Solutions Manual to Accompany
2-1
Essentials of Modern
Business Statistics
With Microsoft Excel
Eighth Edition
David R. Anderson
University of Cincinnati
Dennis J. Sweeney
University of Cincinnati
Thomas A. Williams
Rochester Institute of Technology
South-Western
Cincinnati, Ohio
2-1
Commented [Ma1]:
Contents
Preface
Chapter
1.
Data and Statistics
2.
Descriptive Statistics: Tabular and Graphical Methods
3.
Descriptive Statistics: Numerical Methods
4.
Introduction to Probability
5.
Discrete Probability Distributions
6.
Continuous Probability Distributions
7.
Sampling and Sampling Distributions
8.
Interval Estimation
9.
Hypothesis Testing
10.
Comparisons Involving Means
11.
Comparisons Involving Proportions and A Test of Independence
12.
Simple Linear Regression
13.
Multiple Regression
14.
Statistical Methods for Quality Control
Preface
The purpose of Essentials of Modern Business Statistics with Microsoft Excel is to provide
students, primarily in the fields of business administration and economics, with a sound
conceptual introduction to the field of Statistics and its many applications. The text is
applications-oriented and has been written with the needs of the nonmathematical in mind.
The solutions manual furnishes assistance by identifying learning objectives and providing
detailed solutions for all exercises in the text.
Note: The solutions to the case problems are included in a separate manual.
Acknowledgements
We would like to provide special recognition to Catherine J. Williams for her efforts in
preparing the solutions manual.
David R. Anderson
Dennis J. Sweeney
Thomas A. Williams
2-3
Chapter 1
DATA and STATISTICS
Learning Objectives
1.
Obtain an appreciation for the breadth of statistical applications in business and economics.
2.
Understand the meaning of the terms elements, variables, and observations as they are used in
STATISTICS.
3.
Understand that DATA are obtained using one of the following scales of measurement:
nominal, ordinal, interval, and ratio.
4.
Obtain an understanding of the difference between qualitative, quantitative, crossectional and time
series DATA.
5.
Learn about the sources of DATA for statistical analysis both internal and external to the firm.
6.
Be aware of how errors can arise in DATA.
7.
Know the meaning of descriptive STATISTICS and statistical inference.
8.
Be able to distinguish between a POPULATION and a SAMPLE.
9.
Understand the role a SAMPLE plays in making statistical inferences about the POPULATION.
2-4
Solutions:
1.
2.
3.
4.
5.
STATISTICS can be referred to as numerical facts. In a broader sense, STATISTICS is the
field of study dealing with the collection, analysis, presentation and interpretation of DATA.
a.
9
b.
4
c.
Country and room rate are qualitative variables; number of rooms and the overall score are
quantitative variables.
d.
Country is nominal; room rate is ordinal; number of rooms is ratio and overall score is interval.
a.
Average number of rooms = 808/9 = 89.78 or approximately 90 rooms
b.
2 of 9 are located in England; approximately 22%
c.
4 of 9 have a room rate of $$; approximately 44%
a.
10
b.
All brands are models of minisystems manufactured.
c.
Average price = 3140/10 = $314
d.
$314
a.
5
b.
Price, CD capacity, and the number of tape decks are quantitative. Sound quality and FM tuning
sensitivity and selectivity are qualitative.
c.
Average CD capacity = 30/10 = 3.
d.
7
(100)  70%
10
e.
4
(100)  40%
10
6.
Questions a, c, and d are quantitative.
Questions b and e are qualitative.
7.
8.
a.
The variable is qualitative.
b.
Nominal with four labels or categories.
a.
1005
b.
Qualitative
2-5
9.
c.
Percentages
d.
.29(1005) = 291.45 or approximately 291.
a.
Qualitative
b.
30 of 71; 42.3%
10. a.
Quantitative; ratio
b.
Qualitative; nominal
c.
Qualitative (Note: Rank is a numeric label that identifies the position of a student in the class. Rank
does not indicate how much or how many and is not quantitative.); ordinal
d.
Qualitative; nominal
e.
Quantitative; ratio
11. a.
Quantitative; ratio
b.
Qualitative; ordinal
c.
Qualitative; ordinal (assuming employees can be ranked by classification)
d.
Quantitative; ratio
e.
Qualitative; nominal
12. a.
The POPULATION is all visitors coming to the state of Hawaii.
b.
Since airline flights carry the vast majority of visitors to the state, the use of questionnaires for
passengers during incoming flights is a good way to reach this POPULATION. The
questionnaire actually appears on the back of a mandatory plants and animals declaration form
that passengers must complete during the incoming flight. A large percentage of passengers
complete the visitor information questionnaire.
c.
Questions 1 and 4 provide quantitative DATA indicating the number of visits and the number of days
in Hawaii. Questions 2 and 3 provide qualitative DATA indicating the categories of reason for the
trip and where the visitor plans to stay.
13. a.
Quantitative - Earnings measured in billions of dollars.
b.
Time series with 6 observations
c.
Volkswagen's annual earnings.
d.
Time series shows an increase in earnings. An increase would be expected in 2003, but it appears
that the rate of increase is slowing.
2-6
14. a.
b.
Type of music is a qualitative variable
The graph, based on time series DATA, is shown below.
Percentage of Music Sales
34
32
30
28
26
24
22
20
1995
1996
1997
1998
1999
2000
2001
Year
c.
The bar graph, based on cross-sectional DATA, is shown below.
% of Music Sales in 1998
30.0
25.0
20.0
15.0
10.0
5.0
0.0
Type of Music
15.
Crossectional DATA. The DATA were collected at the same or approximately the same point in time.
16. a.
We would like to see DATA from product taste tests and test marketing the product.
b.
Such DATA would be obtained from specially designed statistical studies.
2-7
17.
Internal DATA on salaries of other employees can be obtained from the personnel
department. External DATA might be obtained from the Department of Labor or industry
associations.
18. a.
(48/120)100% = 40% in the SAMPLE died from some form of heart disease. This can be used as
an estimate of the percentage of all males 60 or older who die of heart disease.
b.
19. a.
The DATA on cause of death is qualitative.
All subscribers of Business Week at the time the 1996 survey was conducted.
b.
Quantitative
c.
Qualitative (yes or no)
d.
Crossectional - 1996 was the time of the survey.
e.
Using the SAMPLE results, we could infer or estimate 59% of the POPULATION of subscribers
have an annual income of $75,000 or more and 50% of the POPULATION of subscribers have
an American Express credit card.
20. a.
56% of market belonged to A.C. Nielsen
$387,325 is the average amount spent per category
b.
3.73
c.
$387,325
21. a.
The two POPULATIONs are the POPULATION of women whose mothers took the drug DES
during pregnancy and the POPULATION of women whose mothers did not take the drug DES
during pregnancy.
b.
It was a survey.
c.
63 / 3.980 = 15.8 women out of each 1000 developed tissue abnormalities.
d.
The article reported “twice” as many abnormalities in the women whose mothers had taken DES
during pregnancy. Thus, a rough estimate would be 15.8/2 = 7.9 abnormalities per 1000 women
whose mothers had not taken DES during pregnancy.
e.
In many situations, disease occurrences are rare and affect only a small portion of the
POPULATION. Large SAMPLEs are needed to collect DATA on a reasonable number of cases
where the disease exists.
22. a.
All adult viewers reached by the Denver, Colorado television station.
b.
The viewers contacted in the telephone survey.
c.
A SAMPLE. It would clearly be too costly and time consuming to try to contact all viewers.
23. a.
Percent of television sets that were tuned to a particular television show and/or total viewing
audience.
b.
All television sets in the United States which are available for the viewing audience. Note this would
not include television sets in store displays.
c.
A portion of these television sets. Generally, individual households would be contacted to determine
2-8
which programs were being viewed.
2-9
d.
24. a.
The cancellation of programs, the scheduling of programs, and advertising cost rates.
This is a statistically correct descriptive statistic for the SAMPLE.
b.
c.
An incorrect generalization since the DATA was not collected for the entire POPULATION.
An acceptable statistical inference based on the use of the word “estimate.”
d.
While this statement is true for the SAMPLE, it is not a justifiable conclusion for the entire POPULATION.
e.
This statement is not statistically supportable. While it is true for the particular SAMPLE observed, it
is entirely possible and even very likely that at least some students will be outside the 65 to 90 range
of grades.
2 - 10
Chapter 2
Descriptive STATISTICS: Tabular and
Graphical Methods
Learning Objectives
1.
Learn how to construct and interpret summarization procedures for qualitative DATA such as :
frequency and relative frequency distributions, bar graphs and pie charts. Be able to use Excel's
COUNTIF function to construct a frequency distribution and the Chart Wizard to construct a bar
graph and pie chart.
2.
Learn how to construct and interpret tabular summarization procedures for quantitative DATA such
as: frequency and relative frequency distributions, cumulative frequency and cumulative relative
frequency distributions. Be able to use Excel's FREQUENCY function to construct a frequency
distribution and the Chart Wizard to construct a histogram.
3.
Learn how to construct a histogram and an ogive as graphical summaries of quantitative DATA.
4.
Be able to use and interpret the exploratory DATA analysis technique of a stem-and-leaf display.
5.
Learn how to construct and interpret cross tabulations and scatter diagrams of bivariate DATA. Be
able to use Excel's Pivot Table report to construct a cross tabulation and the Chart Wizard to
construct a scatter diagram.
2 - 11
Solutions:
1.
Class
A
B
C
2.
a.
1 - (.22 + .18 + .40) = .20
b.
.20(200) = 40
Frequency
60
24
36
120
Relative Frequency
60/120 = 0.50
24/120 = 0.20
36/120 = 0.30
1.00
c/d
Class
A
B
C
D
Total
3.
a.
360° x 58/120 = 174°
b.
360° x 42/120 = 126°
Frequency
.22(200) = 44
.18(200) = 36
.40(200) = 80
.20(200) = 40
200
Percent Frequency
22
18
40
20
100
c.
No
Opinion
16.7%
Yes
48.3%
No
35%
d.
70
60
Frequency
50
40
30
20
10
0
Yes
No
No Opinion
Response
4.
a.
The DATA are qualitative.
b.
TV Show
Millionaire
Frasier
Chicago Hope
Charmed
Total:
Frequency
24
15
7
4
50
12
Percent
Frequency
48
30
14
8
100
c.
30
Frequency
25
20
15
10
5
0
Millionaire
Frasier
Chicago
Charmed
TV Show
Charmed
8%
Chicago
14%
Millionaire
48%
Frasier
30%
d.
5.
Millionaire has the largest market share. Frasier is second.
a.
Name
Brown
Davis
Johnson
Jones
Smith
Williams
Frequency
7
6
10
7
12
8
Relative Frequency
.14
.12
.20
.14
.24
.16
Percent Frequency
14%
12%
20%
14%
24%
16%
50
1.00
b.
14
12
Frequency
10
8
6
4
2
0
Brown
c.
Brown
Davis
Johnson
Jones
Smith
Williams
Davis
Johnson
Jones
Smith
.14 x 360 = 50.4
.12 x 360 = 43.2
.20 x 360 = 72.0
.14 x 360 = 50.4
.24 x 360 = 86.4
.16 x 360 = 57.6


Williams
16%
Smith
24%
Brown
14%
Jones
14%
Davis
12%
Johnson
20%
d.
6.
Most common: Smith, Johnson and Williams
a.
Book
7 Habits
Millionaire
Motley
Dad
Frequency
10
16
9
13
14
Percent Frequency
16.66
26.67
15.00
21.67
Williams
WSJ Guide
Other
Total:
6
6
60
10.00
10.00
100.00
The Ernst & Young Tax Guide 2000 with a frequency of 3, Investing for Dummies with a frequency
of 2, and What Color is Your Parachute? 2000 with a frequency of 1 are grouped in the "Other"
category.
b.
The rank order from first to fifth is: Millionaire, Dad, 7 Habits, Motley, and WSJ Guide.
c.
The percent of sales represented by The Millionaire Next Door and Rich Dad, Poor Dad is 48.33%.
7.
Rating
Outstanding
Very Good
Good
Average
Poor
Frequency
19
13
10
6
2
50
Relative Frequency
0.38
0.26
0.20
0.12
0.04
1.00
Management should be pleased with these results. 64% of the ratings are very good to outstanding.
84% of the ratings are good or better. Comparing these ratings with previous results will show
whether or not the restaurant is making improvements in its ratings of food quality.
8.
a.
Position
Pitcher
Catcher
1st Base
2nd Base
3rd Base
Shortstop
Left Field
Center Field
Right Field
9.
Frequency
17
4
5
4
2
5
6
5
7
55
b.
Pitchers (Almost 31%)
c.
3rd Base (3 - 4%)
d.
Right Field (Almost 13%)
e.
Infielders (16 or 29.1%) to Outfielders (18 or 32.7%)
Relative Frequency
0.309
0.073
0.091
0.073
0.036
0.091
0.109
0.091
0.127
1.000
a/b.
Starting Time
7:00
7:30
8:00
8:30
9:00
Frequency
3
4
4
7
2
Percent Frequency
15
20
20
35
10
20
16
100
c.
Bar Graph
8
7
Frequency
6
5
4
3
2
1
0
7:00
7:30
8:00
8:30
9:00
Starting Time
d.
9:00
10%
7:00
15%
7:30
20%
8:30
35%
8:00
20%
e.
10. a.
The most preferred starting time is 8:30 a.m.. Starting times of 7:30 and 8:00 a.m. are next.
The DATA refer to quality levels from 1 "Not at all Satisfied" to 7 "Extremely Satisfied."
b.
Rating
3
4
5
6
7
Frequency
2
4
12
24
18
60
Relative Frequency
0.03
0.07
0.20
0.40
0.30
1.00
c.
Bar Graph
30
25
Frequency
20
15
10
5
0
3
4
5
6
7
Rating
d.
The survey DATA indicate a high quality of service by the financial consultant. The most
common ratings are 6 and 7 (70%) where 7 is extremely satisfied. Only 2 ratings are below the
middle scale value of 4. There are no "Not at all Satisfied" ratings.
11.
Class
Frequency
Relative Frequency
Percent Frequency
12-14
15-17
18-20
21-23
24-26
2
8
11
10
9
40
0.050
0.200
0.275
0.250
0.225
1.000
5.0
20.0
27.5
25.5
22.5
100.0
Total
12.
Class
less than or equal to 19
less than or equal to 29
less than or equal to 39
less than or equal to 49
less than or equal to 59
Cumulative Frequency
10
24
41
48
50
18
Cumulative Relative Frequency
.20
.48
.82
.96
1.00
13.
18
16
14
Frequency
12
10
8
6
4
2
0
10-19
20-29
30-39
40-49
50-59
1.0
.8
.6
.4
.2
0
10
20
30
40
50
14. a/b.
Class
6.0 - 7.9
8.0 - 9.9
10.0 - 11.9
12.0 - 13.9
14.0 - 15.9
Frequency
4
2
8
3
3
20
PercentFrequency
20
10
40
15
15
100
60
15. a/b.
Waiting Time
0-4
5-9
10 - 14
15 - 19
20 - 24
Totals
Frequency
4
8
5
2
1
20
RelativeFrequency
0.20
0.40
0.25
0.10
0.05
1.00
c/d.
Waiting Time
Less than or equal to 4
Less than or equal to 9
Less than or equal to 14
Less than or equal to 19
Less than or equal to 24
e.
Cumulative Frequency
4
12
17
19
20
Cumulative Relative Frequency
0.20
0.60
0.85
0.95
1.00
12/20 = 0.60
16. a.
Stock Price ($)
10.00 - 19.99
20.00 - 29.99
30.00 - 39.99
40.00 - 49.99
50.00 - 59.99
60.00 - 69.99
Total
Relative
Frequency
0.40
0.16
0.24
0.08
0.04
0.08
1.00
Frequency
10
4
6
2
1
2
25
20
Percent
Frequency
40
16
24
8
4
8
100
12
Frequency
10
8
6
4
2
0
10.0019.99
20.0029.99
30.0039.99
40.0049.99
50.0059.99
60.0069.99
Stock Price
Many of these are low priced stocks with the greatest frequency in the $10.00 to $19.99 range.
b.
Earnings per
Share ($)
-3.00 to -2.01
-2.00 to -1.01
-1.00 to -0.01
0.00 to 0.99
1.00 to 1.99
2.00 to 2.99
Total
Frequency
2
0
2
9
9
3
25
Relative
Frequency
0.08
0.00
0.08
0.36
0.36
0.12
1.00
Percent
Frequency
8
0
8
36
36
12
100
10
Frequency
9
8
7
6
5
4
3
2
1
0
-3.00 to
-2.01
-2.00 to
-1.01
-1.00 to
-0.01
0.00 to
0.99
1.00 to
1.99
2.00 to
2.99
Earnings per Share
The majority of companies had earnings in the $0.00 to $2.00 range. Four of the companies lost
money.
17. a.
Amount
0-99
100-199
200-299
300-399
400-499
b.
Frequency
5
5
8
4
3
25
Histogram
22
Relative Frequency
.20
.20
.32
.16
.12
1.00
9
8
7
Frequency
6
5
4
3
2
1
0
0-99
100-199
200-299
300-399
400-499
Amount ($)
c.
18. a.
The largest group spends $200-$300 per year on books and magazines. There are more in the $0 to
$200 range than in the $300 to $500 range.
Lowest salary: $93,000
Highest salary: $178,000
b.
Salary
($1000s)
91-105
106-120
121-135
136-150
151-165
166-180
Total
Frequency
4
5
11
18
9
3
50
c.
Proportion $135,000 or less: 20/50.
d.
Percentage more than $150,000: 24%
Relative
Frequency
0.08
0.10
0.22
0.36
0.18
0.06
1.00
Percent
Frequency
8
10
22
36
18
6
100
20
18
16
Frequency
14
12
10
8
6
4
2
0
91-105
106-120 121-135 136-150 151-165 166-180
Salary ($1000s)
e.
19. a/b.
Number
140 - 149
150 - 159
160 - 169
170 - 179
180 - 189
190 - 199
Totals
Frequency
2
7
3
6
1
1
20
Relative Frequency
0.10
0.35
0.15
0.30
0.05
0.05
1.00
c/d.
Number
Less than or equal to 149
Less than or equal to 159
Less than or equal to 169
Less than or equal to 179
Less than or equal to 189
Less than or equal to 199
Cumulative Frequency
2
9
12
18
19
20
24
Cumulative Relative Frequency
0.10
0.45
0.60
0.90
0.95
1.00
e.
Frequency
20
15
10
5
140
20. a.
160
180
200
The percentage of people 34 or less is 20.0 + 5.7 + 9.6 + 13.6 = 48.9.
b.
The percentage of the POPULATION over 34 years old is
16.3 + 13.5 + 8.7 + 12.6 = 51.1
c.
The percentage of the POPULATION that is between 25 and 54 years old inclusively is
13.6 + 16.3 + 13.5 = 43.4
d.
The percentage less than 25 years old is 20.0 + 5.7 + 9.6 = 35.3.
So there are (.353)(275) = 97.075 million people less than 25 years old.
e.
An estimate of the number of retired people is (.5)(.087)(275) + (.126)(275) = 46.6125 million.
21. a/b.
Computer
Usage (Hours)
0.0 2.9
3.0 5.9
6.0 8.9
9.0 - 11.9
12.0 - 14.9
Total
Frequency
5
28
8
6
3
50
Relative
Frequency
0.10
0.56
0.16
0.12
0.06
1.00
c.
30
Frequency
25
20
15
10
5
0
0.0 - 2.9
3.0 - 5.9
6.0 - 8.9
9.0 - 11.9 12.0 - 14.9
Computer Usage (Hours)
d.
60
50
Frequency
40
30
20
10
0
3
6
9
12
15
Computer Usage (Hours)
e.
The majority of the computer users are in the 3 to 6 hour range. Usage is somewhat skewed toward
the right with 3 users in the 12 to 15 hour range.
26
22.
23.
24.
5
7 8
6
4 5 8
7
0 2 2 5 5 6 8
8
0 2 3 5
Leaf Unit = 0.1
6
3
7
5 5 7
8
1 3 4 8
9
3 6
10
0 4 5
11
3
Leaf Unit = 10
11
6
12
0 2
13
0 6 7
14
2 2 7
15
5
16
0 2 8
17
0 2 3
25.
26.
9
8 9
10
2 4 6 6
11
4 5 7 8 8 9
12
2 4 5 7
13
1 2
14
4
15
1
Leaf Unit = 0.1
0
4 7 8 9 9
1
1 2 9
2
0 0 1 3 5 5 6 8
3
4 9
4
8
5
6
7
1
28
27.
4
1 3 6 6 7
5
0 0 3 8 9
6
0 1 1 4 4 5 7 7 9 9
7
0 0 0 1 3 4 4 5 5 6 6 6 7 8 8
8
0 1 1 3 4 4 5 7 7 8 9
9
0 2 2 7
or
4
1 3
4
6 6 7
5
0 0 3
5
8 9
6
0 1 1 4 4
6
5 7 7 9 9
7
0 0 0 1 3 4 4
7
5 5 6 6 6 7 8 8
8
0 1 1 3 4 4
8
5 7 7 8 9
9
0 2 2
9
7
28. a.
0
5 8
1
1 1 3 3 4 4
1
5 6 7 8 9 9
2
2 3 3 3 5 5
2
6 8
3
3
6 7 7 9
4
0
4
7 8
5
5
6
0
b.
2000 P/E
Forecast
5-9
10 - 14
15 - 19
20 - 24
25 - 29
30 - 34
35 - 39
40 - 44
45 - 49
50 - 54
55 - 59
60 - 64
Total
Frequency
2
6
6
6
2
0
4
1
2
0
0
1
30
29. a.
30
Percent
Frequency
6.7
20.0
20.0
20.0
6.7
0.0
13.3
3.3
6.7
0.0
0.0
3.3
100.0
y
x
1
2
Total
A
5
0
5
B
11
2
13
C
2
10
12
Total
18
12
30
1
2
Total
A
100.0
0.0
100.0
B
84.6
15.4
100.0
C
16.7
83.3
100.0
b.
y
x
c.
y
x
d.
30. a.
1
2
A
27.8
0.0
B
61.1
16.7
C
11.1
83.3
T otal
100.0
100.0
Category A values for x are always associated with category 1 values for y. Category B values for x
are usually associated with category 1 values for y. Category C values for x are usually associated
with category 2 values for y.
56
40
y
24
8
-8
-24
-40
-40
-30
-20
-10
0
10
20
30
40
x
b.
There is a negative relationship between x and y; y decreases as x increases.
31.
Quality Rating
Good
Very Good
Excellent
Total
Meal Price ($)
20-29
30-39
33.9
2.7
54.2
60.5
11.9
36.8
100.0
100.0
10-19
53.8
43.6
2.6
100.0
40-49
0.0
21.4
78.6
100.0
As the meal price goes up, the percentage of high quality ratings goes up. A positive relationship
between meal price and quality is observed.
32. a.
Sales/Margins/ROE
A
B
C
D
E
Total
0-19
20-39
EPS Rating
40-59
1
1
3
4
1
4
1
2
4
1
6
0-19
20-39
EPS Rating
40-59
60-79
1
5
2
1
80-100
8
2
3
9
13
60-79
11.11
41.67
28.57
20.00
80-100
88.89
16.67
42.86
Total
9
12
7
5
3
36
b.
Sales/Margins/ROE
A
B
C
D
E
8.33
14.29
60.00
33.33
14.29
20.00
66.67
33.33
Total
100
100
100
100
100
Higher EPS ratings seem to be associated with higher ratings on Sales/Margins/ROE. Of those
companies with an "A" rating on Sales/Margins/ROE, 88.89% of them had an EPS Rating of 80 or
32
higher. Of the 8 companies with a "D" or "E" rating on Sales/Margins/ROE, only 1 had an EPS
rating above 60.
33. a.
Sales/Margins/ROE
A
B
C
D
E
Total
A
1
1
1
1
4
Industry Group Relative Strength
B
C
D
2
2
4
5
2
3
3
2
1
1
1
2
11
7
10
E
Total
9
12
7
5
3
36
1
1
2
4
b/c. The frequency distributions for the Sales/Margins/ROE DATA is in the rightmost column of the
crosstabulation. The frequency distribution for the Industry Group Relative Strength DATA is in the
bottom row of the crosstabulation.
d.
Once the crosstabulation is complete, the individual frequency distributions are available in the
margins.
34. a.
80
70
Relative Price Strength
60
50
40
30
20
10
0
0
20
40
60
80
100
120
EPS Rating
b.
One might expect stocks with higher EPS ratings to show greater relative price strength. However,
the scatter diagram using this DATA does not support such a relationship.
The scatter diagram appears similar to the one showing "No Apparent Relationship" in Figure 2.19.
35. a.
The crosstabulation is shown below:
Count of Observation
Position
Guard
Speed
4-4.5
4.5-5
5-5.5
5.5-6
Grand Total
12
1
13
Offensive tackle
2
Wide receiver
6
9
Grand Total
6
11
7
3
19
4
12
15
40
b.
There appears to be a relationship between Position and Speed; wide receivers had faster speeds than
offensive tackles and guards.
c.
The scatter diagram is shown below:
10
9
Rating
8
7
6
5
4
4
4.5
5
5.5
6
Speed
d.
There appears to be a relationship between Speed and Rating; slower speeds appear to be associated
with lower ratings. In other words,, prospects with faster speeds tend to be rated higher than
prospects with slower speeds.
36. a.
Vehicle
F-Series
Silverado
Taurus
Camry
Accord
Frequency
17
12
8
7
6
34
Percent Frequency
34
24
16
14
12
Total
b.
50
100
The two top selling vehicles are the Ford F-Series Pickup and the Chevrolet Silverado.
Accord
12%
F-Series
34%
Camry
14%
Taurus
16%
Silverado
24%
c.
37. a/b.
Industry
Beverage
Chemicals
Electronics
Food
Aerospace
Totals
c.
Frequency
2
3
6
7
2
20
Percent Frequency
10
15
30
35
10
100
8
7
Frequency
6
5
4
3
2
1
0
Beverage
Chemicals Electronics
Food
Aerospace
Industry
38. a.
Response
Accuracy
Approach Shots
Mental Approach
Power
Practice
Putting
Short Game
Strategic Decisions
Total
b.
Frequency
16
3
17
8
15
10
24
7
100
Percent Frequency
16
3
17
8
15
10
24
7
100
Poor short game, poor mental approach, lack of accuracy, and limited practice.
39. a-d.
Sales
0 - 499
500 - 999
1000 - 1499
1500 - 1999
2000 - 2499
Frequency
13
3
0
3
1
Relative
Frequency
0.65
0.15
0.00
0.15
0.05
36
Cumulative
Frequency
13
16
16
19
20
Cumulative
Relative
Frequency
0.65
0.80
0.80
0.95
1.00
Total
20
1.00
e.
14
12
Frequency
10
8
6
4
2
0
0-499
500-999
1000-1499
1500-1999
2000-2499
Sales
40. a.
Closing Price
0 - 9.99
10 - 19.99
20 - 29.99
30 - 39.99
40 - 49.99
50 - 59.99
60 - 69.99
70 - 79.99
Totals
Frequency
9
10
5
11
2
2
0
1
40
Relative Frequency
0.225
0.250
0.125
0.275
0.050
0.050
0.000
0.025
1.000
b.
Closing Price
Less than or equal to 9.99
Less than or equal to 19.99
Less than or equal to 29.99
Less than or equal to 39.99
Less than or equal to 49.99
Cumulative Frequency
9
19
24
35
37
Cumulative Relative Frequency
0.225
0.475
0.600
0.875
0.925
Less than or equal to 59.99
Less than or equal to 69.99
Less than or equal to 79.99
39
39
40
0.975
0.975
1.000
c.
12
10
Frequency
8
6
4
2
0
10
20
30
40
50
60
70
80
Closing Price
d.
Over 87% of common stocks trade for less than $40 a share and 60% trade for less than $30 per
share.
41. a.
Exchange
American
New York
Over the Counter
Frequency
3
2
15
20
Relative
Frequency
0.15
0.10
0.75
1.00
b.
Earnings Per
Share
0.00 - 0.19
0.20 - 0.39
0.40 - 0.59
0.60 - 0.79
0.80 - 0.99
Frequency
7
7
1
3
2
20
Relative
Frequency
0.35
0.35
0.05
0.15
0.10
1.00
Seventy percent of the shadow stocks have earnings per share less then $0.40. It looks like low EPS
should be expected for shadow stocks.
Price-Earning
Ratio
0.00 - 9.9
Frequency
3
38
Relative
Frequency
0.15
10.0 - 19.9
20.0 - 29.9
30.0 - 39.9
40.0 - 49.9
50.0 - 59.9
7
4
3
2
1
20
0.35
0.20
0.15
0.10
0.05
1.00
P-E Ratios vary considerably, but there is a significant cluster in the 10 - 19.9 range.
42.
Income ($)
18,000-21,999
22,000-25,999
26,000-29,999
30,000-33,999
34,000-37,999
Total
Frequency
13
20
12
4
2
51
Relative
Frequency
0.255
0.392
0.235
0.078
0.039
1.000
25
Frequency
20
15
10
5
0
18,000 - 21,999
22,000 - 25,999
26,000 - 29,999
Per Capita Income
43. a.
30,000 - 33,999
34,000 - 37,999
0
8 9
1
0 2 2 2 3 4 4 4
1
5 5 6 6 6 6 7 7 8 8 8 8 9 9 9
2
0 1 2 2 2 3 4 4 4
2
5 6 8
3
0 1 3
b/c/d.
Number Answered
Correctly
5-9
10 - 14
15 - 19
20 - 24
25 - 29
30 - 34
Totals
e.
Relative
Frequency
0.050
0.200
0.375
0.225
0.075
0.075
1.000
Frequency
2
8
15
9
3
3
40
Cumulative
Frequency
2
10
25
34
37
40
Relatively few of the students (25%) were able to answer 1/2 or more of the questions correctly. The
DATA seem to support the Joint Council on Economic Education’s claim. However, the degree of
difficulty of the questions needs to be taken into account before reaching a final conclusion.
44. a/b.
High Temperature
3
4
Low Temperature
3
9
4
3 6 8
5
7
5
0 0 0 2 4 4 5 5 7 9
6
1 4 4 4 4 6 8
6
1 8
7
3 5 7 9
7
2 4 5 5
8
0 1 1 4 6
8
9
0 2 3
9
c.
It is clear that the range of low temperatures is below the range of high temperatures. Looking at the
stem-and-leaf displays side by side, it appears that the range of low temperatures is about 20 degrees
below the range of high temperatures.
d.
There are two stems showing high temperatures of 80 degrees or higher. They show 8 cities with
high temperatures of 80 degrees or higher.
40
e.
Frequency
High Temp.
Low. Temp.
0
1
0
3
1
10
7
2
4
4
5
0
3
0
20
20
Temperature
30-39
40-49
50-59
60-69
70-79
80-89
90-99
Total
Low Temperatur
45. a.
80
75
70
65
60
55
50
45
40
35
30
40
50
60
70
80
90
100
High Temperature
b.
There is clearly a positive relationship between high and low temperature for cities. As one goes up
so does the other.
46. a.
Occupation
Cabinetmaker
Lawyer
Physical Therapist
Systems Analyst
Total
30-39
40-49
1
5
1
2
7
30-39
40-49
10
50
Satisfaction Score
50-59 60-69 70-79
2
4
3
2
1
1
5
2
1
1
4
3
10
11
8
80-89
1
2
3
Total
10
10
10
10
40
b.
Occupation
Cabinetmaker
Lawyer
Physical Therapist
Systems Analyst
20
Satisfaction Score
50-59 60-69 70-79
20
40
30
20
10
10
50
20
10
10
40
30
80-89
10
20
Total
100
100
100
100
c.
Each row of the percent crosstabulation shows a percent frequency distribution for an occupation.
Cabinet makers seem to have the higher job satisfaction scores while lawyers seem to have the
lowest. Fifty percent of the physical therapists have mediocre scores but the rest are rather high.
47. a.
40,000
35,000
Revenue $mil
30,000
25,000
20,000
15,000
10,000
5,000
0
0
10,000
20,000
30,000
40,000
50,000
60,000
70,000
80,000
90,000
Employees
b.
There appears to be a positive relationship between number of employees and revenue. As the
number of employees increases, annual revenue increases.
48. a.
Fuel Type
Year Constructed Elect. Nat. Gas Oil Propane Other
1973 or before
40
183
12
5
7
1974-1979
24
26
2
2
0
1980-1986
37
38
1
0
6
1987-1991
48
70
2
0
1
Total 149
317
17
7
14
Total
247
54
82
121
504
b.
Year Constructed
1973 or before
1974-1979
1980-1986
1987-1991
Total
Frequency
247
54
82
121
504
Fuel Type
Electricity
Nat. Gas
Oil
Propane
Other
Total
42
Frequency
149
317
17
7
14
504
100,000
c.
Crosstabulation of Column Percentages
Fuel Type
Year Constructed Elect. Nat. Gas Oil Propane Other
1973 or before
26.9
57.7
70.5
71.4
50.0
1974-1979
16.1
8.2
11.8
28.6
0.0
1980-1986
24.8
12.0
5.9
0.0
42.9
1987-1991
32.2
22.1
11.8
0.0
7.1
Total 100.0 100.0 100.0 100.0 100.0
d.
Crosstabulation of row percentages.
Year Constructed
1973 or before
1974-1979
1980-1986
1987-1991
e.
Fuel Type
Elect. Nat. Gas Oil Propane Other
16.2
74.1
4.9
2.0
2.8
44.5
48.1
3.7
3.7
0.0
45.1
46.4
1.2
0.0
7.3
39.7
57.8
1.7
0.0
0.8
Total
100.0
100.0
100.0
100.0
Observations from the column percentages crosstabulation
For those buildings using electricity, the percentages have not changes greatly over the years. For
the buildings using natural gas, the majority were constructed in 1973 or before; the second largest
percentage was constructed in 1987-1991. Most of the buildings using oil were constructed in 1973
or before. All of the buildings using propane are older.
Observations from the row percentages crosstabulation
Most of the buildings in the CG&E service area use electricity or natural gas. In the period 1973 or
before most used natural gas. From 1974-1986, it is fairly evenly divided between electricity and
natural gas. Since 1987 almost all new buildings are using electricity or natural gas with natural gas
being the clear leader.
49. a.
Crosstabulation for stockholder's equity and profit.
Stockholders' Equity ($000)
0-1200
1200-2400
2400-3600
3600-4800
4800-6000
Total
b.
0-200
10
4
4
200-400
1
10
3
18
2
16
Profits ($000)
400-600
600-800
3
1
3
6
1
2
800-1000
1000-1200
1
2
1
1
1
2
4
4
Total
12
16
13
3
6
50
800-1000
0.00
12.50
7.69
33.33
0.00
1000-1200
8.33
0.00
7.69
66.67
0.00
Total
100
100
100
100
100
Crosstabulation of Row Percentages.
Stockholders' Equity ($1000s)
0-1200
1200-2400
2400-3600
3600-4800
4800-6000
0-200
83.33
25.00
30.77
0.00
200-400
8.33
62.50
23.08
0.00
33.33
Profits ($000)
400-600
600-800
0.00
0.00
0.00
0.00
23.08
7.69
0.00
0.00
50.00
16.67
c.
50. a.
Stockholder's equity and profit seem to be related. As profit goes up, stockholder's equity goes up.
The relationship, however, is not very strong.
Crosstabulation of market value and profit.
Market Value ($1000s)
0-8000
8000-16000
16000-24000
24000-32000
32000-40000
Total
b.
51. a.
27
Profit ($1000s)
300-600
600-900
4
4
2
2
1
1
2
2
1
13
6
900-1200
4
Total
27
12
4
4
3
50
900-1200
0.00
16.67
25.00
25.00
0.00
Total
100
100
100
100
100
2
1
1
Crosstabulation of Row Percentages.
Market Value ($1000s)
0-8000
8000-16000
16000-24000
24000-32000
32000-40000
c.
0-300
23
4
0-300
85.19
33.33
0.00
0.00
0.00
Profit ($1000s)
300-600
600-900
14.81
0.00
33.33
16.67
50.00
25.00
25.00
50.00
66.67
33.33
There appears to be a positive relationship between Profit and Market Value. As profit goes up,
Market Value goes up.
Scatter diagram of Profit vs. Stockholder's Equity.
1400.0
1200.0
Profit ($1000s)
1000.0
800.0
600.0
400.0
200.0
0.0
0.0
1000.0
2000.0
3000.0
4000.0
5000.0
Stockholder's Equity ($1000s)
b.
Profit and Stockholder's Equity appear to be positively related.
44
6000.0
7000.0
52. a.
Scatter diagram of Market Value and Stockholder's Equity.
45000.0
Market Value ($1000s)
40000.0
35000.0
30000.0
25000.0
20000.0
15000.0
10000.0
5000.0
0.0
0.0
1000.0
2000.0
3000.0
4000.0
5000.0
6000.0
Stockholder's Equity ($1000s)
b.
There is a positive relationship between Market Value and Stockholder's Equity.
7000.0
Chapter 3
Descriptive STATISTICS: Numerical Methods
Learning Objectives
1.
Understand the purpose of measures of location.
2.
Be able to compute the mean, median, mode, quartiles, and various percentiles.
3.
Understand the purpose of measures of variability.
4.
Be able to compute the range, interquartile range, variance, standard deviation, and coefficient of
variation.
5.
Understand how z scores are computed and how they are used as a measure of relative location of a
DATA value.
6.
Know how Chebyshev’s theorem and the empirical rule can be used to determine the percentage of the
DATA within a specified number of standard deviations from the mean.
7.
Learn how to construct a 5-number summary and a box plot.
8.
Be able to compute and interpret covariance and correlation as measures of association between two
variables.
9.
Be able to compute a weighted mean.
46
Solutions:
x
1.
xi
75

n
 15
5
10, 12, 16, 17, 20
Median = 16 (middle value)
x
2.
xi
96

n
 16
6
10, 12, 16, 17, 20, 21
Median =
3.
16 17
 16.5
2
15, 20, 25, 25, 27, 28, 30, 32
20
i
(8)  1.6
2nd position = 20
100
20  25
 22.5
2
25
(8)  2
100
i
65
i
(8)  5.2
6th position = 28
100
i
Mean 
4.
28  30
 29
2
75
(8)  6
100
xi

