value ( p .02) is less than .05 , H0 is rejected. There is sufficient evidence to indicate
the mean response differs for noontime and non-noontime watchers at .05 .
c.
Since the p-value ( p .02) is greater than .01 , H0 is not rejected. There is insufficient evidence
to indicate the mean response differs for noontime and non-noontime watchers at .01 .
Since the two sample means are so close together, there appears to be no “practical” difference
between the two means. Even if there is a statistically significant difference between the two means,
there is no practical difference.
d.
8.104
a.
Let p1 proportion of managers and professionals who are male and p2 proportion of part-time
MBA students who are male. To see if the samples are sufficiently large:
n1 pˆ1 162(.95) 153.9 and n1qˆ1 162(.05) 8.1
n2 pˆ 2 109(.689) 75.101 and n2 qˆ2 109(.311) 33.899
Since n1qˆ1 8.1 15, the normal approximation may not be adequate. We will go ahead and perform
the test.
First, we calculate the overall estimate of the common proportion under H0.
pˆ
n1 pˆ1 n2 pˆ 2 162(.95) 109(.689)
.845
n1 n2
162 109
To determine if the population of managers and professionals consists of more males than the
part-time MBA population, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.95 .689) 0
1
1
.845(.155)
162 109
5.82
The rejection region requires .05 in the upper tail of the z-distribution. From Table II, Appendix
D, z.05 1.645 . The rejection region is z 1.645 .
Since the observed value of the test statistic falls in the rejection ( z 5.82 1.645) , H0 is rejected.
There is sufficient evidence to indicate that population of managers and professionals consists of
more males than the part-time MBA population at .05 .
Copyright © 2014 Pearson Education, Inc.
470
Chapter 8
b.
We had to assume:
1. Both samples were randomly selected
2. Both sample sizes are sufficiently large.
c.
Let p1 proportion of managers and professionals who are married and p2 proportion of part-time
MBA students who are married. To see if the samples are sufficiently large:
n1 pˆ1 162(.912) 147.7 and n1qˆ1 162(.088) 14.3
n2 pˆ 2 109(.534) 58.2 and n2 qˆ2 109(.466) 50.8
Since n1qˆ1 14.3 15, the normal approximation may not be adequate. We will go ahead and perform
the test.
First, we calculate the overall estimate of the common proportion under H0.
pˆ
n1 pˆ1 n2 pˆ 2 162(.912) 109(.534)
.760
n1 n2
162 109
To determine if the population of managers and professionals consists of more married individuals
than the part-time MBA population, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.912 .534) 0
1
1
.760(.240)
162 109
7.14
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic falls in the rejection ( z 7.14 2.33) , H0 is rejected.
There is sufficient evidence to indicate that population of managers and professionals consists of
more married individuals than the part-time MBA population at .01 .
d.
We had to assume:
1. Both samples were randomly selected
2. Both sample sizes are sufficiently large.
8.105
a.
Let p1 proportion of employed individuals who had a routine checkup in the past year and p2
proportion of unemployed individuals who had a routine checkup in the past year. The researchers
are interested in whether there is a difference in these two proportions, so the parameter of interest is
p1 p2 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
b.
471
To determine if there is a difference in the proportions of employed and unemployed individuals who
had a routine checkup in the past year, we test:
H 0 : p1 p2 0
H a : p1 p2 0
c.
Some preliminary calculations are:
pˆ1
x1
642
.563
n1 1,140
pˆ
x1 x2
642 740
1,382
.615
n1 n2 1,140 1,106 2, 246
( pˆ1 pˆ 2 ) 0
8.106
qˆ 1 pˆ 1 .615 .385
(.563 .669) 0
5.16
1
1 1
1
.615(.385)
ˆ ˆ
pq
1,140 1,106
n1 n2
The rejection region requires / 2 .01 / 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
The test statistic is z
d.
x2
740
.669
n2 1,106
pˆ 2
e.
The p-value is p P( z 5.16) P( z 5.16) (.5 .5) (.5 .5) 0 . This agrees with what was
reported.
f.
Since the observed value of the test statistic falls in the rejection region ( z 5.16 2.58) , H0 is
rejected. There is sufficient evidence to indicate a difference in the proportion of employed and
unemployed individuals who had routine checkups in the past year at .01 .
a.
Let 1 mean annual percentage turnover for U.S. plants and 2 mean annual percentage turnover
for Japanese plants. The descriptive statistics are:
Descriptive Statistics: US, Japan
Variable
US
Japan
s 2p
N
5
5
Mean
6.562
3.118
Median
6.870
3.220
StDev
1.217
1.227
Minimum
4.770
1.920
Maximum
8.000
4.910
Q1
5.415
1.970
Q3
7.555
4.215
( n1 1) s12 ( n2 1) s22 (5 1)1.217 2 (5 1)1.227 2
1.4933
n1 n2 2
552
To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants,
we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is t
( x1 x2 ) D0
1 1
s
n1 n2
2
p
(6.562 3.118) 0
1 1
1.4933
5 5
4.456
Copyright © 2014 Pearson Education, Inc.
472
Chapter 8
The rejection region requires .05 in the upper tail of the t-distribution with
df n1 n2 – 2 5 5 – 2 8 . From Table III, Appendix D, t.05 1.860 . The rejection region is
t 1.860 .
Since the observed value of the test statistic falls in the rejection region (t 4.456 1.86) , H0 is
rejected. There is sufficient evidence to indicate the mean annual percentage turnover for U.S. plants
exceeds that for Japanese plants at .05 .
b.
The p-value p P(t 4.456) . Using MINITAB, with df n1 n2 – 2 5 5 – 2 8 ,
Cumulative Distribution Function
Student's t distribution with 8 DF
x
4.456
P( X <= x )
0.998939
p P(t 4.456) 1 .9989 .0011 . Since the p-value is so small, there is evidence to reject H0 for
.0011 .
c.
The necessary assumptions are:
1.
2.
3.
Both sampled populations are approximately normal.
The population variances are equal.
The samples are randomly and independently sampled.
There is no indication that the populations are not normal. The sample sizes are so small, it is hard to
check the assumptions. Both sample variances are similar, so there is no evidence the population
variances are unequal. There is no indication the assumptions are not valid.
8.107
a.
The two populations of interest are all male cell phone users and all female cell phone users.
b.
The estimate of the proportion of men who sometimes do not drive safely while talking or texting on
a cell phone is pˆ1 .32 . The estimate of the proportion of women is pˆ 2 .25 .
c.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D,
z.05 1.645 . A 90% confidence interval for the difference between the proportions of men and
women who sometimes do not drive safely while talking or texting on a cell phone is:
( pˆ1 pˆ 2 ) z.05
pˆ1qˆ1 pˆ 2 qˆ2
.32(.68) .25(.75)
(.32 .25) 1.645
.07 .041 (.029, .111)
n1
n2
643
643
d.
Since the interval does not contain 0, then there is a sufficient evidence to indicate that there is a
difference between the proportions of men and women who sometimes do not drive safely while
talking or texting on a cell phone. Also, the interval contains all positive values so we can conclude
that men are more likely than women to sometimes not drive safely while talking or texting on a cell
phone.
e.
The estimate of the proportion of men who used their cell phone in an emergency is pˆ1 .71 . The
estimate of the proportion of women is pˆ 2 .77 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
f.
473
Let p1 proportion of men who used their cell phone in an emergency and p2 the proportion of
women who used their cell phone in an emergency.
Some preliminary calculations are:
pˆ1
x1
x1 n1 pˆ1 643(.71) 456.53
n1
pˆ
x1 x2 456.53 495.11
.74
643 643
n1 n2
pˆ 2
x2
x2 n2 pˆ 2 643(.77) 495.11
n2
qˆ 1 pˆ 1 .74 .26
To determine whether the proportions of men and women who used their cell phones in an emergency
differ, we test:
H 0 : p1 p2 0
H a : p1 p2 0
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
The test statistic is z
(.71 .77) 0
1
1
.74(.26)
643 643
2.45 .
The rejection region requires / 2 .10 / 2 .05 in each tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
Since the observed value of the test statistic falls in the rejection region ( z 2.45 1.645) , H0 is
rejected. There is sufficient evidence to indicate the proportions of men and women who used their
cell phone in an emergency differ at .10 .
8.108
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table II, Appendix D, z.05 1.645 .
We estimate p1 p2 .5 .
z / 2 ( p1q1 p2 q2 )
2
n1 n2
8.109
a.
ME
2
(1.645) 2 .5(.5) .5(.5)
.052
541.205 542
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Purchasers, Nonpurchasers
Variable
Purchase
Nonpurch
s 2p
N
20
20
Mean
39.80
47.20
Median
38.00
52.00
StDev
10.04
13.62
Minimum Maximum
23.00
59.00
22.00
66.00
Q1
32.25
33.50
( n1 1) s12 ( n2 1) s22 (20 1)13.62 2 (20 1)10.04 2
143.153
n1 n2 2
20 20 2
Let 1 mean age of nonpurchasers and 2 mean age of purchasers.
Copyright © 2014 Pearson Education, Inc.
Q3
48.75
58.75
474
Chapter 8
To determine if there is a difference in the mean age of purchasers and nonpurchasers, we
test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is t
( x1 x2 ) 0
1 1
s
n1 n2
2
p
(47.20 39.80) 0
1
1
143.153
20 20
1.956
The rejection region requires / 2 .10 / 2 .05 in each tail of the t-distribution with
df n1 n2 – 2 20 20 – 2 38 . From Table III, Appendix D, t.05 1.684 . The rejection
region is t 1.684 or t 1.684 .
Since the observed value of the test statistic falls in the rejection region (t 1.956 1.684) , H0
is rejected. There is sufficient evidence to indicate the mean age of purchasers and
nonpurchasers differ at .10 .
b.
The necessary assumptions are:
1.
2.
3.
Both sampled populations are approximately normal.
The population variances are equal.
The samples are randomly and independently sampled.
c.
The p-value is p P(t 1.956) P(t 1.956) (.5 .4748) (.5 .4748) .0504 . The probability of
observing a test statistic of this value or more unusual if H0 is true is .0504. Since this value is less
than .10 , H0 is rejected. There is sufficient evidence to indicate there is a difference in the mean
age of purchasers and nonpurchasers.
d.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
d f 38 , t.05 1.684 . The confidence interval is:
1 1
1
1
( x2 x1 ) t.05 s 2p (39.8 47.2) 1.684 143.153
20 20
n1 n2
7.4 6.37 (13.77, 1.03)
We are 90% confident that the difference in mean ages between purchasers and nonpurchasers is
between 13.77 and 1.03.
8.110
Let p1 proportion of larvae that died in containers containing high carbon dioxide levels and p2
proportion of larvae that died in containers containing normal carbon dioxide levels. The parameter of
interest for this problem is p1 p2 , or the difference in the death rates for the two groups.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
475
Some preliminary calculations are:
pˆ
x1 x2 .10(80) .05(80)
.075
n1 n2
80 80
qˆ 1 pˆ 1 .075 .925
To determine if an increased level of carbon dioxide is effective in killing a higher percentage of leaf-eating
larvae, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
80 80
(.10 .05) 0
1 1
.075(.925)
80 80
1.201
The rejection region requires .01 in the upper tail of the z distribution. From Table II, Appendix D,
z.01 2.33 . The rejection region is z 2.33 .
Since the observed value of the test statistic does not fall in the rejection region ( z 1.201 2.33) , H0 is not
rejected. There is insufficient evidence to indicate that an increased level of carbon dioxide is effective in
killing a higher percentage of leaf-eating larvae at .01 .
8.111
a.
Let p1 proportion of African-American drivers searched by the LAPD and p2 proportion of white
drivers searched by the LAPD.
Some preliminary calculations are:
pˆ1
x1 12, 016
.195
n1 61, 688
pˆ
x1 x2
12, 016 5,312
17,328
.103
n1 n2 61, 688 106,892 168, 580
pˆ 2
x2
5,312
.050
n2 106,892
To determine if the proportions of African-American and white drivers searched differs, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
.195 .050
1
1
.103(.897)
61, 688 106,892
94.35
The rejection region requires / 2 .05 / 2 .025 in each tail of the z-distribution. From Table II,
Appendix D, z.025 1.96 . The rejection region is z 1.96 or z 1.96 .
Since the observed value of the test statistic falls in the rejection region ( z 94.35 1.96) , H0 is
rejected. There is sufficient evidence to indicate the proportions of African-American drivers and
white drivers searched differs at .05 .
Copyright © 2014 Pearson Education, Inc.
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Chapter 8
b.
Let p1 proportion of ‘hits’ for African-American drivers searched by the LAPD and p2
proportion of ‘hits’ for white drivers searched by the LAPD.
Some preliminary calculations are:
pˆ1
x1
5,134
.427
n1 12, 016
pˆ
x1 x2
5,134 3, 006
8,140
.470
n1 n2 12, 016 5, 312 17,328
pˆ 2
x2 3, 006
.566
n2 5,312
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D,
z.025 1.96 . The 95% confidence interval is:
( pˆ1 pˆ 2 ) z.025
pˆ1qˆ1 pˆ 2 qˆ2
.427(.573) .566(.434)
(.427 .566) 1.96
n1
n2
12, 016
5,312
.139 .016 (.155, -.123)
We are 95% confident that the difference in ‘hit’ rates between African-American drivers and white
drivers searched by the LAPD is between .155 and .123.
d
51
8.112
i
155
3.04
51
a.
d i 1
b.
We do not need to estimate anything – we know the parameter’s value.
c.
Using MINITAB, the descriptive statistics are:
51
Descriptive Statistics: MATH2011, MATH2001, DiffMath
Variable
MATH2011
MATH2001
DiffMath
N
51
51
51
Mean
536.84
533.80
3.04
StDev
41.84
33.79
13.40
Minimum
457.00
474.00
-31.00
Q1
501.00
505.00
-6.00
Median
527.00
526.00
2.00
Q3
570.00
561.00
12.00
Maximum
617.00
603.00
32.00
Let 1 mean Math SAT score in 2011 and 2 mean Math SAT score in 2001. Then d 1 2 .
To determine if the true mean Math SAT score in 2011 differs from that in 2001, we test:
H 0 : d 0
H a : d 0
The test statistic is z
d Do 3.04 0
1.62
sd
13.40
51
nd
The rejection region requires / 2 .10 / 2 .05 in each tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 . The rejection region is z 1.645 or z 1.645 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
477
Since the observed value of the test statistic does not fall in the rejection region ( z 1.62 1.645) , H0
is not rejected. There is insufficient evidence to indicate the true mean Math SAT score in 2011 is
different than that in 2001 at .10 .
8.113
For probability .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 . Since we
have no prior information about the proportions, we use p1 p2 .5 to get a conservative estimate.
( z / 2 )2 ( p1q1 p2 q2 ) (1.96) .5(1 .5) .5(1 .5) 1.9208
4,802
.0004
.022
ME 2
2
n1 n2
8.114
For confidence level .95, .05 and / 2 .05 / 2 .025 . From Table II, Appendix D, z.025 1.96 . The
standard deviation can be estimated by dividing the range by 4:
Range 4
1
4
4
z / 2 12 22
2
n1 n2
8.115
(ME )2
1.962 (12 12 )
192.08 193
.22
Some preliminary calculations are:
s12
s22
x
x
2
1
2
1
n1
n1 1
x
x
2
2
2252
5 126 31.5
5 1
4
10, 251
2
2
n2
n2 1
227 2
5 45.2 11.3
5 1
4
10,351
Let 12 variance for instrument A and 22 variance for instrument B. Since we wish to determine if there
is a difference in the precision of the two machines, we test:
H 0 : 12 22
H a : 12 22
The test statistic is F
Larger sample variance s12 31.5
2.79
=
Smaller sample variance s22 11.3
The rejection region requires / 2 .10 / 2 .05 in the upper tail of the F-distribution with
1 n1 1 5 1 4 and 2 n2 1 5 1 4 . From Table VI, Appendix D, F.05 6.39 . The rejection
region is F 6.39 .
Since the observed value of the test statistic does not fall in the rejection region (F 2.79 6.39) , H0 is not
rejected. There is insufficient evidence of a difference in the precision of the two instruments at
.10 .
Copyright © 2014 Pearson Education, Inc.
478
8.116
Chapter 8
Let 1 the mean relational intimacy score for participants in the CMC group and 2 the mean relational
intimacy score for participants in the FTF group.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: CMC, FTF
Variable
CMC
FTF
N Mean
24 3.500
24 3.542
StDev
0.780
0.658
Minimum
2.000
2.000
Q1
3.000
3.000
Median
3.500
4.000
Q3
4.000
4.000
Maximum
5.000
5.000
Some preliminary calculations are:
s 2p
n1 1 s12 n2 1 s22 24 1 .7802 24 1 .6582
n1 n2 2
24 24 2
0.5207
To determine if the mean relational intimacy score for participants in the CMC group is lower than the
mean relational intimacy score for participants in the FTF group, we test:
H 0 : 1 2 0
H a : 1 2 0
The test statistic is t
x1 x2 Do
1 1
s 2p
n1 n2
3.500 3.542 0
1
1
.5207
24 24
0.042
.20
.20831
The rejection region requires .10 in the lower tail of the t-distribution with
df n1 n2 – 2 24 24 – 2 46 . From Table III, Appendix D, t.10 1.303 . The rejection region is
t 1.303 .
Since the observed value of the test statistic does not fall in the rejection region (t .20 1.303) , H0 is
not rejected. There is insufficient evidence to indicate that the mean relational intimacy score for
participants in the CMC group is lower than the mean relational intimacy score for participants in the FTF
group at .10 .
8.117
a.
Let C1 mean relational intimacy score for the CMC group on the first meeting and C3 mean
relational intimacy score for the CMC group on the third meeting. Let Cd difference in mean
relational intimacy score between the first and third meetings for the CMC group. To determine if
the mean relational intimacy score will increase between the first and third meetings, we test:
H 0 : Cd 0
H a : Cd 0
b.
The researchers used the paired t-test because the same individuals participated in each of the three
meeting sessions. Thus, the samples would not be independent.
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
479
c.
Since the p-value is so small ( p .003) , H0 would be rejected. There is sufficient evidence to
indicate that the mean relational intimacy score for participants in the CMC group increased from the
first to the third meeting for any value of .003 .
d.
Let F1 mean relational intimacy score for the FTF group on the first meeting and F3 mean
relational intimacy score for the FTF group on the third meeting. Let Fd difference in mean
relational intimacy score between the first and third meetings for the FTF group. To determine if the
mean relational intimacy score will change between the first and third meetings, we test:
H 0 : Fd 0
H a : Fd 0
8.118
e.
Since the p-value is not small ( p .39) , H0 would be not be rejected. There is insufficient evidence
to indicate that the mean relational intimacy score for participants in the FTF group changed from the
first to the third meeting for any value of .39 .
a.
Let 1 mean scale score for employees who report positive spillover of work skills and 2 mean
scale score for employees who did not report positive work spillover. To determine if the mean scale
score for employees who report positive spillover of work skills differs from the mean scale score for
employees who did not report positive work spillover, we test:
H 0 : 1 2 0
H a : 1 2 0
b.
It is appropriate to apply the large sample z-test because there are 114 workers that have been studied
and divided into two groups.
c.
From the printout, the test statistics is t 8.847 (equal variances not assumed) and the p-value is
p .000 .
Since the p-value is less than ( p .000 .05) , H0 is rejected. There is sufficient evidence to
indicate the mean scale score for employees who report positive spillover of work skills is different
from the mean scale score for employees who did not report positive work spillover at .05 .
d.
We are 95% confident that the difference between the mean use of creative ideas scale scores for the
two groups in between .627 and .988. Since interval does not contain 0, then we can say that there is
a significant difference on the mean scale scores between the two groups. Yes, the inference derived
from the confidence interval agrees with that from the hypothesis test.
e.
Let p1 proportion of male workers who reported positive work spillover and p2 proportion of
male workers who did not report positive spillover of work skills. To determine if the proportions of
male workers in the two groups are significantly different, we test:
H 0 : p1 p2 0
H a : p1 p2 0
From the printout, the test statistic is z .75 and the p-value is p .453 . Since the p-value is not
small, there is no evidence to reject H0. There is insufficient evidence to indicate the proportions of
male workers in the two groups are significantly different at any value of .453 .
Copyright © 2014 Pearson Education, Inc.
480
8.119
Chapter 8
Attitude towards the Advertisement:
The p-value is p .091 . There is no evidence to reject H0 for .05 . There is no evidence to indicate the
first ad will be more effective when shown to males for .05 . There is evidence to reject H0 for .10 .
There is evidence to indicate the first ad will be more effective when shown to males for .10 .
Attitude toward Brand of Soft Drink:
The p-value is p .032 . There is evidence to reject H0 for .032 . There is evidence to indicate the
first ad will be more effective when shown to males for .032 .
Intention to Purchase the Soft Drink:
The p-value is p .050 . There is no evidence to reject H0 for .05 . There is no evidence to indicate the
first ad will be more effective when shown to males for .05 . There is evidence to reject H0 for
.050 . There is evidence to indicate the first ad will be more effective when shown to males for
.050 .
No, I do not agree with the author’s hypothesis. The results agree with the author’s hypothesis for only the
attitude toward the Brand using .05 . If we want to use .10 , then the author’s hypotheses are all
supported.
8.120
a.