657
5.
a.
Median = 57
6th item
Mode = 53
It appears 3 times
x
xi

n
b.
1106.4
 36.88
30
There are an even number of items. Thus, the median is the average of the 15th and 16th items after
the DATA have been placed in rank order.
Median =
c.
 59.727
11
n
36.6  36.7
 36.65
2
Mode = 36.4 This value appears 4 times
d.
First Quartile i 
F25 I30  7.5
G
H100J
K
Rounding up, we see that Q1 is at the 8th position.
Q1 = 36.2
e.
Third Quartile i 
F75 I
G
H100 J
K30  22.5
Rounding up, we see that Q3 is at the 23rd position.
Q3 = 37.9
6.
a.
x
xi

n
1845
 92.25
20
Median is average of 10th and 11th values after arranging in ascending order.
Median 
66  95
2
 80.5
DATA are multimodal
b.
x
xi

n
1334
 66.7
20
Median 
66  70
2
 68
Mode = 70 (4 brokers charge $70)
7.
c.
Comparing all three measures of central location (mean, median and mode), we conclude that it costs
more, on average, to trade 500 shares at $50 per share.
d.
Yes, trading 500 shares at $50 per share is a transaction value of $25,000 whereas trading 1000
shares at $5 per share is a transaction value of $5000.
a.
x
xi
n

1380
 46
30
b.
Yes, the mean here is 46 minutes. The newspaper reported on average of 45 minutes.
c.
Median 
d.
Q1 = 7 (value of 8th item in ranked order)
45  52.9
 48.95
2
Q3 = 70.4 (value of 23rd item in ranked list)
48
e.
Find position i 
40th percentile =
40
30  12; 40th percentile is average of values in 12th and 13th positions.
100
28.8 + 29.1
= 28.95
2
8.
a.
x
xi

n
695
 34.75
20
Mode = 25 (appears three times)
b.
DATA in order: 18, 20, 25, 25, 25, 26, 27, 27, 28, 33, 36, 37, 40, 40, 42, 45, 46, 48, 53, 54
Median (10th and 11th positions) 
33  36
 34.5
2
At home workers are slightly younger
c.
i 
25
(20)  5; use positions 5 and 6
100
Q1 
25  26
2
75
i
 25.5
(20)  15; use positions 15 and 16
100
Q3 
d.
i 
42  45
2
32
 43.5
(20)  6.4; round up to position 7
100
32nd percentile = 27
At least 32% of the people are 27 or younger.
9.
a.
x
xi
n

270, 377
 10,815.08 Median (Position 13) = 8296
25
b.
Median would be better because of large DATA values.
c.
i = (25 / 100) 25 = 6.25
Q1 (Position 7) = 5984
i = (75 / 100) 25 = 18.75
Q3 (Position 19) = 14,330
d.
i = (85/100) 25 = 21.25
85th percentile (position 22) = 15,593. Approximately 85% of the websites have less than 15,593
unique visitors.
10. a.
xi = 435
x
xi

435
n
 48.33
9
DATA in ascending order:
28 42 45 48 49 50 55 58 60
Median = 49
Do not report a mode; each DATA value occurs once.
The index could be considered good since both the mean and median are less than 50.
b.
i
25
100
9  2.25
Q1 (3rd position) = 45
i
75
100
9  6.75
Q3 (7th position) = 55
11.
Using the mean we get xcity =15.58, xcountry = 18.92
For the SAMPLEs we see that the mean mileage is better in the country than in the
city. City
13.2 14.4 15.2 15.3 15.3 15.3 15.9 16 16.1 16.2 16.2 16.7 16.8

Median
Mode: 15.3
Country
17.2 17.4 18.3 18.5 18.6 18.6 18.7 19.0 19.2 19.4 19.4 20.6 21.1

Median
Mode: 18.6, 19.4
The median and modal mileages are also better in the country than in the city.
50
12. a.
xi
x

12, 780
n
b.
xi
x

1976
n
c.
d.
13.
 $639
20
 98.8 pictures
20
xi
2204

 110.2 minutes
n
20
This is not an easy choice because it is a multicriteria problem. If price was the only criterion, the
lowest price camera (Fujifilm DX-10) would be preferred. If maximum picture capacity was the only
criterion, the maximum picture capacity camera (Kodak DC280 Zoom) would be preferred. But, if
battery life was the only criterion, the maximum battery life camera (Fujifilm DX10) would be
preferred. There are many approaches used to select the best choice in a multicriteria situation.
These approaches are discussed in more specialized books on decision analysis.
x
Range 20 - 10 = 10
10, 12, 16, 17, 20
25
i
(5)  1.25
100
Q1 (2nd position) = 12
75
i
(5)  3.75
100
Q3 (4th position) = 17
IQR = Q3 - Q1 = 17 - 12 = 5
14.
xi
x
n
s2 

75
 15
5
( x i  x )2
n 1

64
 16
4
s  16  4
15.
15, 20, 25, 25, 27, 28, 30, 34
i
25
(8)  2
100
Q1 
i
75
(8)  6
100
Q1 
IQR = Q3 - Q1 = 29 - 22.5 = 6.5
x
xi  204  25.5
n
8
Range = 34 - 15 = 19
20  25
2
28  30
2
 22.5
 29
s2 
( x i  x )2

n 1
242
 34.57
7
s  34.57  5.88
16. a.
b.
Range = 190 - 168 = 22
(x i  x )2  376
s 2 = 376 = 75.2
5
c.
s  75.2  8.67
d.
Coefficient of Variation 
17.
8.67
178
100  4.87
Range = 92-67 = 25
IQR = Q3 - Q1 = 80 - 77 = 3
x = 78.4667
∑  x  x   411.7333
2
i
411.7333
 29.4095
s2  ∑  xi  x  
n 1
14
2
s  29.4095  5.4231
18. a.
x
xi
n
115.13 (Mainland); 36.62 (Asia)
Median (7th and 8th position)
Mainland = (110.87 + 112.25) / 2 = 111.56
Median (6th and 7th position)
Asia = (32.98 + 40.41) / 2 = 36.695
b.
Range = High - Low
Range
Standard Deviation
Coefficient of Variation
Mainland
86.24
26.82
23.30
52
Asia
42.97
11.40
31.13
c.
19. a.
b.
Greater mean and standard deviation for Mainland. Greater coefficient of variation for Asia.
Range = 60 - 28 = 32
IQR = Q3 - Q1 = 55 - 45 = 10
435
x
 48.33
9
(x i  x )2  742
s2 
(x  x )2
i
n 1

742
 92.75
8
s  92.75  9.63
c.
20.
The average air quality is about the same. But, the variability is greater in Anaheim.
Dawson Supply: Range = 11 - 9 = 2
s
4.1
 0.67
9
J.C. Clark: Range = 15 - 7 = 8
s
21. a.
60.1
 2.58
9
Winter
Range = 21 - 12 = 9
IQR = Q3 - Q1 = 20-16 = 4
Summer
Range = 38 - 18 = 20
IQR = Q3 - Q1 = 29-18 = 11
b.
Winter
Summer
c.
Variance
8.2333
44.4889
Standard Deviation
2.8694
6.6700
Winter
Coefficient of Variation =
x
s 100 
2.8694
17.7
100  16.21
s 100 
6.6700
25.6
100  26.05
Summer
Coefficient of Variation =
d.
x
More variability in the summer months.
22. a.
500 Shares at $50
Min Value = 34
Max Value = 195
Range = 195 - 34 = 161
Q1 
45  50
 47.5
Q3 
140 140
2
 140
2
Interquartile range = 140 - 47.5 = 92.5
1000 Shares at $5
Min Value = 34
Max Value = 90
Range = 90 - 34 = 56
Q1 
60  60.5
 60.25
Q3 
2
79.5  80
 79.75
2
Interquartile range = 79.75 - 60.25 = 19.5
b.
500 Shares at $50
(x  x )2
51, 402.25
i

 2705.3816
n 1
19
s  2705.3816  52.01
s2 
1000 Shares at $5
(x  x )2
5526.2
i

 290.8526
n 1
19
s  290.8526  17.05
s2 
c.
500 Shares at $50
Coefficient of Variation =
s
(100) 
x
52.01
(100)  56.38
92.25
1000 Shares at $5
Coefficient of Variation =
s
x
d.
(100) 
17.05
(100)  25.56
66.70
The variability is greater for the trade of 500 shares at $50 per share. This is true whether we use the
standard deviation or the coefficient of variation as a measure.
23.
s2 = 0.0021 Production should not be shut down since the variance is less than .005.
24.
Quarter milers
s = 0.0564
54
Coefficient of Variation = (s/ x )100 = (0.0564/0.966)100 = 5.8
Milers
s = 0.1295
Coefficient of Variation = (s/ x )100 = (0.1295/4.534)100 = 2.9
Yes; the coefficient of variation shows that as a percentage of the mean the quarter milers’ times
show more variability.
25.
x
xi
75

n
 15
5
s2 
( x  x) 2

n 1
10
z
10 15
64
4
4
 1.25
4
20
20 15
z
 1.25
4
12
z
12 15
 0.75
4
17
z
17 15
 .50
4
16
z
16 15
 .25
4
26.
z
520  500
 .20
100
z
650  500
 1.50
100
z
500  500
 0.00
100
z
450  500
 0.50
100
z
280  500
 2.20
100
27. a.
z 
40  30
5
 2 1
1
22
 0.75 At least 75%
b.
z 
45  30
5
c.
z 
38  30
5
d.
z 
42  30
5
e.
z 
48  30
5
28. a.
 3 1
 1.6 1
 0.89 At least 89%
1
 0.61 At least 61%
1.62
 2.4 1
 3.6 1
1
2.42
1
 0.83 At least 83%
 0.92 At least 92%
3.62
Approximately 95%
b.
Almost all
c.
Approximately 68%
29. a.
1
32
This is from 2 standard deviations below the mean to 2 standard deviations above the mean.
With z = 2, Chebyshev’s theorem gives:
1
1
z2
 1
1
22
 1
1
4

3
4
Therefore, at least 75% of adults sleep between 4.5 and 9.3 hours per day.
b.
This is from 2.5 standard deviations below the mean to 2.5 standard deviations above the mean.
With z = 2.5, Chebyshev’s theorem gives:
1
1
1
 1 2  1
 .84
z2
6.25
2.5
Therefore, at least 84% of adults sleep between 3.9 and 9.9 hours per day.
1
c.
With z = 2, the empirical rule suggests that 95% of adults sleep between 4.5and 9.3 hours per day.
The probability obtained using the empirical rule is greater than the probability obtained using
Chebyshev’s theorem.
30. a.
2 hours is 1 standard deviation below the mean. Thus, the empirical rule suggests that 68% of the
kids watch television between 2 and 4 hours per day. Since a bell-shaped distribution is symmetric,
approximately, 34% of the kids watch television between 2 and 3 hours per day.
b.
c.
1 hour is 2 standard deviations below the mean. Thus, the empirical rule suggests that 95% of the
kids watch television between 1 and 5 hours per day. Since a bell-shaped distribution is symmetric,
approximately, 47.5% of the kids watch television between 1 and 3 hours per day. In part (a) we
concluded that approximately 34% of the kids watch television between 2 and 3 hours per day; thus,
approximately 34% of the kids watch television between 3 and 4 hours per day. Hence,
approximately 47.5% + 34% = 81.5% of kids watch television between 1 and 4 hours per day.
Since 34% of the kids watch television between 3 and 4 hours per day, 50% - 34% = 16% of the kids
watch television more than 4 hours per day.
56
31. a.
Approximately 68% of scores are within 1 standard deviation from the mean.
b.
Approximately 95% of scores are within 2 standard deviations from the mean.
c.
Approximately (100% - 95%) / 2 = 2.5% of scores are over 130.
d.
Yes, almost all IQ scores are less than 145.
71.00  90.06
32. a.
z
 0.95
b.
z
c.
The z-score in part a indicates that the value is 0.95 standard deviations below the mean. The z-score
in part b indicates that the value is 3.90 standard deviations above the mean.
20
168  90.06
20
 3.90
The labor cost in part b is an outlier and should be reviewed for accuracy.
33. a.
b.
x is approximately 63 or $63,000, and s is 4 or $4000
This is from 2 standard deviations below the mean to 2 standard deviations above the mean.
With z = 2, Chebyshev’s theorem gives:
1
1
z2
 1
1
22
 1
1
4

3
4
Therefore, at least 75% of benefits managers have an annual salary between $55,000 and $71,000.
c.
The histogram of the salary DATA is shown below:
9
8
7
Frequency
6
5
4
3
2
1
0
56-58
58-60
60-62
62-64
64-66
66-68
68-70
70-72
72-74
Salary
Although the distribution is not perfectly bell shaped, it does appear reasonable to assume that the
distribution of annual salary can be approximated by a bell-shaped distribution.
d.
With z = 2, the empirical rule suggests that 95% of benefits managers have an annual salary between
$55,000 and $71,000. The probability is much higher than obtained using Chebyshev’s theorem, but
requires the assumption that the distribution of annual salary is bell shaped.
e. There are no outliers because all the observations are within 3 standard deviations of the mean.
34. a.
x is 100 and s is 13.88 or approximately 14
b.
If the distribution is bell shaped with a mean of 100 points, the percentage of NBA games in which
the winning team scores more than 100 points is 50%. A score of 114 points is z = 1 standard
deviation above the mean. Thus, the empirical rule suggests that 68% of the winning teams will score
between 86 and 114 points. In other words, 32% of the winning teams will score less than 86 points
or more than 114 points. Because a bell-shaped distribution is symmetric, approximately 16% of the
winning teams will score more than 114 points.
c.
For the winning margin, x is 11.1 and s is 10.77. To see if there are any outliers, we will first
compute the z-score for the winning margin that is farthest from the SAMPLE mean of 11.1, a
winning margin of 32 points.
z
xx

s
32 11.1
 1.94
10.77
Thus, a winning margin of 32 points is not an outlier (z = 1.94 < 3). Because a winning margin of 32
points is farthest from the mean, none of the other DATA values can have a z-score that is less than 3
or greater than 3 and hence we conclude that there are no outliers
35. a.
x
xi
n

79.86
 3.99
20
58
Median =
b.
4.17  4.20
 4.185 (average of 10th and 11th values)
2
Q1 = 4.00 (average of 5th and 6th values)
Q3 = 4.50 (average of 15th and 16th values)
12.5080
( x  x )2

 0.8114
19
n 1
c.
s
d.
Allison One: z 
4.12  3.99
 0.16
0.8114
Omni Audio SA 12.3: z 
e.
2.32  3.99
 2.06
0.8114
The lowest rating is for the Bose 501 Series. It’s z-score is:
z 
2.14  3.99
0.8114
 2.28
This is not an outlier so there are no outliers.
36.
15, 20, 25, 25, 27, 28, 30, 34
Smallest = 15
i
25
(8)  2
100
Median 
i
Q1 
20  25
 22.5
2
Q3 
28  30
 29
2
25  27
 26
2
75
(8)  8
100
Largest = 34
37.
15
38.
5, 6, 8, 10, 10, 12, 15, 16, 18
Smallest = 5
20
25
30
35
i
25
100
(9)  2.25 Q1 = 8 (3rd position)
Median = 10
i
75
100
(9)  6.75 Q3 = 15 (7th position)
Largest = 18
5
39.
10
15
20
IQR = 50 - 42 = 8
Lower Limit:
Upper Limit:
Q1 - 1.5 IQR = 42 - 12 = 30
Q3 + 1.5 IQR = 50 + 12 = 62
65 is an outlier
40. a.
b.
Five number summary: 5 9.6 14.5 19.2 52.7
IQR = Q3 - Q1 = 19.2 - 9.6 = 9.6
Lower Limit:
Upper Limit:
c.
Q1 - 1.5 (IQR) = 9.6 - 1.5(9.6) = -4.8
Q3 + 1.5(IQR) = 19.2 + 1.5(9.6) = 33.6
The DATA value 41.6 is an outlier (larger than the upper limit) and so is the DATA value 52.7.
The financial analyst should first verify that these values are correct. Perhaps a typing error has
caused
25.7 to be typed as 52.7 (or 14.6 to be typed as 41.6). If the outliers are correct, the analyst might
consider these companies with an unusually large return on equity as good investment candidates.
d.
*
-10
41. a.
5
20
35
Median (11th position) 4019
i
25
(21)  5.25
100
Q1 (6th position) = 1872
60
*
50
65
i
75
(21)  15.75
100
Q3 (16th position) = 8305
608, 1872, 4019, 8305, 14138
b.
Limits:
IQR = Q3 - Q1 = 8305 - 1872 = 6433
Lower Limit:
Q1 - 1.5 (IQR) = -7777
Upper Limit:
Q3 + 1.5 (IQR) = 17955
c.
There are no outliers, all DATA are within the limits.
d.
Yes, if the first two digits in Johnson and Johnson's sales were transposed to 41,138, sales would
have shown up as an outlier. A review of the DATA would have enabled the correction of the
DATA.
e.
0
42. a.
6,000
3,000
9,000
12,000
15,000
Mean = 105.7933 Median = 52.7
b.
Q1 = 15.7 Q3 = 78.3
c.
IQR = Q3 - Q1 = 78.3 - 15.7 = 62.6
Lower limit for box plot = Q1 - 1.5(IQR) = 15.7 - 1.5(62.6) = -78.2
Upper limit for box plot = Q3 + 1.5 (IQR) = 78.3 + 1.5(62.6) = 172.2
Note: Because the number of shares covered by options grants cannot be negative, the lower limit for
the box plot is set at 0. This, outliers are value in the DATA set greater than 172.2.
Outliers: Silicon Graphics (188.8) and ToysRUs (247.6)
d.
43. a.
Mean percentage = 26.73. The current percentage is much greater.
Five Number Summary (Midsize)
51 71.5 81.5 96.5 128
Five Number Summary (Small)
73 101 108.5 121 140
b.
Box Plots
Midsize
50
60
70
80
90
100
110
120
130
60
70
80
90
100
110
120
130
140
Small Size
50
c.
150
The midsize cars appear to be safer than the small cars.
44. a.
x = 37.48 Median = 23.67
b.
c.
Q1 = 7.91 Q3 = 51.92
IQR = 51.92 - 7.91 = 44.01
Lower Limit:
Q1 - 1.5(IQR) = 7.91 - 1.5(44.01) = -58.11
Upper Limit:
Q3 + 1.5(IQR) = 51.92 + 1.5(44.01) = 117.94
Russia, with a percent change of 125.89, is an outlier.
Turkey, with a percent change of 254.45 is another outlier.
d.
With a percent change of 22.64, the United States is just below the 50th percentile - the median.
45. a.
70
60
50
y
40
30
20
10
0
0
5
10
x
62
15
20
b.
Negative relationship
c/d. xi  40
x
40
8
yi  230
5
( xi  x)( yi  y)  240
y
230
 46
5
( x i  x) 2  118
( y i  y) 2  520
sxy  (xi  x )( yi  y )  240  60
n 1
5 1
sx 
(x  x )2

n 1
118
 5.4314
5 1
sy 
( y  y )2

n 1
520
 11.4018
5 1
rxy 
sxy
sx sy
60

 0.969
(5.4314)(11.4018)
There is a strong negative linear relationship.
46. a.
18
16
14
12
y
10
8
6
4
2
0
0
5
10
15
x
20
25
30
b.
Positive relationship
c/d. xi  80
x
80
 16
5
( xi  x )( yi  y)  106
y
yi  50
50
 10
5
( x i  x) 2  272
( y i  y) 2  86
sxy  (xi  x )( yi  y)  106  26.5
n 1
5 1
sx 
(x  x )2

n 1
272
 8.2462
5 1
86
( y  y)2

 4.6368
n 1
5 1
sxy
26.5
rxy 

 0.693
sx sy (8.2462)(4.6368)
sy 
A positive linear relationship
47. a.
750
700
y = SAT
650
600
550
500
450
400
2.6
2.8
3
3.2
x = GPA
b.
Positive relationship
64
3.4
3.6
3.8
x
c/d. xi  19.8
19.8
 3.3
( xi  x )( yi  y)  143
y
yi  3540
6
3540
 590
6
( x i  x) 2  0.74
( y i  y) 2  36,400
sxy  (xi  x )( yi  y)  143  28.6
n 1
6 1
sx 
(x  x )2

n 1
0.74
 0.3847
6 1
sy 
( y  y)2

n 1
36, 400
 85.3229
6 1
rxy 
sxy
28.6

sx sy
 0.8713
(0.3847)(85.3229)
A positive linear relationship
48.
Let x = driving speed and y = mileage
x
xi  420
420
 42
10
(x  x )( y  y )  475
i
yi  270
y
(x  x )2  1660
i
i
270
 27
10
( y  y)2  164
i
sxy  (xi  x )( yi  y)  475  52.7778
n 1
10 1
sx 
1660
(x  x )2

 13.5810
n 1
10 1
sy 
164
( y  y)2

 4.2687
n 1
10 1
rxy 
sxy
sx sy

52.7778
 .91
(13.5810)(4.2687)
A strong negative linear relationship
49. a.
b.
50. a.
The SAMPLE correlation coefficient is .78.
There is a positive linear relationship between the performance score and the overall rating.
The SAMPLE correlation coefficient is .92.
b.
There is a strong positive linear relationship between the two variables.
51.
The SAMPLE correlation coefficient is .88. This indicates a strong positive linear relationship
between the daily high and low temperatures.
52. a.
x
wi xi
wi
b.

6(3.2)  3(2)  2(2.5)  8(5) 70.2

 3.69
6 3 2 8
19
3.2  2  2.5  5
4

12.7
 3.175
4
53.
fi
4
7
9
5
25
Mi
5
10
15
20
fi Mi
20
70
135
100
325
fi Mi  325  13
n
25
x
s2 
fi
Mi
Mi  x
4
7
9
5
5
10
15
20
-8
-3
+2
+7
fi (M i  x )2 
n 1
600
(M i  x )2
64
9
4
49
 25
24
s  25  5
54. a.
Grade xi
4 (A)
3 (B)
2 (C)
1 (D)
0 (F)
x
b.
wi xi
wi

Weight wi
9
15
33
3
0
60 Credit Hours
9(4) 15(3)  33(2)  3(1) 150

 2.5
9 15  33  3
60
Yes; satisfies the 2.5 grade point average requirement
55.
fi
4
7
Mi
5
10
66
fi Mi
20
70
f i (M i  x )2
256
63
36
245
600
9
5
25
x
135
100
325
fi Mi  325  13
n
s2 
15
20
25
fi
Mi
Mi  x
(Mi  x ) 2
fi (Mi  x ) 2
4
7
9
5
5
10
15
20
-8
-3
+2
+7
64
9
4
49
256
63
36
245
600
600
fi (M i  x )2 
 25
n 1
24
s  25  5
56.
Mi
fi
fi Mi
Mi x
74
192
280
105
23
6
680
2
7
12
17
22
27
148
1,344
3,360
1,785
506
162
7,305
-8.742647
-3.742647
1.257353
6.257353
11.257353
16.257353
(Mi  x ) 2
fi (Mi  x ) 2
76.433877
14.007407
1.580937
39.154467
126.728000
264.301530
5,656.1069
2,689.4221
442.6622
4,111.2190
2,914.7439
1,585.8092
17,399.9630
Estimate of total gallons sold: (10.74)(120) = 1288.8
7305
x
 10.74
680
s2 
17, 399.9630
 25.63
679
s  5.06
57. a.
Class
0
1
2
3
4
Totals
x
i fMi
n

1745
fi
15
10
40
85
350
500
Mi
0
1
2
3
4
 3.49
500
b.
Mi  x
( M i  x )2
f i ( M i  x )2
fi Mi
0
10
80
255
1400
1745
-3.49
-2.49
-1.49
-0.49
+0.51
s2 
58. a.
x
( M i  x )2 f i
n 1
xi

444.95
 0.8917
499
3463
n

12.18
6.20
2.22
0.24
0.26
Total
182.70
62.00
88.80
20.41
91.04
444.95
s  0.8917  0.9443
 138.52
25
Median = 129 (13th value)
Mode = 0 (2 times)
b.
It appears that this group of young adults eats out much more than the average American. The mean
and median are much higher than the average of $65.88 reported in the newspaper.
c.
Q1 = 95 (7th value)
Q3 = 169 (19th value)
d.
Min = 0
Max = 467
Range = 467 - 0 = 467
IQR = Q3 - Q1 = 169 - 95 = 74
e.
s2 = 9271.01
f.
The z - score for the largest value is:
z 
s = 96.29
467  138.52
96.29
 3.41
It is the only outlier and should be checked for accuracy.
59. a.
xi = 760
x
xi

n
760
 38
20
Median is average of 10th and 11th items.
Median 
36  36
2
 36
The modal cash retainer is 40; it appears 4 times.
b.
For Q1,
68
25
i
20  5
100
Since i is integer,
Q1 
28  30
2
 29
For Q3,
i
75
20  15
100
Since i is integer,
Q3 
c
40  50
2
 45
Range = 64 – 15 = 49
Interquartile range = 45 – 29 = 16
d.
2
3318
 174.6316
s2  ∑  xi  x  
n 1
20 1
s
e.
60. a.
 174.6316  13.2148
Coefficient of variation =
x
xi

n
260
s
100 
x
13.2148
100  34.8
38
 18.57
14
Median = 16.5 (Average of 7th and 8th values)
b.
s2 = 53.49
c.
Quantex has the best record: 11 Days
d.
z 
s = 7.31
27  18.57
 1.15
7.31
Packard-Bell is 1.15 standard deviations slower than the mean.
e.
z 
12  18.57
 0.90
7.31
IBM is 0.9 standard deviations faster than the mean.
f.
Check Toshiba:
z 
37  18.57
7.31
 2.52
On the basis of z - scores, Toshiba is not an outlier, but it is 2.52 standard deviations slower than the
mean.
61.
SAMPLE mean = 7195.5
Median = 7019 (average of positions 5 and 6)
SAMPLE variance = 7,165,941
SAMPLE standard deviation = 2676.93
62. a.
b.
The SAMPLE mean is 83.135 and the SAMPLE standard deviation is 16.173.
With z = 2, Chebyshev’s theorem gives:
1
1
 1
z2
1
 1
1
22

4
3
4
Therefore, at least 75% of household incomes are within 2 standard deviations of the mean. Using
the SAMPLE mean and SAMPLE standard deviation computed in part (a), the range within 75%
of household incomes must fall is 83.135  2(16.173) = 83.135  32.346; thus, 75% of household
incomes must fall between 50.789 and 115.481, or $50,789 to $115,481.
c.
With z = 2, the empirical rule suggests that 95% of household incomes must fall between $50,789 to
$115,481. For the same range, the probability obtained using the empirical rule is greater than the
probability obtained using Chebyshev’s theorem.
d.
The z-score for Danbury, CT is 3.04; thus, the Danbury, CT observation is an outlier.
63. a.
Public Transportation: x 
Automobile: x 
320
320
 32
10
 32
10
b.
Public Transportation: s = 4.64
Automobile: s = 1.83
c.
Prefer the automobile. The mean times are the same, but the auto has less variability.
d.
DATA in ascending order:
Public:
25 28 29 29 32 32 33 34 37 41
Auto:
29 30 31 31 32 32 33 33 34 35
Five number Summaries
Public:
25 29 32 34 41
Auto:
29 31 32 33 35
70
Box Plots:
Public:
24
28
32
36
40
28
32
36
40
Auto:
24
The box plots do show lower variability with automobile transportation and support the conclusion in
part c.
64. a.
b.
65. a.
The SAMPLE covariance is 502.67. Because the SAMPLE covariance is positive, there is a positive
linear relationship between income and home price.
The SAMPLE correlation coefficient is .933; this indicates a strong linear relationship between
income and home price.
Let x = media expenditures ($ millions) and y = shipments in barrels (millions)
xi  404.1
x
404.1
 40.41
(x  x )( y  y )  3763.481
i
yi  119.9
10
i
i
A positive relationship
sx 
19, 248.469
(x  x )2

 46.2463
10 1
n 1
939.349
( y  y )2

 10.2163
n 1
10 1
sxy
418.1646

rxy 
 0.885
sx sy (46.2463)(10.2163)
sy 
119.9
(x  x )2  19, 248.469
sxy  (xi  x )( yi  y)  3763.481  418.1646
n 1
10 1
b.
y
 11.99
10
( y  y)2  939.349
i
Note: The same value can also be obtained using Excel's CORREL function
66. a.
b.
The scatter diagram indicates a positive relationship
xi  798
yi  11, 688
x2  71, 306
i
rxy 
xi yi  1, 058, 019
y2  16, 058, 736
i
xi yi  xiyi  / n
1, 058, 019  (798)(11, 688) / 9


 .9856
2
2
x2  x  / n y2  y  / n
71, 306  (798)2 / 9 16, 058, 736  (11, 688)2 / 9
i
i
i
i
Strong positive relationship
67. a.
The scatter diagram is shown below:
72
3.5
3
Earnings
2.5
2
1.5
1
0.5
0
0
5
10
15
20
25
30
Book Value
b.
68. a.
b.
69.
The SAMPLE correlation coefficient is .75; this indicates a linear relationship between book value
and earnings.
(800 + 750 + 900)/3 = 817
Month
January
February
March
Weight
1
2
3
x
wi xi
x
wi xi
wi
wi

1(800)  2(750)  3(900) 5000

 833
1 2  3
6

20(20)  30(12) 10(7) 15(5) 10(6) 965

 11.4 days
20  30 10 15 10
85
70.
a.
x
fi
Mi
fi Mi
Mi  x
( Mi  x ) 2
fi ( Mi  x ) 2
10
40
150
175
75
15
10
475
47
52
57
62
67
72
77
470
2080
8550
10850
5025
1080
770
28,825
-13.68
-8.68
-3.68
+1.32
+6.32
+11.32
+16.32
187.1424
75.3424
13..5424
1.7424
39.9424
128.1424
266.3424
1871.42
3013.70
2031.36
304.92
2995.68
1922.14
2663.42
14,802.64
28, 825
475
 60.68
b.
s2 
14, 802.64
 31.23
474
s  31.23  5.59
71.
x
1030
fi
Mi
fi Mi
Mi  x
( Mi  x ) 2
f i ( Mi  x ) 2
2
6
4
4
2
2
20
29.5
39.5
49.5
59.5
69.5
79.5
59.0
237.0
198.0
238.0
139.0
159.0
1,030.0
-22
-12
-2
8
18
28
484
144
4
64
324
784
968
864
16
256
648
1568
4320
 51.5
20
s
4320
 227.37
19
s = 15.08
74
Chapter 4
Introduction to Probability
Learning Objectives
1.
Obtain an appreciation of the role probability information plays in the decision making process.
2.
Understand probability as a numerical measure of the likelihood of occurrence.
3.
Know the three methods commonly used for assigning probabilities and understand when they should
be used.
4.
Know how to use the laws that are available for computing the probabilities of events.
5.
Understand how new information can be used to revise initial (prior) probability estimates using
Bayes’ theorem.
Solutions:
1.
Number of experimental Outcomes = (3) (2) (4) = 24
F6I 6! 65 4  3 2 1
HGJ3K3!3!  (3 2 1)(3 2 1)  20
2.
ABC
ABD
ABE
ABF
ACD
P36 
3.
6!
(6  3)!
ACE
ACF
ADE
ADF
AEF
BCD
BCE
BCF
BDE
BDF
BEF
CDE
CDF
CEF
DEF
2nd Toss
3rd Toss
 (6)(5)(4)  120
BDF BFD DBF DFB FBD FDB
4.
a.
1st Toss
H
H
T
H
(H,H,H)
T
(H,H,T)
H
(H,T,H)
T
(H,T,T)
H
T
H
(T,H,H)
T
(T,H,T)
T
H
(T,T,H)
T
(T,T,T)
b.
Let: H be head and T be tail
(H,H,H) (T,H,H)
(H,H,T) (T,H,T)
(H,T,H) (T,T,H)
(H,T,T) (T,T,T)
c.
5.
The outcomes are equally likely, so the probability of each outcomes is 1/8.
P(Ei) = 1 / 5 for i = 1, 2, 3, 4, 5
P(Ei)  0 for i = 1, 2, 3, 4, 5
P(E1) + P(E2) + P(E3) + P(E4) + P(E5) = 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 = 1
The classical method was used.
13- 76
6.
P(E1) = .40, P(E2) = .26, P(E3) = .34
The relative frequency method was used.
7.
8.
No. Requirement (4.3) is not satisfied; the probabilities do not sum to 1. P(E1) + P(E2) + P(E3) +
P(E4) = .10 + .15 + .40 + .20 = .85
a.
There are four outcomes possible for this 2-step experiment; planning commission positive - council
approves; planning commission positive - council disapproves; planning commission negative council approves; planning commission negative - council disapproves.
b.
Let
p = positive, n = negative, a = approves, and d = disapproves
Planning Commission
Council
a
(p, a)
d
p
(p, d)
n
a
.
(n, a)
d
(n, d)
F5 0 I 50! 50  49  48  47
GJ
H4 K4!46!  4  3 2 1  230,300
9.
10. a.
Use the relative frequency approach:
P(California) = 1,434/2,374 = .60
b.
Number not from 4 states = 2,374 - 1,434 - 390 - 217 - 112 = 221
P(Not from 4 States) = 221/2,374 = .09
c.
P(Not in Early Stages) = 1 - .22 = .78
d.
Estimate of number of Massachusetts companies in early stage of development - (.22)390  86
13- 77
e.
If we assume the size of the awards did not differ by states, we can multiply the probability an award
went to Colorado by the total venture funds disbursed to get an estimate.
Estimate of Colorado funds = (112/2374)($32.4) = $1.53 billion
Authors' Note: The actual amount going to Colorado was $1.74 billion.
11. a.
No, the probabilities do not sum to one. They sum to .85.
b.
Owner must revise the probabilities so they sum to 1.00.
12. a.
Use the counting rule for combinations:
FG49IJ 49!  (49)(48)(47)(46)(45)  1,906,884
H5 K 5!44! (5)(4)(3)(2)(1)
b.
Very small: 1/1,906,884 = 0.0000005
c.
Multiply the answer to part (a) by 42 to get the number of choices for the six numbers.
No. of Choices = (1,906,884)(42) = 80,089,128
Probability of Winning = 1/80,089,128 = 0.0000000125
13.
Initially a probability of .20 would be assigned if selection is equally likely. DATA does not appear
to confirm the belief of equal consumer preference. For example using the relative frequency
method we would assign a probability of 5 / 100 = .05 to the design 1 outcome, .15 to design 2, .30
to
design 3, .40 to design 4, and .10 to design 5.
14. a.
P (E2) = 1 / 4
b.
P(any 2 outcomes) = 1 / 4 + 1 / 4 = 1 / 2
c.
P(any 3 outcomes) = 1 / 4 + 1 / 4 + 1 / 4 = 3 / 4
15. a.
S = {ace of clubs, ace of diamonds, ace of hearts, ace of spades}
b.
S = {2 of clubs, 3 of clubs, . . . , 10 of clubs, J of clubs, Q of clubs, K of clubs, A of clubs}
c.
There are 12; jack, queen, or king in each of the four suits.
d.
For a: 4 / 52 = 1 / 13 = .08
For b: 13 / 52 = 1 / 4 = .25
For c: 12 / 52 = .23
13- 78
16. a.
(6) (6) = 36 SAMPLE points
b.
Die 2
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Total for Both
Die 1
c.
6 / 36 = 1 / 6
d.
10 / 36 = 5 / 18
e.
No. P(odd) = 18 / 36 = P(even) = 18 / 36 or 1 / 2 for both.
f.
Classical. A probability of 1 / 36 is assigned to each experimental outcome.
17. a.
(4, 6), (4, 7), (4 , 8)
b.
.05 + .10 + .15 = .30
c.
(2, 8), (3, 8), (4, 8)
d.
.05 + .05 + .15 = .25
e.
.15
18. a.
0; probability is .05
b.
4, 5; probability is .10 + .10 = .20
c.
0, 1, 2; probability is .05 + .15 + .35 = .55
19. a.
b.
Yes, the probabilities are all greater than or equal to zero and they sum to one.
P(A) = P(0) + P(1) + P(2) = .08 + .18 + .32
= .58
13- 79
.
c.
P(B) = P(4) = .12
20. a.
P(N) = 56/500 = .112
b.
P(T) = 43/500 = .086
c.
Total in 6 states = 56 + 53 + 43 + 37 + 28 + 28 = 245
P(B) = 245/500 = .49
Almost half the Fortune 500 companies are headquartered in these states.
21. a.
P(A) = P(1) + P(2) + P(3) + P(4) + P(5)
=
20 12 6
3
1
  