To determine if the mean salary of all males with post-graduate degrees exceeds the mean salary of
all females with post-graduate degrees, we test:
H 0 : M F
H a : M F
b.
The test statistic is z
( xM xF ) 0
s
8.121
2
xM
s
2
xF
(61, 340 32, 227)
2,185 2 932 2
12.26
c.
The rejection region requires .01 in the upper tail of the z-distribution. From Table II, Appendix
D, z.01 2.33 . The rejection region is z 2.33 .
d.
Since the observed value of the test statistic falls in the rejection region ( z 12.26 2.33) , H0 is
rejected. There is sufficient evidence to indicate the mean salary of all males with post-graduate
degrees exceeds the mean salary of all females with post-graduate degrees at .01 .
a.
Let p1 proportion of 9th grade boys who gambled weekly or daily in 1992 and p2 proportion of 9th
grade boys who gambled weekly or daily in 1998. The researchers are interested in whether there is a
difference in these two proportions, so the parameter of interest is p1 p2 .
Some preliminary calculations are:
pˆ1
x1
4, 684
.218
n1 21, 484
pˆ
x1 x2
4, 684 5,313
9, 997
.224
n1 n2 21, 484 23,199 44, 683
pˆ 2
x2
5,313
.229
n2 23,199
Copyright © 2014 Pearson Education, Inc.
qˆ 1 pˆ 1 .224 .776
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
481
To determine if there is a difference in the proportions of 9th grade boys who gambled weekly or
daily in 1992 and 1998, we test:
H 0 : p1 p2 0
H a : p1 p2 0
The test statistic is z
( pˆ1 pˆ 2 ) 0
1 1
ˆ ˆ
pq
n1 n2
(.218 .229) 0
1
1
.224(.776)
21, 484 23,199
2.79
The rejection region requires / 2 .01 / 2 .005 in each tail of the z-distribution. From Table II,
Appendix D, z.005 2.58 . The rejection region is z 2.58 or z 2.58 .
Since the observed value of the test statistic falls in the rejection region ( z 2.79 2.58) , H0 is
rejected. There is sufficient evidence to indicate a difference in the proportions of 9th grade boys who
gambled weekly or daily in 1992 and 1998 at .01 .
b.
Yes. If samples sizes are large enough, differences can almost always be found. Suppose we
compute a 99% confidence interval. For confidence coefficient .99, .01 and / 2 .01/ 2 .005 .
From Table II, Appendix D, z.005 2.58 . The 99% confidence interval is:
( pˆ1 pˆ 2 ) z / 2
pˆ1qˆ1 pˆ 2 qˆ2
.218(.782) .229(.771)
(.218 .229) 2.58
n1
n2
21, 484
23,199
.011 .010 (.021, .001)
We are 99% confident that the difference in the proportions of 9th grade boys who gambled weekly or
daily in 1992 and 1998 is between .021 and .001.
8.122
a.
We cannot make inferences about the difference between the mean salaries of male and female
accounting/finance/banking professionals because no standard deviations are provided.
b.
To determine if the mean salary for males is significantly greater than that for females, we test:
H 0 : 1 2 0
H a : 1 2 0
The rejection region requires .05 in the upper tail of the z-distribution. From Table II,
Appendix D, z.05 1.645 .
To make things easier, we will assume that the standard deviations for the 2 groups are the same.
The test statistic is z
x1 x2 Do 69, 484 52, 012 0
2
1
n1
2
2
n2
1
1
1400
1400
2
17,836
(.037796)
Copyright © 2014 Pearson Education, Inc.
471,896.2038
482
Chapter 8
In order to reject H0 this test statistic must fall in the rejection region, or be greater than 1.645.
Solving for we get:
z
471,896.2038
1.645
471,896.2038
286,866.99
1.645
Thus, to reject H0 the average of the two standard deviations has to be less than $286,866.99.
8.123
c.
Yes. In fact, reasonable values for the standard deviation will be around $5,000. which is much
smaller than the required $286,866.99.
d.
These data were collected from voluntary subjects who responded to a Web-based survey. Thus, this
is not a random sample, but a self-selected sample. Generally, subjects who respond to surveys tend
to have very strong opinions, which may not be the same as the population in general. Thus, the
results from this self-selected sample may not reflect the results from the population in general.
Let 1 mean output for Design 1, 2 mean output for Design 2, and d 1 2 .
Some preliminary calculations are:
Difference
(Design 1 - Design 2)
53
271
206
266
213
183
118
87
Working Days
8/16
8/17
8/18
8/19
8/20
8/23
8/24
8/25
d
d 1,397 174.625
nd
8
sd2
d2
d
2
nd
nd 1
(1,397)
8
6,548.839
8 1
289, 793
2
sd s 2d 6,548.839 80.925
To determine if Design 2 is superior to Design 1, we test:
H 0 : d 0
H a : d 0
The test statistic is t
d o
sd
nd
174.625 0
80.925
8
6.103
Since no value was given, we will use .05 . The rejection region requires .05 in the lower tail of
the t-distribution with df nd 1 8 1 7 . From Table III, Appendix D, t.05 1.895 . The rejection
region is t 1.895 .
Since the observed value of the test statistic falls in the rejection region (t 6.103 1.895) , H0 is
rejected. There is sufficient evidence to indicate Design 2 is superior to Design 1 at .05 .
Copyright © 2014 Pearson Education, Inc.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
For confidence coefficient .95, .05 and / 2 .05 / 2 .025 . From Table III, Appendix D, with
df nd 1 8 1 7 , t.025 2.365 . A 95% confidence interval for d is:
d t.025
sd
nd
174.625 2.365
80.925
8
174.625 67.666 (242.29, 106.96)
Since this interval does not contain 0, there is evidence to indicate Design 2 is superior to Design 1.
Copyright © 2014 Pearson Education, Inc.
483
Chapter 9
Design of Experiments and
Analysis of Variance
9.1
Since only one factor is utilized, the treatments are the four levels (A, B, C, D) of the qualitative factor.
9.2
The treatments are the combinations of levels of each of the two factors. There are 2 5 10 treatments.
They are:
(A, 50), (A, 60), (A, 70), (A, 80), (A, 90), (B, 50), (B, 60), (B, 70), (B, 80), (B, 90)
9.3
One has no control over the levels of the factors in an observational experiment. One does have control of
the levels of the factors in a designed experiment.
9.4
a.
College GPA's are measured on college students. The experimental units are college students.
b.
Household income is measured on households. The experimental units are households.
c.
Gasoline mileage is measured on automobiles. The experimental units are the automobiles of a
particular model.
d.
The experimental units are the sectors on a computer diskette.
e.
The experimental units are the states.
a.
This is an observational experiment. The economist has no control over the factor levels or
unemployment rates.
b.
This is a designed experiment. The manager chooses only three different incentive programs to
compare, and randomly assigns an incentive program to each of nine plants.
c.
This is an observational experiment. Even though the marketer chooses the publication, he has no
control over who responds to the ads.
d.
This is an observational experiment. The load on the facility's generators is only observed, not
controlled.
e.
This is an observational experiment. One has no control over the distance of the haul, the goods
hauled, or the price of diesel fuel.
a.
The response variable is QB production score.
b.
There is one factor which is draft position.
c.
The treatments are the three levels of draft position – Top 10, between picks 11-50, and after pick 50.
d.
The experimental units are the drafted quarterbacks.
9.5
9.6
484
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 485
9.7
9.8
a.
The experimental units are the firms with CPAs.
b.
The response variable is the firm’s likelihood of reporting sustainability policies.
c.
There are two factors – firm size and firm type.
d.
There are two levels of firm size – large and small. There are two levels of firm type – public and
private.
e.
The treatments are the combinations of the factor levels. There are 2 2 4 treatments – large/public,
large/private, small/public, and small/private.
a.
The experimental units are the accounting alumni.
b.
The response variable is income.
c.
There are 2 factors in the problem: Mach score classification and Gender.
d.
Mach score classification has 3 levels – high, moderate, and low. Gender has 2 levels – male and
female.
e.
There are a total of 2 3 6 treatments in his experiment. The treatments are all of the
Mach score rating-gender combinations.
9.9
9.10
9.11
9.12
a.
The study is designed because the experimental units (study participants) were randomly assigned to
the treatments (gift givers and gift receivers).
b.
The experimental units are the study participants. The response variable is the level of appreciation
measured on a scale from 1 to 7. There is one factor – role. There are two levels of role and thus,
two treatments. The treatments are gift giver or gift receiver.
a.
The response variable in this problem is the consumer’s opinion on the value of the discount offer.
b.
There are two treatments in this problem: Within-store price promotion and between-store price
promotion.
c.
The experimental units are the consumers.
a.
There are 2 factors in this problem, each with 2 levels. Thus, there are a total of 2 2 4 treatments.
b.
The 4 treatments are: (Within-store, home), (Within-store, in store), (Between-store, home), and
(Between-store, in store).
a.
There are 2 factors in the problem: Type of yeast and Temperature. Type of yeast has
2 levels – Brewer’s yeast and baker’s yeast. Temperature has 4 levels – 45o, 48o, 51o and 54oC.
b.
The response variable is the autolysis yield.
c.
There are a total of 2 4 8 treatments in this experiment. The treatments are all the type of yeasttemperature combinations.
d.
This is a designed experiment.
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Chapter 9
9.13
a.
The experimental units for this study are the students in the introductory psychology class.
b.
The study is a designed experiment because the students are randomly assigned to a particular study
group.
c.
There are 2 factors in this problem: Class standing and study group.
d.
Class standing has 3 levels: Low, Medium, and High. Study group has 2 levels: practice test and
review.
e.
There are a total of 3 2 6 treatments. They are: (Low, Review), (Low, Practice exam), (Medium,
Review), (Medium, Practice exam), (High, Review), and (High, Practice exam).
f.
The response variable is the final exam score.
a.
The dependent variable is the dissolution time.
b.
There are 3 factors in this experiment: Binding agent, binding concentration, and relative density.
Binding agent has 2 levels – khaya gum and PVP. Binding concentration has 2 levels .5% and
4.0%. Relative density has 2 levels – high and low.
c.
There could be a total of 2 2 2 8 treatments for this experiment. They are:
9.14
khaya gum, .5%, high
khaya gum, .5%, low
khaya gum, 4.0%, high
khaya gum, 4.0%, low
9.15
9.16
9.17
PVP, .5%, high
PVP, .5%, low
PVP, 4.0%, high
PVP, 4.0%, low
a.
From Table VI with 1 4 and 2 4 , F.05 6.39 .
b.
From Table VIII with 1 4 and 2 4 , F.01 15.98 .
c.
From Table V with 1 30 and 2 40 , F.10 1.54 .
d.
From Table VII with 1 15 and 2 12 , F.025 3.18 .
a.
P ( F 3.48) 1 .05 .95 using Table VI, Appendix D, with 1 5 and 2 9
b.
P F 3.09 .01 using Table VIII, Appendix D, with 1 15 and 2 20
c.
P F 2.40 .05 using Table VI, Appendix D, with 1 15 and 2 15
d.
P ( F 1.83) 1 .10 .90 using Table V, Appendix D, with 1 8 and 2 40
a.
In the second dot diagram #2, the difference between the sample means is small relative to the
variability within the sample observations. In the first dot diagram #1, the values in each of the
samples are grouped together with a range of 4, while in the second diagram #2, the range of values is
8.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 487
b.
For diagram #1,
x1
x 7 8 9 9 10 11 54 9
1
6
n
6
x2
x 12 13 14 14 15 16 84 14
x2
x 10 10 12 16 18 18 84 14
For diagram #2,
x1
c.
x 5 5 7 11 13 13 54 9
1
6
n
6
2
6
n
6
2
6
n
For diagram #1,
6
x 54 84 11.5
x
12
n
SST ni ( xi x )2 6(9 11.5)2 6(14 11.5)2 75
2
i 1
For diagram #2,
SST ni ( xi x )2 6 9 11.5 6 14 11.5 75
2
2
2
i 1
d.
x
x
For diagram #1,
s12
2
542
496
6 2
6 1
1
2
1
n1
n1 1
s22
x
x
x
x
2
2
2
2
n2
n2 1
842
6 2
6 1
1186
SSE (n1 1) s12 (n2 1) s22 (6 1)2 (6 1)2 20
x
x
For diagram #2,
s12
1
2
1
n1
n1 1
2
542
558
6 14.4
6 1
s22
2
2
2
2
n2
n2 1
842
6 14.4
6 1
1248
SSE (n1 1) s12 (n2 1)s22 (6 1)14.4 (6 1)14.4 144
e.
For diagram #1, SS Total SST SSE 75 20 95
SST is
75
SST
100% 100% 78.95% of SS Total
95
SS (Total )
For diagram #2, SS Total SST SSE 75 144 219
SST is
f.
75
SST
100%
100% 34.25% of SS Total
219
SS (Total )
For diagram #1, MST
75
SST
75 ,
k 1 2 1
MSE
20
SSE
2,
n k 12 2
Copyright © 2014 Pearson Education, Inc.
F
MST 75
37.5
2
MSE
488
Chapter 9
For diagram #2, MST
g.
75
144
75
SST
SSE
MST
75 , MSE
14.4 , F
5.21
k 1 2 1
n k 12 2
MSE 14.4
The rejection region for both diagrams requires .05 in the upper tail of the F-distribution with
1 k 1 2 1 1 and 2 n k 12 2 10 . From Table VI, Appendix D, F.05 4.96 . The
rejection region is F 4.96 .
For diagram #1, since the observed value of the test statistic falls in the rejection region
( F 37.5 4.96) , H0 is rejected. There is sufficient evidence to indicate the samples were drawn
from populations with different means at .05 .
For diagram #2, since the observed value of the test statistic falls in the rejection region
( F 5.21 4.96) , H0 is rejected. There is sufficient evidence to indicate the samples were drawn
from populations with different means at .05 .
h.
9.18
We must assume both populations are normally distributed with common variances.
For each dot diagram, we want to test:
H 0 : 1 2
H a : 1 2
From Exercise 9.17,
Diagram #1
x1 9
Diagram #2
x1 9
x2 14
x2 14
2
1
s 2
s12 14.4
s22 2
s22 14.4
a.
Diagram #1
s 2 s22 2 2
sp2 1
2
2
2
(n1 n2 )
Diagram #2
s 2 s22 14.4 14.4
sp2 1
14.4
2
2
(n1 n2 )
In Exercise 9.17, MSE 2
In Exercise 9.17, MSE 14.4
The pooled variance for the two-sample t-test is the same as the MSE for the F-test.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 489
b.
t=
Diagram #1
x1 x2
9 14
=
1 1
1 1
2 +
sp2
6 6
n1 n2
= 6.12
In Exercise 9.17, F 37.5
t=
Diagram #2
x1 x2
9 14
=
1 1
1 1
14.4 +
sp2
6 6
n1 n2
= 2.28
In Exercise 9.17, F 5.21
The test statistic for the F-test is the square of the test statistic for the t-test.
c.
Diagram #1
For the t-test, the rejection region requires
/ 2 .05 / 2 .025 in each tail of the tdistribution with df n1 n2 2 6 6 2
10 . From Table III, Appendix D,
t.025 2.228 .
Diagram #2
For the t-test, the rejection region
is the same as Diagram #1 since
we are using the same , n1, and
n2 for both tests.
The rejection region is t 2.228 or t 2.228 .
In Exercise 9.17, the rejection region for both diagrams using the F-test is F 4.96 .
The tabled F value equals the square of the tabled t value.
d.
Diagram #1
For the t-test, since the test statistic falls in
the rejection region (t 6.12 2.228) ,
we would reject H0. In Exercise 9.17,
using the F-test, we rejected H0.
e.
Diagram #2
For the t-test, since the test statistic falls in
the rejection region (t 2.28 2.228) ,
we would reject H0. In Exercise 9.17,
using the F-test, we rejected H0.
Assumptions for the t-test:
1.
2.
3.
Both populations have relative frequency distributions that are approximately normal.
The two population variances are equal.
Samples are selected randomly and independently from the populations.
Assumptions for the F-test:
1.
2.
3.
Both population probability distributions are normal.
The two population variances are equal.
Samples are selected randomly and independently from the respective populations.
The assumptions are the same for both tests.
Copyright © 2014 Pearson Education, Inc.
490
9.19
Chapter 9
Refer to Exercise 9.17, the ANOVA table is:
For diagram #1:
Source
Treatment
Error
Total
df
1
10
11
SS
75
20
95
MS
75
2
F
37.5
SS
75
144
219
MS
75
14.4
F
5.21
For diagram #2:
Source
Treatment
Error
Total
9.20
a.
df
1
10
11
SSE SS Total SST 46.5 17.5 29.0
df for Error is 41 6 35
MST
SST 17.5
2.9167
6
k 1
MSE
SSE 29.0
.8286
35
nk
F
MST 2.9167
3.52
.8286
MSE
The ANOVA table is:
Source
Treatment
Error
Total
df
6
35
41
SS
17.5
29.0
46.5
MS
2.9167
.8286
F
3.52
b.
The number of treatments is k. We know k 1 6 k 7 .
c.
The total sample size is n 41 1 42 , where 41 df Total.
d.
First, one would number the 42 experimental units from 1 to 42. Then generate over 100 uniform
random numbers from 1 to 42. The first 6 different random numbers will correspond to treatment 1.
The next 6 different random numbers will correspond to treatment 2. Repeat the process for
treatments, 3, 4, 5, 6, and 7.
e.
To determine if there is a difference among the population means, we test:
H 0 : 1 2 7
H a : At least one of the population means differs from the rest
The test statistic is F 3.52 .
The rejection region requires .10 in the upper tail of the F-distribution with numerator
1 k 1 7 1 6 and denominator 2 n k 42 7 35 . From Table V, Appendix D,
F.10 1.98 . The rejection region is F 1.98 .
Since the observed value of the test statistic falls in the rejection region ( F 3.52 1.98) , H0 is
rejected. There is sufficient evidence to indicate a difference among the population means at .10 .
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 491
f.
The observed significance level is P( F 3.52) . Using MINITAB,
Cumulative Distribution Function
F distribution with 6 DF in numerator and 35 DF in denominator
x
3.52
P( X <= x )
0.992128
P( F 3.52) 1 .992128 .007872 .
g.
H 0 : 1 2
H a : 1 2
x1 x2
1 1
MSE
n1 n2
The test statistic is t
3.7 4.1
1 1
.8286
6 6
.76
The rejection region requires / 2 .10 / 2 .05 in each tail of the t-distribution with df n k 35 .
From Table III, Appendix D, t.05 1.697 . The rejection region is t 1.697 and t 1.697 .
Since the observed value of the test statistic does not fall in the rejection region (t .76 1.697) ,
H0 is not rejected. There is insufficient evidence to indicate that 1 and 2 differ at .10 .
h.
For confidence coefficient .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with
df 35 , t.05 1.697 . The confidence interval is:
1 1
1 1
( x1 x2 ) t.05 MSE (3.7 4.1) 1.697 .8286 .4 .892 1.292, .492
6 6
n1 n 2
i.
The confidence interval is:
x1 t.05
9.21
a.
MSE
.8286
3.7 1.697
3.7 .631 3.069, 4.331
6
6
Using MINITAB, the results are:
One-way ANOVA: T1, T2, T3
Source
Factor
Error
Total
DF
2
9
11
S = 1.449
b.
SS
12.30
18.89
31.19
MS
6.15
2.10
R-Sq = 39.44%
F
2.93
P
0.105
R-Sq(adj) = 25.98%
H 0 : 1 2 3
H a : At least two treatment means differ
The test statistic is F 2.931 and the p-value is p .105 .
Copyright © 2014 Pearson Education, Inc.
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Chapter 9
Since the p-value is not less than ( p .105 .01) , H0 is not rejected. There is insufficient evidence
to indicate a difference in the treatment means at .01 .
9.22
a.
The type of design used was a completely randomized design.
b.
The dependent variable is the decrease in the number of promotional cards sold after implementation
of the pay cuts.
c.
There is one factor in this example – type of pay cut. The factor levels are: unilateral wage cut,
general wage cut, and baseline.
d.
Let 1 mean decrease in cards sold for those receiving the “unilateral wage cut”, 2 mean
decrease in cards sold for those receiving the “general wage cut” and 3 mean decrease in cards
sold for those receiving the “baseline”. To determine if the average decrease in cards sold differs
depending on whether one or more of the workers received a pay cut, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
9.23
e.
Since the p-value is less than ( p .001 .01) , H0 is rejected. There is sufficient evidence to
indicate the average decrease in the number of cards sold differs depending on whether one or more
of the workers received a pay cut.
a.
A completely randomized design was used for this study. The experimental units are the bus
customers. The dependent variable is the performance score. There is one factor which is bus depot
with 3 levels – Depot 1, Depot 2, and Depot 3. These factor levels are the treatments of the
experiment.
b.
Yes. The p-value from the ANOVA F-test was p .0001 . For a 95% confidence level, .05 .
Since the p-value is less than ( p .0001 .05) , H0 is rejected. There is sufficient evidence to
indicate the mean customer performance scores differed across the three bus depots at .05 .
9.24
a.
This is a completely randomized design because the subjects were randomly assigned to one of three
groups.
b.
The response variable was the total WTP (willing to pay) value and the treatments were the 3 types of
instructions given.
c.
To determine if the mean total WTP values differed among the three groups, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
d.