50 50 50 50 50
= .40 + .24 + .12 + .06 + .02
= .84
b.
P(B) = P(3) + P(4) + P(5)
= .12 + .06 + .02
= .20
c.
22. a.
P(2) = 12 / 50 = .24
P(A) = .40, P(B) = .40, P(C) = .60
b.
P(A  B) = P(E1, E2, E3, E4) = .80. Yes P(A  B) = P(A) + P(B).
c.
Ac = {E3, E4, E5} Cc = {E1, E4} P(Ac) = .60 P(Cc) = .40
d.
A  Bc = {E1, E2, E5} P(A  Bc) = .60
e.
P(B  C) = P(E2, E3, E4, E5) = .80
23. a.
P(A) = P(E1) + P(E4) + P(E6) = .05 + .25 + .10 = .40
P(B) = P(E2) + P(E4) + P(E7) = .20 + .25 + .05 = .50
P(C) = P(E2) + P(E3) + P(E5) + P(E7) = .20 + .20 + .15 + .05 = .60
b.
A  B = {E1, E2, E4, E6, E7}
P(A  B) = P(E1) + P(E2) + P(E4) + P(E6) + P(E7)
= .05 + .20 + .25 + .10 + .05
= .65
c.
A  B = {E4}
P(A  B) = P(E4) = .25
13- 80
d.
Yes, they are mutually exclusive.
e.
Bc = {E1, E3, E5, E6}; P(Bc) = P(E1) + P(E3) + P(E5) + P(E6)
= .05 + .20 + .15 + .10
= .50
24.
Let E = experience exceeded expectations
M = experience met expectations
a.
Percentage of respondents that said their experience exceeded expectations
= 100 - (4 + 26 + 65) = 5%
P(E) = .05
b.
P(M  E) = P(M) + P(E) = .65 + .05 = .70
25.
Let Y = high one-year return
M = high five-year return
a.
P(Y) = 15/30 = .50
P(M) = 12/30 = .40
P(Y  M) = 6/30 = .20
b.
P(Y  M) = P(Y) + P(M) - P(Y  M)
= .50 + .40 - .20 = .70
c.
1 - P(Y  M) = 1 - .70 = .30
26.
Let Y = high one-year return
M = high five-year return
a.
P(Y) = 9/30 = .30
P(M) = 7/30 = .23
b.
P(Y  M) = 5/30 = .17
c.
P(Y  M) = .30 + .23 - .17 = .36
P(Neither) = 1 - .36 = .64
27.
Big Ten
Pac-10
a.
Yes
No
P(Neither) =
6823
13, 429
Yes
849
2112
2,961
No
3645
6823
10,468
 .51
13- 81
4494
8935
13,429
b.
P(Either) =
c.
P(Both) =
2961
 4494  849  .05
13, 429 13, 429 13, 429
849
13, 429
28.
 .06
Let: B = rented a car for business reasons
P = rented a car for personal reasons
a.
P(B  P) = P(B) + P(P) - P(B  P)
= .54 + .458 - .30 = .698
b.
P(Neither) = 1 - .698 = .302
29. a.
P(E) =
1033
 .36
2851
P(R) =
854
2851
P(D) =
964
2851
 .30
 .34
b.
Yes; P(E  D) = 0
c.
Probability =
1033
 .43
2375
d.
Yes
e.
P(E  A) = P(E) + P(A) = .36 + .18 = .54
30. a.
b.
c.
31. a.
b.
P(A B) = P(A  B) = .40 = .6667
P(B)
.60
P(B A) = P(A  B) = .40 = .80
P(A)
.50
No because P(A | B)  P(A)
P(A  B) = 0
P(A B) = P(A  B) = 0 = 0
P(B)
.4
c.
No. P(A | B)  P(A);  the events, although mutually exclusive, are not independent.
d.
Mutually exclusive events are dependent.
13- 82
13- 83
32. a.
Single
Married
Total
Under 30
.55
.10
.65
30 or over
.20
.15
.35
Total
.75
.25
1.00
b.
65% of the customers are under 30.
c.
The majority of customers are single: P(single) = .75.
d.
.55
e.
Let: A = event under 30
B = event single
f.
P(A  B) = .55
P(B A) = P(A  B) = .55 = .8462
P(A)
.65
P(A)P(B) = (.65)(.75) = .49
Since P(A  B)  P(A)P(B), they cannot be independent events; or, since P(A | B)  P(B), they
cannot be independent.
33. a.
Reason for Applying
Quality
Cost/Convenience
Other
Total
Full Time
.218
.204
.039
.461
Part Time
.208
.307
.024
.539
.426
.511
.063
1.00
b.
It is most likely a student will cite cost or convenience as the first reason - probability = .511.
School quality is the first reason cited by the second largest number of students - probability = .426.
c.
P(Quality | full time) = .218 / .461 = .473
d.
P(Quality | part time) = .208 / .539 = .386
13- 84
e.
For independence, we must have P(A)P(B) = P(A  B).
From the table, P(A  B) = .218, P(A) = .461, P(B) = .426
P(A)P(B) = (.461)(.426) = .196
Since P(A)P(B)  P(A  B), the events are not independent.
34. a.
P(O) = 0.38 + 0.06 = 0.44
b.
P(Rh-) = 0.06 + 0.02 + 0.01 + 0.06 = 0.15
c.
P(both Rh-) = P(Rh-) P(Rh-) = (0.15)(0.15) = 0.0225
d.
P(both AB) = P(AB) P(AB) = (0.05)(0.05) = 0.0025
e.
P(Rh  O) 
f.
P(Rh+) = 1 - P(Rh-) = 1 - 0.15 = 0.85
P(B Rh+) =
P(Rh  O)
P(O)
P(B  Rh+)
0.06

 0.136
0.44
=
P(Rh+)
35. a.
0.09
= 0.106
0.85
P(Up for January) = 31 / 48 = 0.646
b.
P(Up for Year) = 36 / 48 = 0.75
c.
P(Up for Year  Up for January) = 29 / 48 = 0.604
P(Up for Year | Up for January) = 0.604 / 0.646 = 0.935
d.
They are not independent since
P(Up for Year)  P(Up for Year | Up for January)
0.75  0.935
36. a.
Occupation
Cabinetmaker
Lawyer
Physical Therapist
Systems Analyst
Under 50
.000
.150
.000
.050
Total
.200
50-59
.050
.050
.125
.025
.250
Satisfaction Score
60-69
70-79
.100
.075
.025
.025
.050
.025
.100
.075
.275
.200
b.
P(80s) = .075 (a marginal probability)
c.
P(80s | PT) = .050/.250 = .20 (a conditional probability)
d.
P(L) = .250 (a marginal probability)
13- 85
80-89
.025
.000
.050
.000
.075
Total
.250
.250
.250
.250
1.000
e.
P(L  Under 50) = .150 (a joint probability)
f.
P(Under 50 | L) = .150/.250 = .60 (a conditional probability)
g.
P(70 or higher) = .275 (Sum of marginal probabilities)
37. a.
P(A  B) = P(A)P(B) = (.55)(.35) = .19
b.
P(A  B) = P(A) + P(B) - P(A  B) = .55 + .35 - .19 = .71
c.
P(shutdown) = 1 - P(A  B) = 1 - .71 = .29
38. a.
b.
P(Telephone) 
52
 0.2737
190
This is an intersection of two events. It seems reasonable to assume the next two messages will be
independent; we use the multiplication rule for independent events.
P(E-mail  Fax) = P(E-mail) P(Fax) =
c.
30
15
190 190
This is a union of two mutually exclusive events.
 0.0125
P(Telephone  Interoffice Mail) = P(Telephone) + P(Interoffice Mail)
=
39. a.
b.
52 18
70  0.7368


190 190 190
Yes, since P(A1  A2) = 0
P(A1  B) = P(A1)P(B | A1) = .40(.20) = .08
P(A2  B) = P(A2)P(B | A2) = .60(.05) = .03
c.
P(B) = P(A1  B) + P(A2  B) = .08 + .03 = .11
d.
1
B) = .08 = .7273
.11
2
B) = .03 = .2727
.11
P(A
P(A
40. a.
P(B  A1) = P(A1)P(B | A1) = (.20) (.50) = .10
P(B  A2) = P(A2)P(B | A2) = (.50) (.40) = .20
P(B  A3) = P(A3)P(B | A3) = (.30) (.30) = .09
b.
P(A2 B) =
.20
.10 + .20 + .09
= .51
c.
Events
A1
P(Ai)
.20
P(B | Ai)
.50
13- 86
P(Ai  B)
.10
P(Ai | B)
.26
A2
A3
.50
.40
.20
.51
.30
.30
.09
.23
1.00
.39
1.00
S1 = successful, S2 = not successful and B = request received for additional information.
41.
a.
P(S1) = .50
b.
P(B | S1) = .75
c.
P(S1 B) =
42.
(.50) (.75)
= .375 = .65
(.50) (.75) + (.50) (.40)
.575
M = missed payment
D1 = customer defaults
D2 = customer does not default
P(D1) = .05
P(D2) = .95
P(M | D2) = .2
P(M | D1) = 1
a.
P( D1)P( M D1)
P( D1 M) = P( D )P(M D ) + P( D )P( M D )
1
1
2
2
(.05) (1)
=
(.05) (1) + (.95) (.2)
=
.05
.24
= .21
b.
43.
Yes, the probability of default is greater than .20.
Let: S = small car
Sc = other type of vehicle
F = accident leads to fatality for vehicle occupant
We have P(S) = .18, so P(Sc) = .82. Also P(F | S) = .128 and P(F | Sc) = .05. Using the tabular form
of Bayes Theorem provides:
Events
S
Sc
Prior
Probabilities
.18
.82
1.00
Conditional
Probabilities
.128
.050
Joint
Probabilities
.023
.041
.064
Posterior
Probabilities
.36
.64
1.00
From the posterior probability column, we have P(S | F) = .36. So, if an accident leads to a fatality,
the probability a small car was involved is .36.
13- 87
44.
Let
A1 = Story about Basketball Team
A2 = Story about Hockey Team
W = "We Win" headline
P(A1) = .60
P(W | A1) = .641
P(A2) = .40
P(W | A2) = .462
Ai
A1
A2
P(W | A1)
.641
.462
P(Ai)
.60
.40
P(W  Ai)
.3846
.1848
.5694
The probability the story is about the basketball team is .6754.
45. a.
Let S = person is age 65 or older
P(S) =
b.
34, 991, 753
 .12
281, 421, 906
Let D = takes prescription drugs regularly
P(D) = P(D  S) + P(D  Sc)
= P(D | S)P(S) + P(D | Sc)P(Sc)
= .82(.12) + .49(.88)
= .53
c.
Let D5 = takes 5 or more prescriptions
P(D5  S) = P(D5 | S)P(S)
= .40(.12)
= .048
d.
P(S | D5) =
P(S D5 )
P(D5 )
P(D5) = P(S  D5) + P(Sc  D5)
= P(D5 | S)P(S) + P(D5 | Sc)P(Sc)
= .40(.12) + (.28)(.88)
= .048 + .246
= .294
P(S | D 5) =
46. a.
.048
 .16
.294
P(Excellent) = .18
P(Pretty Good) = .50
P(Pretty Good  Excellent) = .18 + .50 = .68
13- 88
P(Ai | M )
.3846/.5694
.1848/.5694
= .6754
= .3246
1.0000
Note: Events are mutually exclusive since a person may only choose one rating.
b.
1035 (.05) = 51.75
We estimate 52 respondents rated US companies poor.
c.
1035 (.01) = 10.35
We estimate 10 respondents did not know or did not answer.
47. a.
b.
(2) (2) = 4
Let
s = successful
u = unsuccessful
Oil
Bonds
s
E1
u
s
E2
u
s
E3
u
E4
c.
O = {E1, E2}
d.
M = {E1, E3}
O  M = {E1, E2, E3}
e.
O  M = {E1}
f.
No; since O  M has a SAMPLE point.
48. a.
P(satisfied) = 0.61
13- 89
b.
The 18 - 34 year old group (64% satisfied) and the 65 and over group (70% satisfied).
c.
P(not satisfied) = 0.26 + 0.04 = 0.30
49.
Let
I = treatment-caused injury
D = death from injury
N = injury caused by negligence
M = malpractice claim filed
$ = payment made in claim
We are given P(I) = 0.04, P(N | I) = 0.25, P(D | I) = 1/7, P(M | N) = 1/7.5 = 0.1333,
and P($ | M) = 0.50
a.
P(N) = P(N | I) P(I) + P(N | Ic) P(Ic)
= (0.25)(0.04) + (0)(0.96)
= 0.01
b. P(D) = P(D | I) P(I) + P(D | Ic) P(Ic)
= (1/7)(0.04) + (0)(0.96)
= 0.006
c.
P(M) = P(M | N) P(N) + P(M | Nc) P(Nc)
= (0.1333)(0.01) + (0)(0.99)
= 0.001333
P($) = P($ | M) P(M) + P($ | Mc) P(Mc)
= (0.5)(0.001333) + (0)(0.9987)
= 0.00067
50. a.
Probability of the event = P(average) + P(above average) + P(excellent)
=
11 14 13


50 50 50
= .22 + .28 + .26
= .76
b.
Probability of the event = P(poor) + P(below average)
=
51. a.
4
8
 .24
50 50
P(leases 1) = 168 / 932 = 0.18
b.
P(2 or fewer) = 401 / 932 + 242 / 932 + 65 / 932 = 708 / 932 = 0.76
c.
P(3 or more) = 186 / 932 + 112 / 932 = 298 / 932 = 0.32
d.
P(no cars) = 19 / 932 = 0.02
13- 90
52. a.
Yes
No
Total
23 and Under
.1026
.0996
.2022
24 - 26
.1482
.1878
.3360
27 - 30
.0917
.1328
.2245
31 - 35
.0327
.0956
.1283
36 and Over
.0253
.0837
.1090
Total
.4005
.5995
1.0000
b.
.2022
c.
.2245 + .1283 + .1090 = .4618
d.
.4005
53. a.
.
P(24 to 26 | Yes) = .1482 / .4005 = .3700
b.
P(Yes | 36 and over) = .0253 / .1090 = .2321
c.
.1026 + .1482 + .1878 + .0917 + .0327 + .0253 = .5883
d.
P(31 or more | No) = (.0956 + .0837) / .5995 = .2991
e.
No, because the conditional probabilities do not all equal the marginal probabilities. For instance,
P(24 to 26 | Yes) = .3700  P(24 to 26) = .3360
54.
Let
I = important or very important
M = male
F = female
a.
P(I) = .49 (a marginal probability)
b.
P(I | M) = .22/.50 = .44 (a conditional probability)
13- 91
c.
P(I | F) = .27/.50 = .54 (a conditional probability)
d.
It is not independent
P(I) = .49  P(I | M) = .44
and
P(I) = .49  P(I | F) = .54
e.
55. a.
Since level of importance is dependent on gender, we conclude that male and female respondents
have different attitudes toward risk.
P(B S) = P(B  S) = .12 = .30
P(S)
.40
We have P(B | S) > P(B).
Yes, continue the ad since it increases the probability of a purchase.
b.
c.
Estimate the company’s market share at 20%. Continuing the advertisement should increase the
market share since P(B | S) = .30.
P(B S) = P(B  S) = .10 = .333
P(S)
.30
The second ad has a bigger effect.
56. a.
P(A) = 200/800 = .25
b.
c.
P(B) = 100/800 = .125
P(A  B) = 10/800 = .0125
d.
P(A | B) = P(A  B) / P(B) = .0125 / .125 = .10
e.
No, P(A | B)  P(A) = .25
57.
Let
A = lost time accident in current year
B = lost time accident previous year
Given: P(B) = .06, P(A) = .05, P(A | B) = .15
a.
P(A  B) = P(A | B)P(B) = .15(.06) = .009
b.
P(A  B) = P(A) + P(B) - P(A  B)
= .06 + .05 - .009 = .101 or 10.1%
58.
Let: A = return is fraudulent
B = exceeds IRS standard for deductions
Given: P(A | B) = .20, P(A | Bc) = .02, P(B) = .08, find P(A) = .3. Note P(Bc) = 1 - P(B)
= .92
13- 92
P(A) = P(A  B) + P(A  Bc)
= P(B)P(A | B) + P(Bc)P(A | Bc)
= (.08)(.20) + (.92)(.02) = .0344
We estimate 3.44% will be fraudulent.
59. a.
b.
P(Oil) = .50 + .20 = .70
Let S = Soil test results
Events
High Quality (A1)
Medium Quality (A2)
No Oil (A3)
P(Ai)
.50
.20
.30
1.00
P(S | Ai)
.20
.80
.20
P(Ai  S)
.10
.16
.06
P(S) = .32
P(Ai | S)
.31
.50
.19
1.00
P(Oil) = .81 which is good; however, probabilities now favor medium quality rather than high quality
oil.
60. a.
A1 = field will produce oil
A2 = field will not produce oil
W = well produces oil
Events
Oil in Field
No Oil in Field
P(Ai)
.25
.75
1.00
P(Wc | Ai)
.20
1.00
P(Wc  Ai)
.05
.75
.80
P(Ai | Wc)
.0625
.9375
1.0000
The probability the field will produce oil given a well comes up dry is .0625.
b.
Events
Oil in Field
No Oil in Field
P(Ai)
.0625
.9375
1.0000
P(Wc | Ai)
.20
1.00
P(Wc  Ai)
.0125
.9375
.9500
P(Ai | Wc)
.0132
.9868
1.0000
The probability the well will produce oil drops further to .0132.
c.
Suppose a third well comes up dry. The probabilities are revised as follows:
Events
Oil in Field
Incorrect Adjustment
P(Ai)
.0132
.9868
1.0000
P(Wc | Ai)
.20
1.00
P(Wc  Ai)
.0026
.9868
.9894
Stop drilling and abandon field if three consecutive wells come up dry.
13- 93
P(Ai | Wc)
.0026
.9974
1.0000
Chapter 5
Discrete Probability Distributions
Learning Objectives
1.
Understand the concepts of a random variable and a probability distribution.
2.
Be able to distinguish between discrete and continuous random variables.
3.
Be able to compute and interpret the expected value, variance, and standard deviation for a discrete
random variable and understand how an Excel worksheet can be used to ease the burden of the
calculations.
4.
Be able to compute probabilities using a binomial probability distribution and be able to compute
these probabilities using Excel's BINOMDIST function.
5.
Be able to compute probabilities using a Poisson probability distribution and be able to compute
these probabilities using Excel's POISSON function.
6.
Know when and how to use the hypergeometric probability distribution and be able to compute these
probabilities using Excel's HYPGEOMDIST function.
13- 94
Solutions:
1.
a.
Head, Head (H,H)
Head, Tail (H,T)
Tail, Head (T,H)
Tail, Tail (T,T)
b.
x = number of heads on two coin tosses
c.
Outcome
(H,H)
(H,T)
(T,H)
(T,T)
2.
Values of x
2
1
1
0
d.
Discrete. It may assume 3 values: 0, 1, and 2.
a.
Let x = time (in minutes) to assemble the product.
b.
It may assume any positive value: x > 0.
c.
Continuous
3.
Let Y = position is offered
N = position is not offered
a.
S = {(Y,Y,Y), (Y,Y,N), (Y,N,Y), (Y,N,N), (N,Y,Y), (N,Y,N), (N,N,Y), (N,N,N)}
b.
Let N = number of offers made; N is a discrete random variable.
c.
Experimental Outcome
Value of N
4.
5.
(Y,Y,Y) (Y,Y,N) (Y,N,Y) (Y,N,N) (N,Y,Y) (N,Y,N) (N,N,Y) (N,N,N)
3
2
2
1
2
1
1
0
x = 0, 1, 2, . . ., 12.
a.
S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
b.
Experimental Outcome
Number of Steps Required
6.
(1,1)
2
(1,2)
3
a.
values: 0,1,2,...,20
discrete
b. values: 0,1,2,...
discrete
c. values: 0,1,2,...,50
discrete
13- 95
(1,3)
4
(2,1)
3
(2,2)
4
(2,3)
5
7.
d.
values: 0  x  8
continuous
e.
values: x > 0
continuous
a.
f (x)  0 for all values of x.
 f (x) = 1 Therefore, it is a proper probability distribution.
8.
b.
Probability x = 30 is f (30) = .25
c.
Probability x  25 is f (20) + f (25) = .20 + .15 = .35
d.
Probability x > 30 is f (35) = .40
a.
x
1
2
3
4
f (x)
3/20 = .15
5/20 = .25
8/20 = .40
4/20 = .20
Total 1.00
b.
f (x)
.4
.3
.2
.1
x
1
c.
2
f (x)  0 for x = 1,2,3,4.
 f (x) = 1
9.
a.
Age
6
7
8
9
Number of Children
37,369
87,436
160,840
239,719
f(x)
0.018
0.043
0.080
0.119
13- 96
3
4
10
11
12
13
14
286,719
306,533
310,787
302,604
289,168
2,021,175
0.142
0.152
0.154
0.150
0.143
1.001
b.
f(x)
.16
.14
.12
.10
.08
.06
.04
.02
x
6
c.
7
8
9
10 11
12 13
14
f(x)  0 for every x
 f(x) = 1
Note:  f(x) = 1.001 in part (a); difference from 1 is due to rounding values of f(x).
10. a.
x
1
2
3
4
5
f(x)
0.05
0.09
0.03
0.42
0.41
1.00
x
1
2
3
4
5
f(x)
0.04
0.10
0.12
0.46
0.28
1.00
b.
c.
P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83
13- 97
d.
Probability of very satisfied: 0.28
e.
Senior executives appear to be more satisfied than middle managers. 83% of senior executives have
a score of 4 or 5 with 41% reporting a 5. Only 28% of middle managers report being very satisfied.
11. a.
Duration of Call
x
1
2
3
4
f(x)
0.25
0.25
0.25
0.25
1.00
b.
f (x)
0.30
0.20
0.10
x
0
1
2
3
4
c.
f (x)  0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00
d.
f (3) = 0.25
e.
P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50
12. a.
b.
13. a.
Yes; f (x)  0 for all x and  f (x) = .15 + .20 + .30 + .25 + .10 = 1
P(1200 or less) = f (1000) + f (1100) + f (1200)
= .15 + .20 + .30
= .65
Yes, since f (x)  0 for x = 1,2,3 and  f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1
b.
f (2) = 2/6 = .333
c.
f (2) + f (3) = 2/6 + 3/6 = .833
14. a.
f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150)
= 1 - .95 = .05
This is the probability MRA will have a $200,000 profit.
13- 98
b.
P(Profit) = f (50) + f (100) + f (150) + f (200)
= .30 + .25 + .10 + .05 = .70
c.
P(at least 100) = f (100) + f (150) + f (200)
= .25 + .10 +.05 = .40
15. a.
x
3
6
9
f (x)
.25
.50
.25
1.00
x f (x)
.75
3.00
2.25
6.00
E (x) =  = 6.00
b.
x - 
-3
0
3
x
3
6
9
(x - )2
9
0
9
f (x)
.25
.50
.25
(x - )2f (x)
2.25
0.00
2.25
4.50
Var (x) =  2 = 4.50
c.
 =
4.50 = 2.12
16. a.
y
2
4
7
8
f (y)
.20
.30
.40
.10
1.00
E(y) =  = 5.20
b.
y-
-3.20
-1.20
1.80
2.80
y
2
4
7
8
(y - )2
10.24
1.44
3.24
7.84
y f (y)
.40
1.20
2.80
.80
5.20
(y - )2 f (y)
2.048
.432
1.296
.784
4.560
f (y)
.20
.30
.40
.10
Var ( y)  4.56
  4.56  2.14
17. a/b.
x
0
f (x)
.10
x f (x)
.00
x-
-2.45
13- 99
(x - )2
6.0025
(x - )2 f (x)
.600250
1
2
3
4
5
E (x)
2

.15
.30
.20
.15
.10
.15
.60
.60
.60
.50
2.45
2.1025
.2025
.3025
2.4025
6.5025
.315375
.060750
.060500
.360375
.650250
2.047500
=  = 2.45
= 2.0475
= 1.4309
18. a/b.
x
0
1
2
3
4
5
-1.45
- .45
.55
1.55
2.55
f (x)
.01
.23
.41
.20
.10
.05
E (x) =
x f (x)
.00
.23
.82
.60
.40
.25
2.30
x-
-2.3
-1.3
-0.3
0.7
1.7
2.7
(x - )2
5.29
1.69
0.09
0.49
2.89
7.29
Var (x) = 2 =
(x - )2 f (x)
.0529
.3887
.0369
.0980
.2890
.3645
1.2300
The expected value, E (x) = 2.3, of the probability distribution is the same as the average reported
in the 1997 Statistical Abstract of the United States. The variance of the number of television sets
per household is Var (x) = 1.23 television sets squared. The standard deviation is  = 1.11
television sets.
19. a.
E (x) =  x f (x) = 0 (.50) + 2 (.50) = 1.00
b.
E (x) =  x f (x) = 0 (.61) + 3 (.39) = 1.17
c.
The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will
make more points in the long run with the 3 - point shot.
20. a.
x
0
400
1000
2000
4000
6000
f (x)
.90
.04
.03
.01
.01
.01
1.00
x f (x)
0.00
16.00
30.00
20.00
40.00
60.00
166.00
E (x) = 166. If the company charged a premium of $166.00 they would break even.
b.
Gain to Policy Holder
-260.00
140.00
740.00
1,740.00
3,740.00
5,740.00
f (Gain)
.90
.04
.03
.01
.01
.01
13- 100
(Gain) f (Gain)
-234.00
5.60
22.20
17.40
37.40
57.40
-94.00
E (gain) = -94.00. The policy holder is more concerned that the big accident will break him than
with the expected annual loss of $94.00.
21. a.
E (x) =  x f (x) = 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5)
= 4.05
b.
E (x) =  x f (x) = 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5)
= 3.84
c.
Executives:
d.
Middle Managers: 2 =  (x - )2 f(x) = 1.1344
Executives:
 = 1.1169
2 =  (x - )2 f(x) = 1.2475
Middle Managers:  = 1.0651
e.
22. a.
The senior executives have a higher average score: 4.05 vs. 3.84 for the middle managers. The
executives also have a slightly higher standard deviation.
E (x) =  x f (x)
= 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445
The monthly order quantity should be 445 units.
b.
23. a.
Cost:
445 @ $50 = $22,250
Revenue: 300 @ $70 = 21,000
$ 1,250 Loss
Laptop: E (x) = .47(0) + .45(1) + .06(2) + .02(3) = .63
Desktop: E (x) = .06(0) + .56(1) + .28(2) + .10(3) = 1.42
b.
Laptop: Var (x) = .47(-.63)2 + .45(.37)2 + .06(1.37)2 + .02(2.37)2 = .4731
Desktop: Var (x) = .06(-1.42)2 + .56(-.42)2 + .28(.58)2 + .10(1.58)2 = .5636
c.
24. a.
From the expected values in part (a), it is clear that the typical subscriber has more desktop
computers than laptops. There is not much difference in the variances for the two types of
computers.
Medium E (x) =  x f (x)
= 50 (.20) + 150 (.50) + 200 (.30) = 145
Large:
E (x) =  x f (x)
= 0 (.20) + 100 (.50) + 300 (.30) = 140
Medium preferred.
b.
Medium
x
f (x)
x-
(x - )2
13- 101
(x - )2 f (x)
50
150
200
.20
.50
.30
-95
5
55
9025
25
3025
1805.0
12.5
907.5
2
 = 2725.0
Large
y
0
100
300
y - 
-140
-40
160
f (y)
.20
.50
.30
(y - )2
19600
1600
25600
(y - )2 f (y)
3920
800
7680
2 = 12,400
Medium preferred due to less variance.
25. a.
S
F
S
F
S
F
2
1
1
2!

b.
f (1) 
c.
f (0) 
d.
f (2) 
e.
P (x  1) = f (1) + f (2) = .48 + .16 = .64
f.
E (x) = n p = 2 (.4) = .8
1
(.4) (.6)
2
0
0
2
(.4) (.6)
2
2
2
1! 1!
(.4) (.6)

(.4)(.6)  .48
2!
0! 2!
0

2!
2! 0!
(1)(.36)  .36
(.16)(1)  .16
Var (x) = n p (1 - p) = 2 (.4) (.6) = .48
 = .48 = .6928
26. a.
f (0) = .3487
b.
f (2) = .1937
c.
P(x  2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298
d.
P(x  1) = 1 - f (0) = 1 - .3487 = .6513
e.
E (x) = n p = 10 (.1) = 1
13- 102
f.
Var (x) = n p (1 - p) = 10 (.1) (.9) = .9
 = .9 = .9487
27. a.
f (12) = .1144
b.
f (16) = .1304
c.
P (x  16)
d.
P (x  15) = 1 - P (x  16) = 1 - .2374 = .7626
e.
E (x) = n p = 20(.7) = 14
f.
Var (x) = n p (1 - p) = 20 (.7) (.3) = 4.2
 =
28. a.
b.
= f (16) + f (17) + f (18) + f (19) + f (20)
= .1304 + .0716 + .0278 + .0068 + .0008
= .2374
4.2 = 2.0494
f (2) 
6
2
2
4
 .3292
(.33) (.67)
P(at least 2)
= 1 - f(0) - f(1)
6
= 1
0
6
(.33) (.67) 
0
6
1
(.33) (.67)
1
5
= 1 - .0905 - .2673 = .6422
c.
f (0) 
10
0
0
10
(.33) (.67)
 .0182
29.
P(At Least 5)
30. a.
Probability of a defective part being produced must be .03 for each trial; trials must be independent.
b.
= 1 - f (0) - f (1) - f (2) - f (3) - f (4)
= 1 - .0000 - .0005 - .0031 - .0123 - .0350
= .9491
Let: D = defective
G = not defective
13- 103
1st part
2nd part
Experimental
Outcome
D
Number
Defective
(D, D)
2
(D, G)
1
(G, D)
1
(G, G)
0
G
D
.
G
D
G
c.
2 outcomes result in exactly one defect.
d.
P (no defects) = (.97) (.97) = .9409
P (1 defect) = 2 (.03) (.97) = .0582
P (2 defects) = (.03) (.03) = .0009
31.
Binomial n = 10 and p = .05
f ( x) 
10!
x!(10  x)!
(.05) x (.95)10x
a.
Yes. Since they are selected randomly, p is the same from trial to trial and the trials are independent.
b.
f (2) = .0746
c.
f (0) = .5987
d.
P (At least 1) = 1 - f (0) = 1 - .5987 = .4013
32. a.
b.
.90
P (at least 1) = f (1) + f (2)
1
1
2!
f (1) =
(.9) (.1)
1! 1!
= 2 (.9) (.1) = .18
2
0
2!
f (2) =
(.9) (.1)
2! 0!
= 1 (.81) (1) = .81
 P (at least 1) = .18 + .81 = .99
Alternatively
13- 104
P (at least 1) = 1 - f (0)
0
2
2!
f (0) =
(.9) (.1) = .01
0! 2!
Therefore, P (at least 1) = 1 - .01 = .99
c.
P (at least 1) = 1 - f (0)
3!
f (0) =
0
3
(.9) (.1) = .001
0! 3!
Therefore, P (at least 1) = 1 - .001 = .999
d.
Yes; P (at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack
would be catastrophic.
20!
33. a.
f(12) =
(.5)12 (.5)8
12!8!
Using Table 5 in Appendix 8, f(12) = .0708
b.
f(0) + f(1) + f(2) + f(3) + f(4) + f(5)
.0000 + .0000 + .0002 + .0011 + .0046 + .0148 = .0207
c.
E(x) = np = 20(.5) = 10
d.
Var (x) = 2 = np(1 - p) = 20(.5)(.5) = 5
 = 5 = 2.24
34. a.
f (3) = .0634
b.
The answer here is the same as part (a). The probability of 12 failures with p = .60 is the same as the
probability of 3 successes with p = .40.
c.
f (3) + f (4) + · · · + f (15) = 1 - f (0) - f (1) - f (2)
= 1 - .0005 - .0047 - .0219
= .9729
35. a.
f (0) + f (1) + f (2) = .0115 + .0576 + .1369 = .2060
b.
f (4) = .2182
c.
1 - [ f (0) + f (1) + f (2) + f (3) ]
d.
 = n p = 20 (.20) = 4
36.
x
0
f (x)
.343
= 1 - .2060 - .2054 = .5886
x-
-.9
(x - )2
.81
13- 105
(x - )2 f (x)
.27783
1
2
3
37.
.441
.189
.027
1.000
.1
1.1
2.1
.01
1.21
4.41
.00441
.22869
.11907
2
 = .63000
E(x) = n p = 30(0.29) = 8.7
2 = n p (1 - p) = 30(0.29)(0.71) = 6.177
 =
6.177 = 2.485
38. a.
f (x) 
3x e3
x!
b.
f (2) 
32 e3 9(.0498)

 .2241
2!
2
c.
f (1) 
31 e3
 3(.0498)  .1494
1!
d.
P (x  2) = 1 - f (0) - f (1) = 1 - .0498 - .1494 = .8008
39. a.
f (x) 
2x e2
x!
b.
 = 6 for 3 time periods
c.
f (x) 
6x e6
x!
d.
f (2) 
22 e2 4(.1353)

 .2706
2!
2
e.
f (6) 
66 e6
 .1606
6!
f.
f (5) 
45 e4
 .1563
5!
40. a.
 = 48 (5 / 60) = 4
f (3) 
b.
43 e4 (64)(.0183)

 .1952
3!
6
 = 48 (15 / 60) = 12
13- 106
f (10) 
1210 e12
 .1048
10!
c.
 = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes.
f (0) 
40 e4
 .0183
0!
The probability none will be waiting after 5 minutes is .0183.
d.
 = 48 (3 / 60) = 2.4
f (0) 
2.40 e2.4
 .0907
0!
The probability of no interruptions in 3 minutes is .0907.
41. a.
b.
30 per hour
 = 1 (5/2) = 5/2
f (3)  5 / 2 e
3!
3
c.
42. a.
b.
f (0) 
f (0) 
(5 / 2)
 .2138
5 / 20 e(5 / 2)  e(5/ 2)  .0821
0!
7
0
7 e

 e 7  .0009
0!
probability = 1 - [f(0) + f(1)]
f (1) 
71 e7

 7e 7  .0064
1!
probability = 1 - [.0009 + .0064] = .9927
c.
 = 3.5
f (0) 
3.50 e3.5

 e 3.5  .0302
0!
probability = 1 - f(0) = 1 - .0302 = .9698
d.
43. a.
probability = 1 - [f(0) + f(1) + f(2) + f(3) + f(4)]
= 1 - [.0009 + .0064 + .0223 + .0521 + .0912]
= .8271
f (0) 
100 e10

 e 10  .000045
0!
13- 107
b.
f (0) + f (1) + f (2) + f (3)
f (0) = .000045 (part a)
1
f (1) =
10 e
-10
= .00045
1!
Similarly, f (2) = .00225, f (3) = .0075
and f (0) + f (1) + f (2) + f (3) = .010245
c.
2.5 arrivals / 15 sec. period Use  = 2.5
f (0) 
d.
44.
2.50 e2.5
 .0821
0!
1 - f (0) = 1 - .0821 = .9179
Poisson distribution applies
a.
 = 1.25 per month
b.
f (0) 
1.250 e1.25
 0.2865
0!
c.
f (1) 
1.251 e1.25
 0.3581
1!
d.
P (More than 1) = 1 - f (0) - f (1) = 1 - 0.2865 - 0.3581 = 0.3554
45. a.
average per month =
18
 1.5
12
b.
f (0) 
1.50 e1.5

 e 1.5  .2231
0!
c.
probability = 1 - [f(0) + f(1)]
= 1 - [.2231 + .3347]
= .4422
3 10  3
1 4 1
46. a.
f (1) 

10
4
3!
1!2!
7!
3!4!
10!
4!6!

(3)(35)
210
13- 108
 .50
b.
3 10  3
2 22
f (2) 

(3)(1)
10
 .067
45
2
c.
f (0) 
3 10  3
0 20

(1)(21)
10
 .4667
45
2
d.
3 10  3
2 42
f (2) 

(3)(21)
10
 .30
210
4
4 15  4
3 10  3
f (3) 
47.

(4)(330)
15
 .4396
3003
10
48.
Hypergeometric with N = 10 and r = 6
a.
f (2) 
6
2
4
1

10
(15)(4)
 .50
120
3
b.
Must be 0 or 1 prefer Coke Classic.
f (1) 
6
1
4
2

10
(6)(6)
 .30
120
3
f (0) 
6
0
4
3

10
(1)(4)
 .0333
120
3
P (Majority Pepsi) = f (1) + f (0) = .3333
49.
Parts a, b & c involve the hypergeometric distribution with N = 52 and n = 2
a.
r = 20, x = 2
13- 109
20
2
f (2) 
32
0

52
(190)(1)
 .1433
1326
2
b.
r = 4, x = 2
4 48
2 0
f (2) 

52
(6)(1)
 .0045
1326
2
c.
r = 16, x = 2
16
2
f (2) 
36
0
52

(120)(1)
 .0905
1326
2
d.
Part (a) provides the probability of blackjack plus the probability of 2 aces plus the probability of
two 10s. To find the probability of blackjack we subtract the probabilities in (b) and (c) from the
probability in (a).
P (blackjack) = .1433 - .0045 - .0905 = .0483
50.
N = 60 n = 10
a.
r = 20 x = 0
0 IF
4 0I
F2G
F40! JI
G
G
10J
H0 J
KH
K b1gH
10!30!K F40! IF
10 !50 ! I

G
J
G
f (0) =
H10!30!KH60! J
K
60I
60!
F
G
J
10!50!
H1 0 K
=
40  39  38  37  36 35 34  33 32  31
6059 5857 56 5554 5352 51
 .01
b.
r = 20 x = 1
13- 110
20IF
40
F
1K
H
H9 IK F40! IF10!50!I
GJ
GJ
f (1) =
 20G JG
H9!31!KH60! JK
60I
F
GJ
H1 0 K
 .07
c.
1 - f (0) - f (1) = 1 - .08 = .92
d.
Same as the probability one will be from Hawaii. In part b that was found to equal approximately
.07.
51. a.
f (2) 
11 14
2 3

(55)(364)
25
 .3768
53,130
5
b.
f (2) 
14 11
2 3

(91)(165)
25
 .2826
53,130
5
c.
f (5) 
14 11
5 0

(2002)(1)
25
 .0377
53,130
5
d.
f (0) 
14 11
0 5
25

(1)(462)
 .0087
53,130
5
52.
Hypergeometric with N = 10 and r = 2.
Focus on the probability of 0 defectives, then the probability of rejecting the shipment is 1 - f (0).
a.
n = 3, x = 0
f (0) 
2 8
0 3

10
56
 .4667
120
3
P (Reject) = 1 - .4667 = .5333
b.
n = 4, x = 0
13- 111
f (0) 
2
0
8
4

10
70
 .3333
210
4
P (Reject) = 1 - .3333 = .6667
c.
n = 5, x = 0
f (0) 
2 8
0 5

10
56
 .2222
252
5
P (Reject) = 1 - .2222 = .7778
d.
Continue the process. n = 7 would be required with the probability of rejecting = .9333
53. a., b. and c.
x
1
2
3
4
5
f (x)
0.18
0.18
0.03
0.38
0.23
1.00
E(x) =  = 3.30
 =
54. a. and b.
x f (x)
0.18
0.36
0.09
1.52
1.15
3.30
x-
-2.30
-1.30
-0.30
0.70
1.70
(x - )2
5.29
1.69
0.09
0.49
2.89
(x - )2 f (x)
0.9522
0.6084
0.0081
0.7448
3.3235
5.6370
x - 
-2.64
-1.64
-0.64
0.36
1.36
(x - )2
6.9696
2.6896
0.4096
0.1296
1.8496
(x - )2 f (x)
0.139392
0.161376
0.114688
0.069984
0.184960
0.670400
2 = 5.6370
5.6370 = 2.3742
x
1
2
3
4
5
f (x)
0.02
0.06
0.28
0.54
0.10
1.00
x f (x)
0.02
0.12
0.84
2.16
0.50
3.64
f (x)  0 and  f (x) = 1
E(x) =  = 3.64
Var (x) = 2 = 0.6704
c.
People do appear to believe the stock market is overvalued. The average response is slightly over
halfway between “fairly valued” and “somewhat over valued.”
55. a.
x
f (x)
13- 112
9
10
11
12
13
b.
E (x)
.30
.20
.25
.05
.20
=  x f (x)
= 9 (.30) + 10 (.20) + 11 (.25) + 12 (.05) + 13 (.20) = 10.65
Expected value of expenses: $10.65 million
c.
Var (x) =  (x - )2 f (x)
= (9 - 10.65)2 (.30) + (10 - 10.65)2 (.20) + (11 - 10.65)2 (.25)
+ (12 - 10.65)2 (.05) + (13 - 10.65)2 (.20)
= 2.1275
d.
Looks Good: E (Profit) = 12 - 10.65 = 1.35 million
However, there is a .20 probability that expenses will equal $13 million and the college will run a
deficit.
56. a.
n = 20
f (3) 
b.
n = 20
f (0) 
c.
and
20
3
3
17
(0.04) (0.04)
and
20
x = 3
 0.0364
x = 0
0
20
(0.04) (0.96)
 0.4420
0
E (x) = n p = 1200 (0.04) = 48
The expected number of appeals is 48.
d.
 = n p (1 - p) = 1200 (0.04)(0.96) = 46.08
 = 46.08 = 6.7882
57. a.
We must have E (x) = np  10
With p = .4, this leads to:
n(.4)  10
n  25
b.
With p = .12, this leads to:
n(.12)  10
n  83.33
So, we must contact 84 people in this age group to have an expected number of internet users of at
least 10.
13- 113
c.
  25(.4)(.6)  2.45
d.
  84(.12)(.88)  2.97
58.
Since the shipment is large we can assume that the probabilities do not change from trial to trial and
use the binomial probability distribution.
a.
n = 5
5
f (0) 
0
5
0
5
(.01) (.99)
1
4
 .9510
b.
f (1) 
c.
1 - f (0) = 1 - .9510 = .0490
d.
No, the probability of finding one or more items in the SAMPLE defective when only 1% of the
items
in the POPULATION are defective is small (only .0490). I would consider it likely that more than
1% of
the items are defective.
59. a.
b.
1
(.01) (.99)
 .0480
E(x) = np = 100(.041) = 4.1
Var (x) = np(1 - p) = 100(.041)(.959) = 3.9319
  3.9319  1.9829
60. a.
61.
E(x) = 800(.41) = 328
b.
  np(1 p)  800(.41)(.59)  13.91
c.
For this one p = .59 and (1-p) = .41, but the answer is the same as in part (b). For a binomial
probability distribution, the variance for the number of successes is the same as the variance for the
number of failures. Of course, this also holds true for the standard deviation.
 = 15
prob of 20 or more arrivals = f (20) + f (21) + · · ·
= .0418 + .0299 + .0204 + .0133 + .0083 + .0050 + .0029
+ .0016 + .0009 + .0004 + .0002 + .0001 + .0001 = .1249
62.
 = 1.5
prob of 3 or more breakdowns is 1 - [ f (0) + f (1) + f (2) ].
1 - [ f (0) + f (1) + f (2) ]
= 1 - [ .2231 + .3347 + .2510]
13- 114
= 1 - .8088
= .1912
63.
 = 10 f (4) = .0189
64. a.
f (3) 
b.
33 e3
 0.2240
3!
f (3) + f (4) + · · · = 1 - [ f (0) + f (1) + f (2) ]
0
f (0) =
-3
3 e
-3
= e
= .0498
0!
Similarly, f (1) = .1494, f (2) = .2240
 1 - [ .0498 + .1494 + .2241 ] = .5767
65.
Hypergeometric N = 52, n = 5 and r = 4.
F4IF48I
2J
G
H
KGJ
H3 K 6(17296)
a.
F52I  2,598,960 .0399
GJ
H5 K
4IF
48I
F
G
H1JKG
H4 JK 4(194580) .2995
F52I 2,598,960
GJ
H5 K
4IF
48I
F
G
H0JKG
H5 JK 1,712,304
c.
F52I  2,598,960 .6588
GJ
H5 K
b.
d.
66. a.
1 - f (0) = 1 - .6588 = .3412
f (1) 
7 3
1 1

10
(7)(3)
 .4667
45
2
b.
f (2) 
7
2
3
0
10

(21)(1)
 .4667
45
2
13- 115
c.
f (0) 
7
0
3
2
10

(1)(3)
 .0667
45
2
13- 116
Chapter 6
Continuous Probability Distributions
Learning Objectives
1.
Understand the difference between how probabilities are computed for discrete and continuous
random variables.
2.
Know how to compute probability values for a continuous uniform probability distribution and be
able to compute the expected value and variance for such a distribution.
3.
Be able to compute probabilities using a normal probability distribution. Understand the role of the
standard normal distribution in this process.
4.
Be able to use tables for the standard normal probability distribution to compute both standard
normal probabilities and probabilities for any normal distribution.
5.
Given a cumulative probability be able to compute the z-value and x-value that cuts off the
corresponding area in the left tail of a normal distribution.
6.
Be able to use Excel's NORMSDIST and NORMDIST functions to compute probabilities for the
standard normal distribution and any normal distribution.
7.
Be able to use Excel's NORMSINV and NORMINV function to find z and x values corresponding to
given cumulative probabilities.
8.
Be able to compute probabilities using an exponential probability distribution and using Excel's
EXPONDIST function.
9.
Understand the relationship between the Poisson and exponential probability distributions.
13- 117
Solutions:
1.
a.
f (x)
3
2
1
x
.50
2.
1.0
1.5
2.0
b.
P(x = 1.25) = 0. The probability of any single point is zero since the area under the curve above
any single point is zero.
c.
P(1.0  x  1.25) = 2(.25) = .50
d.
P(1.20 < x < 1.5) = 2(.30) = .60
a.
f (x)
.15
.10
.05
x
0
10
b.
P(x < 15) = .10(5) = .50
c.
P(12  x  18) = .10(6) = .60
d.
E(x) 
e.
Var(x) 
10  20
2
20
 15
(20 10)2
12
 8.33
13- 118
30
40
3.
a.
f (x)
3 / 20
1 / 10
1 / 20
x
110
120
130
Minutes
4.
b.
P(x  130) = (1/20) (130 - 120) = 0.50
c.
P(x > 135) = (1/20) (140 - 135) = 0.25
d.
E(x) 
120 140
2
 130 minutes
a.
f (x)
1.5
1.0
.5
x
0
5.
1
2
b.
P(.25 < x < .75) = 1 (.50) = .50
c.
P(x  .30) = 1 (.30) = .30
d.
P(x > .60) = 1 (.40) = .40
a.
Length of Interval = 261.2 - 238.9 = 22.3
1
for 238.9  x  261.2
f (x)  22.3
0
b.
elsewhere
Note: 1 / 22.3 = 0.045
P(x < 250) = (0.045)(250 - 238.9) = 0.4995
13- 119
3
140
Almost half drive the ball less than 250 yards.
c.
P(x  255) = (0.045)(261.2 - 255) = 0.279
d.
P(245  x  260) = (0.045)(260 - 245) = 0.675
e.
P(x  250) = 1 - P(x < 250) = 1 - 0.4995 = 0.5005
The probability of anyone driving it 250 yards or more is 0.5005. With 60 players, the expected
number driving it 250 yards or more is (60)(0.5005) = 30.03. Rounding, I would expect 30 of these
women to drive the ball 250 yards or more.
6.
a.
P(12  x  12.05) = .05(8) = .40
b.
P(x  12.02) = .08(8) = .64
c.
P(x  11.98)  P(x  12.02)
14
4244
3 14
4244
3
.005(8)  .04 .64  .08(8)
Therefore, the probability is .04 + .64 = .68
7.
a.
P(10,000  x < 12,000) = 2000 (1 / 5000) = .40
The probability your competitor will bid lower than you, and you get the bid, is .40.
b.
P(10,000  x < 14,000) = 4000 (1 / 5000) = .80
c.
A bid of $15,000 gives a probability of 1 of getting the property.
d.
Yes, the bid that maximizes expected profit is $13,000.
The probability of getting the property with a bid of $13,000 is
P(10,000  x < 13,000) = 3000 (1 / 5000) = .60.
The probability of not getting the property with a bid of $13,000 is .40.
The profit you will make if you get the property with a bid of $13,000 is $3000 = $16,000 - 13,000.
So your expected profit with a bid of $13,000 is
EP ($13,000) = .6 ($3000) + .4 (0) = $1800.
If you bid $15,000 the probability of getting the bid is 1, but the profit if you do get the bid is only
$1000 = $16,000 - 15,000. So your expected profit with a bid of $15,000 is
EP ($15,000) = 1 ($1000) + 0 (0) = $1,000.
13- 120
 = 10
70
80
90
100
110 120
130
8.
9.
a.
 =5
35
40
45
50
55
60
65
b.
.6826 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50.
c.
.9544 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50.
10. a.
P(z  1.5) = .9332
b.
P(z  1.0) = .8413
c.
P(1.0  z  1.5) = .9332 - .8413 = .0919
d.
P(0 < z < 2.5) = .9938 - .5000 = .4938
11. a.
P(z  -1) = P(z  1) = .8413
b.
P(z  -1) = 1 - P(z  1) = 1 - .8413 = .1587
c.
P(z  -1.5) = P(z  1.5) = .9332
d.
P(-2.5  z) = P(z  2.5) = .9938
e.
P(-3 < z  0) = P(0 < z < 3) = .9986 - .5000 = .4986
12. a.
.7967 - .5000 = .2967
b.
.9418 - .5000 = .4418
c.
1.0000 - .6700 = .3300
13- 121
d.
e.
.5910
.8849
f.
1.0000 - .7611 = .2389
13. a.
.6879 - .0239 = .6640
b.
.8888 - .6985 = .1903
c.
.9599 - .8508 = .1091
14. a.
z = 1.96
b.
z = 1.96
c.
z = .61
d.
Area to left of z is .8686
z = 1.12
e.
z = .44
f.
Area to left of z is .6700
z = .44
15. a.
b.
Compute .9030 / 2 = .4515 so the area to the left of z is .5000 + .4515 = .9515.
c.
Compute .2052 / 2 = .1026 so the area to the left of z is .5000 + .1026 = .6026. z = .26.
d.
Look in the table for an area of .9948; z = 2.56.
e.
Look in the table for an area of .6915. Since the value we are seeking is below the mean, the z value
must be negative. Thus, z = -.50.
16. a.
17.
Look in the table for an area of 1.0000 - .2119 = .7881. Now z = .80 cuts off an area of .2199 in the
upper tail. Thus, for an area of .2119 in the lower tail z = -80.
z = 1.66.
Look in the table for an area of .9900. The area value in the table closest to .9900 provides the
value z = 2.33.
b.
Look in the table for an area of .9750. This corresponds to z = 1.96.
c.
Look in the table for an area of .9500. Since .9500 is exactly halfway between .9495 (z = 1.64) and
.9505 (z = 1.65), we select z = 1.645. However, z = 1.64 or z = 1.65 are also acceptable
answers.
d.
Look in the table for an area of .9000. The area value in the table closest to .9000 provides the value
z = 1.28.
Let x = amount spent
 = 527,  = 160
13- 122
z
700  527
a.
 1.08
160
P(x > 700) = P(z > 1.08) = .5000 - .3599 = .1401
b.
z
100  527
160
 2.67
P(x < 100) = P(z < -2.67) = .5000 - .4962 = .0038
c.
At 700, z = 1.08 from part (a)
At 450, z 
450  527
160
 .48
P(450 < x < 700) = P(-.48 < z < 1.08) = .8599 - .3156 = .5443
d.
z
300  527
160
 1.42
P(x  300) = P(z  -1.42) = .5000 - .4222 = .0778
18. a.
Find P(x  60)
At x = 60
60 - 49
= 0.69
16
z =
P(x < 60) = 0.7549
P(x  60) = 1 - P(x < 60) = 0.2451
b.
Find P(x  30)
At x = 30
z =
30 - 49
= – 1.19
16
P(x  30) = 1.0000 - 0.8830 = 0.1170
c.
Find z-score so that P(z  z-score) = 0.10
z-score = 1.28 cuts off 10% in upper tail
Now, solve for corresponding value of x.
1.28 
x  49
16
x = 49 + (16)(1.28) = 69.48
So, 10% of subscribers spend 69.48 minutes or more reading The Wall Street Journal.
19.
We have  = 3.5 and  = .8.
13- 123
a.
z
5.0  3.5
 1.88
.8
P(x > 5.0) = P(z > 1.88) = 1 - P(z < 1.88) = 1 - .9699 = .0301
The rainfall exceeds 5 inches in 3.01% of the Aprils.
b.
z
3  3.5
 .63
.8
P(x < 3.0) = P(z < -.63) = P(z > .63) = 1 - P(z < .63) = 1 - .7357 = .2643
The rainfall is less than 3 inches in 26.43% of the Aprils.
c.
z = 1.28 cuts off approximately .10 in the upper tail of a normal distribution.
x = 3.5 + 1.28(.8) = 4.524
If it rains 4.524 inches or more, April will be classified as extremely wet.
We use  = 27 and  = 8
20.
a.
z
11 27
8
 2
P(x  11) = P(z  -2) = 1.0000 - .9772 = .0228
The probability a randomly selected subscriber spends less than 11 hours on the computer is .025.
b.
z
40  27
8
 1.63
P(x > 40) = P(z > 1.63) = 1 - P(z  1.63) = 1 - .9484 = .0516
5.16% of subscribers spend over 40 hours per week using the computer.
c.
A z-value of .84 cuts off an area of .20 in the upper tail.
x = 27 + .84(8) = 33.72
A subscriber who uses the computer 33.72 hours or more would be classified as a heavy user.
21.
From the normal probability tables, a z-value of 2.05 cuts off an area of approximately .02 in the
upper tail of the distribution.
x =  + z = 100 + 2.05(15) = 130.75
A score of 131 or better should qualify a person for membership in Mensa.
Use  = 441.84 and  = 90
22.
a.
At 400
13- 124
400  441.84
z
 .46
90
At 500
z
500  441.84
 .65
90
P(0  z < .65) = .2422
P(-.46  z < 0) = .1772
P(400  z  500) = .1772 + .2422 = .4194
The probability a worker earns between $400 and $500 is .4194.
b.
Must find the z-value that cuts off an area of .20 in the upper tail. Using the normal tables, we find
z = .84 cuts off approximately .20 in the upper tail.
So, x =  + z = 441.84 + .84(90) = 517.44
Weekly earnings of $517.44 or above will put a production worker in the top 20%.
c.
At 250, z 
250  441.84
90
 2.13
P(x  250) = P(z  -2.13) = 1.0000 - .9834 = .0166
The probability a randomly selected production worker earns less than $250 per week is .0166.
23. a.
b.
z
60  80
10
 2 Area to left is 1.0000 - .9772 = .0228
At x = 60
z
60  80
 2
Area to left is .0228
 .5
Area to left is .3085
10
At x = 75
z
75  80
10
P(60  x  75) = .3085 - .0228 = .2857
c.
z
90  80
10
1
Area = 1 - .8413 = .1587
Therefore 15.87% of students will not complete on time.
(60) (.1587) = 9.522
We would expect 9.522 students to be unable to complete the exam in time.
13- 125
24. a.
x ∑
 902.75
n
s
b.
xi
∑(x  x )2
 114.185
n 1
We will use x as an estimate of  and s as an estimate of  in parts (b) - (d) below.
Remember the DATA are in thousands of shares.
At 800
z
800  902.75
 .90
114.185
P(x  800) = P(z  -.90) = 1 - P(z  .90) = 1 - .8159 = .1841
The probability trading volume will be less than 800 million shares is .1841
c.
At 1000
z
1000  902.75
 .85
114.185
P(x  1000) = P(z  .85) = 1 - P(z  .85) = 1 - .8023 = .1977
The probability trading volume will exceed 1 billion shares is .1977
d.
A z-value of 1.645 cuts off an area of .05 in the upper tail
x =  + z = 902.75 + 1.645(114.185) = 1,090.584
They should issue a press release any time share volume exceeds 1,091 million.
 = 442.54,  = 65
25.
a.
z
400  442.54
65
 .65
P(x > 400) = P(z > -.65) = .5000 +.2422 = .7422
b.
z
300  442.54
65
 2.19
P(x  300) = P(z  -2.19) = .5000 - .4857 = .0143
c.
At x = 400, z = -.65 from part (a)
At x = 500, z 
500  442.54
65
 .88
13- 126
P(400 < x < 500) = P(-.65 < z < .88) = .8106 - .2578 = .5528
26. a.
P(x  6) = 1 - e-6/8 = 1 - .4724 = .5276
b.
P(x  4) = 1 - e-4/8 = 1 - .6065 = .3935
c.
P(x  6) = 1 - P(x  6) = 1 - .5276 = .4724
d.
27. a.
P(4  x  6) = P(x  6) - P(x  4) = .5276 - .3935 = .1341
P(x  x0 )  1 ex / 3
0
b.
P(x  2) = 1 - e-2/3 = 1 - .5134 = .4866
c.
P(x  3) = 1 - P(x  3) = 1 - (1 - e3/3 ) = e-1 = .3679
d.
P(x  5) = 1 - e-5/3 = 1 - .1889 = .8111
e.
P(2  x  5) = P(x  5) - P(x  2)
= .8111 - .4866
= .3245
28. a.
P(x  10) = 1 - e-10/20 = .3935
b.
P(x  30) = 1 - P(x  30) = 1 - (1 - e-30/20 ) = e-30/20 = .2231
c.
P(10  x  30) = P(x  30) - P(x  10)
= (1 - e-30/20 ) - (1 - e-10/20 )
= e-10/20 - e-30/20
= .6065 - .2231 = .3834
29. a.
f(x)
.09
.08
.07
.06
.05
.04
.03
.02
.01
x
6
12
18
13- 127
24
b.
P(x  12) = 1 - e-12/12 = 1 - .3679 = .6321
c.
P(x  6) = 1 - e-6/12 = 1 - .6065 = .3935
d.
P(x  30) = 1 - P(x < 30)
= 1 - (1 - e-30/12)
= .0821
30. a.
50 hours
b.
P(x  25) = 1 - e-25/50 = 1 - .6065 = .3935
c.
P(x  100) = 1 - (1 - e-100/50)
= .1353
31. a.
P(x  2) = 1 - e-2/2.78 = .5130
b.
P(x  5) = 1 - P(x  5) = 1 - (1 - e-5/2.78 ) = e-5/2.78 = .1655
c.
P(x  2.78) = 1 - P(x  2.78) = 1 - (1 - e-2.78/2.78 ) = e-1 = .3679
This may seem surprising since the mean is 2.78 minutes. But, for the exponential distribution, the
probability of a value greater than the mean is significantly less than the probability of a value less
than the mean.
32. a.
If the average number of transactions per year follows the Poisson distribution, the time between
transactions follows the exponential distribution. So,
=
1
30
and
1
of a year
 1
 1/ 30
 30
then f(x) = 30 e-30x
b.
A month is 1/12 of a year so,
P x
1
12
 1 P x 
1
12
 1 (1 e
30 /12
)e
30 /12
 .0821
The probability of no transaction during January is the same as the probability of no transaction
during any month: .0821
c.
Since 1/2 month is 1/24 of a year, we compute,
P x
1
24
 1 e
30 / 24
 1.2865  .7135
13- 128
33. a.
Let x = sales price ($1000s)
1
f (x)  25
for 200  x  225
0
elsewhere
b.
P(x  215) = (1 / 25) (225 - 215) = 0.40
c.
P(x < 210) = (1 / 25)(210 - 200) = 0.40
d.
E (x) = (200 + 225)/2 = 212,500
If she waits, her expected sale price will be $2,500 higher than if she sells it back to her company
now. However, there is a 0.40 probability that she will get less. It’s a close call. But, the expected
value approach to decision making would suggest she should wait.
34. a.
For a normal distribution, the mean and the median are equal.
  63,000
b.
Find the z-score that cuts off 10% in the lower tail.
z-score = -1.28
Solving for x,
– 1.28 = x – 63,000
15,000
x = 63,000 - 1.28 (15000)
= 43,800
c.
The lower 10% of mortgage debt is $43,800 or less.
Find P(x > 80,000)
At x = 80,000
z =
80,000 – 63,000
= 1.13
15,000
P(x > 80,000) = 1.0000 - .8708 = 0.1292
d.
Find the z-score that cuts off 5% in the upper tail.
z-score = 1.645. Solve for x.
1.645 = x – 63,000
15,000
x = 63,000 + 1.645 (15,000)
= 87,675
The upper 5% of mortgage debt is in excess of $87,675.
35. a.
P(defect) = 1 - P(9.85  x  10.15)
13- 129
= 1 - P(-1  z  1)
= 1 - .6826
= .3174
Expected number of defects = 1000(.3174) = 317.4
b.
P(defect) = 1 - P(9.85  x  10.15)
= 1 - P(-3  z  3)
= 1 - .9972
= .0028
Expected number of defects = 1000(.0028) = 2.8
c.
Reducing the process standard deviation causes a substantial reduction in the number of defects.
 = 6,312
36.
a.
z = -1.645 cuts off .05 in the lower tail
So,
1.645 


b.
1000  6312

1000  6312
 3229
1.645
At 6000, z 
At 4000, z 
6000  6312
3229
4000  6312
3229
 .10
 .72
P(4000 < x < 6000) = P(-.72 < z < -.10) = .4602 - .2358 = .2244
c.
z = 1.88 cuts off approximately .03 in the upper tail
x = 6312 + 1.88(3229) = 12,382.52
The households with the highest 3% of expenditures spent more than $12,382.
 = 10,000
37.
a.
 = 1500
At x = 12,000
13- 130
z
12, 000 10, 000
1500
 1.33 Area to left is .9082
P(x > 12,000) = 1.0000 - .9082 = .0918
b.
At .95
z = 1.645 = x - 10,000
1500
Therefore, x = 10,000 + 1.645(1500) = 12,468.
95%
0.05
10,000
12,468
12,468 tubes should be produced.
38. a.
At x = 200
z
200 150
25
 2 Area = .9772
P(x > 200) = 1 - .9772 = .0228
b.
Expected Profit = Expected Revenue - Expected Cost
= 200 - 150 = $50
39. a.
Find P(80,000  x  150,000)
At x = 150,000
z =
150,000 – 126,681
= 0.78
30,000
z =
80,000 – 126,681
= – 1.56
30,000
P(x  150,000) = 0.7823
At x = 80,000
P(x  80,000) = 1.0000 - .9406 = 0.0594
P(80,000  x  150,000) = 0.7823 - 0.0594 = 0.7229
13- 131
b.
Find P(x < 50,000)
At x = 50,000
z =
50,000 – 126,681
= – 2.56
30,000
P(x < 50,000) = 1.0000 - .9948 = 0.0052
c.
Find the z-score cutting off 95% in the left tail.
z-score = 1.645. Solve for x.
1.645 = x – 126,681
30,000
x = 126,681 + 1.645 (30,000)
= 176,031
The probability is 0.95 that the number of lost jobs will not exceed 176,031.
40. a.
At 400,
z = 400 - 450 = -.500
100
Area to left is .3085
At 500,
z = 500 - 450 = +.500
100
Area to left is .6915
P(400  x  500) = .6915 - .3085 = .3830
38.3% will score between 400 and 500.
b.
At 630,
z = 630 - 450 = 1.80
100
96.41% do worse and 3.59% do better .
c.
At 480,
z = 480 - 450 = .30
100
38.21% are acceptable.
41. a.
At 75,000
13- 132
Area to left is .6179
75, 000  67, 000
z
 1.14
7, 000
P(x > 75,000) = P(z > 1.14) = 1 - P(z  1.14) = 1 - .8729 = .1271
The probability of a woman receiving a salary in excess of $75,000 is .1271
b.
At 75,000
75, 000  65, 500
z
 1.36
7, 000
P(x > 75,000) = P(z > 1.36) = 1 - P(z  1.36) = 1 - .9131 = .0869
c.
The probability of a man receiving a salary in excess of $75,000 is .0869
At x = 50,000
50, 000  67, 000
z
 2.43
7, 000
P(x < 50,000) = P(z < -2.43) = 1 - P(z < 2.43) = 1 - .9925 = .0075
The probability of a woman receiving a salary below $50,000 is very small: .0075
d.
The answer to this is the male copywriter salary that cuts off an area of .01 in the upper tail of the
distribution for male copywriters.
Use z = 2.33
x = 65,500 + 2.33(7,000) = 81,810
A woman who makes $81,810 or more will earn more than 99% of her male counterparts.
42.
 = .6
At 2%
z = -2.05
z = x - 

x = 18
 -2.05 =
18 - 

.6
 = 18 + 2.05 (.6) = 19.23 oz.
0.02
18
 =19.23
The mean filling weight must be 19.23 oz.
13- 133
43. a.
P(x  15) = 1 - e-15/36 = 1 - .6592 = .3408
b.
P(x  45) = 1 - e-45/36 = 1 - .2865 = .7135
Therefore P(15  x  45) = .7135 - .3408 = .3727
c.
P(x  60) = 1 - P(x < 60)
= 1 - (1 - e-60/36) = .1889
44. a.
Mean time between arrivals = 1/7 minutes
b.
f(x) = 7e-7x
c.
d.
P(x > 1) = 1 - P(x < 1) = 1 - [1 - e-7(1)] = e-7 = .0009
12 seconds is .2 minutes
P(x > .2) = 1 - P(x < .2) = 1- [1- e-7(.2)] = e-1.4 = .2466
45. a.
b.
1  x / 36.5
e
 .0274e.0274 x
36.5
P(x < 40) = 1 - e-.0274(40) = 1 - .3342 = .6658
P(x < 20) = 1 - e-.0274(20) = 1 - .5781 = .4219
P(20 < x < 40) = .6658 - .4219 = .2439
c.
From part (b), P(x < 40) = .6658
P(x > 40) = P(x  40) = 1 - P(x < 40) = 1 - .6658 = .3342
46. a.
1

 0.5 therefore  = 2 minutes = mean time between telephone calls

b.
Note: 30 seconds = .5 minutes
P(x  .5) = 1 - e-.5/2 = 1 - .7788 = .2212
c.
P(x  1) = 1 - e-1/2 = 1 - .6065 = .3935
d.
P(x  5)
= 1 - P(x < 5)
= 1 - (1 - e-5/2)
= .0821
13- 134
Chapter 7
Sampling and Sampling Distributions
Learning Objectives
1.
Understand the importance of sampling and how results from SAMPLEs can be used to provide
estimates of POPULATION parameters such as the POPULATION mean, the POPULATION
standard deviation and / or the POPULATION proportion.
2.
Know what simple random sampling is and how simple random SAMPLEs are selected.
3.
Be able to select a simple random SAMPLE using Excel.
4.
Understand the concept of a sampling distribution.
5.
Know the central limit theorem and the important role it plays in sampling.
6.
Know the characteristics of the sampling distribution of the SAMPLE mean ( x ) and the
sampling distribution of the SAMPLE proportion ( p ).
7.
Learn about a variety of sampling methods including stratified random sampling, cluster sampling,
systematic sampling, convenience sampling and judgment sampling.
8.
Know the definition of the following terms:
simple random sampling
factor sampling with replacement
sampling without replacement
sampling distribution
point estimator
finite POPULATION correction
standard error
13- 135
Solutions:
1.
a.
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
b.
With 10 SAMPLEs, each has a 1/10 probability.
c.
E and C because 8 and 0 do not apply.; 5 identifies E; 7 does not apply; 5 is skipped since E is
already in the SAMPLE; 3 identifies C; 2 is not needed since the SAMPLE of size 2 is
complete.
2.
Using the last 3-digits of each 5-digit grouping provides the random numbers:
601, 022, 448, 147, 229, 553, 147, 289, 209
Numbers greater than 350 do not apply and the 147 can only be used once. Thus, the
simple random SAMPLE of four includes 22, 147, 229, and 289.
3.
4.
459, 147, 385, 113, 340, 401, 215, 2, 33, 348
a.
We first number the companies from 1 to 10: 1 AT&T, 2 IBM, , 10 Pfizer.
Random Number
6
8
5
4
1*
Company in SAMPLE
Microsoft
Motorola
Cisco
Johnson & Johnson
AT&T
*Note that the random numbers 5 and 6 were skipped because we are sampling without replacement.
b.
Company
AT&T
IBM
American Online
Johnson & Johnson
Cisco Systems
Microsoft
General Electric
Motorola
Intel
Pfizer
Random Number
Assigned
6
8
5
4
5
6
1
1
3
8
Company
in
SAMPLE