One would number the subjects from 1 to 252. Then, use a random number generator to generate 350
to 400 random numbers from 1 to 252 (We need to generate more than 252 random numbers to
account for duplicates.) The first 84 different random numbers will be assigned to group 1, the next
84 different random numbers will be assigned to groups 2, and the rest will be assigned to group 3.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 493
9.25
a.
To determine if the mean LUST discount percentages across the seven states differ, we test:
H 0 : 1 2 7
H a : At least two treatment means differ
b.
From the ANOVA table, the test statistic is F 1.60 and the p-value is p 0.174 .
Since the observed p-value is not less than ( p .174 .10) , H0 is not rejected. There is insufficient
evidence to indicate a difference in the mean LUST discount percentages among the seven states at
.10 .
9.26
a.
To determine if differences exist in the mean rates of return among the three types of fund groups, we
test:
H 0 : 1 2 3
H a : At least two treatment means differ
b.
The rejection region requires .01 in the upper tail of the F-distribution with1 k 1 3 1 2
and 2 n k 90 3 87 . Using MINITAB,
Inverse Cumulative Distribution Function
F distribution with 2 DF in numerator and 87 DF in denominator
P( X <= x )
0.99
x
4.85777
The rejection region is F 4.86 .
c.
9.27
Since the observed value of the test statistic falls in the rejection region ( F 6.965 4.86) , H0 is
rejected. There is sufficient evidence to indicate differences exist in the mean rates of return among
the three types of fund groups at .01 .
To determine if the mean road rage score differs for the three income groups, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
The test statistic is F 3.90 and the p-value is p .01 . Since the p-value is less than .05 , H0 is
rejected. There is sufficient evidence to indicate the mean road rage score differs for the three income
groups for .01 . Since the sample means increase as the income increases, it appears that road rage
increases as income increases.
9.28
a.
The experimental units are the participants in the study.
b.
The dependent variable is the brand recall score.
c.
There is one factor in this study – TV viewing group. Since there is only one factor, the treatments
correspond to the factor levels of this variable. Thus, the treatments are the same as the three levels of
TV viewer group. These 3 levels are violent content code, sex content code, and neutral TV.
Copyright © 2014 Pearson Education, Inc.
494
Chapter 9
d.
The means given are only sample means. If new samples were selected and sample means computed,
the values and order of the sample means could change. In addition, the variances are not taken into
account.
e.
The test statistic is F 20.45 and the p-value is p 0.000 .
f.
Since the p-value is less than ( p 0.000 .01) , Ho is rejected. There is sufficient evidence to
indicate differences in the mean recall scores among the three viewing groups at .01 . The
researchers can conclude that the content of the TV show affects the recall of imbedded commercials.
g.
Using MINITAB, the histograms of the three viewing groups are:
Histogram of VIOLENT, SEX, NEUTRAL
Normal
VIOLENT
SEX
24
VIO LENT
Mean 2.083
StDev 1.730
N
108
30
18
20
SEX
Mean 1.713
StDev 1.664
N
108
Frequency
12
10
6
0
0
-2
0
2
4
6
-2
0
2
4
6
NEUTRAL
30
NEUTRAL
Mean 3.167
StDev 1.811
N
108
20
10
0
0
2
4
6
The assumptions for ANOVA are that the data are approximately normal and the variances of the
groups are the same. From the legend above, the standard deviations are 1.730, 1.664, and 1.811.
These are all very similar. From the plots, the distributions of the violent group and the neutral group
are fairly normal. The distribution of the sex group is skewed to the right and may not be normal.
9.29
a.
This was a completely randomized design.
b.
The experimental units are the college students. The dependent variable is the attitude toward
tanning score and the treatments are the 3 conditions (view product advertisement with models with a
tan, view product advertisement with models with no tan, and view product advertisement with no
model).
c.
Let 1 mean attitude score for those viewing product advertisement with models with a tan,
2 mean attitude score for those viewing product advertisement with models without a tan, and
3 mean attitude score for those viewing product advertisement with no models. To determine if
the treatment mean scores differ among the three groups, we test:
H 0 : 1 2 3
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 495
d.
These are just sample means. To determine if the population means differ, we have to determine how
many standard deviations are between these sample means. In addition, the next time an experiment
was conducted, the sample means could change.
e.
The hypotheses are:
H 0 : 1 2 3
H a : At least two treatment means differ
The test statistic is F 3.60 and the p-value is p .03 . Since the p-value is less than
( p .03 .05) , H0 is rejected. There is sufficient evidence to indicate a difference in the mean
attitude scores among the three groups at .05 .
9.30
f.
We must assume that we have random samples from approximately normal populations with equal
variances.
a.
To determine if the mean knowledge gain differs among the three groups, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
b.
Using MINITAB, the results are:
One-way ANOVA: NO, CHECK, FULL
Source
Factor
Error
Total
DF
2
72
74
S = 2.706
c.
SS
6.64
527.36
534.00
MS
3.32
7.32
R-Sq = 1.24%
F
0.45
P
0.637
R-Sq(adj) = 0.00%
The test statistic is F 0.45 and the p-value is p 0.637 .
Since the p-value ( p 0.637) is larger than any reasonable significance level, H0 is not rejected.
There is insufficient evidence to indicate a difference in the mean knowledge gained among the three
levels of assistance for any reasonable value of .
Practically speaking, there is not one type of assistance that helps students more than another.
9.31
a.
To determine if the mean level of trust differs among the six treatments, we test:
H 0 : 1 2 6
H a : At least two treatment means differ
b.
The test statistic is F 2.21 .
The rejection region requires .05 in the upper tail of the F-distribution with1 k 1 6 1 5
and 2 n k 230 6 224 . Using MINITAB,
Copyright © 2014 Pearson Education, Inc.
496
Chapter 9
Inverse Cumulative Distribution Function
F distribution with 5 DF in numerator and 231 DF in denominator
P( X <= x )
0.95
x
2.25436
The rejection region is F 2.25 .
Since the observed value of the test statistic does not fall in the rejection region ( F 2.21 2.25) , H0
is not rejected. There is insufficient evidence to indicate that at least two mean trusts differ at
.05 .
9.32
c.
We must assume that all six samples are drawn from normal populations, the six population variances
are the same, and that the samples are independent.
d.
I would classify this experiment as designed. Each subject was randomly assigned to receive one of
the six scenarios.
a.
I would classify this experiment as designed. Each subject was randomly assigned to receive one of
the three dosages (DM, honey, nothing). There are 3 treatments in the study corresponding to the 3
dosages: DM, honey, nothing.
b.
Using MINITAB, the output is:
One-way ANOVA: TotalScore versus Treatment
Source
Treatment
Error
Total
DF
2
102
104
SS
318.51
927.72
1246.23
MS
159.25
9.10
S = 3.016
R-Sq = 25.56%
F
17.51
P
0.000
R-Sq(adj) = 24.10%
To determine if differences exist in the mean improvement scores among the 3 treatment groups, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
The test statistic is F 17.51 and the p-value is p 0.000 .
Since the observed p-value ( p 0.000) is less than any reasonable value of , H0 is rejected. There is
sufficient evidence to indicate a difference in the mean improvement scores among the three levels of
dosage for any reasonable value of .
9.33
To determine if the mean THICKNESS differs among the 4 types of housing, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 11.74 and the p-value is p 0.000 . Since the observed p-value ( p 0.000) is less
than any reasonable value of , H0 is rejected. There is sufficient evidence to indicate a difference in the
mean thickness among the four levels of housing for any reasonable value of .
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 497
To determine if the mean WHIPPING CAPACITY differs among the 4 types of housing, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 31.36 and the p-value is p 0.000 . Since the observed p-value ( p 0.000) is less
than any reasonable value of , H0 is rejected. There is sufficient evidence to indicate a difference in the
mean whipping capacity among the four levels of housing for any reasonable value of .
To determine if the mean STRENGTH differs among the 4 types of housing, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 1.70 and the p-value is p 0.193 . Since the observed p-value ( p 0.193) is higher
than any reasonable value of , H0 is not rejected. There is insufficient evidence to indicate a difference
in the mean strength among the four levels of housing for any reasonable value of .
Thus, the mean thickness and the mean percent overrun differ among the 4 housing systems.
n x
3
i i
9.34
a.
x i 1
76
26(10.5)25(3.9) 25(1.4) 405.5
5.3355
76
76
SST ni ( xi x )2 26(10.5 5.3355)2 25(3.9 5.3355)2 25(1.4 5.3355)2 1132.1941
3
i 1
b.
SSE ( n1 1) s12 ( n2 1) s22 ( n3 1) s32 26 – 1 7.6 25 – 1 7.5 25 – 1 7.5
c.
SS Total SST SSE 1,132.1941 4,144 5, 276.1941
2
2
1,444 1,350 1,350 4,144
MST
F
SST 1132.1942
566.0971
3 1
k 1
MSE
4144
SSE
56.7671
n k 76 3
MST 566.0971
9.97
MSE 56.7671
The ANOVA table is:
Source
Groups
Error
Total
d.
2
df
2
73
75
SS
1132.1941
4144.00
5276.1941
MS
566.0971
56.77
F-value
9.97
To determine if the mean drops in anxiety levels differ among the 3 groups, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
Copyright © 2014 Pearson Education, Inc.
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Chapter 9
The test statistic is F 9.97 .
The rejection region requires .01 in the upper tail of the F-distribution with1 k 1 3 1 2
and 2 n k 76 3 73 . From Table VIII, Appendix D, F.01 4.92 . The rejection region is
F 4.92 .
Since the observed value of the test statistic falls in the rejection region ( F 9.97 4.92) , H0 is
rejected. There is sufficient evidence to indicate a difference in the mean drops in anxiety levels
among the three groups at .01 .
e.
The assumption of constant variance is satisfied since the three sample variances are all very
similar (7.6 2 57.76, 7.52 56.25, and 7.52 56.25) .
We are unable to check the normality assumption since we need the individual drops in anxiety levels
to create a histogram or stem-and-leaf plot.
9.35
The number of pairwise comparisons is equal to k (k 1) / 2 .
a.
For k 3 , the number of comparisons is 3(3 1) / 2 3 .
b.
For k 5 , the number of comparisons is 5(5 1) / 2 10 .
c.
For k 4 , the number of comparisons is 4(4 1) / 2 6 .
d.
For k 10 , the number of comparisons is 10(10 1) / 2 45 .
9.36
The experimentwise error rate is the probability of making a Type I error for at least one of all of the
comparisons made. If the experimentwise error rate is .05 , then each individual comparison is made at
a value of which is less than .05.
9.37
A comparisonwise error rate is the error rate (or the probability of declaring the means different when, in
fact, they are not different, which is also the probability of a Type I error) for each individual comparison.
That is, if each comparison is run using .05 , then the comparisonwise error rate is .05.
9.38
a.
From the diagram, the following pairs of treatments are significantly different because they are not
connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, and E and D. All other
pairs of means are not significantly different because they are connected by lines.
b.
From the diagram, the following pairs of treatments are significantly different because they are not
connected by a line: A and B, A and D, C and B, C and D, E and B, E and D, and B and D. All other
pairs of means are not significantly different because they are connected by lines.
c.
From the diagram, the following pairs of treatments are significantly different because they are not
connected by a line: A and E, A and B, and A and D. All other pairs of means are not significantly
different because they are connected by lines.
d.
From the diagram, the following pairs of treatments are significantly different because they are not
connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, E and D, and B and D.
All other pairs of means are not significantly different because they are connected by lines.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 499
9.39
( 1 2 ) : 2, 15
Since all values in the interval are positive, 1 is significantly greater than 2 .
( 1 3 ) : 4, 7
Since all values in the interval are positive, 1 is significantly greater than 3 .
( 1 4 ) : 10, 3
Since 0 is in the interval, 1 is not significantly different from 4 .
However, since the center of the interval is less than 0, 4 is larger than 1 .
( 2 3 ) : 5, 11
Since 0 is in the interval, 2 is not significantly different from 3 .
However, since the center of the interval is greater than 0, 2 is larger than 3 .
( 2 4 ) : 12, 6 Since all values in the interval are negative, 4 is significantly greater than 2 .
( 3 4 ) : 8, 5
Since all values in the interval are negative, 4 is significantly greater than 3 .
Thus, the largest mean is 4 followed by 1 , 2 ,and 3 .
9.40
a.
The number of pairwise comparisons is c
general baseline , and unilateral baseline .
k (k 1) 3(3 1) 6
3 . These are: general unnilateral ,
2
2
2
b.
A multiple comparison procedure is recommended to keep the experimentwise error rate at the
selected level.
c.
Since the confidence interval contains only positive values, there is evidence of a significant
difference in the average decrease in promotional cards sold. Since the values are positive, this
indicates that the average decrease for the baseline is greater than the average decrease for the general
wage cut.
9.41
Since all confidence intervals contain only positive values, this indicates that there is evidence that all
population means are different. The largest mean is for Depot 1, then next highest is Depot 2, and the
lowest is Depot 3.
9.42
a.
The test statistic is F 22.68 and the p-value is p 0.001 . Since the observed p-value ( p 0.001) is
less than any reasonable level we select (.01, .05, or .10), we reject H0. There is sufficient evidence
to indicate a difference in the mean number of alternatives listed among the three emotional states for
any .001 .
b.
The probability of declaring at least one pair of means different when they are not is .05.
c.
The mean number of alternatives listed under the guilty state is significantly higher than mean
number of alternatives listed under the angry and neutral states. There is no difference in the mean
number of alternatives listed under the angry and neutral states.
a.
Tukey’s multiple comparison method is preferred over other methods because it controls
experimental error at the chosen level. It is more powerful than the other methods.
b.
From the confidence interval comparing large-cap and medium-cap mutual funds, we find that 0 is in
the interval. Thus, 0 is not an unusual value for the difference in the mean rates of return between
large-cap and medium-cap mutual funds. This means we would not reject H0. There is insufficient
evidence of a difference in mean rates of return between large-cap and medium-cap mutual funds at
.05 .
9.43
Copyright © 2014 Pearson Education, Inc.
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Chapter 9
c.
From the confidence interval comparing large-cap and small-cap mutual funds, we find that 0 is not
in the interval. Thus, 0 is an unusual value for the difference in the mean rates of return between
large-cap and small-cap mutual funds. This means we would reject H0. There is sufficient evidence
of a difference in mean rates of return between large-cap and small-cap mutual funds at .05 .
d.
From the confidence interval comparing medium-cap and small-cap mutual funds, we find that 0 is in
the interval. Thus, 0 is not an unusual value for the difference in the mean rates of return between
medium-cap and small-cap mutual funds. This means we would not reject H0. There is insufficient
evidence of a difference in mean rates of return between medium-cap and small-cap mutual funds at
.05 .
e.
From the above, the mean rate of return for large-cap mutual funds is the largest, followed by
medium-cap, followed by small-cap mutual funds. The mean rate of return for large-cap funds is
significantly larger than that for small-cap funds. No other differences exist.
f.
We are 95% confident of this decision.
9.44
The mean attitude score for those viewing the product advertisement with models with no tan was
significantly lower than the mean attitude scores of the other two groups. There is no significant difference
in the mean attitude scores between those viewing the product advertisement with models with tans and
those viewing the product advertisement with no models. This indicates that the type of product
advertisement can influence a consumer’s attitude towards tanning.
9.45
a.
The probability of declaring at least one pair of means different when they are not is .01.
b.
There are a total of
k (k 1) 3(3 1)
3 pair-wise comparisons. They are:
2
2
‘Under $30 thousand’ to ‘Between $30 and $60 thousand’
‘Under $30 thousand’ to ‘Over $60 thousand’
‘Between $30 and $60 thousand’ to ‘Over $60 thousand’
c.
Means for groups in homogeneous subsets are displayed in the table:
Subsets
Income
Group
Under $30,000
$30,000-$60,000
Over $60,000
d.
N
379
392
267
1
4.60
2
5.08
5.15
Two of the comparisons in part b will yield confidence intervals that do not contain 0. They are:
‘Under $30 thousand’ to ‘Between $30 and $60 thousand’
‘Under $30 thousand’ to ‘Over $60 thousand’
9.46
k (k 1) 3(3 1)
3.
2
2
a.
The total number of pairwise comparisons made in the Bonferroni analysis is
b.
The confidence interval for comparing the V and S groups is (.923, .183). (Violence is subtracted
from Sex.) Since the confidence interval contains 0, there is no indication that there is a difference in
mean recall between the V and S groups at .05 .
c.
The confidence interval for comparing the V and N groups is (.530, 1.636). (Violence is subtracted
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 501
from Neutral.) Since the confidence interval does not contain 0, there is evidence to indicate there is
a difference in mean recall between the V and N groups at .05 . Since both endpoints are
positive, there is evidence to indicate the mean recall for the Neutral group is significantly higher
than that of the V group.
The confidence interval for comparing the S and N groups is (.901, 2.007). (Sex is subtracted from
Neutral.) Since the confidence interval does not contain 0, there is evidence to indicate there is a
difference in mean recall between the S and N groups at .05 . Since both endpoints are positive,
there is evidence to indicate the mean recall for the Neutral group is significantly higher than that of
the S group.
d.
Yes. When compared to the Neutral group, the mean recalls for the V and S groups are significantly
lower than the mean recall for the Neutral group.
9.47
The mean level of trust for the "no close" technique is significantly higher than that for "the visual close"
and the "thermometer close" techniques. The mean level of trust for the "impending event" technique is
significantly higher than that for the "thermometer close" technique. No other significant differences exist.
9.48
Using MINITAB, the multiple comparisons of the means is shown below:
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 98.06%
Honey subtracted from:
DM
Control
Lower
-4.120
-5.890
Center
-2.381
-4.201
Upper
-0.642
-2.511
----+---------+---------+---------+----(-----*------)
(------*------)
----+---------+---------+---------+-----5.0
-2.5
0.0
2.5
Upper
-0.104
----+---------+---------+---------+----(------*------)
----+---------+---------+---------+-----5.0
-2.5
0.0
2.5
DM subtracted from:
Control
Lower
-3.535
Center
-1.820
None of the three confidence intervals contain 0:
The confidence interval for the difference in mean improvement scores between DM and Honey is (4.120
and 0.642). Since this confidence interval is strictly below zero, this implies that the improvement scores
for Honey are significantly higher than those of DM.
The confidence interval for the difference in mean improvement scores between the Control group and
Honey is (5.890 and 2.511). Since this confidence interval is strictly below zero, this implies that the
improvement scores for Honey are significantly higher than those of the Control Group.
Compared to the Control group (giving no treatment) and DM, honey is a preferable treatment since it has
significantly higher improvement scores. The state is appropriate.
Copyright © 2014 Pearson Education, Inc.
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9.49
Chapter 9
a.
The confidence interval for ( CAGE BARN ) is (.1250, .0323). Since 0 is not contained in this
interval, there is sufficient evidence of a difference in the mean shell thickness between cage and barn
egg housing systems. Since this interval is negative, this implies that the thickness is larger for the
barn egg housing system.
b.
The confidence interval for ( CAGE FREE ) is (.1233, .0307). Since 0 is not contained in this
interval, there is sufficient evidence of a difference in the mean shell thickness between cage and free
range egg housing systems. Since this interval is negative, this implies that the thickness is larger for
the free range egg housing system.
c.
The confidence interval for ( CAGE ORGANIC ) is (.1050, .0123). Since 0 is not contained in this
interval, there is sufficient evidence of a difference in the mean shell thickness between cage and
organic egg housing systems. Since this interval is negative, this implies that the thickness is larger
for the organic egg housing system.
d.
The confidence interval for ( BARN FREE ) is (.0501, .0535). Since 0 is contained in this interval,
there is insufficient evidence of a difference in the mean shell thickness between barn and free range
egg housing systems. Since the center of the interval is greater than 0, the sample mean for barn is
greater than that for free range.
e.
The confidence interval for ( BARN ORGANIC ) is (.0318, .0718). Since 0 is contained in this
interval, there is insufficient evidence of a difference in the mean shell thickness between barn and
organic egg housing systems. Since the center of the interval is greater than 0, the sample mean for
barn is greater than that for organic.
f.
The confidence interval for ( FREE ORGANIC ) is (.0335, .0701). Since 0 is contained in this
interval, there is insufficient evidence of a difference in the mean shell thickness between free range
and organic egg housing systems. Since the center of the interval is greater than 0, the sample mean
for free range is greater than that for organic.
g.
We rank the housing system means as follows:
Housing System:
Cage < Organic < Free < Barn
We are 95% confident that the mean shell thickness for the cage housing system is significantly less
than the mean thickness for the other three housing systems. There is no significant difference in the
mean shell thicknesses among the barn, free range and organic housing systems.
9.50
a.
There are 3 blocks used since Block df b 1 2 and 5 treatments since the treatment df k 1 4 .
b.
There were 15 observations since the Total df n 1 14 .
c.
H 0 : 1 2 5
H a : At least two treatment means differ
MST
9.109
MSE
d.
The test statistic is F
e.
The rejection region requires .01 in the upper tail of the F distribution with1 k 1 5 1 4
and 2 n k b 1 15 5 3 1 8 . From Table VIII, Appendix D, F.01 7.01 . The rejection
region is F 7.01 .
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Design of Experiments and Analysis of Variance 503
f.
Since the observed value of the test statistic falls in the rejection region ( F 9.109 7.01) , H0 is
rejected. There is sufficient evidence to indicate that at least two treatment means differ at .01 .
g.
The assumptions necessary to assure the validity of the test are as follows:
1.
The probability distributions of observations corresponding to all the block-treatment
combinations are normal.
2.
The variances of all the probability distributions are equal.
B2
SSB i CM
i 1 k
a.
SSB
x 49 266.7778
2
b
9.51
where CM
2
i
9
n
172 152 172
266.7778 .8889
3
3
3
SSE SS Total SST SSB 30.2222 21.5555 .8889 7.7778
MST
SST 21.5555
10.7778
k 1
2
MSE
SSE
7.7778
1.9445
n k b 1
4
FT
MST 10.7778
5.54
MSE 1.9445
MSB
FB
SSB .8889
.4445
b 1
2
MSB .4445
.23
MSE 1.9445
The ANOVA table is:
Source
Treatment
Block
Error
Total
b.
df
2
2
4
8
SS
21.5555
.8889
7.7778
30.2222
MS
10.7778
.4445
1.9445
F
5.54
.23
H 0 : 1 2 3
H a : At least two treatment means differ
MST
5.54
MSE
c.
The test statistic is F
d.
A Type I error would be concluding at least two treatment means differ when they do not.
A Type II error would be concluding all the treatment means are the same when at least two differ.
e.
The rejection region requires .05 in the upper tail of the F distribution with1 k 1 3 1 2 and
2 n k b 1 9 3 3 1 4 . From Table VI, Appendix A, F.05 6.94 . The rejection region is
F 6.94 .
Since the observed value of the test statistic does not fall in the rejection region ( F 5.54 6.94) , H0
is not rejected. There is insufficient evidence to indicate at least two of the treatment means differ at
.05 .
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Chapter 9
9.52
a.
The ANOVA Table is as follows:
Source
Treatment
Block
Error
Total
b.
df
2
3
6
11
SS
12.032
71.749
.708
84.489
MS
6.016
23.916
.118
F
50.958
202.586
To determine if the treatment means differ, we test:
H 0 : A B C
H a : At least two treatment means differ
The test statistic is F
MST
50.958
MSE
The rejection region requires .05 in the upper tail of the F distribution with1 k 1 3 1 2
and 2 n k b 1 12 3 4 1 6 . From Table VI, Appendix D, F.05 5.14 . The rejection
region is F 5.14 .
Since the observed value of the test statistic falls in the rejection region ( F 50.958 5.14) , H0 is
rejected. There is sufficient evidence to indicate that the treatment means differ at .05 .
c.
To see if the blocking was effective, we test:
H 0 : 1 2 3 4
H a : At least two block means differ
The test statistic is F
MSB
202.586
MSE
The rejection region requires .05 in the upper tail of the F distribution with1 b 1 4 1 3
and 2 n k b 1 12 3 4 1 6 . From Table VI, Appendix D, F.05 4.76 . The rejection
region is F 4.76 .
Since the observed value of the test statistic falls in the rejection region ( F 202.586 4.76) , H0 is
rejected. There is sufficient evidence to indicate that blocking was effective in reducing the
experimental error at .05 .
d.
From the printouts, we are given the differences in the sample means. The difference between
Treatment B and both Treatments A and C are positive (1.125 and 2.450), so Treatment B has the
largest sample mean. The difference between Treatment A and C is positive (1.325), so Treatment A
has a larger sample mean than Treatment C. So Treatment B has the largest sample mean, Treatment
A has the next largest sample mean and Treatment C has the smallest sample mean.
From the printout, all the means are significantly different from each other.
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Design of Experiments and Analysis of Variance 505
e.
The assumptions necessary to assure the validity of the inferences above are:
1.
2.
9.53
a.
The probability distributions of observations corresponding to all the block-treatment
combinations are normal.
The variances of all the probability distributions are equal.
SST .2(500) 100
SSB .3(500) 150
SSE SS Total SST SSB 500 100 150 250
MST
SST 100
33.3333
k 1 4 1
MSE
SSE
250
250
10.4167
n k b 1 36 4 9 1 24
FT
MST 33.3333
3.20
MSE 10.4167
MSB
FB
SSB 150
18.75
b 1 9 1
MSB
18.75
1.80
MSE 10.4167
To determine if differences exist among the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 3.20 .
The rejection region requires .05 in the upper tail of the F distribution with1 k 1 4 1 3 and
2 n k b 1 36 4 9 1 24 . From Table VI, Appendix D, F.05 3.01 . The rejection region is
F 3.01 .
Since the observed value of the test statistic falls in the rejection region ( F 3.20 3.01) , H0 is
rejected. There is sufficient evidence to indicate differences among the treatment means at .05 .
To determine if differences exist among the block means, we test:
H 0 : 1 2 9
H a : At least two block means differ
The test statistic is F 1.80 .
The rejection region requires .05 in the upper tail of the F distribution with1 b 1 9 1 8
and 2 n k b 1 36 4 9 1 24 . From Table VI, Appendix D, F.05 2.36 . The rejection
region is F 2.36 .
Since the observed value of the test statistic does not fall in the rejection region
( F 1.80 2.36) , H0 is not rejected. There is insufficient evidence to indicate differences among
the block means at .05 .
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Chapter 9
b.
SST .5(500) 250
SSB .2(500) 100
SSE SS Total SST SSB 500 250 100 150
MST
SST 250
83.3333
k 1 4 1
MSE
SSE
150
6.25
n k b 1 36 4 9 1
FT
MST 83.3333
13.33
MSE
6.25
MSB
FB
SSB 100
12.5
b 1 9 1
MSB 12.5
2
MSE 6.25
To determine if differences exist among the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 13.33 .
The rejection region is F 3.01 (same as above).
Since the observed value of the test statistic falls in the rejection region ( F 13.33 3.01) , H0 is
rejected. There is sufficient evidence to indicate differences exist among the treatment means at .05 .
To determine if differences exist among the block means, we test:
H 0 : 1 2 9
H a : At least two block means differ
The test statistic is F 2.00 .
The rejection region is F 2.36 (same as above).
Since the observed value of the test statistic does not fall in the rejection region ( F 2.00 2.36) , H0
is not rejected. There is insufficient evidence to indicate differences exist among the block means at
.05 .
c.
SST .2(500) 100
SSB .5(500) 250
SSE SS Total SST SSB 500 100 250 150
MST
SST 100
33.3333
k 1 4 1
MSE
SSE
150
6.25
n k b 1 36 4 9 1
FT
MST 33.3333
5.33
MSE
6.25
MSB
FB
SSB 250
31.25
b 1 9 1
MSB 31.25
5.00
MSE
6.25
To determine if differences exist among the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
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Design of Experiments and Analysis of Variance 507
The test statistic is F 5.33 .
The rejection region is F 3.01 (same as above).
Since the observed value of the test statistic falls in the rejection region ( F 5.33 3.01) , H0 is
rejected. There is sufficient evidence to indicate differences exist among the treatment means at
.05 .
To determine if differences exist among the block means, we test:
H 0 : 1 2 9
H a : At least two block means differ
The test statistic is F 5.00 .
The rejection region is F 2.36 (same as above).
Since the observed value of the test statistic falls in the rejection region ( F 5.00 2.36) , H0 is
rejected. There is sufficient evidence to indicate differences exist among the block means at .05 .
d.
SST .4(500) 200
SSB .4(500) 200
SSE SS Total SST SSB 500 200 200 100
MST
SST 200
66.6667
k 1 4 1
MSE
SSE
100
4.1667
n k b 1 36 4 9 1
FT
MST 66.6667
16.0
MSE
4.1667
MSB
FB
SSB 200
25
b 1 9 1
MSB
25
6.00
MSE 4.1667
To determine if differences exist among the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 16.0 .
The rejection region is F 3.01 (same as above).
Since the observed value of the test statistic falls in the rejection region ( F 16.0 3.01) , H0 is
rejected. There is sufficient evidence to indicate differences among the treatment means at .05 .
To determine if differences exist among the block means, we test:
H 0 : 1 2 9
H a : At least two block means differ
The test statistic is F 6.00 .
The rejection region is F 2.36 (same as above).
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Chapter 9
Since the observed value of the test statistic falls in the rejection region ( F 6.00 2.36) , H0 is
rejected. There is sufficient evidence to indicate differences exist among the block means at .05 .
e.
SST .2(500) 100
SSB .2(500) 100
SSE SS Total SST SSB 500 100 100 300
MST
SST 100
33.3333
k 1 4 1
MSE
SSE
300
12.5
n k b 1 36 4 9 1
FT
MST 33.3333
2.67
MSE
12.5
MSB
FB
SSB 100
12.5
b 1 9 1
MSB 12.5
1.00
MSE 12.5
To determine if differences exist among the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 2.67 .
The rejection region is F 3.01 (same as above).
Since the observed value of the test statistic does not fall in the rejection region ( F 2.67 3.01) , H0
is not rejected. There is insufficient evidence to indicate differences exist among the treatment means
at .05 .
To determine if differences exist among the block means, we test:
H 0 : 1 2 9
H a : At least two block means differ
The test statistic is F 1.00 .
The rejection region is F 2.36 (same as above).
Since the observed value of the test statistic does not fall in the rejection region ( F 1.00 2.36) , H0
is not rejected. There is insufficient evidence to indicate differences among the block means at
.05 .
9.54
a.
This experimental design is a randomized block design because in part B, the same subjects provided
WTP amounts for insuring both a sculpture and a painting. Each subject had 2 responses.
b.
The dependent (response) variable is the WTP amount. The treatments are the two scenarios
(sculpture and painting). The blocks are the 84 subjects.
c.
To determine if there is a difference in the mean WTP amounts between sculptures and paintings, we
test:
H 0 : 1 2
H a : 1 2
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Design of Experiments and Analysis of Variance 509
9.55
a.
A randomized block design should be used to analyze the data because the same employees were
measured at all three time periods. Thus, the blocks are the employees and the treatments are the
three time periods.
b.
There is still enough information in the table to make a conclusion because the p-values are given.
c.
To determine if there are differences in the mean competence levels among the three time periods, we
test:
H 0 : 1 2 3
H a : At least two treatment means differ
9.56
d.
The p-value is p 0.001 . At a significance level > .001, we reject H0. There is sufficient evidence to
conclude that there is a difference in the mean competence levels among the three time periods for
any value of .001 .
e.
With 90% confidence, the mean competence before the training is significantly less than the mean
competence 2-days after and 2-months after. There is no significant difference in the mean
competence between 2-days after and 2-months after.
a.
This was a randomized complete block design. The blocks are the months and the treatments were
the 3 types of measures of electrical consumption.
b.
df Method k 1 3 1 2 , df Error n k b 1 12 3 4 1 6 ,
SST (k 1) MST (3 1)(.195) .390 ,
FMonth
MSB 10.780
159.23
MSE
.069
Source
Forecast Method
Month
Error
Total
c.
SSB (b 1) MSB (4 1)(10.780) 32.340 ,
df
2
3
6
11
SS
.390
32.340
.414
33.144
MS
.195
10.780
.069
F-value
2.83
156.23
p-value
.08
< .01
To determine if there is a difference in the mean electrical consumption values among the three
methods, we test:
H 0 : 1 2 3
H a : At least 2 of the treatment means differs
The test statistic is F 2.83 and the p-value is p .08 . Since the p-value is not less than
( p .08 .05) , H0 is not rejected. There is insufficient evidence to indicate a difference in mean
electrical consumption values among the three methods at .05 .
9.57
a.
The treatments were the 8 different activities.
b.
The blocks were the 15 adults who participated in the study.
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9.58
Chapter 9
c.
Since the p-value is less than ( p .001 .01) , H0 is rejected. There is sufficient evidence to
indicate a difference in mean heart rate among the 8 activities at .01 .
d.
The treadmill jogging had the highest mean heart rate. It was significantly greater than the mean
heart rates of all the other activities. Brisk treadmill walking had the second highest mean heart rate.
It was significantly less than the mean heart rate of treadmill jogging, but significantly greater than
the mean heart rates of the other 6 activities. There was no significant difference in the mean heart
rates among the treatments Wii aerobics, Wii muscle conditioning, Wii yoga, and Wii balance. The
mean heart rates for these activities were significantly less than the mean heart rates for treadmill
jogging and brisk treadmill walking, but greater than the mean heart rates of handheld gaming and
rest. There was no significant difference in the mean heart rate between handheld gaming and rest.
The mean heart rate for these two activities were significantly less than those for the other 6
activities.
a.
The time of the year (month) could affect the number of rigs running, so a
randomized complete block design was used to “block” out the month to month variation.
b.
There are 3 treatments in this experiment. They are the three states – California, Utah, and Alaska.
c.
There are 3 blocks in this experiment – the three months selected: Month 1, Month 2, and Month 3.
d.
To determine if there is a difference in the mean number of rigs running among the three states, we
test:
H 0 : 1 2 3
9.59
e.
From the printout, the test statistic is F 38.0685 and the p-value is p .0025 . Since the p-value is
so small, we would reject H0 for any value of .0025 . There is sufficient evidence to indicate a
difference in the mean number of oil rigs running among the three states.
f.
From the XLSTAT printout, there is no significant difference in the mean number of oil rigs running
in Alaska and Utah. However, both of these states have a significantly smaller number of rigs
running than does California. Thus, California has the largest mean number of oil rigs running.
a.
To compare the mean item scores, we test:
H 0 : 1 2 5
H a : At least 2 of the treatment means differs
b.
Each of the 11 items were reviewed by each of the 5 systematic reviews. Since all reviews were
made on each item, the observations are not independent. Thus, the randomized block ANOVA is
appropriate.
c.
The p-value for Review is p 0.319. Since the p-value is not small, H0 would not be rejected for any
reasonable value of . There is insufficient evidence to indicate a difference in the mean review
scores among the 5 systematic reviews.
The p-value for Item is p 0.000. Since the p-value is small, H0 would be rejected for any
reasonable value of . There is sufficient evidence to indicate a difference in the mean scores
among the 5 reviews.
d.
None of the means are significantly different because all means are connected with the letter ‘a’.
This agrees with the conclusion drawn in part c about the treatment Review.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 511
e.
9.60
The experiment-wise error rate is .05. This means that the probability of declaring at least 2 means
different when they are not different is .05.
Using SAS, the ANOVA Table is:
The ANOVA Procedure
Dependent Variable: temp
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
11
18.53700000
1.68518182
0.52
0.8634
Error
18
58.03800000
3.22433333
Corrected Total
29
76.57500000
R-Square
Coeff Var
Root MSE
temp Mean
0.242076
1.885189
1.795643
95.25000
Source
DF
Anova SS
Mean Square
F Value
Pr > F
STUDENT
PLANT
9
2
18.41500000
0.12200000
2.04611111
0.06100000
0.63
0.02
0.7537
0.9813
To determine if there are differences among the mean temperatures among the three treatments, we test:
H 0 : 1 2 3
H a : At least 2 of the treatment means differs
The test statistic is F 0.02 . The associated p-value is p .9813 . Since the p-value is very large, there is
no evidence of a difference in mean temperature among the three treatments for any reasonable value of .
Since there is no difference, we do not need to compare the means. It appears that the presence of plants or
pictures of plants does not reduce stress.
9.61
Using MINITAB, the ANOVA table is:
Two-way ANOVA: Rate versus Week, Day
Analysis of Variance for Rate
Source
DF
SS
Week
8
575.2
Day
4
94.2
Error
32
376.9
Total
44
1046.4
Day
1
2
3
4
5
Mean
8.8
4.6
5.8
5.4
6.4
MS
71.9
23.5
11.8
F
6.10
2.00
P
0.000
0.118
Individual 95% CI
-+---------+---------+---------+---------+
(--------*---------)
(--------*---------)
(--------*--------)
(--------*---------)
(---------*--------)
-+---------+---------+---------+---------+
2.5
5.0
7.5
10.0
12.5
Copyright © 2014 Pearson Education, Inc.
512
Chapter 9
To determine if there is a difference in mean rate of absenteeism among the 5 days of the week, we test:
H 0 : 1 2 3 4 5
H a : At least 2 of the treatment means differs
The test statistic is F 2.00 and the p-value is p .118 .
Since the p-value is not small, H0 is not rejected. There is insufficient evidence to indicate a difference in
mean rate of absenteeism among the 5 days of the week for any value of .118 .
To test for the effectiveness of blocking, we test:
H 0 : 1 2 9
H a : At least 2 of the block means differs
The test statistic is F 6.10 and the p-value is p .000 .
Since the p-value is so small, H0 is rejected. There is sufficient evidence to indicate blocking was effective
at any reasonable value of .
9.62
a.
The treatments are the 4 pre-slaughter phases. The blocks are the 8 cows.
b.
Using SPSS, the output is:
Tests of Between-Subjects Effects
Dependent Variable:Rate
Type III Sum of
Source
Squares
df
Mean Square
F
Sig.
a
10
244.400
5.108
.001
341551.125
1
341551.125
7137.777
.000
Cow
1922.875
7
274.696
5.741
.001
Phase
521.125
3
173.708
3.630
.030
Error
1004.875
21
47.851
Total
345000.000
32
3448.875
31
Corrected Model
Intercept
Corrected Total
2444.000
a. R Squared = .709 (Adjusted R Squared = .570)
The ANOVA table is simpler form is:
Source
Phase
Cow
Error
Total
df
3
7
21
31
SS
521.125
1922.875
1004.875
3448.875
MS
173.708
274.696
47.851
F-value
3.630
5.741
Copyright © 2014 Pearson Education, Inc.
p-value
.030
.001
Design of Experiments and Analysis of Variance 513
c.
To determine if there are differences among the mean heart rates of cows in the four pre-slaughter
phases, we test:
H 0 : 1 2 3 4
H a : At least 2 of the treatment means differs
The test statistic is F 3.63 and the p-value is p .030 . Since the p-value is less than
( p .030 .05) , H0 is rejected. There is sufficient evidence to indicate a difference in heart rates
of cows among the four pre-slaughter phases at .05 .
d.
Since we rejected H0 in part c, the multiple comparison procedure is warranted. Using SPSS, the
results are:
Homogeneous Subsets
Rate
a,b
Tukey HSD
Subset
Phase
N
1
2
2.00
8
97.0000
3.00
8
103.1250
103.1250
4.00
8
105.1250
105.1250
1.00
8
Sig.
108.0000
.119
.508
Means for groups in homogeneous subsets are
displayed.
Based on observed means.
The error term is Mean Square(Error) = 47.851.
a. Uses Harmonic Mean Sample Size = 8.000.
b. Alpha = 0.05.
There is a significant difference in the mean heart rates between the first phase and the second phase.
The mean heart rate at the first phase is significantly greater than the mean heart rate at the second
phase. No other differences exist.
Copyright © 2014 Pearson Education, Inc.
514
9.63
Chapter 9
Using MINITAB, the ANOVA table is:
Two-way ANOVA: Corrosion versus Time, System
Source
Time
System
Error
Total
DF
2
3
6
11
S = 0.3060
System
1
2
3
4
SS
63.1050
9.5833
0.5617
73.2500
MS
31.5525
3.1944
0.0936
R-Sq = 99.23%
Mean
9.0667
9.7333
11.0667
8.7333
F
337.06
34.12
P
0.000
0.000
R-Sq(adj) = 98.59%
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----*-----)
(-----*----)
(----*-----)
(----*-----)
------+---------+---------+---------+--8.80
9.60
10.40
11.20
To determine if there is a difference in mean corrosion rates among the 4 systems, we test:
H 0 : 1 2 3 4
H a : At least 2 of the treatment means differs
The test statistic is F 34.12 and the p-value is p .000 .
Since the p-value is so small, H0 is rejected. There is sufficient evidence to indicate a difference in mean
corrosion rates among the 4 systems at any reasonable value of .
Using SAS, Tukey’s multiple comparison results are:
Tukey's Studentized Range (HSD) Test for CORROSION
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher
Type II error rate than REGWQ.
Alpha
0.05
Error Degrees of Freedom
6
Error Mean Square
0.093611
Critical Value of Studentized Range 4.89559
Minimum Significant Difference
0.8648
Means with the same letter are not significantly different.
Tukey Grouping
Mean
N
SYSTEM
A
11.0667
3
3
B
B
B
9.7333
3
2
9.0667
3
1
8.7333
3
4
C
C
C
The mean corrosion rate for system 3 is significantly larger than all of the other mean corrosion rates. The
mean corrosion rate of system 2 is significantly larger than the mean for system 4. If we want the system
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 515
(epoxy coating) with the lowest corrosion rate, we would pick either system 1 or system 4. There is no
significant difference between these two groups and they are in the lowest corrosion rate group.
9.64
9.65
a.
There are two factors.
b.
No, we cannot tell whether the factors are qualitative or quantitative.
c.
Yes. There are four levels of factor A and three levels of factor B.
d.
A treatment would consist of a combination of one level of factor A and one level of factor B. There
are a total of 4 3 = 12 treatments.
e.
One problem with only one replicate is there are no degrees of freedom for error. This is overcome
by having at least two replicates.
a.
The ANOVA table is:
Source
A
B
AB
Error
Total
df
2
3
6
12
23
SS
.8
5.3
9.6
1.3
17.0
MS
.4000
1.7667
1.6000
.1083
F
3.69
16.31
14.77
df for A is a 1 3 1 2
df for B is b 1 4 1 3
df for AB is ( a 1)(b 1) 2 3 6
df for Error is n ab 24 3 4 12
df for Total is n 1 24 1 23
SSE SS Total SST SSB 17.0 .8 5.3 9.6 1.3
MSB
SS B 5.3
1.7667
b 1 4 1
MSE
1.3
SSE
.1083
n ab 24 3(4)
FAB
b.