Note that both American Online and Cisco were assigned a random number of 5. We broke the tie
by including the first to receive a 5 in the SAMPLE.
c.
d.
10!
(10)(9)(8)(7)(6)
Number of SAMPLEs of Size 5 =

 252
5!(10  5)!
(5)(4)(3)(2)(1)
Use Excel's RAND() function to assign a random number between 0 and 1 to each of the companies,
then proceed as in part (b) above. The five with the smallest random numbers can be found by using
Excel's SORT tool.
13- 136
5.
a.
283, 610, 39, 254, 568, 353, 602, 421, 638, 164
b.
Generate a random number for each of the 645 students. Include the students associated with the 50
smallest random numbers in the SAMPLE.
6.
2782, 493, 825, 1807, 289
7.
Use the DATA disk accompanying the book and the EAI file. Generate a random number using
the RAND() function for each of the 2500 managers. Then sort the list of managers with respect to
the random numbers. The first 50 managers are the SAMPLE.
8.
a.
21 random numbers were needed. The teams selected are Wisconsin, Clemson, Washington, USC,
Oklahoma, and Colorado.
b.
Use Excel to generate 25 random numbers - one for each team. Then sort the list of teams with
respect to the list of random numbers.
We can also use the same first two digits in column 9 of Table 7.1. Using the random numbers in
Table 7.1, the following 6 teams are used in the SAMPLE: Nebraska, Florida State, Michigan, Texas,
Washington, and TCU. These are the teams with the six smallest random numbers. (There is a tie
between TCU and Colorado for 6th smallest.)
9.
511, 791, 99, 671, 152, 584, 45, 783, 301, 568, 754, 750
10.
finite, infinite, infinite, infinite, finite
11. a.
x  xi / n 
b.
b.
9
6
( x  x) 2
n 1
( x i  x )2 = (-4)2 + (-1)2 + 12 (-2)2 + 12 + 52 = 48
s 
s=
12. a.
54
48
 3.1
61
p = 75/150 = .50
p = 55/150 = .3667
13. a.
Totals
x  xi / n 
465
xi
( xi  x)
94
100
85
94
92
465
+1
+7
-8
+1
-1
0
 93
5
13- 137
( xi  x )2
1
49
64
1
1
116
b.
14. a.
( x  x )2

n 1
149/784 = 0.19
s
116
 5.39
4
b.
251/784 = 0.32
c.
Total receiving cash = 149 + 219 + 251 = 619
619/784 = 0.79
15. a.
x  x / n 
70
 7 years
i
10
b.
s
20.2
(x  x )2

 1.5 years
n 1
10 1
16.
p = 1117/1400 = 0.80
17. a.
595/1008 = .59
b.
332/1008 = .33
c.
81/1008 = .08
18. a.
Use the DATA disk accompanying the book and the EAI file. Generate a random number for
each manager and select managers associated with the 50 smallest random numbers as the
SAMPLE.
b.
Use Excel's AVERAGE function to compute the mean for the SAMPLE.
c.
Use Excel's STDEV function to compute the SAMPLE standard deviation.
d.
Use the SAMPLE proportion as a point estimate of the POPULATION proportion.
19. a.
The sampling distribution is normal with
E ( x ) =  = 200
 x   / n  50 / 100  5
For +5, ( x -  ) = 5
z 
x  
x

5
1
5
Probability of being within  5 is .6826
b.
For + 10, (x  ) = 10
z 
x  
x

Probability of being within  10 is .9544
13- 138
10
5
2
x  / n
20.
 x  25 / 50  3.54
 x  25 / 100  2.50
 x  25 / 150  2.04
 x  25 / 200  1.77
The standard error of the mean decreases as the SAMPLE size increases.
21. a.
b.
 x   / n  10 / 50  1.41
n / N = 50 / 50,000 = .001
Use  x   / n  10 / 50  1.41
c.
n / N = 50 / 5000 = .01
Use  x   / n  10 / 50  1.41
d.
n / N = 50 / 500 = .10
Use  x 


500  50 10
N  n 
 1.34

N 1 n
500  1 50
Note: Only case (d) where n /N = .10 requires the use of the finite POPULATION correction factor.
22. a.
b.
Using the central limit theorem, we can approximate the sampling distribution of x with a normal
probability distribution provided n  30.
n = 30
 x   / n  50 / 30  9.13
x
400
n = 40
 x   / n  50 / 40  7.91
x
13 - 139
400
23. a.
 x   / n  16 / 50  2.26
For , ( x  )  2
z 
x  
x
2

 0.88
2.26
P(0  z  0.88) = .3106
For  2, the probability is 2(.3106) = .6212
b.
x 
16
100
 1.60
z 
x  
x

2
 1.25
1.60
P(0  z  1.25) = .3944
For  2, the probability is 2(.3944) = .7888
c.
x 
16
200
 1.13
z 
x  
x

2
 1.77
1.13
P(0  z  1.77) = .4616
d.
For  2, the probability is 2(.4616) = .9232
16
 0.80
x 
400
2
x  

 2.50
z 
0.80
x
P(0  z  2.50) = .4938
For  2, the probability is 2(.4938) = .9876
e.
The larger SAMPLE provides a higher probability that the SAMPLE mean will be within  2 of .
24. a.
 x   / n  4000 / 60  516.40
x
51,800
E( x )
The normal distribution is based on the Central Limit Theorem.
13- 140
b.
For n = 120, E ( x ) remains $51,800 and the sampling distribution of x can still be approximated
by a normal distribution. However,  x is reduced to 4000 / 120 = 365.15.
c.
As the SAMPLE size is increased, the standard error of the mean,  x , is reduced. This appears
logical from the point of view that larger SAMPLEs should tend to provide SAMPLE means that are
closer to the POPULATION mean. Thus, the variability in the SAMPLE mean, measured in terms
of  x , should decrease as the SAMPLE size is increased.
 x   / n  4000 / 60  516.40
51,300 51,800
52,300
25. a.
z
52, 300  51,800
 .97
516.40
P(0  z  .97) = .3340
b.
For  500, the probability is 2(.3340) = .6680
 x   / n  4000 / 120  365.15
z = 52,300 - 51,800 = +1.37
365.15
P(0  z  1.37) = .4147
For  500, the probability is 2(.4147) = .8294
26. a.
A normal distribution
E ( x )  1.20
 x   / n  0.10 / 50  0.014
b.
z 
z 
1.22  1.20
 1.41
0.014
1.18  1.20
0.014
 1.41
P(0  z  1.41) = .4207
P(-1.41  z  0) = .4207
probability = 0.4207 + 0.4207 = 0.8414
13- 141
x
c.
z 
z 
1.21 1.20
 0.71
0.014
1.19  1.20
0.014
P(0  z  .71) = .2612
 0.71
P(-.71  z  0) = .2612
probability = 0.2612 + 0.2612 = 0.5224
27. a.
E( x ) = 1017
 x   / n  100 / 75  11.55
z
1027 1017
 0.87
P(0  z  .87) = .3078
11.55
z
1007 1017
11.55
 0.87 P(-.87  z  0) = .3078
probability = 0.3078 + 0.3078 = 0.6156
b.
z
z
1037 1017
11.55
997 1017
11.55
 1.73
P(0  z  1.73) = .4582
 1.73 P(-1.73  z  0) = .4582
probability = 0.4582 + 0.4582 = 0.9164
28. a.

z 
x  34, 000
/ n
Error = x - 34,000 = 250
n = 30
z
=
250
= .68
.2518 x 2 = .5036
= .88
.3106 x 2 = .6212
= 1.25
.3944 x 2 = .7888
= 1.77
.4616 x 2 = .9232
= 2.50
.4938 x 2 = .9876
2000 / 30
n = 50
z
=
250
2000 / 50
n = 100
z =
250
2000 / 100
n = 200
z =
250
2000 / 200
n = 400
z =
250
2000 / 400
b. A larger SAMPLE increases the probability that the SAMPLE mean will be within a
specified distance from the POPULATION mean. In the salary example, the probability
of being within
250 of  ranges from .5036 for a SAMPLE of size 30 to .9876 for a SAMPLE of size 400.
13- 142
29. a.
E( x ) = 982
 x   / n  210 / 40  33.2
z
x  
/ n

100
210 / 40
 3.01
.4987 x 2 = .9974
b.
z
x 
/ n

25
210 / 40
 .75
.2734 x 2 = .5468
c.
30. a.
The SAMPLE with n = 40 has a very high probability (.9974) of providing a SAMPLE mean within 
$100. However, the SAMPLE with n = 40 only has a .5468 probability of providing a SAMPLE
mean within  $25. A larger SAMPLE size is desirable if the  $25 is needed.
Normal distribution,
E( x ) = 166,500
 x   / n  42,000 / 100  4200
b.
z 
x    10,000  2.38
 / n 4,200
P(-2.38  z  2.38) = .9826
c.
$5000 z = 5000/4200 = 1.19
P(-1.19  z  1.19) = .7660
$2500 z = 2500/4200 = .60
P(-.60  z  .60) = .4514
$1000 z = 1000/4200 = .24
P(-.24  z  .24) = .1896
d.
Increase SAMPLE size to improve precision of the estimate. SAMPLE size of 100 only has a
.4514 probability of being within  $2,500.
 = 1.46  = .15
31.
a.
n = 30
z
x  
/ n

.03
.15 / 30
 1.10
P(1.43  x  1.49) = P(-1.10  z  1.10) = .3643(2) = .7286
b.
n = 50
z
x  
/ n

.03
.15 / 50
 1.41
13- 143
P(1.43  x  1.49) = P(-1.41  z  1.41) = .4207(2) = .8414
c.
n = 100
z
x  
/ n

.03
.15 / 100
 2.00
P(1.43  x  1.49) = P(-2  z  2) = .4772(2) = .9544
d.
32. a.
b.
A SAMPLE size of 100 is necessary.
n / N = 40 / 4000 = .01 < .05; therefore, the finite POPULATION correction factor is not necessary.
With the finite POPULATION correction factor
 x 
4000  40 8.2
N  n 
 1.29

4000  1 40
N 1 n
Without the finite POPULATION correction factor
 x   / n  1.30
Including the finite POPULATION correction factor provides only a slightly different value for  x
than when the correction factor is not used.
c.
z 
x  
1.30
2

 1.54
1.30
P(-1.54  z  1.54) = .8764
33. a.
E ( p ) = p = .40
0.40(0.60)
p(1 p)

 0.0490
100
n
b.
 p 
c.
Normal distribution with E ( p ) = .40 and  p = .0490
d.
It shows the probability distribution for the SAMPLE proportion p .
34. a.
E ( p ) = .40
 p 
0.40(0.60)
p(1 p)

 0.0346
n
200
z 
p p
p

0.03
0.0346
P(-.87  z  .87) = .6156
13- 144
 0.87
b.
z 
p p
p

0.05
 1.45
0.0346
P(-1.45  z  1.45) = .8530
35.

 p 
p(1  p)
n
p 
(0.55)(0.45)
 0.0497
100
p 
(0.55)(0.45)
 0.0352
200
p 
(0.55)(0.45)
 0.0222
500
p 
(0.55)(0.45)
 0.0157
1000
 p decreases as n increases
36. a.
p 
z 
(0.30)(0.70)
 0.0458
100
p p
p

0.04
 0.87
0.0458
P(-.87  z  .87) = 2(.3078) = .6156
Area = 0.3078 x 2 = 0.6156
b.
p 
z 
(0.30)(0.70)
 0.0324
200
p p
p

0.04
 1.23
0.0324
Area = 0.3907 x 2 = 0.7814
c.
p 
z 
(0.30)(0.70)
 0.0205
500
p p
p

0.04
 1.95
0.0205
Area = 0.4744 x 2 = 0.9488
13- 145
d.
p 
z 
(0.30)(0.70)
 0.0145
1000
p p
p
0.04

 2.76
0.0145
Area = 0.4971 x 2 = 0.9942
e.
With a larger SAMPLE, there is a higher probability p will be within  .04 of the
POPULATION proportion p.
37. a.
 
0.30(0.70)
p(1  p)

 0.0458
100
n
p
0.30
The normal distribution is appropriate because n p = 100 (.30) = 30 and n (1 - p ) = 100 (.70) =
70 are both greater than 5.
b.
P (.20  p  .40) = ?
z = .40 - .30 = 2.18
.0458
P(0  z  2.18) = .4854
Probability sought is 2(.4854) = .9708
c.
P (.25  p  .35) = ?
z = .35 - .30 = 1.09
.0458
P(-1.09  z  1.09) = .7242
38. a.
E ( p ) = .76
 p 
0.76(1 0.76)
p(1 p)
 0.0214

400
n
The normal distribution is appropriate because np = 400(.76) = 304 and n(1-p) = 400 (.24) = 96 are
both greater than 5.
13- 146
b.
z 
z 
0.79  0.76
 1.40
0.0214
0.73  0.76
0.0214
 1.40
P(0  z  1.40) = .4192
P(-1.40  z  0) = .4192
probability = 0.4192 + 0.4192 = 0.8384
c.
 p 
z 
0.76(1 0.76)
p(1 p)
 0.0156

750
n
0.79  0.76
0.0156
z 
0.73  0.76
0.0156
 1.92
P(0  z  1.92) = .4726
 1.92
P(-1.92  z  0) = .4726
probability = 0.4726 + 0.4726 = 0.9452
39. a.
Normal distribution
E ( p ) = .50
p 
b.
z 
p(1 p)

n
(.50)(1.50)
 .0206
589
p  p  .04  1.94
.0206
p
.4738 x 2 = .9476
c.
z 
p  p  .03  1.46
.0206
p
.4279 x 2 = .8558
d.
z 
p  p  .02  .97
.0206
p
.3340 x 2 = .6680
40. a.
Normal distribution
E ( p ) = 0.25
13- 147
 p 
b.
z 
(0.25)(0.75)
p(1 p)

 0.0306
n
200
0.03
 0.98
0.0306
P(0  z  .98) = .3365
probability = 0.3365 x 2 = 0.6730
c.
z 
0.05
 1.63
0.0306
P(0  z  1.63) = .4484
probability = 0.4484 x 2 = 0.8968
41. a.
Normal distribution with E( p ) = p = .25 and
p 
b.
z 
p(1 p)

n
.25(1.25)
 .0137
1000
p  p  .03  2.19
.0137
p
P(.22  p  .28) = P(-2.19  z  2.19) = .4857(2) = .9714
c.
z
p p
.25(1 .25)
500

.03
 1.55
.0194
P(.22  p  .28) = P(-1.55  z  1.55) = .4394(2) = .8788
42. a.
 
p(1 p)  0.15(0.85)

 0.0505
n
50
p
0.15
b.
P (.12  p  .18) = ?
z = .18 - .15 = .59
.0505
P(-.59  z  .59) = .4448
c.
P ( p  .10) = ?
13- 148
z = .10 - .15 = -.99
.0505
P(z  -.99) = .3389 + .5000 = .8389
43. a.
E ( p ) = 0.17
 p 
(0.17)(1  0.17)
p(1 p)
 0.01328

800
n
Normal distribution
b.
z 
z 
0.19  0.17
 1.51
0.01328
0.34  0.37
0.01328
 1.51
P(0  z  1.51) = .4345
P(-1.51  z  0) = .4345
probability = 0.4345 + 0.4345 = 0.8690
c.
 p 
z 
(0.17)(1 0.17)
p(1 p)
 0.0094

1600
n
0.19  0.17
0.0094
z 
0.15  0.17
0.0094
 2.13
P(0  z  2.13) = .4834
 2.13
P(-2.13  z  0) = .4834
probability = 0.4834 + 0.4834 = 0.9668
44.
112, 145, 73, 324, 293, 875, 318, 618
45. a.
Normal distribution
E(x ) = 3
 x 
b.
z


1.2
n
x  
/ n
 .17
50

.25
1.2 / 50
 1.47
.4292 x 2 = .8584
46. a.
Normal distribution
E ( x ) = 31.5
13- 149
 x 
b.


n
12
50
 1.70
1
 0.59
1.70
z 
P(0  z  .59) = .2224
probability = 0.2224 x 2 = 0.4448
c.
3
 1.77
1.70
z 
P(0  z  1.77) = .4616
probability = 0.4616 x 2 = 0.9232
47. a.
E ( x ) = $24.07
 x 

0.50
z

n
4.80
120
 0.44
 1.14
0.44
P(0  z  1.14) = .3729
probability = 0.3729 x 2 = 0.7458
b.
1.00
z
 2.28
0.44
P(0  z  2.28) = .4887
probability = 0.4887 x 2 = 0.9774
 = 41,979  = 5000
48.
a.
 x  5000 / 50  707
b.
z 

x    0  0
707
x

P( x > 41,979) = P(z > 0) = .50
c.
z 
x  

x


1000
 1.41
707
P(40,979  x  42,979) = P(-1.41  z  1.41) = (.4207)(2) = .8414
d.
 x  5000 / 100  500
z 
x  
x

1000
 2.00
500
P(40,979  x  42,979) = P(-2  z  2) = (.4772)(2) = .9544
13- 150
49. a.
 x 

N  n 
N 1 n
N = 2000
 x 
N = 5000
 x 
N = 10,000




 x 




2000  50 144
 20.11
2000  1 50
5000  50 144
 20.26
5000  1 50
10,000  50 144
 20.31
10,000  1 50
Note: With n / N  .05 for all three cases, common statistical practice would be to ignore
144

 20.36 for each case.
50
the finite POPULATION correction factor and
use  x
b.
N = 2000
z =
25
= 1.24
20.11
z
25
 1.23
20.26
z =
25
= 1.23
20.31
P(-1.24  z  1.24) = .7850
N = 5000
P(-1.23  z  1.23) = .7814
N = 10,000
P(-1.23  z  1.23) = .7814
All probabilities are approximately .78
50. a.
x 

n
500

n
 20
n = 500 / 20 = 25 and n = (25)2 = 625
b.
For  25,
z = 25 = 1.25
20
P(-1.25  z  1.25) = .7888
51.
Sampling distribution of x
13- 151
 
0.05

n


30
0.05


1.9

2.1
x
1.9 + 2.1 = 2
2
The area below  = 2.1 must be .95. An area of .95 in the standard normal table shows
z = 1.645.
 =

Thus,

2.1 2.0

(0.1) 30  0.33
1.645
 / 30
 1.645
Solve for 
52.
p = .305
a.
Normal distribution with E( p ) = p = .305 and
p 
b.
z 
p(1 p)

n
.305(1.305)
 .0326
200
p  p  .04  1.23
.0326
p
P(.265  p  .345) = P(-1.23  z  1.23) = .3907(2) = .7814
c.
z 
p  p  .02  .61
.0326
p
P(.285  p  .325) = P(-.61  z  .61) = .2291(2) = .4582
53.
 p 
p(1 p)

n
(0.40)(0.60)
 0.0245
400
P ( p  .375) = ?
z = .375 - .40 = -1.02
.0245
P(z  -1.02) = P(z  1.02) = .8461
13- 152
P ( p  .375) = .8461
54. a.
p(1 p)

n
p 
z 
p p
p
.05

(.71)(1 .71)
 .0243
350
 2.06
.0243
.4803 x 2 = .9606
b.
z 
p p
p

.75 .71
 1.65
.0243
Area = .4505
P ( p  .75) = 1.0000 - .9505 = .0495
55. a.
Normal distribution with E ( p ) = .15 and
 p 
b.
(0.15)(0.85)
p(1 p)

 0.0292
150
n
P (.12  p  .18) = ?
z = .18 - .15 = 1.03
.0292
P(-1.03  z  1.03) = 2(.3485) = .6970
56. a.
 p 
.25(.75)
p(1 p)

.0625
n
n
Solve for n
n
.25(.75)
 48
(.0625)2
b.
Normal distribution with E ( p ) = .25 and  x = .0625
c.
P ( p  .30) = ?
z
.30 .25
.0625
P(z  .8) = 1 - P(z  .8) = 1 - .7881 = .2119
Thus P ( p  .30) = .2119
13- 153
 .8
Chapter 8
Interval Estimation
Learning Objectives
1.
Be able to construct and interpret an interval estimate of a POPULATION mean and / or a
POPULATION proportion.
2.
Understand the concept of a sampling error.
3.
Be able to use knowledge of a sampling distribution to make probability statements about the
sampling error.
4.
Understand and be able to compute the margin of error.
5.
Learn about the t distribution and when it should be used in constructing an interval estimate for a
POPULATION mean.
6.
Be able to use the worksheets presented in the chapter as templates for constructing interval estimates.
7.
Be able to determine the size of a simple random SAMPLE necessary to estimate a POPULATION
mean and a POPULATION proportion with a specified level of precision.
8.
Know the definition of the following terms:
confidence interval
confidence coefficient
confidence level
precision
sampling error
margin of error
degrees of freedom
13- 154
Solutions:
1.
2.
a.
 x   / n  5 / 40  0.79
b.
At 95%, z / n  1.96(5 / 40)  1.55
a.
32  1.645 (6 / 50 )
32  1.4
b.
(30.6 to 33.4)
32  1.96 (6 / 50 )
32  1.66
c.
(30.34 to 33.66)
32  2.576 (6 / 50 )
32  2.19
3.
a.
(29.81 to 34.19)
80  1.96 (15 / 60 )
80  3.8
b.
(76.2 to 83.8)
80  1.96 (15 / 120 )
80  2.68
c.
(77.32 to 82.68)
Larger SAMPLE provides a smaller margin of error.
126  1.96 (s / n )
4.
1.96
16.07
n
4
n
1.96(16.07)
 7.874
4
n  62
5.
6.
a.
 x   / n  5.00 / 49  .7143
b.
1.96 / n  1.96(5.00 / 49 )  1.4
c.
34.80  1.4 or (33.40 to 36.20)
a.
x  369
b.
s = 50
13- 155
c.
369  1.96 (50/ 250 )
369  6.20 (362.8 to 375.2)
x  z.025 ( / n )
7.
3.37  1.96 (.28 / 120 )
3.37  .05
8.
a.
x  z / 2
(3.32 to 3.42)

n
12,000  1.645 (2, 200 / 245)
12,000  231
b.
(11,769 to 12,231)
12,000  1.96 (2, 200 / 245)
12,000  275
c.
(11,725 to 12,275)
12,000  2.576 (2, 200 / 245)
12,000  362
9.
(11,638 to 12,362)
d.
Interval width must increase since we want to make a statement about  with greater confidence.
a.
x
b.
s
c.
Margin of Error =1.96
xi
 13.75
n
(x i  x )2
n 1
 4.8969
s
n
 1.96
4.8969
 1.24
60
95% Confidence Interval: 13.75  1.24 or $12.51 to $14.99
10.
x  z.025
s
n
7.75  1.96
3.45
180
7.75  .50
11. a.
(7.25 to 8.25)
Using Excel we obtained a SAMPLE mean of x = 6.34 and a SAMPLE standard deviation of 2.163.
The confidence interval is shown below:
6.34  1.96 (2.163 / 50 )
13- 156
6.34  .60
The 95% confidence interval estimate is 5.74 to 6.94.
12. a.
x
xi

114
n
b.
s
 3.8 minutes
30
(x  x )2
 2.26 minutes
n 1
s
Margin of Error = z.025
c.
x  z.025
.95
b.
.90
c.
.01
d.
.05
e.
.95
f.
.85
2.26
30
 .81 minutes
(2.99 to 4.61)
14. a.
1.734
b.
-1.321
c.
3.365
d.
-1.761 and +1.761
e.
-2.048 and +2.048
15. a.
 1.96
s
n
3.8  .81
13. a.
n
x  xi / n 
80
 10
8
84
( x  x )2
 3.464

n 1
8 1
b.
s 
c.
With 7 degrees of freedom, t.025 = 2.365
x  t.025 (s / n )
13- 157
10  2.365 (3.464 / 8 )
10  2.90
16. a.
(7.10 to 12.90)
17.25  1.729 (3.3 / 20 )
17.25  1.28
b.
(15.97 to 18.53)
17.25  2.09 (3.3 / 20 )
17.25  1.54
c.
(15.71 to 18.79)
17.25  2.861 (3.3 / 20 )
17.25  2.11
(15.14 to 19.36)
At 90% , 80  t.05 (s / n ) with df = 17 t.05 = 1.740
17.
80  1.740 (10 / 18 )
80  4.10
At 95%, 80  2.11 (10 /
(75.90 to 84.10)
18) with df = 17 t.05 = 2.110
80  4.97
18. a.
x
xi  18.96  $1.58
n
12
b.
s
.239
(x  x )2

 .1474
12 1
n 1
c.
t.025 = 2.201
(75.03 to 84.97)
x  t.025 (s / n )
1.58  2.201 (.1474 / 12 )
1.58  .09
19.
(1.49 to 1.67)
x  xi / n  6.53 minutes
s
( x  x )2
 0.54 minutes
n 1
x  t.025 (s / n )
13- 158
6.53  2.093 (0.54 / 20 )
6.53  .25
20. a.
(6.28 to 6.78)
22.4  1.96 (5 / 61)
22.4  1.25
b.
(21.15 to 23.65)
With df = 60, t.025 = 2.000
22.4  2 (5 / 61)
22.4  1.28
c.
Confidence intervals are essentially the same regardless of whether z or t is used.
x
21.
(21.12 to 23.68)
xi
864

n
s
 $108
8
654
(x  x )2

 9.6658
8 1
n 1
t.025 = 2.365
x  t.025 (s / n )
108  2.365 (9.6658 / 8)
108  8.08
22. a.
b.
(99.92 to 116.08)
Using Excel, x = 6.86 and s = 0.78
x  t.025 (s / n ) t.025 = 2.064
df = 24
6.86  2.064 (0.78 / 25)
6.86  0.32
(6.54 to 7.18)
(1.96)2 (25) 2
23.
n  z.025 
E2
24. a.
Planning value of  = Range/4 = 36/4 = 9
2
n  z.025 
E2
2
b.
2
2
52
(1.96)2 (9)2
32
 96.04 Use n  97
 34.57
Use n  35
13- 159
c.
n
25. a.
n
b.
n
(1.96)2 (9)2
22
(1.96)2 (6.82)2
(1.5)2
(1.645)2 (6.82)2
22
n z 
E2
2
26. a.
b.
n
c.
n
27. a.
n
b.
n
c.
n
2
(1.96)2 (9400)2
(200)2
(1.96)2 (2,000)2
(500)2
(1.96)2 (2,000)2
(200) 2
(1.96)2 (2,000)2
(100)2
2
 79.41
Use n  80
 31.47
Use n  32
 339.44 Use 340
(1000)2
(500)2
n z 
E2
Use n  78
(1.96)2 (9400)2
(1.96)2 (9400)2
2
28. a.
 77.79
 1357.78
Use 1358
 8486.09
Use 8487
 61.47
Use n  62
 384.16
Use n  385
 1536.64
Use n  1537
(1.645)2 (220)2
 52.39 Use 53
(50)2
(1.96)2 (220)2
b.
n
c.
n
d.
Must increase SAMPLE size to increase confidence.
29. a.
n
b.
n
(50)2
 74.37
(2.576)2 (220)2
(50)2
(1.96)2 (6.25)2
22
(1.96)2 (6.25)2
12
Use 75
 128.47 Use 129
 37.52
Use n  38
 150.06
Use n  151
13- 160
(1.96)2 (7.8)2
30.
n
31. a.
p = 100/400 = 0.25
b.
c.
 58.43
22
Use n  59
0.25(0.75)
p(1 p)

 0.0217
n
400
p  z.025
p(1  p)
n
.25  1.96 (.0217)
.25  .0424
32. a.
.70  1.645
(.2076 to .2924)
0.70(0.30)
800
.70  .0267
b.
.70  1.96
(.6733 to .7267)
0.70(0.30)
800
.70  .0318
n
33.
2
z.025
p(1 p)
(.6682 to .7318)

2
(1.96) (0.35)(0.65)
E2
34.
 349.59
Use n  350
(0.05)2
Use planning value p = .50
n
35. a.
(1.96)2 (0.50)(0.50)
(0.03)2
0.6904(1 0.6904)
p(1 p)
 1.645
 0.0267
n
814
1.645
c.
0.6904  0.0267
b.
Use n  1068
p = 562/814 = 0.6904
b.
36. a.
 1067.11
(0.6637 to 0.7171)
p = 152/346 = .4393
p 
p(1 p)

n
.4393(1.4393)
 .0267
346
13- 161
p  z.025 p
.4393  1.96(.0267)
.4393  .0523
p(1 p)
, p = 182/650 = .28
n
p  1.96
37.
(.3870 to .4916)
.28  1.96
(0.28)(0.72)
650
0.28  0.0345
38. a.
(0.2455 to 0.3145)
(0.26)(0.74)
p(1 p)
 1.96
 0.0430
n
400
1.96
b.
0.26  0.0430
c.
n
39. a.
n
(0.2170 to 0.3030)
1.962 (0.26)(0.74)
(0.03)2
2
z.025
p(1 p)

 821.25
(1.96)2 (.33)(1.33)
E2
b.
n
2
z.005
p(1 p)

(2.576)2 (.33)(1.33)
41.
 1630.19
(.03)2
E
b.
 943.75 Use 944
(.03)2
2
40. a.
Use n  822
p = 255/1018 = 0.2505
1.96
(0.2505)(1 0.2505)
= 0.0266
1018
p 
p(1 p)

n
.16(1.16)
 .0102
1285
Margin of Error = 1.96  p = 1.96(.0102) = .02
.16  1.96  p
.16  .02
z
42.
n
2
.025
(.14 to .18)
p(1 p)
E2
13- 162
Use 1631
September
n
October
n
November
n
Pre-Election n 
43. a.
n 
1.962 (.50)(1.50)
.042
1.962 (.50)(1.50)
.032
1.962 (.50)(1.50)
1.962 (.50)(1.50)
p = 445/601 = 0.7404
c.
0.7404  1.96
b.
z.025
s
n
 1.96
(0.7054 to 0.7755)
20, 500
400
 2009
x  z.025 (s / n )
50,000  2009
45. a.
Use n  601
(0.7404)(0.2596)
601
0.7404  0.0350
44. a.
 9604
.012
b.
 1067.11 Use 1068
 2401
.022
1.962 (0.5)(1 0.5)
 600.25
(0.04) 2
 600.25 Use 601
(47,991 to 52,009)
x  z.025 (s / n )
252.45  1.96 (74.50 / 64 )
252.45  18.25 or $234.20 to $270.70
b.
Yes. the lower limit for the POPULATION mean at Niagara Falls is $234.20 which is greater
than
$215.60.
46. a.
Using Excel, x = 49.8 minutes
b.
Using Excel, s = 15.99 minutes
c.
x  1.96 (s / n )
49.8  1.96 (15.99 / 200 )
49.8  2.22
47. a.
(47.58 to 52.02)
Using Excel, we find x = 16.8 and s = 4.25
13- 163
With 19 degrees of freedom, t.025 = 2.093
x  2.093 (s / n )
16.8  2.093 (4.25 / 20 )
16.8  1.99
b.
(14.81 to 18.79)
Using Excel, we find x = 24.1 and s = 6.21
24.1  2.093 (6.21 / 20 )
24.1  2.90
c.
48. a.
(21.2 to 27.0)
16.8 / 24.1 = 0.697 or 69.7% or approximately 70%
x  xi / n 
132
 13.2
10
547.6
( x  x )2

 7.8
9
n 1
b.
s
c.
With d f = 9, t.025 = 2.262
x  t.025 (s / n )
13.2  2.262 (7.8 / 10 )
13.2  5.58
d.
The  5.58 shows poor precision. A larger SAMPLE size is desired.
49.
n
50.
n
51.
n
n
52.
(7.62 to 18.78)
n
1.962 (45)2
102
 77.79
(2.33)2 (2.6)2
 36.7
Use n  37
 61.47
Use n  62
12
(1.96)2 (8) 2
22
(2.576)2 (8) 2
22
(1.96)2 (675)2
1002
Use n  78
 106.17
Use n  107
 175.03
Use n  176
13- 164
53. a.
p  1.96
p(1 p)
, p = 212/450 = .47
n
(0.47)(0.53)
450
0.47  1.96
0.47  0.0461
b.
0.47  2.576
(0.4239 to 0.5161)
(0.47)(0.53)
450
0.47  0.06
c.
54. a.
b.
c.
55. a.
The margin of error becomes larger.
p = 200/369 = 0.5420
(0.5420)(0.4580)
p(1 p)
 1.96
 0.0508
n
369
0.5420  0.0508
(0.4912 to 0.5928)
1.96
p = 504 / 1400 = .36
b.
1.96
56. a.
n
b.
n
57. a.
(0.41 to 0.53)
(0.36)(0.64)
 0.0251
1400
(2.33)2 (0.70)(0.30)
(0.03)2
(2.33)2 (0.50)(0.50)
(0.03)2
 1266.74
Use n  1267
 1508.03
Use n  1509
p = 110 / 200 = 0.55
0.55  1.96
(0.55)(0.45)
200
.55  .0689
b.
58. a.
b.
n
(.4811 to .6189)
(1.96)2 (0.55)(0.45)
(0.05) 2
 380.32
Use n  381
p = 340/500 = .68
p 
p(1 p)

n
.68(1.68)
 .0209
500
13- 165
p  z.025 p
.68  1.96(.0209)
.68  .0409
59. a.
n
(.6391 to .7209)
(1.96)2 (0.3)(0.7)
(0.02)2
 2016.84
b.
p = 520/2017 = 0.2578
c.
p  1.96
p(1 p)
n
(0.2578)(0.7422)
2017
0.2578  1.96
0.2578  0.0191
60. a.
b.
c.
Use n  2017
(0.2387 to 0.2769)
p = 618 / 1993 = .3101
p  1.96
p(1 p)
1993
0.3101  1.96
(0.3101)(0.6899)
1993
.3101  .0203
(.2898 to .3304)
n 
z
z 2 p(1 p)
E2
(1.96)2 (0.3101)(0.6899)
(0.01)2
 8218.64
Use n  8219
No; the SAMPLE appears unnecessarily large. The .02 margin of error reported in part (b) should
provide adequate precision.
13- 166
Chapter 9
Hypothesis Testing
Learning Objectives
1.
Learn how to formulate and test hypotheses about a POPULATION mean and/or a POPULATION proportion.
2.
Understand the types of errors possible when conducting a hypothesis test.
3.
Be able to determine the probability of making various errors in hypothesis tests.
4.
Know how to compute and interpret p-values.
5.
Be able to use the Excel worksheets presented in the chapter as templates for conducting hypothesis
tests about POPULATION means and proportions.
6.
Know the definition of the following terms:
null hypothesis
alternative hypothesis
type I error
type II error
critical value
level of significance
one-tailed test
two-tailed test
p-value
13- 167
Solutions:
1.
a.
H0:   600
Manager’s claim.
Ha:  > 600
2.
b.
We are not able to conclude that the manager’s claim is wrong.
c.
The manager’s claim can be rejected. We can conclude that  > 600.
a.
H0:   14
Ha:  > 14
3.
4.
b.
There is no statistical evidence that the new bonus plan increases sales volume.
c.
The research hypothesis that  > 14 is supported. We can conclude that the new bonus plan
increases the mean sales volume.
a.
H0:  = 32
Specified filling weight
Ha:   32
Overfilling or underfilling exists
b.
There is no evidence that the production line is not operating properly. Allow the production process
to continue.
c.
Conclude   32 and that overfilling or underfilling exists. Shut down and adjust the production
line.
a.
H0:   220
Ha:  < 220
5.
Research hypothesis
Research hypothesis to see if mean cost is less than $220.
b.
We are unable to conclude that the new method reduces costs.
c.
Conclude  < 220. Consider implementing the new method based on the conclusion that it lowers
the mean cost per hour.
a. The Type I error is rejecting H0 when it is true. In this case, this error occurs if the researcher
concludes that the mean newspaper-reading time for individuals in management positions is greater
than the national average of 8.6 minutes when in fact it is not.
b. The Type II error is accepting H0 when it is false. In this case, this error occurs if the researcher
concludes that the mean newspaper-reading time for individuals in management positions is less than
or equal to the national average of 8.6 minutes when in fact it is greater than 8.6 minutes.
6.
a.
H0:   1
The label claim or assumption.
Ha:  > 1
b.
Claiming  > 1 when it is not. This is the error of rejecting the product’s claim when the claim is
true.
13- 168
c.
7.
a.
Concluding   1 when it is not. In this case, we miss the fact that the product is not meeting its
label specification.
H0:   8000
Ha:  > 8000
8.
Research hypothesis to see if the plan increases average sales.
b.
Claiming  > 8000 when the plan does not increase sales. A mistake could be implementing the
plan when it does not help.
c.
Concluding   8000 when the plan really would increase sales. This could lead to not
implementing a plan that would increase sales.
a.
H0:   220
Ha:  < 220
9.
b.
Claiming  < 220 when the new method does not lower costs. A mistake could be implementing
the method when it does not help.
c.
Concluding   220 when the method really would lower costs. This could lead to not
implementing a method that would lower costs.
a.
z = -1.645
Reject H0 if z < -1.645
b.
z 
x   9.46  10