MSAB
FA
MSA
SS A
.8
.40
a 1 3 1
9.6
SS AB
1.60
(a 1)(b 1) (3 1)(4 1)
MS A .4000
=
3.69
MSE .1083
FB
MS AB 1.6000
=
14.77
MSE
.1083
Sum of Squares for Treatment SSA SSB SSAB .8 5.3 2.6 15.7
MST
15.7
SST
1.4273
ab 1 3(4) 1
FT
MST 1.4273
=
13.18
MSE .1083
To determine if the treatment means differ, we test:
Copyright © 2014 Pearson Education, Inc.
MS B 1.7667
=
16.31
MSE .1083
516
Chapter 9
H 0 : 1 2 12
H a : At least 2 of the treatment means differs
The test statistic is F 13.18 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 3(4) 1 11 and 2 n ab 24 3(4) 12 . From Table VI, Appendix D, F.05 2.75 .
The rejection region is F 2.75 .
Since the observed value of the test statistic falls in the rejection region ( F 13.18 2.75) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
c.
d.
e.
Yes. We need to partition the Treatment Sum of Squares into the Main Effects and Interaction Sum
of Squares. Then we test whether factors A and B interact. Depending on the conclusion of the test
for interaction, we either test for main effects or compare the treatment means.
Two factors are said to interact if the effect of one factor on the dependent variable is not the same at
different levels of the second factor. If the factors interact, then tests for main effects are not
necessary. We need to compare the treatment means for one factor at each level of the second.
To determine if the factors interact, we test:
H0: Factors A and B do not interact to affect the response mean
Ha: Factors A and B do interact to affect the response mean
The test statistic is F
MS AB
14.77
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (3 1)(4 1) 6 and 2 n ab 24 3(4) 12 . From Table VI, Appendix D,
F.05 3.00 . The rejection region is F 3.00 .
Since the observed value of the test statistic falls in the rejection region ( F 14.77 3.00) , H0 is
rejected. There is sufficient evidence to indicate the two factors interact to affect the response mean
at .05 .
9.66
f.
No. Testing for main effects is not warranted because interaction is present. Instead, we compare the
treatment means of one factor at each level of the second factor.
a.
Factor A has 3 1 4 levels and factor B has 1 1 2 levels.
b.
There are a total of 23 1 24 observations and 4 2 8 treatments. Therefore, there were 24 / 8 3
observations for each treatment.
c.
AB
Error
MSA
df (a 1)(b 1) (4 1)(2 1) 3
df n ab 24 4(2) 16
SSA
SSA ( a 1) MSA (4 1)(.75) 2.25
a 1
MSB
Copyright © 2014 Pearson Education, Inc.
SS B .95
.95
b 1 2 1
Design of Experiments and Analysis of Variance 517
MSAB
SSAB
SSAB (a 1)(b 1) MSAB (4 1)(2 1)(.30) .9
(a 1)(b 1)
SSE SS Total SSA SSB SSAB 6.5 2.25 .95 .9 2.4
Treatment df ab 1 4(2) 1 7
FA
MS A .75
5.00
MSE .15
FB
MST
MSE
SST
4.1
.5857
ab 1 7
MS B .95
6.33
MSE .15
FAB
FT
2.4
SSE
.15
n ab 24 4(2)
MST .5857
3.90
MSE
.15
MS AB .30
2.00
MSE
.15
The ANOVA table is:
Source
Treatments
A
B
AB
Error
Total
d.
df
7
3
1
3
16
23
SS
4.1
2.25
.95
.90
2.40
6.50
MS
.59
.75
.95
.30
.15
F
3.90
5.00
6.33
2.00
To determine whether the treatment means differ, we test:
H 0 : 1 2 8
H a : At least two treatment means differs
The test statistic is F
MST
3.90
MSE
The rejection region requires .10 in the upper tail of the F-distribution with
1 ab 1 4(2) 1 7 and 2 n ab 24 4(2) 16 . From Table V, Appendix D, F.10 2.13 .
The rejection region is F 2.13 .
Since the observed value of the test statistic falls in the rejection region ( F 3.90 2.13) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .10 .
e.
To determine if the factors interact, we test:
H0: Factors A and B do not interact to affect the response mean
Ha: Factors A and B do interact to affect the response mean
The test statistic is F 2.00 .
The rejection region requires .10 in the upper tail of the F-distribution with
1 (a 1)(b 1) (4 1)(2 1) 3 and 2 n ab 24 4(2) 16 . From Table V, Appendix D,
F.10 2.46 . The rejection region is F 2.46 .
Copyright © 2014 Pearson Education, Inc.
518
Chapter 9
Since the observed value of the test statistic does not fall in the rejection region ( F 2.00 2.46) , H0
is not rejected. There is insufficient evidence to indicate factors A and B interact at .10 .
To determine if the four means of factor A differ, we test:
H0: There is no difference in the four means of factor A
Ha: At least two of the factor A means differ
The test statistic is F 5.00 .
The rejection region requires .10 in the upper tail of the F-distribution with1 a 1 4 1 3
and 2 n ab 24 4(2) 16 . From Table V, Appendix D, F.10 2.46 . The rejection region is
F 2.46 .
Since the observed value of the test statistic falls in the rejection region ( F 5.00 2.46) , H0 is
rejected. There is sufficient evidence to indicate at least two of the four means of factor A differ at
.10 .
To determine if the 2 means of factor B differ, we test:
H0: There is no difference in the two means of factor B
Ha: At least two of the factor B means differ
The test statistic is F 6.33 .
The rejection region requires .10 in the upper tail of the F-distribution with1 b 1 2 1 1
and 2 n ab 24 4(2) 16 . From Table V, Appendix D, F.10 3.05 . The rejection region is
F 3.05 .
Since the observed value of the test statistic falls in the rejection region ( F 6.33 3.05) , H0 is
rejected. There is sufficient evidence to indicate the two means of factor B differ at .10 .
All of the tests performed are warranted because interaction was not significant.
9.67
a.
The treatments are the combinations of the levels of factor A and the levels of factor B. There are
2 3 6 treatments. The treatment means are:
x11
x 3.1 4.0 3.55
x12
x21
x 5.9 5.3 5.6
x22
11
2
2
21
2
2
x 4.6 4.2 4.4
x13
x 6.4 7.1 6.75
x 2.9 2.2 2.55
x23
x 3.3 2.5 2.9
12
2
2
22
2
2
Copyright © 2014 Pearson Education, Inc.
13
2
2
23
2
2
Design of Experiments and Analysis of Variance 519
Using MINITAB, the graph is:
Scatterplot of A1, A2 vs B
7
Variable
A1
A2
Y-Data
6
5
4
3
2
1
b.
2
B
3
The treatment means appear to be different because the sample means are quite different. The factors
appear to interact because the lines are not parallel.
SST SSA SSB SSAB 4.4408 4.1267 18.0667 26.5742
MST
26.5742
SST
5.315
ab 1 2(3) 1
FT
MST 5.315
21.62
MSE .246
To determine whether the treatment means differ, we test:
H 0 : 1 2 6
H a : At least two treatment means differs
The test statistic is F
MST
21.62
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 2(3) 1 5 and 2 n ab 12 2(3) 6 . From Table VI, Appendix D, F.05 4.39 .
The rejection region is F 4.39 .
Since the observed value of the test statistic falls in the rejection region ( F 21.62 4.39) , H0 is
rejected. There is sufficient evidence to indicate that the treatment means differ at .05 . This
supports the plot in a.
c.
Yes. Since there are differences among the treatment means, we test for interaction. To determine
whether the factors A and B interact, we test:
H0: Factors A and B do not interact to affect the mean response
Ha: Factors A and B do interact to affect the mean response
The test statistic is F
MSAB 9.0033
36.62
=
MSE .24583
The rejection region requires .05 in the upper tail of the F-distribution with
Copyright © 2014 Pearson Education, Inc.
520
Chapter 9
1 (a 1)(b 1) (2 1)(3 1) 2 and 2 n ab 12 2(3) 6 . From Table VI, Appendix D,
F.05 5.14 . The rejection region is F 5.14 .
Since the observed value of the test statistic falls in the rejection region ( F 36.62 5.14) , H0 is
rejected. There is sufficient evidence to indicate that factors A and B interact to affect the response
mean at .05 .
No. Because interaction is present, the tests for main effects are not warranted.
e.
The results of the tests in parts b and c support the visual interpretation in part a.
a.
The treatments are the combinations of the levels of factor A and the levels of factor B. There are
2 2 4 treatments. The treatment means are:
x11
x 29.6 35.2 32.4
x12
11
2
x21
2
12.9 17.6
x21
15.25
2
2
Using MINITAB, the graph is:
x22
x 47.3 42.1 44.7
12
2
x22
2
2
28.4 22.7
25.55
2
Scatterplot of A1, A2 vs B
Variable
A1
A2
45
40
35
Y-Data
9.68
d.
30
25
20
15
1
2
B
The factors do not appear to interact—the lines are almost parallel. The treatment means do appear
to differ because the sample means range from 15.25 to 44.7.
x 235.8 6,950.205
CM
2
b.
2
i
n
8
SS (Total ) x 2 CM 7922.92 6950.205 972.715
SSA
A CM 154.2 81.6 6950.205 7609.05 6950.205 658.845
SSB
B CM 95.3 140.5 6950.205 7205.585 6950.205 255.38
2
i
2(2)
br
2
i
ar
2
2
2(2)
2
2(2)
2
2(2)
Copyright © 2014 Pearson Education, Inc.
SSAB
Design of Experiments and Analysis of Variance 521
AB SSA SSB CM
2
ij
r
2
64.8 89.4 2 30.52 51.12
658.845 255.38 6950.205 7866.43 7864.43 2
2
2
2
2
SSE SS Total SSA SSB SSAB 972.715 658.845 255.38 2 56.49
A df a 1 2 1 1
B df b 1 2 1 1
AB df (a 1)(b 1) (2 1)(2 1) 1
Error df n ab 8 2 2 4
Total df n 1= 8 1 7
MSA
SSA 658.845
SS B 255.38
658.845 MSB
255.38
a 1
b 1
1
1
MSE
SS E 56.49
14.1225
n ab
4
FA
MS A 658.845
46.65
MSE 14.1225
FB
MS B 255.38
18.08
MSE 14.1225
MSAB
FAB
2
SS AB
2
(a 1)(b 1) 1
MS AB
2
.14
MSE 14.1225
The ANOVA table is:
Source
A
B
AB
Error
Total
c.
df
1
1
1
4
7
SS
658.845
255.380
2.000
56.490
972.715
MS
658.845
255.380
2.000
14.1225
F
46.65
18.08
.14
SST SSA SSB SSAB 658.845 255.380 2.000 916.225
MST
SST
916.225
305.408
ab 1
3
FT
df ab 1 2 2 1 3
MST 305.408
=
21.63
MSE 14.1225
To determine whether the treatment means differ, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differs
The test statistic is F 21.63 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 2(2) 1 3 and 2 n ab 8 2(2) 4 . From Table VI, Appendix D, F.05 6.59 .
The rejection region is F 6.59 .
Copyright © 2014 Pearson Education, Inc.
522
Chapter 9
Since the observed value of the test statistic falls in the rejection region ( F 21.63 6.59) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
This agrees with the conclusion in part a.
d.
Since there are differences among the treatment means, we test for the presence of interaction:
H0: Factors A and B do not interact to affect the response means
Ha: Factors A and B do interact to affect the response means
The test statistic is F .14 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (2 1)(2 1) 1 and 2 n ab 8 2(2) 4 . From Table VI, Appendix D,
F.05 7.71 . The rejection region is F 7.71 .
Since the observed value of the test statistic does not fall in the rejection region ( F .14 7.71) ), H0
is not rejected. There is insufficient evidence to indicate the factors interact at .05 .
e.
Since the interaction was not significant, we test for main effects.
To determine whether the two means of factor A differ, we test:
H 0 : 1 2
H a : 1 2
The test statistic is F 46.65 .
The rejection region requires .05 in the upper tail of the F-distribution with1 a 1 2 1 and
2 n ab 8 2(2) 4 . From Table VI, Appendix D, F.05 7.71 . The rejection region is
F 7.71 .
Since the observed value of the test statistic falls in the rejection region ( F 46.65 7.71) , H0 is
rejected. There is sufficient evidence to indicate the two means of factor A differ at .05 .
To determine whether the two means of factor B differ, we test:
H 0 : 1 2
H a : 1 2
The test statistic is F 18.08 .
The rejection region requires .05 in the upper tail of the F-distribution with 1 = b 1 = 2 1 = 1
and 2 n ab 8 2(2) 4 . From Table VI, Appendix D, F.05 7.71 . The rejection region is
F 7.71 .
Since the observed value of the test statistic falls in the rejection region ( F 18.08 7.71) , H0 is
rejected. There is sufficient evidence to indicate the two means of factor B differ at .05 .
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 523
f.
The results of all the tests agree with those in part a.
g.
Since no interaction is present, but the means of both factors A and B differ, we compare the two
means of factor A and compare the two means of factor B. Since there are only two means to
compare for each factor, the higher population mean corresponds to the higher sample mean.
Factor A:
x1
x 29.6 35.2 47.3 42.1 38.55
x2
x 12.9 17.6 28.4 22.7 20.4
1
2(2)
br
2
2(2)
br
The mean for level 1 of factor A is significantly higher than the mean for level 2.
Factor B:
x1
x 29.6 35.2 12.9 17.6 23.825
x2
x 47.3 42.1 28.4 22.7 35.125
1
2(2)
ar
2
2(2)
ar
The mean for level 2 of factor B is significantly higher than the mean for level 1.
9.69
a.
SSA .2 1000 200 ,
SSB .11000 100 ,
SSAB .1 1000 100
SSE SS Total SSA SSB SSAB 1000 200 100 100 600
SST SSA SSB SSAB 200 100 100 400
MSB
SS B 100
50
b 1 3 1
MSE
600
SSE
33.333
n ab 27 3(3)
FA
MSAB
MSA
SSA 200
100
a 1 3 1
100
SSAB
25
(a 1)(b 1) (3 1)(3 1)
MST
400
SST
50
ab 1 3(3) 1
MSA
100
=
= 3.00
MSE 33.333
FB
MS B
50
1.50
MSE 33.333
MSAB
25
.75
MSE 33.333
FT
MST
50
1.50
MSE 33.333
FAB
Source
A
B
AB
Error
Total
df
2
2
4
18
26
SS
200
100
100
600
1000
MS
100
50
25
33.333
F
3.00
1.50
.75
Copyright © 2014 Pearson Education, Inc.
524
Chapter 9
To determine whether the treatment means differ, we test:
H 0 : 1 2 9
H a : At least two treatment means differs
The test statistic is F
MST
1.50
MSE
Suppose .05 . The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 3(3) 1 8 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D, F.05 2.51 .
The rejection region is F 2.51 .
Since the observed value of the test statistic does not fall in the rejection region ( F 1.50 2.51) , H0 is
not rejected. There is insufficient evidence to indicate the treatment means differ at .05 . Since
there are no treatment mean differences, we have nothing more to do.
b.
SSA .1 1000 100 ,
SSB .1 1000 100 ,
SSAB .5 1000 500
SSE SS Total SSA SSB SSAB 1000 100 100 500 300
SST SSA SSB SSAB 100 100 500 700
MSB
SS B 100
50
b 1 3 1
MSAB
MSE
300
SSE
16.667
n ab 27 3(3)
MST
FA
MSA
SSA 100
50
a 1 3 1
500
SSAB
125
(a 1)(b 1) (3 1)(3 1)
SST
700
87.5
ab 1 9 1
MS A
50
=
= 3.00
MSE 16.667
FB
MS B
50
3.00
MSE 16.667
MS AB
125
=
= 7.50
MSE 16.667
FT
MST
87.5
=
= 5.25
MSE 16.667
FAB
Source
A
B
AB
Error
Total
df
2
2
4
18
26
SS
100
100
500
300
1000
MS
50
50
125
16.667
F
3.00
3.00
7.50
To determine if the treatment means differ, we test:
H 0 : 1 2 9
H a : At least two treatment means differs
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 525
The test statistic is F
MST
5.25
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 3(3) 1 8 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D, F.05 2.51 .
The rejection region is F 2.51 .
Since the observed value of the test statistic falls in the rejection region ( F 5.25 2.51) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
Since the treatment means differ, we next test for interaction between factors A and B. To determine
if factors A and B interact, we test:
H0: Factors A and B do not interact to affect the mean response
Ha: Factors A and B do interact to affect the mean response
The test statistic is F
MS AB
7.50
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (3 1)(3 1) 4 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D,
F.05 2.93 . The rejection region is F 2.93 .
Since the observed value of the test statistic falls in the rejection region ( F 7.50 2.93) , H0 is
rejected. There is sufficient evidence to indicate the factors A and B interact at .05 . Since
interaction is present, no tests for main effects are necessary.
c.
SSA .4 1000 400 ,
SSB .1 1000 100 ,
SSAB .2 1000 200
SSE SS Total SSA SSB SSAB 1000 400 100 200 300
SST SSA SSB SSAB 400 100 200 700
MSA
MSB
SS B 100
50
b 1 3 1
MSAB
MSE
300
SSE
16.667
n ab 27 3(3)
MST
SSA 400
50
a 1 3 1
200
MSAB
50
(a 1)(b 1) (3 1)(3 1)
700
SST
87.5
ab 1 3(3) 1
FA
MSA
200
=
= 12.00
MSE 16.667
FB
MSB
50
=
= 3.00
MSE 16.667
FAB
MSAB
50
=
= 3.00
MSE 16.667
FT
MST
87.5
= 5.25
MSE 16.667
Copyright © 2014 Pearson Education, Inc.
526
Chapter 9
Source
A
B
AB
Error
Total
df
SS
400
100
200
300
1000
2
2
4
18
26
MS
200
50
50
16.667
F
12.00
3.00
3.00
To determine if the treatment means differ, we test:
H 0 : 1 2 9
H a : At least two treatment means differs
The test statistic is F
MST
5.25
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 3(3) 1 8 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D, F.05 2.51 .
The rejection region is F 2.51 .
Since the observed value of the test statistic falls in the rejection region ( F 5.25 2.51) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
Since the treatment means differ, we next test for interaction between factors A and B. To determine
if factors A and B interact, we test:
H0: Factors A and B do not interact to affect the mean response
Ha: Factors A and B do interact to affect the mean response
The test statistic is F
MS AB
3.00
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (3 1)(3 1) 4 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D,
F.05 2.93 . The rejection region is F 2.93 .
Since the observed value of the test statistic falls in the rejection region ( F 3.00 2.93) , H0 is
rejected. There is sufficient evidence to indicate the factors A and B interact at .05 . Since
interaction is present, no tests for main effects are necessary.
d.
SSA .4 1000 400 ,
SSB .4 1000 400 ,
SSAB .1 1000 100
SSE SS Total SSA SSB SSAB 1000 400 400 100 100
SST SSA SSB SSAB 400 400 100 900
MSB
SS B 400
200
b 1 3 1
MSAB
MSA
SSA 400
200
a 1 3 1
100
SSAB
25
(a 1)(b 1) (3 1)(3 1)
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 527
MSE
FA
100
SSE
5.556
n ab 27 3(3)
MST
900
SST
112.5
ab 1 3(3) 1
MSA
200
=
= 36.00
MSE 5.556
FB
MSB
200
=
= 36.00
MSE 5.556
MSAB
25
=
= 4.50
MSE 5.556
FT
MST 112.5
= 20.25
MSE 5.556
FAB
Source
A
B
AB
Error
Total
df
2
2
4
18
26
SS
400
400
100
100
1000
MS
200
200
25
5.556
F
36.00
36.00
4.50
To determine if the treatment means differ, we test:
H 0 : 1 2 9
H a : At least two treatment means differs
The test statistic is F
MST
20.25
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 3(3) 1 8 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D, F.05 2.51 .
The rejection region is F 2.51 .
Since the observed value of the test statistic falls in the rejection region ( F 20.25 2.51) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
Since the treatment means differ, we next test for interaction between factors A and B. To determine
if factors A and B interact, we test:
H0: Factors A and B do not interact to affect the mean response
Ha: Factors A and B do interact to affect the mean response
The test statistic is F
MSAB
4.50
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (3 1)(3 1) 4 and 2 n ab 27 3(3) 18 . From Table VI, Appendix D,
F.05 2.93 . The rejection region is F 2.93 .
Since the observed value of the test statistic falls in the rejection region ( F 4.50 2.93) , H0 is
rejected. There is sufficient evidence to indicate the factors A and B interact at .05 . Since
interaction is present, no tests for main effects are necessary.
Copyright © 2014 Pearson Education, Inc.
528
Chapter 9
9.70
a.
The experimental design used was a factorial design.
b.
The two factors are diet and age. There are 2 levels of diet – fine limestone (FL) and coarse
limestone (CL). There are 2 levels of age – young and old. There are 2 2 4 treatments: FL/young,
FL/old, CL/young, and CL/old.
c.
The experimental units are the hens.
d.
The dependent variable is egg shell thickness.
e.
If diet and age do not interact, then the effect of diet on the egg shell thickness is the same at each
level of age.
f.
This indicates that there is no significant difference in egg shell thickness between the young and old
hens.
g.