 1.91
2 / 50
s/ n
Reject H0; conclude Ha is true.
10. a.
z = 2.05
Reject H0 if z > 2.05
x   16.5  15

 1.36
7 / 40
s/ n
b.
z 
c.
Using the cumulative normal probability table, the area to the right of z = 1.36 is 1 - .9131 = .0869.
Thus, the p-value is .0869
d.
Do not reject H0
11.
Reject H0 if z < -1.645
a.
z 
x    22  25

 2.50
s / n 12 / 100
Reject H0
b.
z 
x    24  25

 .83
s / n 12 / 100
Do Not Reject H0
13- 169
c.
z 
x    23.5  25

 1.25
s / n 12 / 100
Do Not Reject H0
d.
z 
x    22.8  25

 1.83
s / n 12 / 100
Reject H0
12. a.
p-value = 1 - .9656 = .0344
Reject H0
b.
p-value = 1 - .6736 = .3264
Do not reject H0
c.
p-value = 1 - .9332 = .0668
Do not reject H0
d.
z = 3.09 is the largest table value with 1 - .999 = .001 area in tail. For z = 3.30, the p-value is less
than .001 or approximately 0. Reject H0.
e.
Since z is to the left of the mean and the rejection region is in the upper tail, the p-value is the area to
the right of z = -1.00. Because the standard normal distribution is symmetric, the area to the right of
z = -1.00 is the same as the area to the left of z = 1.00. Thus, the p-value = .8413. Do not reject H0.
13. a.
H0:   1056
Ha:  < 1056
b.
Reject H0 if z < -1.645
c.
910 1056
x 
 1.83
z  s / n0  1600 / 400
d.
Reject H0 and conclude that the mean refund of “last minute” filers is less than $1056.
e.
p-value = 1.0000 - .9664 = .0336
14. a.
z.01 = 2.33
Reject H0 if z > 2.33
x 
z
c.
Reject H0; conclude the mean television viewing time per day is greater than 6.70.
15. a.
s/ n

7.25  6.70
b.
2.5 / 200
 3.11
A summary of the SAMPLE DATA is shown below:
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
100
$9300
$4500
H0:   10,192
Ha:  < 10,192
Reject H0 if z < –1.645.
13- 170
z 
x  
s/ n

9300  10,192
 1.98
4500 / 100
b.
The area to the left of z = -1.98 is the same as the area to the right of z = 1.98. Using the cumulative
normal probability table, the area to the right of z = 1.98 is 1 - .9761 = .0239. Thus, the p-value =
.0239.
c.
The manager can use the SAMPLE results to conclude that the mean sales price of used cars at the
dealership is less than the mean sales price of used cars using the national average. The manager may
want to explore the possible reasons for the lower prices at the dealership. Perhaps sales personnel
are making excessive price concessions to close the sales. Perhaps the dealership is missing out on a
portion of the late model used car market that might warrant used cars with higher prices. The
manager’s judgment and insight might suggest other reasons the dealership is experiencing the lower
mean sales prices.
16.
A summary of the SAMPLE DATA is shown below:
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
30
27,500
1000
H0:   28,000
Ha:  < 28,000
Reject H0 if z < -1.645
z
x   0 27,500  28,000

 2.74
s/ n
1000 / 30
Reject H0; Tires are not meeting the at least 28,000 design specification.
Because the standard normal distribution is symmetric, the area to the left of z = -2.74 is the same as
the area to the right of z = 2.74. Using the cumulative normal probability table, the area to the right
of z = 2.74 is 1 - .9969 = .0031. Thus, the p-value = .0031.
17. a.
H0:   
Ha:  < 13
b.
z.01 = 2.33
Reject H0 if z < -2.33
x 
z
d.
Reject H0; conclude Canadian mean internet usage is less than 13 hours per month.
s/ n

10.8 13
c.
9.2 / 145
 2.88
Note: p-value = .002
18. a.
H0:   5.72
13- 171
Ha:  > 5.72

5.98  5.72
c.
d.
p-value < ; reject H0. Conclude teens in Chicago have a mean expenditure greater than 5.72.
19. a.
z
x
 2.12
s / n 1.24 / 102
p-value = 1.0000 - .9830 = .0170
b.
H0:   181,900
Ha:  < 181,900
x
z
c.
p-value = 1.0000 - .9983 = .0017
d.
p-value < ; reject H0. Conclude mean selling price in South is less than the national mean selling
price.
20. a.
s/ n

166, 400 181, 900
b.
33, 500 / 40
 2.93
H0:   37,000
Ha:  > 37,000
x
z
c.
p-value = 1.0000 - .9292 = .0708
d.
p-value > ; do not reject H0. Cannot conclude POPULATION mean salary has increased in June 2001.
21. a.
b.
s/ n

38,100  37, 000
b.
5200 / 48
 1.47
Reject H0 if z < -1.96 or z > 1.96
z
x   10.8  10

 2.40
s / n 2.5 / 36
Reject H0; conclude Ha is true.
22. a.
Reject H0 if z < -2.33 or z > 2.33
x   14.2  15

 1.13
5 / 50
s/ n
b.
z
c.
p-value = (2) (1 - .8708) = .2584
d.
Do not reject H0
23.
Reject H0 if z < -1.96 or z > 1.96
a.
z 
x    22  25

 2.68
s / n 10 / 80
Reject H0
13- 172
b.
c.
d.
x    27  25

 1.79
s / n 10 / 80
x   23.5  25

z 
 1.34
s / n 10 / 80
x    28  25

z 
 2.68
s / n 10 / 80
z 
Do not reject H0
Do not reject H0
Reject H0
24. a.
p-value = 2(1 - .9641) = .0718
Do not reject H0
b.
p-value = 2(1 - .6736) = .6528
Do not reject H0
c.
p-value = 2(1 - .9798) = .0404
Reject H0
d.
approximately 0 Reject H0
e.
p-value = 2(1 - .8413) = .3174
25. a.
Do not reject H0
z.025 = 1.96
Reject H0 if z < -1.96 or z > 1.96
x 
z
c.
Do not reject H0. Cannot conclude a change in the POPULATION mean has occurred.
d.
p-value = 2(1.000 - .9382) = .1236
26. a.
s/ n

38.5  39.2
b.
4.8 / 112
 1.54
H0:  = 8
Ha:   8
Reject H0 if z < -1.96 or if z > 1.96
z
x   0  7.5  8

 1.71
s / n 3.2 / 120
Do not reject H0; cannot conclude the mean waiting time differs from eight minutes.
b.
27. a.
Using the cumulative normal probability table, the area to the left of z = -1.71 is 1 - .9564 = .0436.
Thus, the p-value = 2 (.0436) = .0872.
H0:  = 16
Continue production
Ha:   16
Shut down
Reject H0 if z < -1.96 or if z > 1.96
b.
z
x  0 16.32  16

 2.19
s/ n
.8 / 30
Reject H0 and shut down for adjustment.
13- 173
c.
z
x   0 15.82  16

 1.23
s/ n
.8 / 30
Do not reject H0; continue to run.
d.
For x = 16.32, p-value = 2 (1 - .9857) = .0286
For x = 15.82, p-value = 2 (1 - .8907) = .2186
28.
A summary of the SAMPLE DATA is shown below:
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
45
2.39
.20
H0:  = 2.2
Ha:   2.2
Reject H0 if z < -2.33 or if z > 2.33
z
x   0 2.39  2.20

 6.37
s/ n
.20 / 45
Reject H0 and conclude 2.2 - minute standard is not being met.
H0:  = 15.20
29.
Ha:   15.20
Reject H0 if z < -1.96 or if z > 1.96
z
x  0 14.30  15.20

 1.06
s/ n
5 / 35
Do not reject H0; the SAMPLE does not provide evidence to conclude that there has been a change.
p-value = 2 (1 - .8554) = .2892
30. a.
H0:  = 1075
Ha:   1075
x 
z
c.
p-value = 2(1.0000 - .9236) = .1528
d.
Do not reject H0. Cannot conclude a change in mean amount of charitable giving.
31. a.
s/ n

1160 1075
b.
840 / 200
 1.43
With 15 degrees of freedom, t.05 = 1.753
Reject H0 if t > 1.753
13- 174
b.
32. a.
t
x  0
s/ n

11 10
3 / 16
 1.33
Do not reject H0
x  ∑ xi / n = 108 / 6 = 18
10
∑( x  x)2
 1.414

n 1
61
b.
s
b.
d.
Reject H0 if t < -2.571 or t > 2.571
x  0
18  20
t

 3.46
s / n 1.414 / 6
e.
Reject H0; conclude Ha is true.
33.
Reject H0 if t < -1.721
a.
t 
13  15
 1.17
8 / 22
Do not reject H0
b.
t 
11.5  15
 2.05
8 / 22
Reject H0
c.
t 
15  15
0
8 / 22
Do not reject H0
d.
t 
19  15
 2.35
8 / 22
Do not reject H0
34.
Excel's TDIST function with 15 degrees of freedom was used to determine each p-value.
a.
p-value = .01
Reject H0
b.
p-value = .10
Do not reject H0
c.
p-value = .03
Reject H0
d.
p-value = .15
Do not reject H0
e.
p-value = .003
Reject H0
35. a.
H0:   3.00
Ha:   3.00
b.
t.025 = 2.262
Reject H0 if t < -2.262 or if t > 2.262
13- 175
c.
x
xi
28

n
 2.80
10
.44
(x  x )2

 .70
n 1
10 1
d.
s
e.
t
f.
Do not reject H0; cannot conclude the POPULATION mean earnings per share has changed.
g.
t.10 = 1.383
x
s/ n

2.80  3.00
.70 / 10
 .90
p-value is greater than .10 x 2 = .20
Actual p-value = .3916
36. a.
A summary of the SAMPLE DATA is shown below:
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
25
84.5
14.5
H0:  = 90
Ha:   90
Degrees of freedom = 24
t.025 = 2.064
Reject H0 if z < -2.064 or if z > 2.064
t
x  0  84.5  90

 1.90
s / n 14.5 / 25
Do not reject H0; we cannot conclude the mean household expenditure in Corning differs from the
U.S. mean expenditure.
b.
37. a.
Using Excel's TDIST function, the p-value corresponding to t = -1.90 is approximately .07.
H0:   55
Ha:  > 55
With 7 degrees of freedom, reject H0 if t < 1.895.
x  ∑ xi / n = 475 / 8 = 59.38
s
∑( x  x)2
123.87

 4.21
7
n 1
13- 176
t
x   0 59.38  55

 2.94
s/ n
4.21 / 8
Reject H0; the mean number of hours worked per week exceeds 55.
b.
38. a.
Using Excel's TDIST function, the p-value corresponding to t = 2.94 is approximately .011.
H0:   4000
Ha:   4000
b.
t.05 = 2.160
13 degrees of freedom
Reject H0 if t < -2.160 or if t > 2.160
x 
t
d.
Do not reject H0; Cannot conclude that the mean cost in New City differs from $4000.
e.
With 13 degrees of freedom
s/ n

4120  4000
c.
275 / 14
 1.63
t.05 = 1.771
t.10 = 1.350
1.63 is between 1.350 and 1.771. Therefore the p-value is between .10 and .20.
39. a.
H0:   280
Ha:  > 280
b.
286.9 - 280 = 6.9 yards
c.
t.05 = 1.860 with 8 degrees of freedom
d.
t
e.
Reject H0; The POPULATION mean distance of the new driver is greater than the USGA
approved driver..
f.
t.05 = 1.860
x 
s/ n

286.9  280
10 / 9
 2.07
t.025 = 2.306
p-value is between .025 and .05
Actual p-value = .0361
40.
H0:   2
13- 177
Ha:  > 2
With 9 degrees of freedom, reject H0 if t > 1.833
x = 2.4 s = .5164
x    2.4  2
t  s / n0  .5164 / 10  2.45
Using Excel's TDIST function, the p-value corresponding to t = 2.45 is approximately .02.
Reject H0 and claim  is greater than 2 hours. For cost estimating purposes, consider using more
than 2 hours of labor time.
41. a.
b.
Reject H0 if z > 1.645
p 
z 
42. a.
b.
.50(.50)
.0354
200
p p
p

.57.50
 1.98
.0354
Reject H0 if z < -1.96 or z > 1.96
p 
z 
.20(.80)
.02
400
p p
p

.175.20
 1.25
.02
c.
p-value = 2(1 - .8944) = .2122
d.
Do not reject H0.
43.
Reject H0
Reject H0 if z < -1.645
a.
p 
z 
.75(.25)
.0250
300
p p
p

.68.75
 2.80
.025
p-value = 1 - .8974 = .0026
Reject H0.
b.
z 
p p
p

.72.75
 1.20
.025
13- 178
p-value = 1 - .8849 = .1151
Do not reject H0.
c.
z 
p p
p

.70.75
 2.00
.025
p-value = 1 - .8772 = .0228
Reject H0.
d.
z 
p p
p

.77.75
.80
.025
In this case, the p-value is the area to the left of z = .80. Thus, the p-value = .7881.
Do not reject H0.
44. a.
H0: p  .40
Ha: p > .40
b.
Reject H0 if z > 1.645
c.
p = 188/420 = .4476
p(1 p)
.40(1.40)
p 

 .0239
n
420
z 
d.
45.
p p
p

.4476 .40
 1.99
.0239
Reject H0. Conclude that there has been an increase in the proportion of users receiving more than
ten e-mails per day.
H0: p  .64
Ha: p < .64
Reject H0 if z < –1.645.
p = 52/100 = .2667
z 
.52.64
.64(.36)
100
 2.5
Reject H0; conclude that less than 64% of the shoppers believe that the supermarket ketchup is as
good as the national name brand ketchup.
46. a.
p = 285/460 = .62
13- 179
b.
H0: p  0.50
Ha: p > 0.50
Reject H0 if z > 2.33
p  p0

p0 (1 p0 )
n
z
.57.50
 3.13
.50(1.50)
500
Reject H0; a Burger King taste preference should be expressed by over 50% of the consumers.
c.
47.
Yes; the statistical evidence shows Burger King fries are preferred. The give-away was a good way
to get potential customers to try the new fries.
A summary of the SAMPLE DATA is shown below:
SAMPLE Size
200
Number of College
Students
42
H0: p = .25
Ha: p  .25
Reject H0 if z < -1.645 or if z > 1.645
p = 42/200 = .21
p 
z
.25(.75)
.0306
200
p  p0
p

.21.25
 1.31
.0306
Do not reject H0; the magazine’s claim of 25% cannot be rejected.
p-value = 2 (1 - .9049) = .1902
48. a.
b.
p = 67/105 = .6381 (about 64%)
p(1 p)

n
p 
z 
p p
p

.50(1.50)
 .0488
105
.6381.50
 2.83
.0488
c.
p-value = 2(1.0000 - .9977) = .0046
d.
p-value < .01, reject H0. Conclude preference is for the four ten-hour day schedule.
13- 180
49. a.
H0: p = .44
Ha: p  .44
b.
p = 205/500 = .41
p(1 p)

n
p 
z 
p p
p

.44(1.44)
 .0222
500
.41.44
 1.35
.0222
p-value = 2(1.0000 - .9115) = .1770
Do not reject H0. Cannot conclude that there has been a change in the proportion of repeat customers.
c.
p = 245/500 = .49
z 
p p
p

.49 .44
 2.25
.0222
p-value = 2(1.0000 - .9878) = .0244
Reject H0. conclude that the proportion of repeat customers has changed. The point estimate of the
percentage of repeat customers is now 49%.
50. a.
p(1 p)

n
p 
z 
p p
p

.75(1.75)
 .025
300
.72 .75
 1.20
.025
b.
p-value = 1.0000 - .8849 = .1151
c.
Do not reject H0. Cannot conclude the manager's claim is wrong based on this SAMPLE evidence.
51. a.
H0: p  .047
Ha: p < .047
b.
p = 35/1182 = .0296
c.
p 
z 
d.
.047(1.047)
 .0062
1182
p p
p

.0296 .047
 2.82
.0062
p-value = 1.0000 - .9976 = .0024
13- 181
e.
52. a.
p-value < , reject H0. The error rate for Brooks Robinson is less than the overall error rate.
H0:   45,250
Ha:  > 45,250
x
47, 000  45, 250
b.
z
c.
p-value = 1.0000 - .9966 = .0034
d.
53. a.
s/ n

6300 / 95
 2.71
p-value < ; reject H0. New York City school teachers must have a higher mean annual salary.
H0:   30
Ha:  < 30
Reject H0 if z < –2.33
z
x   0 29.5  30

 1.96
s / n 1.8 / 50
Do not reject H0; the SAMPLE evidence does not support the conclusion that the Buick LeSabre
provides less than 30 miles per gallon.
b.
p-value = 1 - .9963 = .0037
c.
x  z.05
x  z.05

n
1.5
45
 27.6.44
Interval is 27.16 to 28.04
54.
H0:   25,000
Ha:  > 25,000
Reject H0 if z > 1.645
z
x  0 26, 000  25, 000

 2.26
s/ n
2, 500 / 32
p-value = 1.0000 - .9881 = .0119
Reject H0; the claim should be rejected. The mean cost is greater than $25,000.
55.
H0:  = 120
Ha:   120
13- 182
With n = 10, use a t distribution with 9 degrees of freedom.
Reject H0 if t < -2.262 or of t > 2.262
x
xi
 118.9
n
s
(x  x )2
 4.93
n 1
x 
118.9 120
t  s / n0  4.93 / 10  .71
Do not reject H0; the results do not permit rejection of the assumption that  = 120.
56. a.
H0:  = 550
Ha:   550
Reject H0 if z < -1.96 or if z > 1.96
z
x  0 562  550

 1.80
s/ n
40 / 36
Do not reject H0; the claim of $550 per month cannot be rejected.
b.
c.
p-value = 2(1 - .9641) = .0718
s
x  z.025
n
x  1.96
40
36
 562  13
Interval is 549 to 575
Do not reject H0 since 550 is in the above interval.
57. a.
A summary of the SAMPLE DATA is shown below:
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
30
80
20
H0:   72
Ha:  > 72
x  72 80  72
z  s / n  20 / 30  2.19
13- 183
p-value = 1 - .9857 = .0143
b.
Since p-value < .05, reject H0; the mean idle time exceeds 72 minutes per day.
H0: p  .79
58.
Ha: p < .79
Reject H0 if z < -1.645
p = 360/500 = .72
z
p  p0  .72.79  3.84
 (.79)(.21)
p
500
Reject H0; conclude that the proportion is less than .79 in 1995.
59.
A summary of the SAMPLE DATA is shown below:
SAMPLE Size
400
Number that Work
with Coworkers
304
H0: p  .72
Ha: p > .72
Reject H0 if z > 1.645
p = 304/400 = .76
p  p0  .76.72  1.87

(.76)(.24)
p
400
Reject H0: conclude that the proportion of workers at Trident is greater.
z
60. a.
b.
The research is attempting to see if it can be concluded that less than 50% of the working
POPULATION hold jobs that they planned to hold.
.50(.50)
.0136
1350
.41.50
z 
 6.62
.0136
p 
Reject H0 if z < -2.33
Reject H0; it can be concluded that less than 50% of the working POPULATION hold jobs that they
planned to hold. The majority hold jobs due to chance, lack of choice, or some other unplanned
reason.
13- 184
.75(.25)
.0229
356
p = 313/356 = .88
 p 
61.
z 
.88.75
.0229
 5.68
Reject H0; conclude p 0. DATA suggest that 88% of women wear shoes that are at least one size too
small.
62. a.
b.
p = 355/546 = .6502
p(1 p)

n
p 
z 
p p
p

.67(1.67)
 .0201
546
.6502 .67
 .98
.0201
c.
p-value = 2(1.0000 - .8365) = .3270
d.
p-value  , do not reject H0. The assumption of two-thirds cannot be rejected.
63. a.
b.
p = 330/400 = .825
p(1 p)

n
p 
z 
p p
p

.78(1.78)
 .0207
400
.825 .78
 2.17
.0207
c.
p-value = 2(1.0000 - .9850) = .03
d.
p-value < , reject H0. Arrival rate has changed from 78%. Service appears to be improving.
64. a.
b.
p = 44/125 = .352
p(1 p)

n
p 
z 
p p
p

.47(1.47)
 .0446
125
.352 .47
 2.64
.0446
c.
p-value = 1.0000 - .9959 = .0041
d.
Reject H0; conclude that the proportion of food containing pesticide residues has been reduced.
13- 185
Chapter 10
Comparisons Involving Means
Learning Objectives
1.
Be able to develop interval estimates and conduct hypothesis tests about the difference between the
means of two POPULATIONs.
2.
Know the properties of the sampling distribution of the difference between two means x1  x2 .
3.
Be able to use the t distribution to conduct statistical inferences about the difference between the
means of two normal POPULATIONs with equal variances.
4.
Understand the concept and use of a pooled variance estimate.
5.
Learn how to analyze the difference between the means of two POPULATIONs when the
SAMPLEs are independent and when the SAMPLEs are matched.
6.
Understand how the analysis of variance procedure can be used to determine if the means of more
than two POPULATIONs are equal.
7.
Know the assumptions necessary to use the analysis of variance procedure.
8.
Understand the use of the F distribution in performing the analysis of variance procedure.
9.
Know how to set up an ANOVA table and interpret the entries in the table.
10.
Be able to use the Excel worksheets and tools presented to conduct comparisons involving means.
13- 186
Solutions:
1.
a.
x1  x2 = 13.6 - 11.6 = 2
b.
sx  x 
1
2
s2
s2
(2.2)2 (3)2
 0.595

 2 
50
35
n1 n2
1
2  1.645(.595)
2  .98 or 1.02 to 2.98
c.
2  1.96(.595)
2  1.17 or 0.83 to 3.17
2.
a.
x1  x2 = 22.5 - 20.1 = 2.4
b.
s2 
c.
sx  x  s
(n  1)s2  (n  1)s2
1
1
2
2
n1  n 2  2
2
1
2
9(2.5)2  7(2)2

10  8  2
 5.27
F1 1 I
F1 1I
 J
 5.27G  J 1.09
G
Hn n K H10 8 K
1
2
16 degrees of freedom, t.025 = 2.12
2.4  2.12(1.09)
2.4  2.31 or .09 to 4.71
3.
a.
x1  ∑ xi / n  54 / 6  9
x2  ∑ xi / n  42 / 6  7
b.
s1 
∑(x  x )2

n1 1
18
 1.90
6 1
s2 
∑(x  x )2

n2 1
16
 1.79
6 1
c.
x1  x2 = 9 - 7 = 2
d.
s2 
e.
With 10 degrees of freedom, t.025 = 2.228
(n  1)s2  (n  1)s2
1
1
2
n1  n2  2
2
5(1.90)2  5(1.79)2

662
13- 187
 3.41
2
sx  x  s
1
2
F1 1 I
F1 1I
 J
 3.41G J 1.07
G
Hn n K H6 6 K
1
2
2  2.228(1.07)
2  2.37 or -0.37 to 4.37
4.
a.
x1  x2 = 1.58 - 0.98 = $0.60
b.
s
x1  x2

s2
1
n1
s2

2
n2
x1  x2  z  / 2 s x  x
1

.122 .082

 .021
50
42
2
.60 ± 1.96(.021)
.60 ± .04 or .56 to .64
5.
a.
22.5 - 18.6 = 3.9 miles per day
b.
x1  x2  z  / 2 s x  x
1
sx  x 
1
2
s2
1
n1
2
s2


2
n2
(8.4)2 (7.4)2

 1.58
50
50
22.5 - 18.6  1.96(1.58)
3.9  3.1 or 0.6 to 7.0
6.
LA
6.72
2.374
x
s
x1  x2  z  / 2 s x  x
1
sx  x 
1
2
Miami
6.34
2.163
2
(2.374)2 (2.163)2
s2

 0.454
 2 
50
50
n1 n2
s2
1
6.72 - 6.34  1.96(.454)
.38  .89 or -.51 to 1.27
7.
a.
x1  x2 = 14.9 - 10.3 = 4.6 years
b.
s
x1  x2

s2
1
n1

s2
2
n2

5.22 3.82

 .66
100
85
13- 188
z.025 sx  x = 1.96(.66) = 1.3
1
c.
2
x1  x2  z.025 sx  x
1
2
4.6  1.3 or 3.3 to 5.9
8.
a.
x1  x2 = 45,700 - 44,500 = 1,200
b.
Pooled variance
s2 
7(700)2 11(850)2
 632, 083
18
sx  x 
1
 362.88

2
With 18 degrees of freedom t.025 = 2.101
1200  2.101(362.88)
1200  762 or 438 to 1962
9.
c.
POPULATIONs are normally distributed with equal variances.
a.
n1 = 10
n2 = 8
x1 = 21.2
x2 = 22.8
s1 = 2.70
s2 = 3.55
x1  x2 = 21.2 - 22.8 = -1.6
Kitchens are less expensive by $1,600.
b.
x1  x2  z  / 2 s x  x
1
2
Degrees of freedom = n1 + n2 - 2 = 16
t.05 = 1.746
s2 
9(2.70)2  7(3.55)2
10  8  2
sx  x 
1
2
 9.63
 1.47
-1.6  1.746(1.47)
13- 189
-1.6  2.57 or -4.17 to +.97
10. a.
b.
c.
x1 = 17.54
x2 = 15.36
x1  x2 = 17.54 - 15.36 = $2.18 per hour greater for union workers.
(n  1)s2  (n  1)s2 14(2.24)2  19(1.99)2
1
2
2
 4.41
s2  1

15  20  2
n1  n 2  2
x1  x2  t / 2 sx  x
1
2
sx  x 
1
 0.72
2
17.54 15.36  t / 2 (.72) = 2.18  t / 2 (.72)
Note: Using Excel's TINV function, t.025 = 2.035.
2.18  2.035(.72)
2.18  1.47 or 0.71 to 3.65
d.
11. a.
There does appear to be a difference in the mean wage rate for these two groups.
sx  x 
1
z
2
s2
s2
(5.2)2 (6)2
 1.18

 2 
40
50
n1 n2
1
(25.2  22.8)
 2.03
1.18
Reject H0 if z > 1.645
Reject H0; conclude Ha is true and    > 0.
b.
12. a.
p-value = 1.0000 - .9788 = .0212
sx  x 
1
z
2
s2
1
n1
s2
(8.4)2 (7.6)2
 1.31

 2 
80
70
n2
( x1  x2 )  (1   2 ) (104  106)  0

 1.53
sx  x
1.31
1
2
Reject H0 if z < -1.96 or z > 1.96
Do not reject H0
b.
13. a.
p-value = 2(1.0000 - .9370) = .1260
x1  x2 =1.4 – 1.0 = 0.4
13- 190
(n  1)s2  (n  1)s2
s2 
1
1
2
n1  n 2  2
sx  x 
1
2

7(.4)2  6(.6)2
872
 0.2523
 0.26
2
With 13 degrees of freedom. t.025 = 2.16
Reject H0 if t < -2.16 or t > 2.16
t
( x1  x2 )  (1   2 )  0.4  1.54
 0.26
sx  x
1
2
Do not reject H0
14. a.
H0: µ1 - µ2 = 0
Ha:     0
b.
Reject H0 if z < -1.96 or if z > 1.96
c.
s
x1  x2
z
s2

1
n1
s2
2

n2
(16.8)2 (15.2)2

 1.79
150
175
 x1  x2   0 39.3  35.4  0
sx  x
1
15.

2

 2.18
1.79
d.
Reject H0; conclude the POPULATION means differ.
e.
p-value = 2(1.0000 - .9854) = .0292
A summary of the SAMPLE DATA is shown below:
Airport
Miami
Los Angeles
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
50
6.34
2.163
50
6.72
2.374
We will treat Los Angeles as POPULATION 1
H0:   


Ha:   
z
( x1  x2 )  (1   2 )  (6.72  6.34)  0

 0.84
s2 s2
(2.374)2  (2.163)2

50
50
Since 0.84 < z.05 = 1.64 we cannot reject H0
13- 191
16.
H0: µ1 - µ2 = 0
Ha:     0
Reject H0 if z < -1.96 or if z > 1.96
z
(x1  x2 )  0 



2
2
(40  35)
(9)2 (10)2

36
49
 2.41
Reject H0; customers at the two stores differ in terms of mean ages.
p-value = 2(1.0000 - .9920) = .0160
17. a.
POPULATION 1 is
supplier A.
POPULATION 2 is
supplier B.
b.
H0:     0
Stay with supplier A
Ha:    > 0
Change to supplier B
Reject H0 if z > 1.645
z
(x1  x2 )  (1  2 )
2
2

(14 12.5)  0
 2.68
(3)2 (2)2

50
30
p-value = 1.0000 - .9963 = .0037
Reject H0; change to supplier B.
18.
A summary of the SAMPLE DATA is shown below:
Employees
Male
Female
SAMPLE
SAMPLE Size SAMPLE Mean Standard Deviation
44
$12.34
$0.92
32
$11.59
$0.76
We will treat the male employees as POPULATION 1.
H0:     0
Ha:    > 0
Reject H0 if z > 2.33
z
(x1  x2 )  (1  2 ) (12.34 11.59)  0

 3.88
s2 s2
(.92)2 (.76)2

44  1332 192
Reject H0; wage discrimination appears to exist.
19. a.
H0:    = 0
Ha:     0
Degrees of freedom = n1 + n2 - 2 = 24
t.025= 2.064
Reject H0 if t < -2.064 or if t > 2.064
x1 = 30.6
x2 = 27
s1 = 3.35
s2 = 2.64
s2
sx  x 
1
s2

2
t 
(30.6  27)  0
1.20
(3.35)2

(2.64) 2
12
14
 1.20
 3.0
Reject H0; the POPULATION means differ.
b.
Public Accountants have a higher mean.
x1  x2 = 30.6 - 27 = 3.6, or $3,600.
20. a.
H0:     0
Ha:     0
2
 x  x  
1
z
2
2
2
 2  2.5  2.5  .36
112
 x1  x2   0

 x x

1

84
69.95  69.56
.36
 1.08
2
b.
p-value = 2(1.0000 - .8599) = .2802
c.
Do no reject H0. Cannot conclude that there is a difference between the POPULATION mean
scores for the two golfers.
21. a.
H0:     0
Ha:     0
b.
t.025 = 2.021
df = n1 + n2 - 2 = 22 + 20 - 2 = 40
Reject H0 if t < -2.021 or if t > 2.021
13- 193
c.
s2 
n1 1 s12  n2 1 s22
n 1  n2  2
sx  x 
1
t
 x1  x2   0
sx  x
1

(22 1)(.8)2  (20 1)(1.1)2
22  20  2
2.5  2.1
 .9108
 .2948


2

 1.36
.2948
2
d.
Do not reject H0. Cannot conclude that a difference between POPULATION mean exists.
e.
Using Excel's TDIST function, p-value = .18.
22. a.
H0:     0
Ha:    > 0
b.
t.05 = 1.711
df = n1 + n2 - 2 = 16 + 10 - 2 = 24
Reject H0 if t > 1.711
c.
s2 
n1 1 s12  n2 1 s22
n 1  n2  2
sx  x 
1
t

2
 x1  x2   0
sx  x
1


(16 1)(.64)2  (10 1)(.75)2
16 10  2
 .2755

6.82  6.25
 .4669
 2.07
.2755
2
d.
Reject H0. Conclude that the consultant with the more experience has the higher POPULATION
mean rating.
e.
Using Excel's TDIST function, p-value = .025.
23. a.
1, 2, 0, 0, 2
b.
d  ∑ di / n  5 / 5  1
c.
sd 
d.
With 4 degrees of freedom, t.05 = 2.132
∑(d  d )2

n 1
4
1
51
Reject H0 if t > 2.132
t
d  d
sd / n

1 0
 2.24
1/ 5
13- 194
Using Excel's TDIST function, p-value = .04.
Reject H0; conclude d > 0.
24. a.
3, -1, 3, 5, 3, 0, 1
b.
d  ∑ di / n  14 / 7  2
c.
sd 
d.
d =2
e.
With 6 degrees of freedom t.025 = 2.447
26
∑(d  d )2
 2.082

7 1
n 1

2  2.447 2.082 / 7

2  1.93 or .07 to 3.93
25.
Difference = rating after - rating before
H0: d  0
Ha: d > 0
With 7 degrees of freedom, reject H0 if t > 1.895
d = .625 and sd = 1.3025
t
d  d
.625  0

 1.36
sd / n 1.3025 / 8
p-value is greater than .10
Do not reject H0; we cannot conclude that seeing the commercial improves the mean potential to
purchase.
26.
Differences: .20, .29, .39, .02, .24, .20, .20, .52, .29, .20
d  ∑ di / n  2.55 /10  .255
sd 
∑(d  d )2
 .1327
n 1
With df = 9, t.025 = 2.262
13- 195
d  t.025
sd
n
.1327
.255  2.262
10
.255  .095 or .16 to .35
27.
Differences: 8, 9.5, 6, 10.5, 15, 9, 11, 7.5, 12, 5
d = 93.5/10 = 9.35 and sd = 2.954
t.025 = 2.262
e
j
9.35  2.262 2.954 / 10  9.35  2.11
Interval estimate is 7.24 to 11.46
28.
H0: d = 0
Ha: d  0
Reject H0 if t < -2.365 or if t > 2.365
df = 7
Differences -.01, .03, -.06, .16, .21, .17, -.09, .11
d  ∑ di / n  .52 / 8  .065
sd 
t
∑(d  d )2
 .1131
n 1
d 0

sd
n
.065
 1.63
.1131
8
Do not reject H0. Cannot conclude that the POPULATION means differ.
29.
Using matched SAMPLEs, the differences are as follows: 4, -2, 8, 8, 5, 6, -4, -2, -3, 0, 11, -5, 5, 9, 5
H0: d  0
Ha: d > 0
d = 3 and sd = 5.21
t
d  d
sd / n

3 0
 2.23
5.21 / 15
Using Excel's TDIST function, p-value = .02.
13- 196
With 14 degrees of freedom, reject H0 if t > 1.761 or if p-value <  = .05.
Reject H0. Conclude that the POPULATION of readers spends more time, on average, watching
television than reading.
30. a.
Difference = Price deluxe - Price Standard
H0: d = 10
Ha: d  10
With 6 degrees of freedom, reject H0 if t < -2.447 or if t > 2.447; alternatively, reject H0
p-value <  = .05.
d = 8.86 and sd = 2.61
t
d  d 8.86  10