This indicates that there is a significant difference in the mean egg shell thickness due to diet. The
mean egg shell thickness for eggs produced by hens on the CL diet is greater than the mean egg shell
thickness for eggs produced by hens on the FL diet.
a.
The two factors are type of statement and order of information. There are 2 2 4 treatments:
concrete/statement first, concrete/behavior first, abstract/statement first, and abstract/behavior first.
b.
This indicates that the effect of type of statement on the level of hypocrisy depends on the order of the
information.
c.
Using MINITAB, a plot of the means is:
9.71
Scatterplot of Hypocrisy vs Order
6.00
Ty pe
Abstract
Concrete
5.75
Hypocrisy
5.50
5.25
5.00
4.75
4.50
Statement
Behavior
Order
9.72
d.
Since the interaction between the type of statement and the order of information was significant, then
the tests for main effects should not be performed. Multiple comparisons on some or all of the pairs
of treatments should be performed next.
a.
There are a total of 2 4 8 treatments.
b.
The interaction between temperature and type was significant. This means that the effect of type of
yeast on the mean autolysis yield depends on the level of temperature.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 529
c.
To determine if the main effect of type of yeast is significant, we test:
H 0 : Ba Br
H a : Ba Br
To determine if the main effect of temperature is significant, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differs
d.
The tests for the main effects should not be run since the test for interaction was significant. If
interaction is significant, then these interaction effects could cover up the main effects. Thus, the
main effect tests would not be informative.
e.
Baker’s yeast:
The mean yield for temperature 54o is significantly lower than the mean yields for the other 3
temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o.
Brewer’s yeast:
The mean yield for temperature 54o is significantly lower than the mean yields for the other 3
temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o.
9.73
a.
If justice reparation potential and producer need interact, then the effect of justice reparation
potential on intension depends on the level of producer need.
b.
To determine if interaction exists, we test:
H0: Justice reparation potential and producer need do not interact
Ha: Justice reparation potential and producer need do interact
The test statistic is F 20.55 and the p-value is p 0.000 . Since the p-value is less than
( p 0.000 .01) , H0 is rejected. There is sufficient evidence to indicate reparation justice
potential and producer need interact to affect intension at .01 .
9.74
c.
No. Since the test for interaction was significant, then the tests for the main effects are not necessary.
d.
This plot indicates that for high reparation justice potential, as producer need changes from High to
Moderate, the mean intension decreases. However, for low reparation justice potential, as producer
need changes from High to Moderate, the mean intension increases. This indicates that the effect of
reparation justice potential on intension depends on the level of producer need.
e.
Yes. This is exactly what the graph shows.
a.
There are a total of 2 4 8 treatments for this study. They include all combinations of Insomnia
status and Education level. The 8 treatments are:
Normal sleeper, College graduate
Normal sleeper, Some college
Normal sleeper, High school graduate
Normal sleeper, High school dropout
Chronic insomnia, College graduate
Chronic insomnia, Some college
Chronic insomnia, High school graduate
Chronic insomnia, High school dropout
Copyright © 2014 Pearson Education, Inc.
530
Chapter 9
b.
Since Insomnia and Education did not interact, this means that the effect of Insomnia on the Fatigue
Severity Scale does not depend on the level of Education. In a graph, the lines will be parallel. A
possible graph of this situation is:
Scatterplot of FSS vs Insomnia
Education
1
2
3
4
11
10
9
FSS
8
7
6
5
4
3
2
1.0
9.75
1.2
1.4
1.6
Insomnia
1.8
2.0
c.
This means that the researchers can infer that the population mean FSS for people who had insomnia
is higher than the population mean FSS for normal sleepers.
d.
This means that at least one level of education had a mean FSS score that differed from the rest.
There may be more than one difference, but there is at least one.
e.
With 95% confidence, we can conclude that the mean FSS value for high school dropouts is
significantly higher than the mean FSS values for the 3 other education levels. There is no significant
difference in the mean FSS values for college graduates, those with some college, and high school
graduates.
a.
dfOrder a 1 2 1 1 , df Menu b 1 2 1 1 , df OxM (a 1)(b 1) (2 1)(2 1) 1 ,
df Error n ab 180 2(2) 176
Source
Order
Menu
Order x Menu
Error
Total
df
1
1
1
176
179
F-value
----11.25
p-value
----<.001
b.
Since the p-value is less than ( p 0.001 .05) , H0 is rejected. There is sufficient evidence to
indicate order and menu interact to affect the amount willing to pay at .05 .
c.
No, these results are not required to complete the analysis. Since the test for interaction was
significant, there is no need to run the main effect tests.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 531
d.
Using MINITAB, a graph of the means is:
Interaction Plot for WillingPay
Order
Vice
Virtue
17
16
Mean
15
14
13
12
11
Homogeneous
Mixed
Menu
9.76
a.
There are two factors for this experiment, housing system and weight class. There are a total of
2 4 8 treatments. The treatments are:
Cage, M
Barn, M
b.
Cage, L
Barn, L
Free, M
Organic, M
Free, L
Organic, L
Using SAS, the results are:
The GLM Procedure
Dependent Variable: OVERRUN
c.
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
7
11364.52381
1623.50340
14.93
<.0001
Error
20
2175.33333
108.76667
Corrected Total
27
13539.85714
R-Square
Coeff Var
Root MSE
OVERRUN Mean
0.839339
2.061383
10.42913
505.9286
Source
DF
Type I SS
Mean Square
F Value
Pr > F
HOUSING
WTCLASS
HOUSING*WTCLASS
3
1
3
10787.79048
329.14286
247.59048
3595.93016
329.14286
82.53016
33.06
3.03
0.76
<.0001
0.0973
0.5303
Source
DF
Type III SS
Mean Square
F Value
Pr > F
HOUSING
WTCLASS
HOUSING*WTCLASS
3
1
3
10787.79048
320.47407
247.59048
3595.93016
320.47407
82.53016
33.06
2.95
0.76
<.0001
0.1015
0.5303
To determine if interaction between housing system and weight class exists, we test:
H0: Housing system and weight class do not interact
Ha: Housing system and weight class do interact
Copyright © 2014 Pearson Education, Inc.
532
Chapter 9
The test statistic is F 0.76 and the p-value is p .5303 . Since the p-value is not less than
( p .5303 .05) , H0 is not rejected. There is insufficient evidence to indicate that housing system
and weight class interact at .05 .
d.
To determine if there is a difference in mean whipping capacity among the 4 housing systems, we
test:
H 0 : 1 2 3 4
H a : At least two treatment means differs
The test statistic is F 33.06 and the p-value is p .0001 . Since the p-value is less than
( p .0001 .05) , H0 is rejected. There is sufficient evidence to indicate a difference in mean
whipping capacity among the 4 housing systems at .05 .
e.
To determine if there is a difference in mean whipping capacity between the 2 weight classes, we test:
H 0 : 1 2
H a : 1 2
The test statistic is F 2.95 and the p-value is p .1015 . Since the p-value is not less than
( p .1015 .05) , H0 is not rejected. There is insufficient evidence to indicate a difference in
mean whipping capacity between the 2 weight classes at .05 .
Yes. Using MINITAB, a plot of the data is:
Scatterplot of Time vs Density
Agent
Gum
PVP
8
7
6
M ean T ime
9.77
5
4
3
2
1
0
0
Low
High
Density
Since the lines are not parallel, this indicates interaction is present. The increase in mean time when
density is increased from low to high for PVP is not as great as the increase in mean time when density is
increased from low to high for GUM.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 533
9.78
Using MINITAB, the results of the ANOVA are:
General Linear Model: NUMBER versus GROUP, SET
Factor
GROUP
SET
Type
fixed
fixed
Levels
3
3
Values
3, 6, 12
FIRST, LAST, MIDDLE
Analysis of Variance for NUMBER, using Adjusted SS for Tests
Source
GROUP
SET
GROUP*SET
Error
Total
DF
2
2
4
81
89
S = 1.00308
Seq SS
15.267
62.600
7.133
81.500
166.500
Adj SS
15.267
62.600
7.133
81.500
R-Sq = 51.05%
Adj MS
7.633
31.300
1.783
1.006
F
7.59
31.11
1.77
P
0.001
0.000
0.142
R-Sq(adj) = 46.22%
Means
SET
FIRST
LAST
MIDDLE
N
30
30
30
NUMBER
3.0000
1.1000
1.4000
GROUP
3
6
12
N
30
30
30
NUMBER
2.4000
1.4333
1.6667
Tukey 95.0% Simultaneous Confidence Intervals
Response Variable NUMBER
All Pairwise Comparisons among Levels of GROUP
GROUP = 3 subtracted from:
GROUP
6
12
GROUP =
Lower
-1.586
-1.352
GROUP
12
6
Center
-0.9667
-0.7333
Upper
-0.3477
-0.1143
---+---------+---------+---------+--(--------*--------)
(--------*-------)
---+---------+---------+---------+---1.40
-0.70
0.00
0.70
subtracted from:
Lower
-0.3857
Center
0.2333
Upper
0.8523
---+---------+---------+---------+--(--------*--------)
---+---------+---------+---------+---1.40
-0.70
0.00
0.70
Copyright © 2014 Pearson Education, Inc.
534
Chapter 9
Tukey 95.0% Simultaneous Confidence Intervals
Response Variable NUMBER
All Pairwise Comparisons among Levels of SET
SET = FIRST subtracted from:
SET
LAST
MIDDLE
Lower
-2.519
-2.219
SET = LAST
SET
MIDDLE
Center
-1.900
-1.600
Upper
-1.281
-0.981
-----+---------+---------+---------+(-----*-----)
(-----*-----)
-----+---------+---------+---------+-2.0
-1.0
0.0
1.0
subtracted from:
Lower
-0.3190
Center
0.3000
Upper
0.9190
-----+---------+---------+---------+(-----*-----)
-----+---------+---------+---------+-2.0
-1.0
0.0
1.0
To determine if group size and photo set interact to affect the number of selections, we test:
H0: Group size and Photo set do not interact to affect the number of selections
Ha: Group size and Photo set interact to affect the number of selections
The test statistic is F 1.77 and the p-value is p .142 . Since the p-value is not small, H0 is not rejected.
There is insufficient evidence to indicate that group size and photo set interact to affect the number of
selections for any reasonable level of .
Since there is no evidence of an interaction, we will next test for the main effects. To determine if group
size had an effect on the mean number of selections, we test:
H 0 : 1 2 3
H a : At least two group size means differs
The test statistic is F 7.59 and the p-value is p .001 . Since the p-value is so small, H0 is rejected. There
is sufficient evidence to indicate that group size has an effect on the mean number of selections for any
level of greater than .001.
To determine if photo set had an effect on the mean number of selections, we test:
H 0 : 1 2 3
H a : At least two photo set means differs
The test statistic is F 31.11 and the p-value is p .000 . Since the p-value is so small, H0 is rejected.
There is sufficient evidence to indicate that photo set has an affect the mean number of selections for any
level of greater than .000.
Since both main effects are significant, we will run Tukey’s multiple comparison procedure on each main
effect to find where the differences exist.
The mean number of selections made for the different group sizes are:
__________________
Means:
1.433
1.667
2.400
Groups:
6
12
3
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 535
The confidence interval comparing size 3 to size 6 is (-1.586, -.3477). Since both endpoints of the interval
are negative, the mean number of selections for size 3 is significantly greater than the mean number of
selections for size 6. The confidence interval comparing size 3 to size 12 is (-1.352, -.1143). Since both
endpoints of the interval are negative, the mean number of selections for size 3 is significantly greater than
the mean number of selections for size 12. The confidence interval comparing size 6 to size 12 is (-.3857,
.8523). Since 0 is contained in the interval, there is no difference in the mean number of selections
between sizes 6 and 12. Thus, there are significantly more selections made for group size 3 than for the
other two sizes.
The mean number of selections made for the different photo sets are:
__________________
Means:
1.10
1.40
3.00
Groups:
Last
Middle
First
The confidence interval comparing the first photo set to the last photo set is (-2.519, -1.281). Since both
endpoints of the interval are negative, the mean number of selections for the first photo set is significantly
greater than the mean number of selections for the last photo set. The confidence interval comparing the
first photo set to the middle photo set is (-2.219, -.981). Since both endpoints of the interval are negative,
the mean number of selections for the first photo set is significantly greater than the mean number of
selections for the middle photo set. The confidence interval comparing the middle photo set to the last
photo set is (-.3190, .9190). Since 0 is contained in the interval, there is no difference in the mean number
of selections between the last photo set and the middle photo set. Thus, there are significantly more
selections made for the first photo set than for the other two photo sets.
9.79
Using MINITAB, a complete factorial design was fit to the data:
General Linear Model: RECALL versus CONTENT, BEFORE
Factor
CONTENT
BEFORE
Type
fixed
fixed
Levels
3
2
Values
NEUTRAL, SEX, VIOLENT
NO, YES
Analysis of Variance for RECALL, using Adjusted SS for Tests
Source
CONTENT
BEFORE
CONTENT*BEFORE
Error
Total
S = 1.73153
DF
2
1
2
318
323
Seq SS
123.265
6.458
7.472
953.421
1090.617
R-Sq = 12.58%
Adj SS
120.004
6.393
7.472
953.421
Adj MS
60.002
6.393
3.736
2.998
F
20.01
2.13
1.25
P
0.000
0.145
0.289
R-Sq(adj) = 11.21%
Grouping Information Using Tukey Method and 95.0% Confidence
CONTENT
NEUTRAL
VIOLENT
SEX
N
108
108
108
Mean
3.167
2.090
1.731
Grouping
A
B
B
Means that do not share a letter are significantly different.
Copyright © 2014 Pearson Education, Inc.
536
Chapter 9
Grouping Information Using Tukey Method and 95.0% Confidence
BEFORE
NO
YES
N
162
162
Mean
2.470
2.188
Grouping
A
A
Means that do not share a letter are significantly different.
First, we test for the interaction term. To determine if content group and whether one had watched the
commercial before interact to affect recall, we test:
H 0 : Content and whether one watched commercial before do not interact
H a : Content and whether one watched commercial before do interact
The test statistic is F 1.25 and the p-value is p .289 . Since the p-value is not small, H0 is not rejected.
There is no evidence to indicate content and whether the commercial was viewed before interact to affect
recall for any reasonable value of .
Next, we test for the main effects.
To determine if the mean recall differs among the content groups, we test:
H 0 : 1 2 3
H a : At least two means differ
The test statistic is F 20.01 and the p-value is p .000 . Since the p-value is very small, H0 is rejected.
There is evidence to indicate the mean recall differs among the different content groups for any reasonable
value of .
Tukey’s multiple comparison on the content means yielded the following. The mean recall for those in the
neutral content group was significantly higher than the mean recall of the other 2 groups. No other
differences existed.
To determine if the mean recall differs between whether one watched the ad before or not, we test:
H 0 : 1 2
H a : 1 2
The test statistic is F 2.13 and the p-value is p .145 . Since the p-value is not small, H0 is not rejected.
There is no evidence to indicate the mean recall differs between whether one watched the ad before or not
for any reasonable value of .
These results agree with the researchers’ conclusions.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 537
9.80
Using MINITAB, the ANOVA results are:
General Linear Model: Deviation versus Group, Trail
Factor
Group
Trail
Type Levels Values
fixed
4 F G M N
fixed
2 C E
Analysis of Variance for Deviation, using Adjusted SS for Tests
Source
Group
Trail
Group*Trail
Error
Total
DF
3
1
3
112
119
Seq SS
16271.2
46445.5
2245.2
82131.7
147093.6
Adj SS
13000.6
46445.5
2245.2
82131.7
Adj MS
4333.5
46445.5
748.4
733.3
F
5.91
63.34
1.02
P
0.001
0.000
0.386
First, we must test for treatment effects.
SST SS Group SS Trail SS GxT 16, 271.2 46, 445.5 2, 245.2 64, 961.9 .
The df 3 1 3 7 .
MST
64,961.9
SST
9, 280.2714
ab 1 4(2) 1
F
MST 9, 280.2714
12.66
MSE
733.3
To determine if there are differences in mean ratings among the 8 treatments, we test:
H0: All treatment means are the same
Ha: At least two treatment means differ
The test statistic is F 12.66 .
Since no was given, we will use .05 . The rejection region requires .05 in the upper tail of the F
distribution with1 ab 1 4(2) 1 7 and 2 n – ab 120 – 4 2 112 . From Table VI, Appendix D,
F.05 2.09 . The rejection region is F 2.09 .
Since the observed value of the test statistic falls in the rejection region ( F 12.66 2.09) , H0 is rejected.
There is sufficient evidence that differences exist among the treatment means at .05 . Since differences
exist, we now test for the interaction effect between Trail and Group.
To determine if Trail and Group interact, we test:
H0: Trail and Group do not interact
Ha: Trail and Group do interact
The test statistic is F 1.02 and the p-value is p .386 .
Since the p-value is greater than ( p .386 .05) , H0 is not rejected. There is insufficient evidence that
Trail and Group interact at .05 . Since the interaction does not exist, we test for the main effects of
Trail and Group.
Copyright © 2014 Pearson Education, Inc.
538
Chapter 9
To determine if there are differences in the mean trail deviations between the two levels of Trail, we test:
H 0 : 1 2
H a : 1 2
The test statistics is F 63.34 and the p-value is p .000 .
Since the p-value is less than ( p .000 .05) , H0 is rejected. There is sufficient evidence that the mean
trail deviations differ between the fecal extract trail and the control trail at .05 .
To determine if there are differences in the mean trail deviations between the four levels of Group, we test:
H 0 : 1 2 3 4
H a : At least two means differ
The test statistics is F 5.91 and p .001 .
Since the p-value is less than ( p .001 .05) , H0 is rejected. There is sufficient evidence that the mean
trail deviations differ among the four groups at .05 .
9.81
a.
Low Load, Ambiguous: Total1 n1 x1 25(18) 450
High Load, Ambiguous: Total2 n2 x2 25(6.1) 152.5
Low Load, Common: Total3 n3 x3 25(7.8) 195
High Load, Common: Total4 n4 x4 25(6.3) 157.5
(sum of all observations)2 (450 152.5 195 157.5)2 9552
9,120.25
n
100
100
b.
CM
c.
Low Load total is 450 195 645 . High Load total is 152.5 157.5 310 .
A
a
2
i
6452 3102
SS( Load ) i 1
CM
9,120.25 10, 242.5 9,120.25 1,122.25
2(25) 2(25)
br
Ambiguous total is 450 152.5 602.5 . Common total is 195 157.5 352.5
B
b
SS( Name)
j 1
ar
2
j
CM
AB
a
SS(Load Name)
602.52 352.52
7, 700.0625 9, 745.25 9,120.25 625
2(25)
2(25)
b
i 1 j 1
r
2
ij
SS (Load) SS(Name) CM
450 2 152.52 1952 157.52
1,122.25 625 9,120.25
25
25
25
25
11,543.5 1,122.25 625 9,120.25 676
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 539
d.
Low Load, Ambiguous: s12 152 225
(n1 1) s12 (25 1)225 5, 400
High Load, Ambiguous: s22 9.52 90.25
(n2 1) s22 (25 1)90.25 2,166
Low Load, Common: s32 9.52 90.25
(n3 1)s32 (25 1)90.25 2,166
High Load, Common: s42 102 100
(n4 1)s42 (25 1)100 2, 400
e.
SSE (n1 1) s12 (n2 1) s22 (n3 1) s32 (n4 1) s42 5, 400 2,166 2,166 2, 400 12,132
f.
SS Total = SS Load SS Name SS Load x Name SSE
g.
The ANOVA table is:
1,122.25 625 676 12,132 14, 555.25
Source
Load
Name
Load x Name
Error
Total
df
1
1
1
96
99
SS
1,122.25
625.00
676.00
12,132.00
14,555.25
MS
1,122.25
625.00
676.00
126.375
F
8.88
4.95
5.35
h.
Yes. We computed 5.35, which is almost the same as 5.34. The difference could be due to round-off
error.
i.
To determine if interaction between Load and Name is present, we test:
H0: Load and Name do not interact
Ha: Load and Name class do interact
The test statistic is F 5.35 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (2 1)(2 1) 1 and 2 n – ab 100 – 2 2 96 . From Table VI, Appendix D,
F.05 3.96 . The rejection region is F 3.96 .
Since the observed value of the test statistic falls in the rejection region ( F 5.35 3.96) , H0 is
rejected. There is sufficient evidence to indicate that Load and Name interact at .05 .
Copyright © 2014 Pearson Education, Inc.
540
Chapter 9
Using MINITAB, a graph of the results is:
Scatterplot of Mean vs Load
Name
17.5
1
2
Mean
15.0
12.5
10.0
7.5
5.0
Low
High
Load
From the graph, the interaction is quite apparent. For Low load, the mean number of jelly beans
taken for the ambiguous name is much higher than the mean number taken for the common name.
However, for High load, there is essentially no difference in the mean number of jelly beans taken
between the two names.
j.
We must assume that:
1.
The response distributions for each Load-Name combination (treatment) is normal.
2.
The response variance is constant for all Load-Name combinations.
3.
Random and independent samples of experimental units are associated with each Load-Name
combination.
9.82
A one-way ANOVA has only one factor with 2 or more levels. A two-way ANOVA has 2 factors, each at
2 or more levels.