 1.16
sd / n 2.61 / 7
Using Excel's TDIST function, p-value = .29.
Do not reject H0; we cannot reject the hypothesis that a $10 price differential exists.
b.
d  t / 2
sd
n
8.86  2.447
2.61
7
8.86  2.41 or 6.45 to 11.27
31. a.
H0: 1 - 2 = 0
Ha: 1 - 2  0
With df = 11, t.025 = 2.201
Reject H0 if t < -2.201 or if t > 2.201; alternatively, reject H0 if p-value <  = .05.
Calculate the difference, di, for each stock.
d  ∑ di / n  85 / 12  7.08
sd 
∑(d  d )2
 3.34
n 1
13- 197
if
t
x  
sd / n
 7.34
p-value  0
Reject H0; a decrease in P/E ratios is being projected for 1998.
b.
d  t.025
sd
n
7.08  2.201
3.34
12
7.08  2.12 or 4.96 to 9.21
32. a.
x = (30 + 45 + 36)/3 = 37
k

SSTR  ∑ nj x j  x
j 1
 = 5(30 - 37) + 5(45 - 37) + 5(36 - 37) = 570
2
2
2
2
MSTR = SSTR /(k - 1) = 570/2 = 285
k
b.
SSE  ∑(n j 1)s j = 4(6) + 4(4) + 4(6.5) = 66
2
j 1
MSE = SSE /(nT - k) = 66/(15 - 3) = 5.5
c.
F = MSTR /MSE = 285/5.5 = 51.82
F.05 = 3.89 (2 degrees of freedom numerator and 12 denominator)
Since F = 51.82 > F.05 = 3.89, we reject the null hypothesis that the means of the three
POPULATIONs are equal.
d.
Source of Variation
Treatments
Error
Total
33. a.
Sum of Squares
570
66
636
Degrees of Freedom
2
12
14
Mean Square
285
5.5
x = (153 + 169 + 158)/3 = 160
k

SSTR  ∑ nj x j  x
j 1
 = 4(153 - 160) + 4(169 - 160) + 4(158 - 160) = 536
2
2
MSTR = SSTR /(k - 1) = 536/2 = 268
13- 198
2
2
F
51.82
k
b.
SSE  ∑(n j 1)s j = 3(96.67) + 3(97.33) +3(82.00) = 828.00
2
j 1
MSE = SSE /(nT - k) = 828.00 /(12 - 3) = 92.00
c.
F = MSTR /MSE = 268/92 = 2.91
F.05 = 4.26 (2 degrees of freedom numerator and 9 denominator)
Since F = 2.91 < F.05 = 4.26, we cannot reject the null hypothesis.
d.
Source of Variation
Treatments
Error
Total
34. a.
x
Sum of Squares
536
828
1364
4(100)  6(85)  5(79)
Degrees of Freedom
2
9
11
Mean Square
268
92
F
2.91
 87
15
k

SSTR  ∑ nj x j  x
j 1
 = 4(100 - 87) + 6(85 - 87) + 5(79 - 87) = 1,020
2
2
2
2
MSTR = SSB /(k - 1) = 1,020/2 = 510
k
b.
SSE  ∑(n j 1)s j = 3(35.33) + 5(35.60) + 4(43.50) = 458
2
j 1
MSE = SSE /(nT - k) = 458/(15 - 3) = 38.17
c.
F = MSTR /MSE = 510/38.17 = 13.36
F.05 = 3.89 (2 degrees of freedom numerator and 12 denominator)
Since F = 13.36 > F.05 = 3.89 we reject the null hypothesis that the means of the three POPULATIONs
are equal.
d.
Source of Variation
Treatments
Error
Total
Sum of Squares
1020
458
1478
Degrees of Freedom
2
12
14
Mean Square
510
38.17
F
13.36
Source of Variation
Treatments
Error
Total
Sum of Squares
1200
300
1500
Degrees of Freedom
3
60
63
Mean Square
400
5
F
80
35. a.
b.
F.05 = 2.76 (3 degrees of freedom numerator and 60 denominator)
Since F = 80 > F.05 = 2.76 we reject the null hypothesis that the means of the 4 POPULATIONs are equal.
13- 199
36. a.
Source of Variation
Treatments
Error
Total
b.
Sum of Squares
120
216
336
Degrees of Freedom
2
72
74
Mean Square
60
3
F
20
F.05 = 3.12 (2 numerator degrees of freedom and 72 denominator)
Since F = 20 > F.05 = 3.12, we reject the null hypothesis that the 3 POPULATION means are equal.
37.
Manufacturer 1
23
6.67
SAMPLE Mean
SAMPLE Variance
Manufacturer 2
28
4.67
Manufacturer 3
21
3.33
x = (23 + 28 + 21)/3 = 24
k

SSTR  ∑ nj x j  x
j 1
 = 4(23 - 24) + 4(28 - 24) + 4(21 - 24) = 104
2
2
2
2
MSTR = SSTR /(k - 1) = 104/2 = 52
k
SSE  ∑(n j 1)s j = 3(6.67) + 3(4.67) + 3(3.33) = 44.01
2
j 1
MSE = SSE /(nT - k) = 44.01/(12 - 3) = 4.89
F = MSTR /MSE = 52/4.89 = 10.63
F.05 = 4.26 (2 degrees of freedom numerator and 9 denominator)
Since F = 10.63 > F.05 = 4.26 we reject the null hypothesis that the mean time needed to mix a batch
of material is the same for each manufacturer.
38.
Superior
5.75
1.64
SAMPLE Mean
SAMPLE Variance
Peer
5.5
2.00
Subordinate
5.25
1.93
x = (5.75 + 5.5 + 5.25)/3 = 5.5
k

SSTR  ∑ nj x j  x
j 1
 = 8(5.75 - 5.5) + 8(5.5 - 5.5) + 8(5.25 - 5.5) = 1
2
2
2
MSTR = SSTR /(k - 1) = 1/2 = .5
k
SSE  ∑(n j 1)s j = 7(1.64) + 7(2.00) + 7(1.93) = 38.99
2
j 1
MSE = SSE /(nT - k) = 38.99/21 = 1.86
13- 200
2
F = MSTR /MSE = 0.5/1.86 = 0.27
F.05 = 3.47 (2 degrees of freedom numerator and 21 denominator)
Since F = 0.27 < F.05 = 3.47, we cannot reject the null hypothesis that the means of the three
POPULATIONs are equal; thus, the source of information does not significantly affect the
dissemination of the information.
39.
Marketing
Managers
5
.8
SAMPLE Mean
SAMPLE Variance
Marketing
Research
4.5
.3
Advertising
6
.4
x = (5 + 4.5 + 6)/3 = 5.17
k

SSTR  ∑ nj x j  x
j 1
 = 6(5 - 5.17) + 6(4.5 - 5.17) + 6(6 - 5.17) = 7.00
2
2
2
2
MSTR = SSTR /(k - 1) = 7.00/2 = 3.5
k
SSE  ∑(n j 1)s j = 5(.8) + 5(.3) + 5(.4) = 7.50
2
j 1
MSE = SSE /(nT - k) = 7.50/(18 - 3) = .5
F = MSTR /MSE = 3.5/.50 = 7.00
F.05 = 3.68 (2 degrees of freedom numerator and 15 denominator)
Since F = 7.00 > F.05 = 3.68, we reject the null hypothesis that the mean perception score is the same
for the three groups of specialists.
40.
Real Estate
Agent
67.73
117.72
SAMPLE Mean
SAMPLE Variance
Architect
61.13
180.10
Stockbroker
65.80
137.12
x = (67.73 + 61.13 + 65.80)/3 = 64.89
k

SSTR  ∑ nj x j  x
j 1
 = 15(67.73 - 64.89) + 15(61.13 - 64.89) + 15(65.80 - 64.89) = 345.47
2
2
2
MSTR = SSTR /(k - 1) = 345.47/2 = 172.74
k
SSE  ∑(n j 1)s j = 14(117.72) + 14(180.10) + 14(137.12) = 6089.16
2
j 1
13- 201
2
MSE = SSE /(nT - k) = 6089.16/(45-3) = 144.98
F = MSTR /MSE = 172.74/144.98 = 1.19
F.05 = 3.22 (2 degrees of freedom numerator and 42 denominator)
Since F = 1.19 < F.05 = 3.22, we cannot reject the null hypothesis that the job stress ratings are the
same for the three occupations.
41.
The Excel output is shown below:
SUMMARY
Groups
Count
Sum
Average
Variance
Banking
12
183
15.25
29.8409
Financial Services
7
128
18.2857
16.5714
Insurance
10
163
16.3
15.1222
ANOVA
Source of Variation
SS
df
MS
Between Groups
40.7732
2
20.3866
Within Groups
563.7786
26
21.6838
Total
604.5517
28
F
P-value
0.9402
0.4034
Since the p-value = 0.4034 >  = 0.05, we cannot reject the null hypothesis that that the mean
price/earnings ratio is the same for these three groups of firms.
42.
x x z
1
2
s2
s2
.05
45, 000  35, 000 1.645
(4000)2

60
(3500)2
80
10,000  1066 or 8,934 to 11,066
43.
H0: 1 - 2 = 0
Ha: 1 - 2  0
Reject H0 if z < -1.96 or if z > 1.96
13- 202
F crit
3.3690
z
(x1  x2 )  (1  2 )  (4.27  3.38)  0  3.99

(1.85)2 (1.46)2
2 2

120  100
Reject H0; a difference exists with system B having the lower mean checkout time.
44. a.
H0: 1 - 2  0
Ha: 1 - 2 > 0
Reject H0 if z > 1.645
b.
n1= 30
n2 = 30
x1 = 16.23
x2 = 15.70
s1 = 3.52
s2 = 3.31
(3.52)2
sx  x 
1
z
2
30

(3.31)2
30
 0.88
( x1  x2 )  0 (16.23  15.70)

 0.59
sx  x
0.88
1
2
Do not reject H0; cannot conclude that the mutual funds with a load have a greater mean rate of
return.
Load funds 16.23% ; no load funds 15.7%
c.
At z = 0.59, Area = 0.2224
p-value = 1.0000 - .7224 = 0.2776
45.
Difference = before - after
H0: d  0
Ha: d > 0
With 5 degrees of freedom, reject H0 if t > 2.015
d = 6.167 and sd = 6.585
t
d   d  6.167  0  2.29

sd / n 6.585 / 6
Using Excel's TDIST function, p-value = .035.
Reject H0; conclude that the program provides weight loss.
46. a.
POPULATION 1 - 1996
13- 203
POPULATION 2 - 1997
H0: 1 - 2  0
Ha: 1 - 2 > 0
b.
d  ∑ di / n  1.74 / 14  0.12
sd 
∑(d  d )2
 0.33
n 1
Degrees of freedom = 13; t.05 = 1.771
Reject H0 if t > 1.771 or if p-value <  = .05
t 
d 0
0.12

 1.42
sd / n 0.33 / 14
Using Excel's TDIST function, p-value = .09.
Do not reject H0. The SAMPLE of 14 companies shows earnings are down in the fourth quarter
by a mean of 0.12 per share. However, DATA does not support the conclusion that mean
earnings for all companies are down in 1997.
47. a.
Area 1
96
50
SAMPLE Mean
SAMPLE Variance
Area 2
94
40
s2  s2 50  40
2
pooled estimate = 1

 45 
2
2


estimate of standard deviation of x1  x2 
t
 4.74
x1  x2 96  94

 .42
4.74
4.74
t.025 = 2.447 (6 degrees of freedom)
Since t = .42 < t.025 = 2.477, the means are not significantly different.
b.
x = (96 + 94)/2 = 95
k

SSTR  ∑ nj x j  x
j 1
 = 4(96 - 95) + 4(94 - 95) = 8
2
2
2
MSTR = SSTR /(k - 1) = 8 /1 = 8
13- 204
k
SSE  ∑(n j 1)s j = 3(50) + 3(40) = 270
2
j 1
MSE = SSE /(nT - k) = 270 /(8 - 2) = 45
F = MSTR /MSE = 8 /45 = .18
F.05 = 5.99 (1 degree of freedom numerator and 6 denominator)
Since F = .18 < F.05 = 5.99 the means are not significantly different.
c.
Area 1
96
50
SAMPLE Mean
SAMPLE Variance
Area 2
94
40
Area 3
83
42
x = (96 + 94 + 83)/3 = 91
k

SSTR  ∑ nj x j  x
j 1
 = 4(96 - 91) + 4(94 - 91) + 4(83 - 91) = 392
2
2
2
2
MSTR = SSTR /(k - 1) = 392 /2 = 196
k
SSE  ∑(n j 1)s j = 3(50) + 3(40) + 3(42) = 396
2
j 1
MSTR = SSE /(nT - k) = 396 /(12 - 3) = 44
F = MSTR /MSE = 196 /44 = 4.45
F.05 = 4.26 (2 degrees of freedom numerator and 6 denominator)
Since F = 4.45 > F.05 = 4.26 we reject the null hypothesis that the mean asking prices for all three
areas are equal.
48.
The Excel output for these DATA is shown below:
SUMMARY
Groups
Count
Sum
Average
Variance
Sport Utility
Small Pickup
10
10
586
488
58.6
48.8
20.9333
17.7333
Full-Size Pickup
10
601
60.1
22.1
ANOVA
Source of Variation
SS
df
13- 205
MS
F
P-value
F crit
Between Groups
753.2667
2
376.6333
Within Groups
546.9
27
20.2556
1300.167
29
Total
18.5941
8.37E-06
3.3541
Because the p-value = .000 <  = .05, we can reject the null hypothesis that the mean resale value is
the same. It appears that the mean resale value for small pickup trucks is much smaller than the mean
resale value for sport utility vehicles or full-size pickup trucks.
49.
Food
52.25
22.25
SAMPLE Mean
SAMPLE Variance
Personal Care
62.25
15.58
Retail
55.75
4.92
x = (52.25 + 62.25 + 55.75)/3 = 56.75
k

SSTR  ∑ nj x j  x
j 1
 = 4(52.25 - 56.75) + 4(62.25 - 56.75) + 4(55.75 - 56.75) = 206
2
2
2
2
MSTR = SSTR /(k - 1) = 206 /2 = 103
k
SSE  ∑(n j 1)s j = 3(22.25) + 3(15.58) + 3(4.92) = 128.25
2
j 1
MSE = SSE /(nT - k) = 128.25 /(12 - 3) = 14.25
F = MSTR /MSE = 103 /14.25 = 7.23
F.05 = 4.26 (2 degrees of freedom numerator and 9 denominator)
Since F = 7.23 exceeds the critical F value, we reject the null hypothesis that the mean age of
executives is the same in the three categories of companies.
50.
Physical
Therapist
63.7
164.68
Lawyer
50.0
124.22
SAMPLE Mean
SAMPLE Variance
Cabinet
Maker
69.1
105.88
Systems
Analyst
61.2
136.62
x 50.0  63.7  69.1 61.2  61
4
k

SSTR  ∑ nj x j  x
j 1
 = 10(50.0 - 61) + 10(63.7 - 61) + 10(69.1 - 61) + 10(61.2 - 61) = 1939.4
2
2
13- 206
2
2
2
MSTR = SSTR /(k - 1) = 1939.4 /3 = 646.47
k
SSE  ∑(n j 1)s j = 9(124.22) + 9(164.68) + 9(105.88) + 9(136.62) = 4,782.60
2
j 1
MSE = SSE /(nT - k) = 4782.6 /(40 - 4) = 132.85
F = MSTR /MSE = 646.47 /132.85 = 4.87
F.05 = 2.87 (3 degrees of numerator and 36 denominator)
Since F = 4.87 > F.05 = 2.87, we reject the null hypothesis that the mean job satisfaction rating is the
same for the four professions.
51.
The Excel output for these DATA is shown
below: Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
West
South
10
10
1080
917
108
91.7
565.5556
384.9
Northeast
10
1211
121.1
826.3222
ANOVA
MS
F
Between Groups
Source of Variation
4338.867
2
2169.4333
3.6630
Within Groups
15991
27
592.2593
20329.87
29
Total
SS
df
P-value
0.0391
F crit
3.3541
Because the p-value = .0391 <  = .05, we can reject the null hypothesis that the mean rate for the
three regions is the same.
52.
The Excel output is shown below:
SUMMARY
Groups
Count
Sum
Average
Variance
West
10
600
60
52.0933
South
10
454
45.4
57.9067
North Central
10
473
47.3
45.9444
Northeast
10
521
52.1
37.8511
ANOVA
13- 207
MS
F
Between Groups
Source of Variation
SS
1271
df
3
423.6667
8.7446
Within Groups
1744.16
36
48.4489
Total
3015.16
39
P-value
0.0002
F crit
2.8663
Since the p-value = 0.0002 <  = 0.05, we can reject the null hypothesis that that the mean base
salary for art directors is the same for each of the four regions.
53.
The Excel output for these DATA is shown below:
SUMMARY
Groups
Count
Sum
Average
Variance
Wide Receiver
Guard
15
13
111.2
79.4
7.4133
6.1077
0.7841
0.5474
Offensive Tackle
12
84.7
7.0583
0.6408
ANOVA
MS
F
Between Groups
Source of Variation
12.4020
SS
df
2
6.2010
9.3283
Within Groups
24.5957
37
0.6647
Total
36.9978
39
P-value
0.0005
F crit
3.2519
Because the p-value = .0005 <  = .05, we can reject the null hypothesis that the mean rating for the
three positions is the same. It appears that wide receivers and tackles have a higher mean rating than
guards.
54.
The output obtained using Excel's Anova: Single factor tool is shown.
13- 208
SUMMARY
Groups
UK
US
Europe
Count
22
22
22
Sum
265.14
329.05
442.3
Average
Variance
12.0518
1.9409
14.9568
3.4100
20.1045
6.4308
df
MS
365.8766
3.9272
ANOVA
Source of Variation
Between Groups
Within Groups
SS
731.7533
247.4164
Total
979.1696
2
63
F
93.1637
P-value
0.0000
F crit
3.1428
65
Since the p-value = 0.0000 is less than  = .05, we can reject the null hypothesis that the mean
download time is the same for Websites located in the United kingdom, United States and Europe.
13- 209
Chapter 11
Comparisons Involving Proportions and
A Test of Independence
Learning Objectives
1.
Know the properties of the sampling distribution of the difference between two proportions ( p1  p2 ) .
2.
Be able to develop interval estimates and conduct hypothesis tests about the difference between the proportions
of two POPULATIONs.
3.
Be able to conduct a goodness of fit test when the POPULATION is hypothesized to have a multinomial
probability distribution.
4.
For a test of independence, be able to set up a contingency table, determine the observed and expected
frequencies, and determine if the two variables are independent.
5.
Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence.
6.
Be able to use the Excel worksheets presented as templates for interval estimates and hypothesis tests involving
proportions.
13- 210
Solutions:
1.
a.
p1  p2 = .48 - .36 = .12
b.
s
p1  p2

p1 (1 p1 )
p2 (1 p2 )


0.48(0.52) 0.36(0.64)
 0.0373

400
300
0.12  1.645(0.0373)
0.12  0.0614 or 0.0586 to 0.1814
c.
0.12  1.96(0.0373)
0.12  0.0731 or 0.0469 to 0.1931
2.
a.
p
n1 p 1  n 2 p 2
n1  n2

200(0.22)  300(0.16)
 0.184
200  300
F1 1 I
G
H200  300JK 0.0354
sp  p  (0.184)(0.816)
1
2
Reject H0 if z > 1.645
z
(.22 .16)  0
 1.69
.0354
Reject H0
b.
3.
p-value = (1.0000 - .9545) = .0455
p1 = 220/400 = 0.55
p2 = 192/400 = 0.48
sp  p 
1
2
0.55(0.45) 0.48(0.52)
 0.0353

400
400
p1  p2 ± 1.96 sp  p
1
2
0.55 - 0.48  1.96(0.0353)
0.07  0.0691 or 0.0009 to 0.1391
7% more executives are predicting an increase in full-time jobs. The confidence interval shows the
difference may be from 0% to 14%.
13- 211
4.
a.
p1 = 682/1082 = .6303 (63%)
p2 = 413/1008 = .4097 (41%)
p1  p2 = .6303 - .4097 = .2206 (22%)
b.


p1  p2

p1 (1 p1 )
p1  p2 1.96 p  p
1

p2 (1 p2 )

.6303(1.6303) .4097(1.4097)

 .0213
1082
1008

(0.58)(0.42) (0.43)(0.57)

2
.2206  1.96(.0213)
.2206  .0418 or .1788 to .2624
p1  p2  z / 2 sp  p
5.
1
p1 (1 p1 )
sp  p 
1

2

2
p2 (1 p2 )
n1  0.57(1710)  975
 sp  p 
1
2
n2  0.08(1710)  137
(0.58)(0.42) (0.43)(0.57)
 0.045

975
137
0.58 - 0.43 ± 1.96(0.045)
0.15 ± 0.09 or 0.07 to 0.24
6.
a.
p1 = 279/300 = 0.93
p2 = 255/300 = 0.85
b.
H0: p1 - p2 = 0
Ha: p1 - p2  0
Reject H0 if z < -1.96 or if z > 1.96
p 
279  255
300  300
 0.89
sp  p  (0.89)(0.11)
1
2
z
F1 1 I
G
H300  300JK 0.0255
p1  p2  0 0.93  0.85

 3.13
sp  p
0.0255
1
2
13- 212
Using Excel's NORMSDIST function, p-value = .002.
c.
Reject H0; women and men differ on this question.
p1  p2  1.96sp  p
1
sp  p 
1
2
2
(0.93)(0.07) (0.85)(0.15)
 0.0253

300
300
0.93 - 0.85  1.96(0.0253)
0.08  0.05 or 0.03 to 0.13
95% confident that 3% to 13% more women than men agree with this statement.
H0: p1  p2
7.
Ha: p1 > p2
z 
b
( p1  p2 )  p 1  p2
sp  p
1
g
2
n p n p
1545(0.675)  1691(0.608)
p 1 1 2 2 
 0.64
n1  n2
1545  1691
1 1
1
1
sp  p  p(1 p)

 ( 0.64 )( 0.36)

n n
1545 1691
1
2
1
F I
G
H JK
2
z
F
G
H
IJ 0.017
K
(0.675  0.608)  0
 3.94
0.017
Since 3.94 > z.05 = 1.645, we reject H0
p-value  0
Conclusion: The proportion of men that feel that the division of housework is fair is greater than the
proportion of women that feel that the division of housework is fair.
8.
a.
A summary of the SAMPLE DATA is shown below:
Respondents
Men
Women
SAMPLE Size
200
300
Number
Cooperating
110
210
H0: p1 - p2 = 0
Ha: p1 - p2  0
Reject H0 if z < -1.96 or if z > 1.96
13- 213
p 
110  210
 0.64
200  300
sp  p  (0.64)(0.36)
1
2
F1 1 I
G
H200  300JK 0.0438
p1  110 / 200  0.55
z
p2  210 / 300  0.70
b
( p1  p2 )  p 1  p2
sp  p
1
g (0.55  0.70)  0  3.42
0.0438
2
Reject H0; there is a difference between response rates for men and women.
b.
0.15  1.96
0.55(0.45) 0.70(0.30)

200
300
.15  .0863 or .0637 to .2363 Greater response rate for women.
9.
a.
H0: p1 - p2 = 0
Ha: p1 - p2  0
Reject H0 if z < -1.96 or if z > 1.96
p 
63  60
 0.3514
150  200
F1 1 I
G
H150  200JK 0.0516
sp  p  (0.3514)(0.6486)
1
2
p1  63 / 150  0.42
z
b
( p1  p2 )  p 1  p2
sp  p
1
p2  60 / 200  0.30
g (0.42  0.30)  0  2.33
0.0516
2
p-value = 2(1.0000 - .9901) = .0198
Reject H0; there is a difference between the recall rates for the two commercials.
b.
(0.42  0.30)  1.96
0.42(58) 0.30(0.70)

150
200
.12  .10 or .02 to .22
10.
p
n 1 p 1  n2 p2
n1  n2

sp  p  p(1 p)
1
2
232(.815)  210(.724)
 .7718
232  210

1

13- 214

1
 .04
z
 p1  p2   0
sp  p
1

.815 .724
 2.28
.04
2
p-value = 2(1.0000 - .9887) = .0226
p-value < .05, reject H0. The POPULATION proportions differ. NYSE is showing a greater
proportion of stocks below their 1997 highs.
11.
H0: p1 - p2  0
Ha: p1 - p2  0
p
n1 p1  n2 p2
n1  n2

240(.40)  250(.32)
 .3592
240  250


sp  p  p(1 p)

 p1  p2   0
.40 .32
1
z
2
sp  p
1

2
1

1
 .0434
 1.85
.0434
p-value = 1.0000 - .9678 = .0322
p-value < .05, reject H0. The proportion of users at work is greater in Washington D.C.
12.
Expected frequencies:
e1 = 200 (.40) = 80, e2 = 200 (.40) = 80
e3 = 200 (.20) = 40
Actual frequencies:
f1 = 60, f2 = 120, f3 = 20
(60 - 80)2 (120 - 80) 2 (20 - 40)2
+
+
2 =
80
80
40
=
400
80
+
1600
+
80
400
40
= 5 + 20 + 10
= 35
2
 .01
= 9.21034 with k - 1 = 3 - 1 = 2 degrees of freedom
Since  2 = 35 > 9.21034 reject the null hypothesis. The POPULATION proportions are not as stated
in the null hypothesis.
13- 215
13.
Expected frequencies:
e1 = 300 (.25) = 75, e2 = 300 (.25) = 75
e3 = 300 (.25) = 75, e4 = 300 (.25) = 75
Actual frequencies:
f1 = 85, f2 = 95, f3 = 50, f4 = 70
2 =
(85 - 75)2
(95 - 75) 2
+
75
100
=
75
(50 - 75)2
+
75
400
+
75
625
+
75
(70 - 75)2
+
75
75
25
+
75
1150
=
75
= 15.33
2
 .05
= 7.81473 with k - 1 = 4 - 1 = 3 degrees of freedom
Since 2 = 15.33 > 7.81473 reject H0
We conclude that the proportions are not all equal.
14.
H0 = pABC = .29, pCBS = .28, pNBC = .25, pOther = .18
Ha = The proportions are not pABC = .29, pCBS = .28, pNBC = .25, pOther = .18
Expected frequencies:
300 (.29) = 87, 300 (.28) = 84
300 (.25) = 75, 300 (.18) = 54
e1 = 87, e2 = 84, e3 = 75, e4 = 54
Actual frequencies:
f1 = 95, f2 = 70, f3 = 89, f4 = 46
2
 .05
= 7.81 (3 degrees of freedom)

2
=
(95 - 87)2
87
= 6.87
+
(70 - 84)2
84
+
(89 - 75)2
+
(46 - 54)2
75
54
Do not reject H0; there is no significant change in the viewing audience proportions.
13- 216
15.
Observed
Frequency
(fi)
177
135
79
41
36
38
506
Hypothesized
Proportion
0.30
0.20
0.20
0.10
0.10
0.10
Totals:
Category
Brown
Yellow
Red
Orange
Green
Blue
Expected
Frequency
(ei)
151.8
101.2
101.2
50.6
50.6
50.6
(fi - ei)2 / ei
4.18
11.29
4.87
1.82
4.21
3.14
29.51
2
 .05
= 11.07 (5 degrees of freedom)
Since 29.51 > 11.07, we conclude that the percentage figures reported by the company have changed.
16.
Observed
Frequency
(fi)
264
255
229
748
Hypothesized
Proportion
1/3
1/3
1/3
Totals:
Category
Full Service
Discount
Both
Expected
Frequency
(ei)
249.33
249.33
249.33
(fi - ei)2 / ei
0.86
0.13
1.66
2.65
2
= 4.61 (2 degrees of freedom)
.10
Since 2.65 < 4.61, there is no significant difference in preference among the three service choices.
17.
Category
News and Opinion
General Editorial
Family Oriented
Business/Financial
Female Oriented
African-American
Observed
Frequency
(fi)
20
15
30
22
16
12
115
Hypothesized
Proportion
1/6
1/6
1/6
1/6
1/6
1/6
Totals:
Expected
Frequency
(ei)
19.17
19.17
19.17
19.17
19.17
19.17
(fi - ei)2 / ei
.04
.91
6.12
.42
.52
2.68
10.69
2
 .10
= 9.24 (5 degrees of freedom)
Since 10.69 > 9.24, we conclude that there is a difference in the proportion of ads with guilt appeals
among the six types of magazines.
18.
Expected frequencies:
2
 2 = (43 - 45) +
45
ei = (1 / 3) (135) = 45
(53 - 45)2
45
+
(39 - 45)2
= 2.31
45
13- 217
19.
With 2 degrees of freedom,  2.05 = 5.99
Do not reject H0; there is no justification for concluding a difference in preference exists.
H0: p1 = .03, p2 = .28, p3 = .45, p4 = .24
df = 3
2
= 11.34
.01
Reject H0 if 2 > 11.34
Rating
Excellent
Good
Fair
Poor
Observed
24
124
172
80
400
(fi - ei)2 / ei
12.00
1.29
.36
2.67
2
 = 16.31
Expected
.03(400) = 12
.28(400) = 112
.45(400) = 180
.24(400) = 96
400
Reject H0; conclude that the ratings differ. A comparison of observed and expected frequencies show
telephone service is slightly better with more excellent and good ratings.
20.
H0 = The column variable is independent of the row variable
Ha = The column variable is not independent of the row variable
Expected Frequencies:
A
28.5
21.5
P
Q
2
2
 =
(20 - 28.5)
+
(44 - 39.9)
28.5
2
+
B
39.9
30.1
(50 - 45.6)
39.9
2
+
C
45.6
34.4
(30 - 21.5)
45.6
2
+
(26 - 30.1)
21.5
30.1
= 7.86
2
 .025
= 7.37776 with (2 - 1) (3 - 1)= 2 degrees of freedom
Since 2 = 7.86 > 7.37776 Reject H0
Conclude that the column variable is not independent of the row variable.
21.
H0 = The column variable is independent of the row variable
Ha = The column variable is not independent of the row variable
Expected Frequencies:
P
Q
R
A
17.5000
28.7500
13.7500
B
30.6250
50.3125
24.0625
13- 218
C
21.8750
35.9375
17.1875
2
+
(30 - 34.4)
34.4
2

2
=
(20 - 17.5000)2
+
17.5000
= 19.78
(30 - 30.6250)2
++
(30 - 17.1875)2
30.6250
17.1875
2
 .05
= 9.48773 with (3 - 1) (3 - 1)= 4 degrees of freedom
Since 2 = 19.78 > 9.48773 Reject H0 Conclude that the column variable is not independent of f the
row variable.
22.
H0 : Type of ticket purchased is independent of the type of flight
Ha: Type of ticket purchased is not independent of the type of flight.
Expected Frequencies:
e11 = 35.59
e21 = 150.73
e31 = 455.68
Ticket
First
First
Business
Business
Full Fare
Full Fare
e12 = 15.41
e22 = 65.27
e32 = 197.32
Flight
Domestic
International
Domestic
International
Domestic
International
Totals:
Observed
Frequency
(fi)
29
22
95
121
518
135
920
Frequency
(ei)
35.59
15.41
150.73
65.27
455.68
197.32
(fi - ei)2 / ei
1.22
2.82
20.61
47.59
8.52
19.68
100.43
2
 .05
= 5.99 with (3 - 1)(2 - 1) = 2 degrees of freedom
Since 100.43 > 5.99, we conclude that the type of ticket purchased is not independent of the type of
flight.
23. a.
Observed Frequency (fij)
Same
Different
Total
Domestic
125
140
265
European
55
105
160
Asian
68
107
175
Total
248
352
600
Domestic
109.53
155.47
265
European
66.13
93.87
160
Asian
72.33
102.67
175
Total
248
352
600
Expected Frequency (eij)
Same
Different
Total
Chi Square (fij - eij)2 / eij
Same
Domestic
2.18
European
1.87
13- 219
Asian
0.26
Total
4.32
Different
1.54
0.18
3.04
2 = 7.36
2
 .05
= 5.99
Degrees of freedom = 2
b.
1.32
Reject H0; conclude brand loyalty is not independent of manufacturer.
Brand Loyalty
Domestic 125/265 = .472 (47.2%)  Highest
European 55/160 = .344 (34.4%)
Asian
68/175 = .389 (38.9%)
24.
Major
Business
Engineering
Oil
30
30
Chemical
22.5
22.5
Industry
Electrical
17.5
17.5
Computer
30
30
Note: Values shown above are the expected frequencies.
2
 .01
= 11.3449 (3 degrees of freedom: 1 x 3 = 3)
2 = 12.39
Reject H0; conclude that major and industry not independent.
25.
Expected Frequencies:
e11 =
e21 =
e31 =
e41 =
e51 =
e61 =
31.0
29.5
13.0
5.5
7.0
14.0
Most Difficult
Spouse
Spouse
Parents
Parents
Children
Children
Siblings
Siblings
In-Laws
In-Laws
Other Relatives
Other Relatives
e12 =
e22 =
e32 =
e42 =
e52 =
e62 =
31.0
29.5
13.0
5.5
7.0
14.0
Gender
Men
Women
Men
Women
Men
Women
Men
Women
Men
Women
Men
Women
Totals:
Observed
Expected
Frequency
(fi)
37
25
28
31
7
19
8
3
4
10
16
12
200
Frequency
(ei)
31.0
31.0
29.5
29.5
13.0
13.0
5.5
5.5
7.0
7.0
14.0
14.0
2
 .05
= 11.0705 with (6 - 1) (2 - 1) = 5 degrees of freedom
13- 220
(fi - ei)2 / ei
1.16
1.16
0.08
0.08
2.77
2.77
1.14
1.14
1.29
1.29
0.29
0.29
13.43
Since 13.43 > 11.0705. we conclude that gender is not independent of the most difficult person to
buy for.
26.
Expected Frequencies:
e11 =
e21 =
e31 =
e41 =
e51 =
e61 =
17.16
14.88
28.03
22.31
17.16
15.45
e12 =
e22 =
e32 =
e42 =
e52 =
e62 =
Magazine
News
News
General
General
Family
Family
Business
Business
Female
Female
African-American
African-American
12.84
11.12
20.97
16.69
12.84
11.55
Observed
Frequency
(fi)
20
10
15
11
30
19
22
17
16
14
12
15
201
Appeal
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Guilt
Fear
Totals:
Frequency
(ei)
17.16
12.84
14.88
11.12
28.03
20.97
22.31
16.69
17.16
12.84
15.45
11.55
(fi - ei)2 / ei
0.47
0.63
0.00
0.00
0.14
0.18
0.00
0.01
0.08
0.11
0.77
1.03
3.41
2
 .01
= 15.09 with (6 - 1) (2 - 1) = 5 degrees of freedom
Since 3.41 < 15.09, the hypothesis of independence cannot be rejected.
27. a.
Observed Frequency (fij)
Correct
Incorrect
Total
Pharm
207
3
210
Consumer
136
4
140
Computer
151
9
160
Telecom
178
12
190
Total
672
28
700
Consumer
134.4
5.6
140
Computer
153.6
6.4
160
Telecom
182.4
7.6
190
Total
672
28
700
Expected Frequency (eij)
Correct
Incorrect
Total
Pharm
201.6
8.4
210
Chi Square (fij - eij)2 / eij
13- 221
Correct
Incorrect
Pharm
.14
3.47
Degrees of freedom = 3
Consumer
.02
.46
Computer
.04
1.06
Telecom
.11
2.55
Total
.31
7.53
2
 = 7.85
2
= 7.81473
.05
Reject H0; conclude that order fulfillment is not independent of industry.
b.
28.
The pharmaceutical industry is doing the best with 207 of 210 (98.6%) correctly filled orders.
Expected Frequencies:
Supplier
A
B
C
Good
88.76
173.09
133.15
Part Quality
Minor Defect
6.07
11.83
9.10
Major Defect
5.14
10.08
7.75
2 = 7.96
2
 .05
= 9.48773 (4 degrees of freedom: 2 x 2 = 4)
Do not reject H0; conclude that the assumption of independence cannot be rejected
29.
Expected Frequencies:
Education Level
Did not complete high school
High school degree
College degree
Democratic
28
32
40
Party Affiliation
Republican
28
32
40
Independent
14
16
20
2 = 13.42
2
 .01
= 13.2767 (4 degrees of freedom: 2 x 2 = 4)
Reject H0; conclude that party affiliation is not independent of education level.
30.
Expected Frequencies:
e11 = 11.81
e21 = 8.40
e31 = 21.79
Siskel
Con
Con
Con
Mixed
Mixed
Mixed
Pro
Pro
e12 = 8.44
e22 = 6.00
e32 = 15.56
Ebert
Con
Mixed
Pro
Con
Mixed
Pro
Con
Mixed
e13 = 24.75
e23 = 17.60
e33 = 45.65
Observed
Frequency
(fi)
24
8
13
8
13
11
10
9
13- 222
Expected
Frequency
(ei)
11.81
8.44
24.75
8.40
6.00
17.60
21.79
15.56
(fi - ei)2 / ei
12.57
0.02
5.58
0.02
8.17
2.48
6.38
2.77
Pro
Pro
Totals:
64
160
45.65
7.38
45.36
2
 .01
= 13.28 with (3 - 1) (3 - 1) = 4 degrees of freedom
Since 45.36 > 13.28, we conclude that the ratings are not independent.
31.
A summary of the SAMPLE DATA is shown below:
Region
I
II
SAMPLE Size
500
800
p1 = 175 / 500 = .35
p2 = 360 / 800 = .45
0.35(0.65) 0.45(0.55)
 0.0276