9.83
In a completely randomized design, independent random selection of treatments to be assigned to
experimental units is required. In a randomized block design, the experimental units are first grouped into
blocks such that within the blocks the experimental units are homogeneous and between the blocks the
experimental units are heterogeneous. Once the experimental units are grouped into blocks, the treatments
are randomly assigned to the experimental units within each block so that each treatment appears one time
in each block.
9.84
There are 3 2 6 treatments. They are A1B1, A1B2, A2B1, A2B2, A3B1, and A3B2.
9.85
When the overall level of significance of a multiple comparisons procedure is , the level of significance
for each comparison is less than . This is because the comparisons within the experiment are not
independent of each other.
9.86
a.
SSE SSTotal SST 62.55 36.95 25.60
df Treatment k 1 4 1 3
MST
SST 36.95
12.32
3
df
df Error n k 20 4 16
MSE
SSE 25.60
1.60
16
df
Copyright © 2014 Pearson Education, Inc.
df Total n 1 20 1 19
F
MST 12.32
7.70
MSE 1.60
Design of Experiments and Analysis of Variance 541
The ANOVA table:
Source
Treatment
Error
Total
b.
df
3
16
19
SS
36.95
25.60
62.55
MS
12.32
1.60
F
7.70
To determine if there is a difference in the treatment means, we test:
H 0 : 1 2 3 4
H a : At least two means differ
where the i represents the mean for the ith treatment.
The test statistic is F
MST
7.70
MSE
The rejection region requires .10 in the upper tail of the F-distribution with1 k 1 4 1 3
and 2 n – k 20 4 16 . From Table V, Appendix D, F.10 2.46 . The rejection region is
F 2.46 .
Since the observed value of the test statistic falls in the rejection region ( F 7.70 2.46) , H0 is
rejected. There is sufficient evidence to conclude that at least two of the means differ at .10 .
c.
x4
x 57 11.4
4
5
n4
For confidence level .90, .10 and / 2 .10 / 2 .05 . From Table III, Appendix D, with df 16 ,
t.05 1.746 . The confidence interval is:
x4 t.05
9.87
a.
1.6
MSE
11.4 1.746
11.4 .99 10.41, 12.39
5
n4
SST SS Tot – SS Block – SSE 22.31 – 10.688 .288 11.334
MST
SST 11.334
3.778 ,
k 1
4 1
MS ( Block )
MSE
FT
df k – 1 4 – 1 3
SS ( Block ) 10.688
2.672 , df b – 1 5 – 1 4
b 1
5 1
SSE
.288
.024 , df n – k – b 1 20 – 4 – 5 1 12
n k b 1 20 4 5 1
MST 3.778
157.42
MSE .024
FB
MS ( Block ) 2.672
111.33
MSE
.024
Copyright © 2014 Pearson Education, Inc.
542
Chapter 9
The ANOVA Table is:
Source
Treatment
Block
Error
Total
b.
df
3
4
12
19
SS
11.334
10.688
0.288
22.310
MS
3.778
2.672
0.024
F
157.42
111.33
To determine if there are differences among the treatment means, we test:
H 0 : A B C D
H a : At least two treatment means differ
The test statistic is F
MST
157.42
MSE
The rejection region requires .05 in the upper tail of the F-distribution with1 k 1 4 1 3
and 2 n – k b 1 20 4 5 1 12 . From Table VI, Appendix D, F.05 3.49 . The rejection
region is F 3.49 .
Since the observed value of the test statistic falls in the rejection region ( F 157.42 3.49) ,
H0 is rejected. There is sufficient evidence to indicate differences among the treatment means at
.05 .
c.
Since there is evidence of differences among the treatment means, we need to compare the treatment
k (k 1) 4(4 1)
means. The number of pairwise comparisons is
6.
2
2
d.
To determine if there are differences among the block means, we test:
H0: All block means are the same
Ha: At least two block means differ
The test statistic is F
MS ( Block )
111.33
MSE
The rejection region requires .05 in the upper tail of the F distribution with1 b 1 5 1 4
and 2 n – k b 1 20 4 5 1 12 . From Table VI, Appendix D, F.05 3.26 . The rejection
region is F 3.26 .
Since the observed value of the test statistic falls in the rejection region ( F 111.33 3.26) , H0 is
rejected. There is sufficient evidence that the block means differ at .05 .
9.88
a.
df AB ( a 1) b 1 3 5 15
df Error n ab 48 4 6 24
SSAB MSAB df 3.1 15 46.5
SS Total SSA SSB SSAB SSE 2.6 9.2 46.5 18.7 77
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 543
SS A 2.6
.8667
a 1 3
MSB
MS A .8667
1.11
MSE .7792
FB
MSA
FA
Source
A
B
AB
Error
Total
b.
df
3
5
15
24
47
SS B 9.2
1.84
b 1 5
MSE
MS B 1.84
2.36
MSE .7792
FAB
SS
2.6
9.2
46.5
18.7
77.0
SSE 18.7
.7792
n ab 24
MS AB
3.1
3.98
MSE .7792
F
1.11
2.36
3.98
MS
.8667
1.84
3.1
.7792
Factor A has a 3 1 4 levels and factor B has b 5 1 6 levels. The number of treatments is
ab 4 6 24 . The total number of observations is n 47 1 48 . Thus, two replicates were
performed.
c.
SST SSA SSB SSAB 2.6 9.2 46.5 58.3
MST
58.3
SST
2.5347
ab 1 4(6) 1
F
MST 2.5347
3.25
MSE .7792
To determine whether the treatment means differ, we test:
H 0 : 1 2 24
H a : At least two treatment means differ
The test statistic is F
MST
3.25
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 4(6) 1 23 and 2 n – ab 48 – 4 6 24 . From Table VI, Appendix D, F.05 2.03
. The rejection region is F 2.03 .
Since the observed value of the test statistic falls in the rejection region ( F 3.25 2.03) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .05 .
d.
Since there are differences among the treatment means, we test for the presence of interaction:
H0: Factor A and factor B do not interact to affect the response mean
Ha: Factor A and factor B do interact to affect the response mean
The test statistic is F
MS AB
3.98
MSE
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (4 1)(6 1) 15 and 2 n – ab 48 – 4 6 24 . From Table VI, Appendix D,
F.05 2.11 . The rejection region is F 2.11 .
Copyright © 2014 Pearson Education, Inc.
544
Chapter 9
Since the observed value of the test statistic falls in the rejection region ( F 3.98 2.11) , H0 is
rejected. There is sufficient evidence to indicate factors A and B interact to affect the response means
at .05 .
Since the interaction is significant, no further tests are warranted. Multiple comparisons need to be
performed.
9.89
a.
A completely randomized design was used.
b.
There are 4 treatments: 3 robots/colony, 6 robots/colony, 9 robots/colony, and 12 robots/colony.
c.
To determine if there were differences in the mean energy expended (per robot) among the 4 colony
sizes, we test:
H 0 : 1 2 3 4
H a : At least two means differ
d.
Since the p-value is less than p .001 .05 , H0 is rejected. There is sufficient evidence to
indicate differences in mean energy expended per robot among the 4 colony sizes at .05 .
9.90
9.91
k k –1
The total number of comparisons conducted is c
f.
The mean energy expended by robots in the 12 robot colony is significantly smaller than the mean
energy expended by robots in any of the other size colonies. There are no differences in the mean
energy expended by robots in the 3 robot colony, the 6 robot colony, and the 9 robot colony.
a.
The response is the evaluation by the undergraduate student of the ethical behavior of the
salesperson.
b.
There are two factors—type of sales job at two levels (high tech. vs. low tech.) and sales task at two
levels (new account development vs. account maintenance).
c.
The treatments are the 2 2 4 combinations of type of sales job and sales task. The treatments are:
high tech./new account development, low tech./new account development,
high tech./account maintenance, and low tech./account maintenance.
d.
The experimental units are the college students.
a.
This is a complete 2 2 factorial design. The 2 factors are Color and Question. There are two levels
of color – Blue and Red. There are two levels of question – difficult and simple. The 4 treatments
are: blue/difficult, blue/simple, red/difficult, red/simple.
b.
Since the p-value is so small ( p .03) , H0 is rejected. There is a significant interaction between color
and question. The effect of color on the mean score is different at each level of question.
2
4 4 – 1
e.
Copyright © 2014 Pearson Education, Inc.
2
6.
Design of Experiments and Analysis of Variance 545
c.
Using MINITAB, the graph is:
Scatterplot of Blue, Red vs Quest
Variable
Blue
Red
80
Y-Data
70
60
50
40
Difficult
Simple
Quest
Since the lines are not parallel, it indicates that there is significant interaction between color and
question.
9.92
a.
This is a randomized block design. The blocks are the 12 plots of land. The treatments are the three
methods used on the shrubs: fire, clipping, and control. The response variable is the mean number of
flowers produced. The experimental units are the 36 shrubs.
b.
Plot
c.
To determine if there is a difference in the mean number of flowers produced among the three
treatments, we test:
H 0 : 1 2 3
H a : The mean number of flowers produced differ for at least two of the methods
The test statistic is F 5.42 and the p-value is p .009 .
Since the p-value is so small ( p .009) , H0 is rejected. There is sufficient evidence that there are
differences in the mean number of flowers produced among the three treatments at .009 .
Copyright © 2014 Pearson Education, Inc.
546
9.93
9.94
Chapter 9
d.
There is no difference in the mean number of flowers produced by Clipping and Burning. The mean
number of flowers produced by Control is significantly less than the mean number for Clipping and
Burning.
a.
The experimental design used in this example was a randomized block design.
b.
The experimental units in this problem are the electronic commerce and internet-based companies.
The response variable is the rate of return for the stock of the companies. The treatments are the 4
categories of companies: e-companies, internet software and service, internet hardware, and internet
communication. The blocks are the 3 age categories: 1 year-old, 3 year-old, and 5 year-old.
a.
The experimental unit in the study is the college tennis coach. The dependent variable is the response
to the statement “the Prospective Student-Athlete Form on the web site contributes very little to the
recruiting process” on a scale from 1 to 7. There is one factor in the study and it is the NCAA
division of the college tennis coach. There are 3 levels of this factor, and thus, there are 3
treatments: Division I, Division II, and Division III.
b.
To determine if the mean responses of tennis coaches from the different divisions differ, we test:
H 0 : 1 2 3
H a : At least two means differ
9.95
c.
Since the observed p-value of the test ( p .003) is less than .05 , H0 is rejected. There is
sufficient evidence to indicate differences in mean response among coaches of the 3 divisions at
.05 .
d.
The mean response for Division I coaches is significantly higher than the mean responses for the
Division II and Division III coaches. There is no difference in the mean responses between Division
II and Division III coaches.
a.
To determine if leadership style affects behavior of subordinates, we test:
H 0 : 1 2 3 4
H a : At least two treatment means differ
The test statistic is F 30.4 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 ab 1 2(2) 1 3 and 2 n – ab 257 – 2 2 253 . From Table VI, Appendix D,
F.05 2.60 . The rejection region is F 2.60 .
Since the observed value of the test statistic falls in the rejection region ( F 30.4 2.60) , H0 is
rejected. There is sufficient evidence to indicate that leadership style affects behavior of subordinates
at .05 .
b.
From the table, the mean response for High control, low consideration is significantly higher than that
for any other three treatments. The mean response for Low control, low consideration is significantly
higher than that for High control, high consideration and for Low control, high consideration. No
other significant differences exist.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 547
c.
The assumptions for Bonferroni's method are the same as those for the ANOVA. Thus, we must
assume that:
i. The populations sampled from are normal.
ii. The population variances are the same.
iii. The samples are independent.
9.96
From the printout, the p-value for treatments or Decoy is p .589 . Since the p-value is not small, we
cannot reject H0. There is insufficient evidence to indicate a difference in mean percentage of a goose flock
to approach to within 46 meters of the pit blind among the three decoy types. This conclusion is valid for
any reasonable value of .
9.97
a.
This is a complete 6 6 factorial design.
b.
There are 2 factors – Coagulant and pH level. There are 6 levels of coagulant: 5, 10, 20, 50, 100, and
200 mg / liter. There are 6 levels of pH: 4.0, 5.0, 6.0, 7.0, 8.0, and 9.0.
There are 6 6 36 treatments. In the pairs, let the coagulant level be the first number and pH level
the second. The 36 treatments are:
(5, 4.0)
(10, 4.0)
(20, 4.0)
(50, 4.0)
(100, 4.0)
(200, 4.0)
9.98
a.
(5, 5.0)
(10, 5.0)
(20, 5.0)
(50, 5.0)
(100, 5.0)
(200, 5.0)
(5, 6.0)
(10, 6.0)
(20, 6.0)
(50, 6.0)
(100, 6.0)
(200, 6.0)
(5, 8.0)
(10, 8.0)
(20, 8.0)
(50, 8.0)
(100, 8.0)
(200, 8.0)
(5, 9.0)
(10, 9.0)
(20, 9.0)
(50, 9.0)
(100, 9.0)
(200, 9.0)
The response is the weight of a brochure. There is one factor and it is carton. The treatments are the
five different cartons, while the experimental units are the brochures.
y .75005 .01406437506
CM
2
b.
(5, 7.0)
(10, 7.0)
(20, 7.0)
(50, 7.0)
(100, 7.0)
(200, 7.0)
n
2
40
SS Total y 2 CM .014066537 .01406437506 .00000216264
T2
.14767 2 .15028 2 .14962 2 .15217 2 .150312
SST i CM
.01406437506
ni
8
8
8
8
8
.01406568209 .01406437506 .00000130703
SSE SS Total SST .00000216264 .00000130703 .00000085561
MST
SST .00000130703
.000000326756
k 1
5 1
MSE
SSE .00000085561
.000000024446
40 5
nk
F
MST .000000326756
13.37
MSE .000000024446
Copyright © 2014 Pearson Education, Inc.
548
Chapter 9
Source
Treatments
Error
Total
df
4
35
39
SS
.00000130703
.00000085561
.00000216264
MS
.000000326756
.000000024446
F
13.37
To determine whether there are differences in mean weight per brochure among the five cartons, we
test:
H 0 : 1 2 3 4 5
H a : At least two treatment means differ
The test statistic is F 13.37 .
The rejection region requires .05 in the upper tail of the F-distribution with1 k 1 5 1 4
and 2 n – k 40 5 35 . From Table VI, Appendix D, F.05 2.61 . The rejection region is
F 2.61 .
Since the observed value of the test statistic falls in the rejection region ( F 13.37 2.61) , H0 is
rejected. There is sufficient evidence to indicate a difference in mean weight per brochure among the
five cartons at .05 .
c.
We must assume that the distributions of weights for the brochures in the five cartons are normal, that
the variances of the weights for the brochures in the five cartons are equal, and that random and
independent samples were selected from each of the cartons.
d.
Using MINITAB, the results of Tukey’s multiple comparison procedure are:
Level
Carton1
Carton2
Carton3
Carton4
Carton5
N
8
8
8
8
8
Mean
0.018459
0.018785
0.018703
0.019021
0.018789
Individual 95% CIs For Mean Based on
Pooled StDev
StDev ---+---------+---------+---------+----0.000105 (-----*-----)
0.000101
(----*-----)
0.000109
(----*-----)
0.000232
(-----*-----)
0.000188
(----*-----)
---+---------+---------+---------+-----0.01840
0.01860
0.01880
0.01900
Pooled StDev = 0.000156
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 549
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 99.32%
Carton1 subtracted from:
Carton2
Carton3
Carton4
Carton5
Carton2
Carton3
Carton4
Carton5
Lower
0.0001013
0.0000188
0.0003375
0.0001050
Center
0.0003262
0.0002437
0.0005625
0.0003300
Upper
0.0005512
0.0004687
0.0007875
0.0005550
------+---------+---------+---------+--(-----*------)
(-----*-----)
(-----*-----)
(-----*------)
------+---------+---------+---------+---0.00035
0.00000
0.00035
0.00070
Carton2 subtracted from:
Carton3
Carton4
Carton5
Carton3
Carton4
Carton5
Lower
-0.0003075
0.0000113
-0.0002212
Center
-0.0000825
0.0002363
0.0000037
Upper
0.0001425
0.0004612
0.0002287
------+---------+---------+---------+--(------*-----)
(------*-----)
(-----*------)
------+---------+---------+---------+---0.00035
0.00000
0.00035
0.00070
Carton3 subtracted from:
Carton4
Carton5
Carton4
Carton5
Lower
0.0000938
-0.0001387
Center
0.0003187
0.0000862
Upper
0.0005437
0.0003112
------+---------+---------+---------+--(-----*------)
(-----*------)
------+---------+---------+---------+---0.00035
0.00000
0.00035
0.00070
Carton4 subtracted from:
Carton5
Carton5
Lower
-0.0004575
Center
-0.0002325
Upper
-0.0000075
------+---------+---------+---------+--(-----*------)
------+---------+---------+---------+---0.00035
0.00000
0.00035
0.00070
Copyright © 2014 Pearson Education, Inc.
550
Chapter 9
The means arranged in order are:
Carton 1
.018459
Carton 3
.018703
Carton 2
.018785
Carton 5
.018789
Carton 4
.019021
The interpretation of the Tukey results are:
The mean weight for carton 4 is significantly higher than the mean weights of all the other cartons.
The mean weights of cartons 2, 3, and 5 are not significantly different from each other, but they are
significantly higher than the mean weight of carton 1.
Since there are differences among the cartons, management should sample from many cartons.
a.
There is one factor in this problem which is Group. There are 5 treatments in this problem,
corresponding to the 5 levels of Group: Casualties, Survivors, Implementers/casualties,
Implementers/survivors, and Formulators. The response variable is the ethics score. The
experimental units are the employees enrolled in an Executive MBA program.
b.
To determine if there are any differences among the mean ethics scores for the five groups, we test:
H 0 : 1 2 3 4 5
H a : At least two means differ
c.
The test statistic is F 9.85 and the p-value is p .000 . Since the p-value (0.000) is less than any
reasonable significance level , H0 is rejected. There is sufficient evidence to indicate a difference
in the mean ethics scores among the five groups of employees for any reasonable value of .
d.
We will check the assumptions of normality and equal variances. Using MINITAB, the histograms are:
Histogram of CASUAL, SURVIVE, IMPCAS, IMPSUR, FORMUL
Normal
0
C ASUAL
1
2
3
4
5
SURVIVE
IMPC AS
30
20
Frequency
9.99
e.
10
IMPSUR
FORMUL
30
0
0
1
2
3
4
5
C ASUAL
Mean
1.787
StDev 0.8324
N
47
SURVIVE
Mean 1.845
StDev 1.023
N
71
IMPC AS
Mean
1.593
StDev 0.6360
N
27
IMPSUR
Mean 2.545
StDev 1.301
N
33
20
10
0
0
1
2
3
4
5
FORMUL
Mean 2.871
StDev 1.176
N
31
The data for some of the 5 groups do not look particularly mound-shaped, so the assumption of
normality is probably not valid.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 551
Using MINITAB, the boxplots are:
Boxplot of CASUAL, SURVIVE, IMPCAS, IMPSUR, FORMUL
5
Data
4
3
2
1
CASUAL
SURVIVE
IMPCAS
IMPSUR
FORMUL
The spreads of responses do not appear to be about the same. The groups Implementers/survivors
and Formulators have more variability than the other three groups. Thus, the assumption of constant
variance is probably not valid.
The assumptions required for the ANOVA F-test do not appear to be reasonably satisfied.
9.100
e.
The Bonferroni method is preferred over other multiple comparisons methods because it does not
require equal sample sizes. The five groups of employees do not have the same sample sizes. In
addition, it is more powerful than Scheffe’s method.
f.
The number of pairwise comparisons for this analysis is c
g.
The mean ethics scores for both Groups 4 and 5 are significantly higher than the mean ethics scores
for Groups 1, 2, and 3. There is no difference in the mean ethics scores between Group 4 and Group
5. There is no difference in the mean ethics scores among Groups 1, 2 and 3.
a.
This is a randomized block design.
Response:
Factor:
Factor type:
Treatments:
Experimental units:
k ( k 1) 5(5 1) 20
10 .
2
2
2
the length of time required for a cut to stop bleeding
drug
qualitative
drugs A, B, and C
subjects
Copyright © 2014 Pearson Education, Inc.
552
Chapter 9
b.
Using MINITAB, the results are:
General Linear Model: Y versus Drug, Person
Factor
Drug
Person
Type Levels Values
fixed
3 A B C
fixed
5 1 2 3 4 5
Analysis of Variance for Y, using Adjusted SS for Tests
Source
Drug
Person
Error
Total
DF
2
4
8
14
Seq SS
156.4
7645.8
160.1
7962.3
Adj SS
156.4
7645.8
160.1
Adj MS
78.2
1911.5
20.0
F
3.91
95.51
P
0.066
0.000
Tukey 90.0% Simultaneous Confidence Intervals
Response Variable Y
All Pairwise Comparisons among Levels of Drug
Drug = A subtracted from:
Drug
B
C
Lower
-11.56
-3.72
Center
-4.820
3.020
Upper
1.922
9.762
-----+---------+---------+---------+(-------*-------)
(--------*-------)
-----+---------+---------+---------+-8.0
0.0
8.0
16.0
Upper
14.58
-----+---------+---------+---------+(--------*-------)
-----+---------+---------+---------+-8.0
0.0
8.0
16.0
Drug = B subtracted from:
Drug
C
Lower
1.098
Center
7.840
Let 1 , 2 , and 3 represent the mean clotting times for the three drugs. To determine if there is a
difference in mean clotting time among the 3 drugs, we test:
H 0 : 1 2 3
H a : At least two treatment means differ
The test statistic is F
MS ( Drug )
3.91 and the p-value is p .066 .
MSE
Since the p-value is less than ( p .066 .10) , H0 is rejected. There is sufficient evidence to
indicate differences in the mean clotting times among the three drugs at .10 .
c.
The observed level of significance is given as p .066 .
d.
To determine if there is a significant difference in the mean response over blocks, we test:
H 0 : 1 2 3 4 5
H a : At least two block means differ
The test statistic is F
MS ( Person)
95.51 and the p-value is p .000 .
MSE
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 553
Since the p-value is less than ( p .000 .10) , H0 is rejected. There is sufficient evidence to
indicate differences in the mean clotting times among the five people at .10 .
e.
The confidence interval to compare drugs A and B is (-11.56, 1.922). Since 0 is in the interval, there
is no evidence of a difference in mean clotting times between drugs A and B.
The confidence interval to compare drugs A and C is (-3.72, 9.762). Since 0 is in the interval, there
is no evidence of a difference in mean clotting times between drugs A and C.
The confidence interval to compare drugs B and C is (1.098, 14.58). Since 0 is not in the interval,
there is evidence of a difference in mean clotting times between drugs B and C. Since the numbers
are positive, the mean clotting time for drug C is greater than that for drug B.
In summary, the mean clotting time for drug C is greater than that for drug B. No other differences
exist.
9.101
9.102
a.
The response is the quality of the steel ingot.
b.
There are two factors: temperature and pressure. They are quantitative factors since they are
numerical.
c.
The treatments are the 3 5 15 factor-level combinations of temperature and pressure.
d.
The steel ingots are the experimental units.
a.
The degrees of freedom for “Type of message retrieval system” is a 1 2 1 1 . The degrees of
freedom for “Pricing option” is b 1 2 1 1 . The degrees of freedom for the interaction of Type
of message retrieval system and Pricing option is ( a 1) b – 1 (2 1)(2 1) 1 . The degrees of
freedom for error is n ab 120 2 2 116 .
Source
Type of message retrieval system
Pricing Option
Type of system pricing option
Error
Total
b.
Df
1
1
1
116
119
SS
-
MS
-
F
2.001
5.019
4.986
To determine if “Type of system” and “Pricing option” interact to affect the mean willingness to buy,
we test:
H0: “Type of system” and “Pricing option” do not interact
Ha: “Type of system” and “Pricing option” interact
c.
The test statistic is F
MSAB
4.986
MSE
The rejection region requires .05 in the upper tail of the F distribution with
1 (a 1)(b 1) (2 1)(2 1) 1 and 2 n – ab 120 – 2 2 116 . From Table VI, Appendix D,
F.05 3.92 . The rejection region is F 3.92 .
Copyright © 2014 Pearson Education, Inc.
554
Chapter 9
Since the observed value of the test statistic falls in the rejection region ( F 4.986 3.92) , H0 is
rejected. There is sufficient evidence to indicate “Type of system” and “Pricing option” interact to
affect the mean willingness to buy at .05 .
9.103
d.
No. Since the test in part c indicated that interaction between “Type of system” and “Pricing option”
is present, we should not test for the main effects. Instead, we should proceed directly to a multiple
comparison procedure to compare selected treatment means. If interaction is present, it can cover up
the main effects.
a.
We will select size as the quantitative variable and color as the qualitative variable. To determine if
the mean size of diamonds differ among the 6 colors, we test:
H 0 : 1 2 3 4 5 6
H a : At least two means differ
b.
Using MINITAB, the ANOVA table is:
One-way ANOVA: Carats versus Color
Analysis of Variance for Carats
Source
DF
SS
MS
Color
5
0.7963
0.1593
Error
302
22.7907
0.0755
Total
307
23.5869
Level
D
E
F
G
H
I
N
16
44
82
65
61
40
Mean
0.6381
0.6232
0.5929
0.5808
0.6734
0.7310
Pooled StDev =
0.2747
F
2.11
P
0.064
Individual 95% CIs For Mean
Based on Pooled StDev
StDev ----------+---------+---------+-----0.3195 (-------------*------------)
0.2677
(-------*-------)
0.2648
(-----*-----)
0.2792
(------*------)
0.2643
(------*------)
0.2918
(-------*--------)
----------+---------+---------+-----0.60
0.70
0.80
The test statistic is F 2.11 and the p-value is p .064 .
Since the p-value is less than ( p .064 .10) , H0 is rejected. There is sufficient evidence to
indicate the mean sizes of diamonds differ among the 6 colors at .10 .
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 555
We will check the assumptions of normality and equal variances. Using MINITAB, the histograms
are:
Histogram of CARAT
Normal
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
D
E
F
D
12
9
Frequency
6
3
G
H
0.6381
0.3195
16
Mean
StDev
N
E
Mean
StDev
N
0
I
0.6232
0.2677
44
F
Mean
StDev
N
12
9
0.5929
0.2648
82
G
6
Mean
StDev
N
3
0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
0.5808
0.2792
65
H
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
Mean
StDev
N
CARAT
Panel variable: COLOR
0.6734
0.2643
61
I
The data for the 6 colors do not look particularly mound-shaped, so the assumption of normality is
probably not valid. However, departures from this assumption often do not invalidate the ANOVA
results.
Using MINITAB, the box plots are:
Boxplot of CARAT vs COLOR
1.1
1.0
0.9
0.8
CARAT
c.
0.7
0.6
0.5
0.4
0.3
0.2
D
E
F
G
H
I
COLOR
The spreads of all the colors appear to be about the same, so the assumption of constant variance is
probably valid.
Copyright © 2014 Pearson Education, Inc.
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Chapter 9
d.
Using MINITAB, the Tukey confidence intervals are:
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of COLOR
Individual confidence level = 99.53%
COLOR = D subtracted from:
COLOR
E
F
G
H
I
Lower
-0.2435
-0.2591
-0.2758
-0.1846
-0.1387
Center
-0.0149
-0.0452
-0.0574
0.0353
0.0929
Upper
0.2136
0.1688
0.1611
0.2552
0.3244
---------+---------+---------+---------+
(-------------*-------------)
(------------*-------------)
(------------*-------------)
(-------------*-------------)
(--------------*-------------)
---------+---------+---------+---------+
-0.16
0.00
0.16
0.32
COLOR = E subtracted from:
COLOR
F
G
H
I
Lower
-0.1765
-0.1952
-0.1046
-0.0632
Center
-0.0303
-0.0424
0.0503
0.1078
Upper
0.1160
0.1104
0.2051
0.2788
---------+---------+---------+---------+
(--------*--------)
(--------*---------)
(---------*---------)
(----------*---------)
---------+---------+---------+---------+
-0.16
0.00
0.16
0.32
COLOR = F subtracted from:
COLOR
G
H
I
Lower
-0.1422
-0.0518
-0.0129
Center
-0.0122
0.0805
0.1381
Upper
0.1178
0.2129
0.2890
---------+---------+---------+---------+
(-------*-------)
(-------*-------)
(---------*--------)
---------+---------+---------+---------+
-0.16
0.00
0.16
0.32
COLOR = G subtracted from:
COLOR
H
I
Lower
-0.0469
-0.0071
Center
0.0927
0.1502
Upper
0.2322
0.3075
---------+---------+---------+---------+
(--------*--------)
(--------*---------)
---------+---------+---------+---------+
-0.16
0.00
0.16
0.32
COLOR = H subtracted from:
COLOR
I
Lower
-0.1017
Center
0.0576
Upper
0.2168
---------+---------+---------+---------+
(---------*---------)
---------+---------+---------+---------+
-0.16
0.00
0.16
0.32
All of the confidence intervals contain 0. Thus, at 95% confidence, there is no evidence that the
mean sizes of the diamonds are different among the different colors. This disagrees with the test of
hypothesis because the test was run using .10 .
9.104
a.
The experimenters expected there to be much variation in the number of participants from week to
week (more participants at the beginning and fewer as time goes on). Thus, by blocking on weeks,
this extraneous source of variation can be controlled.
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 557
b.
df(Week) b 1 6 1 5
MS ( Prompt )
SST 1185.0
296.25
4
df
F ( Prompt )
MST 296.25
39.87
MSE
7.43
MS
296.25
77.28
7.43
F
39.87
10.40
The ANOVA table is:
Source
Prompt
Week
Error
Total
c.
df
4
5
20
29
SS
1185.0
386.4
148.6
1720.0
p
0.0001
0.0001
To determine if a difference exists in the mean number of walkers per week among the five walker
groups, we test:
H 0 : 1 2 3 4 5
H a : At least two treatment means differ
where i represents the mean number of walkers in group i.
The test statistic is F 39.87 .
The rejection region requires .05 in the upper tail of the F-distribution with1 k 1 5 1 4
and 2 n – k b 1 30 5 6 1 20 . From Table VI, Appendix D, F.05 2.87 . The rejection
region is F 2.87 .
Since the observed value of the test statistic falls in the rejection region ( F 39.87 2.87) , H0 is
rejected. There is sufficient evidence to indicate differences exist among the mean number of
walkers per week among the 5 walker groups at .05 .
d.
The following conclusions are drawn:
There is no significant difference in the mean number of walkers per week in the "Frequent/High"
group and the "Frequent/Low group". The means for these two groups are significantly higher than
the means for the other three groups. There is no significant difference in the mean number of
walkers per week in the "Infrequent/Low" group and the "Infrequent/High" group. The means for
these two groups are significantly higher than the mean for the "Control group.
e.
9.105
In order for the above inferences to be valid, the following assumptions must hold:
1)
The probability distributions of observations corresponding to all block-treatment conditions
are normal.
2)
The variances of all the probability distributions are equal.
a.
The treatments are the 3 3 9 combinations of PES and Trust. The nine treatments are: (BC, Low),
(PC, Low), (NA, Low), (BC, Med), (PC, Med), (NA, Med), (BC, High), (PC, High), and (NA, High).
b.
df(Trust) 1 3 1 2 ;
SSE SSTot SS PES SS Trust SSPT 301.55 4.35 15.20 3.50 278.50
Copyright © 2014 Pearson Education, Inc.
Chapter 9
SS ( PES ) 4.35
2.175
2
df ( PES )
MS (Trust )
SS ( PT ) 3.50
.875
4
df ( PT )
MSE
278.50
SSE
1.458
191
df ( Error )
FPES
MS ( PES ) 2.175
1.49
MSE
1.458
FTrust
MS (Trust ) 7.600
5.21
MSE
1.458
FPT
MS ( PT ) .875
.600
MSE
1.458
MS ( PES )
MS ( PT )
SS (Trust ) 15.20
7.600
2
df (Trust )
The ANOVA table is:
Source
PES
Trust
PES Trust
Error
Total
c.
df
2
2
4
191
199
SS
4.35
15.20
3.50
278.50
301.55
MS
2.175
7.600
.875
1.458
F
1.49
5.21
0.60
To determine if PES and Trust interact, we test:
H0: PES and Trust do not interact to affect the mean tension
Ha: PES and Trust do interact to affect the mean tension
The test statistic is F 0.60 .
The rejection region requires .05 in the upper tail of the F-distribution with
1 (a 1)(b 1) (3 1)(3 1) 4 and 2 n – ab 215 – 3 3 191 . From Table VI, Appendix D,
F.05 2.37 . The rejection region is F 2.37 .
Since the observed value of the test statistic does not fall in the rejection region ( F 0.60 2.37) , H0
is not rejected. There is insufficient evidence to indicate that PES and Trust interact at .05 .
d.
The plot of the treatment means is:
Interaction Plot for Mean
Trust
High
Low
Medium
6.5
6.0
Mean
558
5.5
5.0
4.5
BC
NA
Perform
PC
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 559
The three lines corresponding to the Trust levels are almost parallel. This indicates that PES and
Trust do not interact. This agrees with the result in part c.
9.106
e.
Since the interaction is not significant, the tests for the main effects should be run.
a.
There are a total of a b 3 3 9 treatments in this study.
b.
Using MINITAB, the ANOVA results are:
ANOVA: Y versus Display, Price
Factor
Display
Price
Type Levels Values
fixed
3 1 2 3
fixed
3 1 2 3
Analysis of Variance for Y
Source
Display
Price
Display*Price
Error
Total
S = 22.2428
DF
2
2
4
18
26
SS
1691393
3089054
510705
8905
5300057
R-Sq = 99.83%
MS
F
845696 1709.37
1544527 3121.89
127676 258.07
495
P
0.000
0.000
0.000
R-Sq(adj) = 99.76%
To get the SS for Treatments, we must add the SS for Display, SS for Price, and the SS for Interaction.
Thus, SST 1, 691, 393 3, 089, 054 510, 705 5, 291,152 . The df 2 2 4 8 .
MST
SST
5, 291,152
661,394
3(3) 1
ab 1
F
MST 661,394
1336.15
MSE
495
To determine whether the treatment means differ, we test:
H 0 : 1 2 9
H a : At least two treatment means differ
The test statistic is F
MST
1, 336.15 .
MSE
The rejection region requires .10 in the upper tail of the F-distribution with
1 ab 1 3(3) 1 8 and 2 n – ab 27 – 3 3 18 . From Table V, Appendix D, F.10 2.04 .
The rejection region is F 2.04 .
Since the observed value of the test statistic falls in the rejection region ( F 1,336.15 2.04) , H0 is
rejected. There is sufficient evidence to indicate the treatment means differ at .10 .
c.
Since there are differences among the treatment means, we next test for the presence of interaction.
H0: Factors A and B do not interact to affect the response means
Ha: Factors A and B do interact to affect the response means
The test statistic is F
MSAB
258.07 and the p-value is p .000 .
MSE
Copyright © 2014 Pearson Education, Inc.
560
Chapter 9
Since the p-value is less than ( p .000 .10) , H0 is rejected. There is sufficient evidence to
indicate the two factors interact at .10 .
9.107
9.108
d.
The main effect tests are not warranted since interaction is present in part c.
e.
The nine treatment means need to be compared.
a.
This is a 2 2 factorial experiment.
b.
The two factors are the tent type (treated or untreated) and location (inside or outside). There are
2 2 4 treatments. The four treatments are (treated, inside), (treated, outside), (untreated, inside),
and (untreated, outside).
c.
The response variable is the number of mosquito bites received in a 20 minute interval.
d.
There is sufficient evidence to indicate interaction is present. This indicates that the effect of the tent
type on the number of mosquito bites depends on whether the person is inside or outside.
a.
This is a completely randomized design with a complete four-factor factorial design.
b.
There are a total of 2 2 2 2 16 treatments.
c.
Using SAS, the output is:
Analysis of Variance Procedure
Dependent Variable: Y
Sum of
Mean
Source
DF
Squares
Square
F Value
Pr > F
Model
15
546745.50
36449.70
5.11
0.0012
Error
16
114062.00
7128.88
Corrected Total
31
660807.50
R-Square
C.V.
Root MSE
Y Mean
0.827390
41.46478
84.433
203.63
Copyright © 2014 Pearson Education, Inc.
Design of Experiments and Analysis of Variance 561
d.
Source
DF
Anova SS
Mean Square
F Value
Pr > F
SPEED
1
56784.50
56784.50
7.97
0.0123
FEED
1
21218.00
21218.00
2.98
0.1037
SPEED*FEED
1
55444.50
55444.50
7.78
0.0131
COLLET
1
165025.13
165025.13
23.15
0.0002
SPEED*COLLET
1
44253.13
44253.13
6.21
0.0241
FEED*COLLET
1
142311.13
142311.13
19.96
0.0004
SPEED*FEED*COLLET
1
54946.13
54946.13
7.71
0.0135
WEAR
1
378.13
378.13
0.05
0.8208
SPEED*WEAR
1
1540.13
1540.13
0.22
0.6483
FEED*WEAR
1
946.13
946.13
0.13
0.7204
SPEED*FEED*WEAR
1
528.13
528.13
0.07
0.7890
COLLET*WEAR
1
1682.00
1682.00
0.24
0.6337
SPEED*COLLET*WEAR
1
512.00
512.00
0.07
0.7921
FEED*COLLET*WEAR
1
72.00
72.00
0.01
0.9212
SPEE*FEED*COLLE*WEAR
1
1104.50
1104.50
0.15
0.6991
To determine if the interaction terms are significant, we must add together the sum of squares for all
interaction terms as well as the degrees of freedom.
SS Interaction 55, 444.50 44, 253.13 142,311.13 54, 946.13 1,540.13 946.13
528.13 1, 682.00 512.00 72.00 1,104.50 303, 339.78
df(Interaction) = 11
MS ( Interaction)
FInteraction
SS ( Interacton) 303,339.78
27,576.34364
11
df ( Interaction)
MS ( Interaction) 27,576.34364
3.87
MSE
7128.88
To determine if interaction effects are present, we test:
H0: No interaction effects exist
Ha: Interaction effects exist
The test statistic is F 3.87 .
The rejection region requires .05 in the upper tail of the F-distribution with1 11 and 2 16 .
From Table VI, Appendix D, F.05 2.49 . The rejection region is F 2.49 .
Since the observed value of the test statistic falls in the rejection region ( F 3.87 2.49) , H0 is
rejected. There is sufficient evidence to indicate that interaction effects exist at .05 .
Since the sums of squares for a balanced factorial design are independent of each other, we can look
at the SAS output to determine which of the interaction effects are significant. The three-way
interaction between speed, feed, and collet is significant ( p .0135) . There are three two-way
interactions with p-values less than .05. However, all of these two-way interaction terms are
imbedded in the significant three-way interaction term.
Copyright © 2014 Pearson Education, Inc.
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Chapter 9
e.
Yes. Since the significant interaction terms do not include wear, it would be necessary to perform the
main effect test for wear. All other main effects are contained in a significant interaction term.
To determine if the mean finish measurements differ for the different levels of wear, we test:
H0: The mean finish measurements for the two levels of wear are the same
Ha: The mean finish measurements for the two levels of wear are different
The test statistic is F 0.05 and the p-value is p .8280 .
Since the p-value is not less than ( p .8280 .05) , H0 is not rejected. There is insufficient
evidence to indicate that the mean finish measurements differ for the different levels of wear at
.05 .
f.
We must assume that:
i. The populations sampled from are normal.
ii. The population variances are the same.
iii. The samples are random and independent.
9.109
Using MINITAB, the ANOVA Table is:
ANOVA: Rating versus Prep, Standing
Factor
Prep
Standing
Type Levels Values
fixed
2 PRACTICE REVIEW
fixed
3 HI LOW MED
Analysis of Variance for Rating
Source
Prep
Standing
Prep*Standing
Error
Total
DF
1
2
2
126
131
SS
54.735
16.500
13.470
478.955
563.659
S = 1.94967
R-Sq = 15.03%
MS
54.735
8.250
6.735
3.801
F
14.40
2.17
1.77
P
0.000
0.118
0.174
R-Sq(adj) = 11.66%
Tukey 95.0% Simultaneous Confidence Intervals
Response Variable Rating
All Pairwise Comparisons among Levels of Prep
Prep = PRACTICE subtracted from:
Prep
REVIEW
Lower
-1.960
Center
-1.288
Upper
-0.6162
---+---------+---------+---------+--(-----------*----------)
---+---------+---------+---------+---1.80
-1.20
-0.60
0.00
First, we must test for treatment effects.
SST SSP SSS SSPS 54.735 16.500 13.470 84.705 .
The df 1 2 2 5 .
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Design of Experiments and Analysis of Variance 563
MST
SST
84.705
16.941
ab 1 2(3) 1
F
MST 16.941
4.46
MSE 3.801
To determine if there are differences in mean ratings among the 6 treatments, we test:
H0: All treatment means are the same
Ha: At least two treatment means differ
The test statistic is F 4.46 .
Since no was given, we will use .05 . The rejection region requires .05 in the upper tail of the F
distribution with1 ab 1 2(3) 1 5 and 2 n – ab 132 – 2 3 126 . From Table VI, Appendix D,
F.05 2.29 . The rejection region is F 2.29 .
Since the observed value of the test statistic falls in the rejection region ( F 4.46 2.29) , H0 is rejected.
There is sufficient evidence that differences exist among the treatment means at .05 . Since differences
exist, we now test for the interaction effect between Preparation and Class Standing.
To determine if Preparation and Class Standing interact, we test:
H0: Preparation and Class Standing do not interact
Ha: Preparation and Class Standing do interact
The test statistic is F 1.77 and the p-value is p .174 .
Since the p-value is greater than ( p .174 .05) , H0 is not rejected. There is insufficient evidence that
Preparation and Class Standing interact at .05 . Since the interaction does not exist, we test for the main
effects of Preparation and Class standing.
To determine if there are differences in the mean rating between the three levels of Class standing, we test:
H 0 : L M H
H a : At leaset two treatment means differ
The test statistics is F 2.17 and the p-value is p 0.118 .
Since the p-value is greater than ( p .118 .05) , H0 is not rejected. There is insufficient evidence that
the mean ratings differ among the 3 levels of Class Standing at .05 .
To determine if there are differences in the mean rating between the two levels of Preparation, we test:
H 0 : P R
H a : P R
The test statistics is F 14.40 and the p-value is p 0.000 .
Since the p-value is less than ( p .000 .05) , H0 is rejected. There is sufficient evidence that the mean
ratings differ between the two levels of preparation at .05 .
There are only 2 levels of Preparation. The mean rating for Practice is higher than the mean rating Review.
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