500
800
sp  p 
1
Number Indicating
An Intent to Purchase
175
360
2
.10  2.575(.0276)
.10  .071 or .029 to .171
32. a.
H0: p1 - p2  0
Ha: p1 - p2 > 0
b.
p1 = 704/1035 = .6802 (68%)
p2 = 582/1004 = .5797 (58%)
p1  p2 = .6802 - .5797 = .1005
p
n 1 p 1  n 2 p2
n1  n2

1035(0.6802) 1004(0.5797)
 .6307
1035 1004


sp  p  p(1 p)
1
z
2



( p1  p2 )  0 .6802 .5797

 4.70
sp  p
.0214
1
2
p-value  0
c.
33. a.
Reject H0; proportion indicating good/excellent increased.
H0: p1 - p2 = 0
13- 223
 .0214
Ha: p1 - p2  0
Reject H0 if z < -1.96 or if z > 1.96
p 
76  90
400  900
 0.1277
F1 1 I
 J  0.02
G
H
400 900 K
sp  p  (0.1277)(0.8723)
1
2
p1  76 / 400  0.19
z
p2  90 / 900  0.10
( p1  p2 )  ( p1  p2 ) (0.19  0.10)  0

 4.50
sp  p
0.02
1
2
p-value  0
Reject H0; there is a difference between claim rates.
b.
0.09  1.96
0.19(0.81) 0.10(0.90)

400
900
.09  .0432 or .0468 to .1332
p 
34.
9  5  14  0.0341
142  268 410
sp  p  (0.0341)(0.9659)
1
2
F1 1 I
G
H142  268JK 0.0188
p1  9 / 142  0.0634
p2  5 / 268  0.0187
p1  p2  0.0634  0.0187  0.0447
z 
0.0447  0
0.0188
 2.38
p-value = 2(1.0000 - .9913) = 0.0174
Reject H0; There is a significant difference in drug resistance between the two states. New Jersey has
the higher drug resistance rate.
35. a.
b.
.38(430) = 163.4
Estimate: 163
.23(285) = 65.55
Estimate: 66
p1  p2  .38 .23  .15
sp  p 
1
2
.38(1.38) .23(1.23)

 .064
163
66
Confidence interval: .15  1.96(.064) or .15  .125(.025 to .275)
13- 224
c.
36. a.
Yes, since the confidence interval in part (b) does not include 0, I would conclude that the Kodak
campaign is more effective than most.
p1  .38
p2  .22
Point estimate = p1  p2  .38 .22  .16
b.
H0: p1 - p2  0
Ha: p1 - p2  0
c.
p
n 1 p 1  n 2 p2
n1  n2

(200)(.38)  (200)(.22)
 .30
200  200
1
sp  p 
1
z
1
2
.38 .22
2

 .0458
 3.49
.0458
z.01 = 2.33
With z = 3.49 > 2.33 we reject H0 and conclude that expectations for future inflation have diminished.
37.
Observed
Expected
60
50
45
50
59
50
36
50
2 = 8.04
2
 .05
= 7.81473 (3 degrees of freedom)
Reject H0; conclude that the order potentials are not the same in each sales territory.
38.
Observed
Expected

2
=
(48 – 37.03)2
37.03
48
37.03
+
323
306.82
79
126.96
16
21.16
63
37.03
(323 – 306.82)2
(63 – 37.03)2
+   +
306.82
37.03
= 41.69
2
 .01 = 13.2767 (4 degrees of freedom)
Since 41.69 > 13.2767, reject H0.
Mutual fund investors' attitudes toward corporate bonds differ from their attitudes toward corporate
stock.
13- 225
39.
Observed
Expected

2
=
(20 – 35) 2
35
20
35
+
(20 – 35)2
35
+
20
35
40
35
(40 – 35)2
+ (60 – 35)
35
35
60
35
2
= 31.43
2
 .05 = 7.81473 (3 degrees of freedom)
Since 31.43 > 7.81473, reject H0.
The park manager should not plan on the same number attending each day. Plan on a larger staff for
Sundays and holidays.
40.
Observed
Expected
13
18
16
18
28
18
17
18
16
18
2 = 7.44
2
 .05
= 9.48773
Do not reject H0; the assumption that the number of riders is uniformly distributed cannot be rejected.
41.
Observed
Frequency
(fi)
105
235
55
90
15
500
Hypothesized
Proportion
0.28
0.46
0.12
0.10
0.04
Totals:
Category
Very Satisfied
Somewhat Satisfied
Neither
Somewhat Dissatisfied
Very Dissatisfied
Expected
Frequency
(ei)
140
230
60
50
20
(fi - ei)2 / ei
8.75
0.11
0.42
32.00
1.25
42.53
2
 .05
= 9.49 (4 degrees of freedom)
Since 42.53 > 9.49, we conclude that the job satisfaction for computer programmers is different than
42.
Expected Frequencies:
Quality
Shift
1st
2nd
3rd
Good
368.44
276.33
184.22
13- 226
Defective
31.56
23.67
15.78
the
2 = 8.11
2
 .05
= 5.99147 (2 degrees of freedom)
43.
Reject H0; conclude that shift and quality are not independent.
Expected Frequencies:
e11 =
e21 =
e31 =
e41 =
1046.19
28.66
258.59
516.55
e12 =
e22 =
e32 =
e42 =
Employment
Full-Time
Full-time
Part-Time
Part-Time
Self-Employed
Self-Employed
Not Employed
Not Employed
632.81
17.34
156.41
312.45
Observed
Frequency
(fi)
1105
574
31
15
229
186
485
344
2969
Region
Eastern
Western
Eastern
Western
Eastern
Western
Eastern
Western
Totals:
Frequency
(ei)
1046.19
632.81
28.66
17.34
258.59
156.41
516.55
312.45
(fi - ei)2 / ei
3.31
5.46
0.19
0.32
3.39
5.60
1.93
3.19
23.37
2
 .05
= 7.81 with (4 - 1) (2 - 1) = 3 degrees of freedom
Since 23.37 > 7.81, we conclude that employment status is not independent of region.
44.
Expected frequencies:
Loan Approval Decision
Approved
Rejected
24.86
15.14
18.64
11.36
31.07
18.93
12.43
7.57
Loan Offices
Miller
McMahon
Games
Runk
2 = 2.21
2
 .05
= 7.81473 (3 degrees of freedom)
Do not reject H0; the loan decision does not appear to be dependent on the officer.
45. a.
Observed Frequency (fij)
Men
Women
Total
Never Married
234
216
450
Married
106
168
274
Divorced
10
16
26
Total
350
400
750
Married
Divorced
Total
Expected Frequency (eij)
Never Married
13- 227
Men
Women
Total
210
240
450
127.87
146.13
274
12.13
13.87
26
350
400
750
Married
3.74
3.27
Divorced
.38
.33
Total
6.86
6.00
2 = 12.86
Chi Square (fij - eij)2 / eij
Never Married
2.74
2.40
Men
Women
2
= 9.21
.01
Degrees of freedom = 2
Reject H0; conclude martial status is not independent of gender.
b.
Martial Status
Never Married
66.9%
54.0%
Men
Women
Married
30.3%
42.0%
Divorced
2.9%
4.0%
Men
100 - 66.9 = 33.1% have been married
Women 100 - 54.0 = 46.0% have been married
46.
Expected Frequencies:
e11 
2 
(50)(18)
(50)(24)
(50)(12)
 9, e12 
 12,    , e25 
6
100
100
100
(4  9)2
9

(10 12)2
  
12
(4  6)2
 9.76
6
2
= 9.48773 (4 degrees of freedom)
.05
Since 9.76 > 9.48773, reject H0.
Banking tends to have lower P/E ratios. We can conclude that industry type and P/E ratio are related.
47.
Expected Frequencies:
County
Urban
Rural
Total
Sun
56.7
11.3
68
Mon
47.6
9.4
57
Days of the Week
Tues Wed
Thur
55.1 56.7
60.1
10.9 11.3
11.9
66
68
72
2 = 6.20
2
 .05
= 12.5916 (6 degrees of freedom)
13- 228
Fri
72.6
14.4
87
Sat
44.2
8.8
53
Total
393
78
471
Do not reject H0; the assumption of independence cannot be rejected.
48.
Expected Frequencies:
Occupied
Vacant
Total
Los Angeles
165.7
34.3
200.0
2
San Diego
124.3
25.7
150.0
San Francisco
186.4
38.6
225.0
2
San Jose
165.7
34.3
200.0
Total
642
133
775
2
(160 - 165.7)
(26 - 34.3)
(116 - 124.3)
++
+
34.3
165.7
124.3
= 7.78
2 =
2
= 7.81473 with 3 degrees of freedom
.05
Since 2 = 7.78  7.81473 Do not reject H0.
We cannot conclude that office vacancies are dependent on metropolitan area, but it is close: the p-value
is slightly larger than .05.
13- 229
Chapter 12
Simple Linear Regression
Learning Objectives
1.
Understand how regression analysis can be used to develop an equation that estimates
mathematically how two variables are related.
2.
Understand the differences between the regression model, the regression equation, and the estimated
regression equation.
3.
Know how to fit an estimated regression equation to a set of SAMPLE DATA based upon the
least- squares method.
4.
Be able to determine how good a fit is provided by the estimated regression equation and compute
the SAMPLE correlation coefficient from the regression analysis output.
5.
Understand the assumptions necessary for statistical inference and be able to test for a significant
relationship.
6.
Learn how to use a residual plot to make a judgement as to the validity of the regression assumptions,
recognize outliers, and identify influential observations.
7.
Know how to develop confidence interval estimates of y given a specific value of x in both the case
of a mean value of y and an individual value of y.
8.
Be able to compute the SAMPLE correlation coefficient from the regression analysis output.
9.
Know the definition of the following terms:
independent and dependent variable
simple linear regression
regression model
regression equation
estimated regression equation
scatter diagram
coefficient of determination
standard error of the estimate
confidence interval
prediction interval
residual plot
standardized residual plot
outlier
influential observation
leverage
13- 230
Solutions:
a.
16
14
12
10
y
1
8
6
4
2
0
0
1
2
3
4
5
6
x
b.
There appears to be a linear relationship between x and y.
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part d we will determine the equation of a straight line that
“best” represents the relationship according to the least squares criterion.
d.
Summations needed to compute the slope and y-intercept are:
x  15
y  40
i
b1 
i
(x  x )( y  y )  26
i
i
i
(xi  x )( yi  y) 26

 2.6
(x i  x )2
10
b0  y  b1 x  8  (2.6)(3)  0.2
yˆ  0.2  2.6x
e.
(x  x )2  10
y$  0.2  2.6(4)  10.6
13- 231
a.
35
30
25
20
y
2.
15
10
5
0
0
2
4
6
8
10
x
b.
There appears to be a linear relationship between x and y.
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part d we will determine the equation of a straight line that
“best” represents the relationship according to the least squares criterion.
d.
Summations needed to compute the slope and y-intercept are:
x  19
i
b1 
y  116
(x  x )( y  y)  57.8
i
i
i
i
(xi  x )( yi  y) 57.8

 1.8766
(x i  x )2
30.8
b0  y  b1 x  23.2  (1.8766)(3.8)  30.3311
y$  30.33  1.88x
e.
(x  x )2  30.8
y$  30.33  1.88(6)  19.05
13- 232
3.
a.
7
6
5
y
4
3
2
1
0
0
2
4
6
8
x
b.
Summations needed to compute the slope and y-intercept are:
x  26
i
b1 
y  17
(x  x )( y  y)  11.6
i
i
i
i
(xi  x )( yi  y) 11.6

 0.5088
(x i  x )2
22.8
b0  y  b1 x  3.4  (0.5088)(5.2)  0.7542
y$  0.75  0.51x
c.
(x  x )2  22.8
y$  0.75  0.51(4)  2.79
13- 233
10
4. a.
135
130
125
y
120
115
110
105
100
61
62
63
64
65
66
67
68
69
x
b.
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part d we will determine the equation of a straight line that
“best” represents the relationship according to the least squares criterion.
d.
Summations needed to compute the slope and y-intercept are:
x  325
i
b1 
y  585
i
(x  x )( y  y)  110
i
i
i
(xi  x )( yi  y) 110

 5.5
(x i  x )2
20
b0  y  b1 x  117  (5.5)(65)  240.5
y$  240.5  5.5x
e.
(x  x )2  20
y$  240.5  5.5x  240.5  5.5(63)  106 pounds
13- 234
There appe
a.
2100
1900
1700
1500
1300
y
5.
1100
900
700
500
300
100
0
20
40
60
80
100
120
140
x
b.
There appears to be a linear relationship between x = media expenditures (millions of dollars) and
= case sales (millions).
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part d we will determine the equation of a straight line that
“best” represents the relationship according to the least squares criterion.
Summations needed to compute the slope and y-intercept are:
x  420.6
i
b1 
y  5958.7
i
(x  x )( y  y)  142, 040.3443
i
i
(x  x )2  9847.6486
i
(xi  x )( yi  y) 142, 040.3443

 14.4238
(x i  x )2
9847.6486
b0  y  b1x  851.2429  (14.4238)(60.0857)  15.42
y$  15.42  14.42x
d.
A one million dollar increase in media expenditures will increase case sales by approximately 14.42
million.
e.
y$  15.42  14.42x  15.42  14.42(70)  993.98
13- 235
y
a.
1.4
1.2
1
0.8
y
6.
0.6
0.4
0.2
0
66
68
70
72
74
76
78
80
82
84
x
b.
There appears to be a linear relationship between x = percentage of flights arriving on time and
= number of complaints per 100,000 passengers.
c.
Summations needed to compute the slope and y-intercept are:
x  667.2
y  7.18
i
b1 
i
(x  x )( y  y)  9.0623
i
i
(x  x )2  128.7
i
(xi  x )( yi  y) 9.0623

 0.0704
(x i  x )2
128.7
b0  y  b1 x  0.7978  (0.0704)(74.1333)  6.02
y$  6.02  0.07x
d.
A one percent increase in the percentage of flights arriving on time will decrease the number of
complaints per 100,000 passengers by 0.07.
e
y$  6.02  0.07x  6.02  0.07(80)  0.42
13- 236
y
1550
1500
S&P
1450
1400
1350
1300
9600
9800
10000
10200
10400
10600
10800
11000
11200
DJIA
7.
a.
b.
Let x = DJIA and y = S&P. Summations needed to compute the slope and y-intercept are:
x  104,850
i
b 
1
y  14, 233
i
(x  x )( y  y )  268, 921
i
(x  x )2  1,806, 384
i
(xi  x )( yi  y)  268, 921  0.14887

(x i  x )2
1,806, 384
b0  y  b1x  1423.3  (.14887)(10, 485)  137.629
yˆ  137.63  0.1489x
c.
yˆ  137.63  0.1489(11, 000)  1500.27 or approximately 1500
13- 237
i
a.
1800
1600
1400
1200
Price
8.
1000
800
600
400
200
0
0
1
2
3
4
5
6
Sidetrack Capability
b.
There appears to be a linear relationship between x = sidetrack capability and y = price, with higher
priced models having a higher level of handling.
c.
Summations needed to compute the slope and y-intercept are:
x  28
i
b1 
y  10, 621
i
(x  x )( y  y)  4003.2
i
i
(xi  x )( yi  y) 4003.2

 204.2449
(x i  x )2
19.6
b0  y  b1 x  1062.1 (204.2449)(2.8)  490.21
yˆ  490.21 204.24x
d.
yˆ  490.21 204.24x  490.21 204.24(4)  1307
13- 238
(x  x )2  19.6
i
a.
Let x = years of experience and y = annual sales ($1000s)
150
140
130
120
110
y
9.
100
90
80
70
60
50
0
2
4
6
8
10
x
b.
Summations needed to compute the slope and y-intercept are:
x  70
i
b1 
y  1080
(x  x )( y  y)  568
i
i
i
i
(xi  x )( yi  y) 568

4
(x i  x )2
142
b0  y  b1 x  108  (4)(7)  80
y$  80  4 x
c.
(x  x )2  142
y$  80  4x  80  4(9)  116
13- 239
12
14
95
Overall Rating
90
85
80
75
70
65
60
100
150
200
250
Performance Score
10. a.
b.
Let x = performance score and y = overall rating. Summations needed to compute the slope and yintercept are:
x  2752
i
b 
1
y  1177
i
(x  x )( y  y)  1723.73
i
i
(xi  x )( yi  y)  1723.73  0.1452

(x i  x )2
11, 867.73
b0  y  b1x  78.4667  (.1452)(183.4667)  51.82
yˆ  51.82  0.145x
c.
yˆ  51.82  0.145(225)  84.4 or approximately 84
13- 240
(x  x )2  11,867.73
i
11. a.
Let x = hotel revenue and y = gaming revenue
900.0
800.0
700.0
600.0
y
500.0
400.0
300.0
200.0
100.0
0.0
0.0
100.0
200.0
300.0
400.0
500.0
600.0
700.0
800.0
x
b.
There appears to be a linear relationship between the variables.
c.
The summations needed to compute the slope and the y-intercept are:
x  2973.3
y  3925.6
i
b1 
i
(x  x )( y  y)  453, 345.042
i
i
(xi  x )( yi  y) 453, 345.042

 0.9385
(x i  x )2
483, 507.581
b0  y  b1 x  392.56  (0.9385)(297.33)  113.52
y$  113.52  0.94x
d.
y$  113.52  0.94x  113.52  0.94(500)  583.5
13- 241
(x  x )2  483, 507.581
i
12. a.
40000
35000
Revenue
30000
25000
20000
15000
10000
5000
0
0
20000
40000
60000
80000
100000
Number of Employees
b.
There appears to be a positive linear relationship between the number of
employees and the revenue.
b.
Let x = number of employees and y = revenue. Summations needed to compute the slope and yintercept are:
x  4200
i
b 
1
y  1669
i
(x  x )( y  y)  4, 658, 594,168
i
i
(xi  x )( yi  y)  4, 658, 594,168  0.316516

(x i  x )2
14, 718, 343,803
b0  y  b1 x  14, 048  (.316516)(40, 299)  1293
yˆ  1293  0.3165x
d.
yˆ  1293  .3165(75, 000)  25, 031
13- 242
(x  x )2  14, 718, 343, 803
i
13. a.
Let x = adjusted gross income ($1000s) and y = total itemized deductions ($1000s)
30.0
25.0
y
20.0
15.0
10.0
5.0
0.0
0.0
20.0
40.0
60.0
80.0
100.0
120.0
140.0
x
b.
The summations needed to compute the slope and the y-intercept are:
x  399
y  97.1
i
b1 
i
(x  x )( y  y)  1233.7
i
i
(x  x )2  7648
i
(xi  x )( yi  y) 1233.7

 0.16131
(x i  x )2
7648
b0  y  b1 x  13.87143  (0.16131)(57)  4.67675
y$  4.68  0.16x
c.
y$  4.68  0.16x  4.68  0.16(52.5)  13.08 or approximately $13,080.
The agent's request for an audit appears to be justified.
13- 243
14. a.
Let x = average room rate ($) and y = occupancy rate (%)
85
80
y
75
70
65
60
60
70
80
90
100
110
x
b.
The summations needed to compute the slope and the y-intercept are:
x  1677.25
i
b 
1
y  1404.3
(x  x )( y  y)  897.9493
i
i
i
(x  x )2  3657.4568
i
(xi  x )( yi  y)  897.9493  0.2455

(x i  x )2
3657.4568
b0  y  b1 x  70.215  (0.2455)(83.8625)  49.63
y$  49.63.2455x
c.
15. a.
y$  49.63.2455x  49.63.2455(80)  69.3%
The estimated regression equation and the mean for the dependent variable are:
y$i  0.2  2.6xi
y 8
The sum of squares due to error and the total sum of squares are
SSE  ∑( y i y$i )2  12.40
SST  ∑( yi  y)2  80
Thus, SSR = SST - SSE = 80 - 12.4 = 67.6
b.
r2 = SSR/SST = 67.6/80 = .845
The least squares line provided a very good fit; 84.5% of the variability in y has been explained by
the least squares line.
c.
rxy  .845  .9192
13- 244
16. a.
The estimated regression equation and the mean for the dependent variable are:
yˆi  30.33 1.88x
y  23.2
The sum of squares due to error and the total sum of squares are
SSE  ∑( yi  yˆi )2  6.33
SST  ∑( yi  y)2  114.80
Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47
b.
r2 = SSR/SST = 108.47/114.80 = .945
The least squares line provided an excellent fit; 94.5% of the variability in y has been explained by
the estimated regression equation.
c.
rxy  .945  .9721
Note: the sign for rxy is negative because the slope of the estimated regression equation is negative.
(b1 = -1.88)
17.
The estimated regression equation and the mean for the dependent variable are:
yˆi  .75  .51x
y  3.4
The sum of squares due to error and the total sum of squares are
SSE  ∑( yi  yˆi )2  5.3
SST  ∑( yi  y)2  11.2
Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9
r2 = SSR/SST = 5.9/11.2 = .527
We see that 52.7% of the variability in y has been explained by the least squares line.
rxy  .527  .7259
18. a.
The estimated regression equation and the mean for the dependent variable are:
yˆ  1790.5  581.1x
y  3650
The sum of squares due to error and the total sum of squares are
SSE  ∑( y i  yˆi )2  85,135.14
SST  ∑( y i  y )2  335, 000
Thus, SSR = SST - SSE = 335,000 - 85,135.14 = 249,864.86
b.
r2 = SSR/SST = 249,864.86/335,000 = .746
We see that 74.6% of the variability in y has been explained by the least squares line.
13- 245
c.
19. a.
rxy  .746  .8637
The estimated regression equation and the mean for the dependent variable are:
yˆ  137.63 .1489x
y  1423.3
The sum of squares due to error and the total sum of squares are
SSE  ∑( y i  yˆi )2  7547.14
SST  ∑( yi  y)2  47, 582.10
Thus, SSR = SST - SSE = 47,582.10 - 7547.14 = 40,034.96
b.
r2 = SSR/SST = 40,034.96/47,582.10 = .84
We see that 84% of the variability in y has been explained by the least squares line.
c.
20. a.
rxy  .84  .92
Let x = income and y = home price. Summations needed to compute the slope and y-intercept are:
x  1424
i
b 
1
y  2455.5
i
(x  x )( y  y)  4011
i
(x  x )2  1719.618
i
i
(xi  x )( yi  y)  4011  2.3325

(x i  x )2
1719.618
b0  y  b1x  136.4167  (2.3325)(79.1111)  48.11
yˆ  48.11 2.3325x
b.
The sum of squares due to error and the total sum of squares are
SSE  ∑( yi  yˆi )2  2017.37
SST  ∑( y i  y)2  11, 373.09
Thus, SSR = SST - SSE = 11,373.09 – 2017.37 = 9355.72
r2 = SSR/SST = 9355.72/11,373.09 = .82
We see that 82% of the variability in y has been explained by the least squares line.
rxy  .82  .91
c.
21. a.
yˆ  48.11 2.3325(95)  173.5 or approximately $173,500
The summations needed in this problem are:
x  3450
i
b1 
y  33, 700
i
(x  x )( y  y )  712, 500
i
i
(xi  x )( yi  y) 712, 500

 7.6
(x i  x )2
93, 750
13- 246
(x  x )2  93, 750
i
b0  y  b1 x  5616.67  (7.6)(575)  1246.67
y$  1246.67  7.6x
b.
$7.60
c.
The sum of squares due to error and the total sum of squares are:
SSE  ∑( y i  yˆi )2  233, 333.33
SST  ∑( y i  y)2  5, 648, 333.33
Thus, SSR = SST - SSE = 5,648,333.33 - 233,333.33 = 5,415,000
r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587
We see that 95.87% of the variability in y has been explained by the estimated regression equation.
d.
22. a.
y$  1246.67  7.6x  1246.67  7.6(500)  $5046.67
The summations needed in this problem are:
x  613.1
i
b 
1
y  70
(x  x )( y  y)  5766.7
i
i
(x  x )2  45,833.9286
i
i
(xi  x )( yi  y)  5766.7
 0.1258

(x i  x )2
45,833.9286
b0  y  b1 x  10  (0.1258)(87.5857)  1.0183
yˆ  1.0183  0.1258x
b.
The sum of squares due to error and the total sum of squares are:
SSE  ∑( yi  yˆi )2  1272.4495
SST  ∑( yi  y)2  1998
Thus, SSR = SST - SSE = 1998 - 1272.4495 = 725.5505
r2 = SSR/SST = 725.5505/1998 = 0.3631
Approximately 37% of the variability in change in executive compensation is explained by the twoyear change in the return on equity.
c.
rxy  0.3631  0.6026
It reflects a linear relationship that is between weak and strong.
23. a.
b.
s2 = MSE = SSE / (n - 2) = 12.4 / 3 = 4.133
s  MSE  4.133  2.033
13- 247
c.
(x i  x )2  10
s
sb 
(x  x )2
1
d.
t

2.033
10
 0.643
b1
2.6

 4.04
sb .643
1
t.025 = 3.182 (3 degrees of freedom)
Since t = 4.04 > t.05 = 3.182 we reject H0: 1 = 0
e.
MSR = SSR / 1 = 67.6
F = MSR / MSE = 67.6 / 4.133 = 16.36
F.05 = 10.13 (1 degree of freedom numerator and 3 denominator)
Since F = 16.36 > F.05 = 10.13 we reject H0: 1 = 0
Source of Variation
Regression
Error
Total
24. a.
Sum of Squares
67.6
12.4
80.0
Mean Square
67.6
4.133
F
16.36
Mean Square
F
s2 = MSE = SSE / (n - 2) = 6.33 / 3 = 2.11
b.
s  MSE  2.11  1.453
c.
(x i  x )2  30.8
s
sb 
(x  x )2
1
d.
Degrees of Freedom
1
3
4
t

1.453
30.8
 0.262
b1  1.88  7.18
sb
.262
1
t.025 = 3.182 (3 degrees of freedom)
Since t = -7.18 < -t.025 = -3.182 we reject H0: 1 = 0
e.
MSR = SSR / 1 = 8.47
F = MSR / MSE = 108.47 / 2.11 = 51.41
F.05 = 10.13 (1 degree of freedom numerator and 3 denominator)
Since F = 51.41 > F.05 = 10.13 we reject H0: 1 = 0
Source of Variation
Sum of Squares
Degrees of Freedom
13- 248
108.47
6.33
114.80
Regression
Error
Total
25. a.
1
3
4
108.47
2.11
51.41
s2 = MSE = SSE / (n - 2) = 5.30 / 3 = 1.77
s  MSE  1.77  1.33
b.
(x i  x )2  22.8
s
sb 
(x  x )2
1
t
b1
sb

.51

1.33
22.8
 0.28
 1.82
.28
1
t.025 = 3.182 (3 degrees of freedom)
Since t = 1.82 < t.025 = 3.182 we cannot reject H0: 1 = 0; x and y do not appear to be related.
c.
MSR = SSR/1 = 5.90 /1 = 5.90
F = MSR/MSE = 5.90/1.77 = 3.33
F.05 = 10.13 (1 degree of freedom numerator and 3 denominator)
Since F = 3.33 < F.05 = 10.13 we cannot reject H0: 1 = 0; x and y do not appear to be related.
26. a.
s2 = MSE = SSE / (n - 2) = 85,135.14 / 4 = 21,283.79
s  MSE  21,283.79  145.89
(x i  x )2  0.74
s
sb 
(x  x )2
1
t
b1
sb
1


145.89
0.74
 169.59
581.08
 3.43
169.59
t.025 = 2.776 (4 degrees of freedom)
Since t = 3.43 > t.025 = 2.776 we reject H0: 1 = 0
b.
MSR = SSR / 1 = 249,864.86 / 1 = 249.864.86
F = MSR / MSE = 249,864.86 / 21,283.79 = 11.74
F.05 = 7.71 (1 degree of freedom numerator and 4 denominator)
13- 249
Since F = 11.74 > F.05 = 7.71 we reject H0: 1 = 0
c.
Source of Variation
Regression
Error
Total
27. a.
Sum of Squares
249864.86
85135.14
335000
Degrees of Freedom
1
4
5
Mean Square
249864.86
21283.79
F
11.74
Summations needed to compute the slope and y-intercept are:
x  37
i
b1 
y  1654
i
(x  x )( y  y)  315.2
i
(x  x )2  10.1
i
i
(xi  x )( yi  y) 315.2

 31.2079
(x i  x )2
10.1
b0  y  b1 x  165.4  (31.2079)(3.7)  19.93
yˆ  19.93  31.21x
b.
SSE = ( y i  yˆi )2  2487.66 SST = ( yi  y)2 = 12,324.4
Thus, SSR = SST - SSE = 12,324.4 - 2487.66 = 9836.74
MSR = SSR/1 = 9836.74
MSE = SSE/(n - 2) = 2487.66/8 = 310.96
F = MSR / MSE = 9836.74/310.96 = 31.63
F.05 = 5.32 (1 degree of freedom numerator and 8 denominator)
Since F = 31.63 > F.05 = 5.32 we reject H0: 1 = 0.
Upper support and price are related.
c.
r2 = SSR/SST = 9,836.74/12,324.4 = .80
The estimated regression equation provided a good fit; we should feel comfortable using the
estimated regression equation to estimate the price given the upper support rating.
d.
28.
yˆ = 19.93 + 31.21(4) = 144.77
SST = 411.73 SSE = 161.37 SSR = 250.36
MSR = SSR / 1 = 250.36
MSE = SSE / (n - 2) = 161.37 / 13 = 12.413
13- 250
F = MSR / MSE = 250.36 / 12.413= 20.17
F.05 = 4.67 (1 degree of freedom numerator and 13 denominator)
Since F = 20.17 > F.05 = 4.67 we reject H0: 1 = 0.
29.
SSE = 233,333.33 SST = 5,648,333.33 SSR = 5,415,000
MSE = SSE/(n - 2) = 233,333.33/(6 - 2) = 58,333.33
MSR = SSR/1 = 5,415,000
F = MSR / MSE = 5,415,000 / 58,333.25 = 92.83
Source of Variation
Regression
Error
Total
Sum of Squares
5,415,000.00
233,333.33
5,648,333.33
Degrees of Freedom
1
4
5
Mean Square
5,415,000
58,333.33
F
92.83
F.05 = 7.71 (1 degree of freedom numerator and 4 denominator)
Since F = 92.83 > 7.71 we reject H0: 1 = 0. Production volume and total cost are related.
30.
Using the computations from Exercise 22,
SSE = 1272.4495 SST = 1998 SSR = 725.5505
s  254.4899  15.95
∑(x i  x )2 = 45,833.9286
s
sb 
(x  x )2
1
t
b1
sb
1

0.1258
15.95

 0.0745
45, 833.9286
 1.69
0.0745
t.025 = 2.571
Since t = 1.69 < 2.571, we cannot reject H0: 1 = 0
There is no evidence of a significant relationship between x and y.
31.
SST = 11,373.09 SSE = 2017.37 SSR = 9355.72
MSR = SSR / 1 = 9355.72
MSE = SSE / (n - 2) = 2017.37/ 16 = 126.0856
F = MSR / MSE = 9355.72/ 126.0856 = 74.20
13- 251
F.01 = 8.53 (1 degree of freedom numerator and 16 denominator)
Since F = 74.20 > F.01 = 8.53 we reject H0: 1 = 0.
32. a.
b.
y$  6.1092  0.8951x
t
b1
sb
1

0.8951
 6.01
0.149
t.025 = 2.306 (8 degrees of freedom)
Since t = 6.01 > t.025 = 2.306 we reject H0: 1 = 0; monthly maintenance expense is related to
usage.
c.
33. a.
r2 = SSR/SST = 1575.76/1924.90 = 0.82. A good fit.
9
b.
y$  20.0  7.21x
c.
t = 5.29 > t.025 = 2.365 we reject H0: 1 = 0
d.
SSE = SST - SSR = 51,984.1 - 41,587.3 = 10,396.8
MSE = 10,396.8 / 7 = 1,485.3
F = MSR / MSE = 41,587.3 / 1,485.3 = 28.00
F.05 = 5.59 (1 degree of freedom numerator and 7 denominator)
Since F = 28 > F.05 = 5.59 we reject H0: 1 = 0.
e.
34. a.
b.
y$  20.0  7.21x  20.0  7.21(50)  380.5 or $380,500
y$ = 80.0 + 50.0 x
F = MSR / MSE = 6828.6 / 82.1 = 83.17
F.05 = 4.20 (1 degree of freedom numerator and 28 denominator)
Since F = 83.17 > F.05 = 4.20 we reject H0: 1 = 0.
Branch office sales are related to the salespersons.
c.
t=
50
= 9.12
5.482
13- 252
t.025 = 2.048 (28 degrees of freedom)
Since t = 9.12 > t.05 = 2.048 we reject H0: 1 = 0
d.
35.
p-value = .000
A portion of the Excel Regression tool output for this problem follows:
Regression STATISTICS
Multiple R
0.7379
R Square
0.5444
Adjusted R Square
0.5094
Standard Error
4.1535
Observations
15
ANOVA
SS
MS
F
Regression
df
1
268.0118
268.0118
15.5357
Residual
13
224.2682
17.2514
Total
14
492.28
Coefficients Standard Error
36.
t Stat
P-value
Intercept
11.3332
2.7700
4.0914
0.0013
Gross Profit Margin (%)
0.6361
0.1614
3.9415
0.0017
a.
$y = 11.3332 + .6361x where x = Gross Profit Margin (%)
b.
Significant relationship: Significance F = .0017 <  = .05
c.
Significant relationship: P-value = .0017 <  = .05
d.
r2 = 0.5444; Not a good fit
A portion of the Excel Regression tool output for this problem follows:
Regression STATISTICS
Multiple R
0.6502
R Square
0.4228
Adjusted R Square
0.3907
Standard Error
11.5925
Observations
20
ANOVA
13- 253
Significance F
0.0017
df
Regression
Residual
Total
Intercept
Age
1
18
19
SS
MS
1771.982016 1771.982
2418.967984 134.3871
4190.95
Coefficients Standard Error
-42.7965
19.3816
1.0043
0.2766
t Stat
P-value
-2.2081
0.0405
3.6312
0.0019
a.
yˆ = 42.7965 + 1.0043x where x = Age
b.
Significant relationship: Significance F = .0019 <  = .05
c.
r2 = 0.4228; Not a good fit
37.
F
Significance F
13.1857
0.0019
A portion of the Excel Regression tool output for this problem follows:
Regression STATISTICS
Multiple R
0.9277
R Square
0.8606
Adjusted R Square
0.8467
Standard Error
6.6343
Observations
12
ANOVA
df
SS
MS
Regression
1
2717.8625
2717.8625
Residual
10
440.1375
44.0137
Total
11
3158
Coefficients Standard Error
Intercept
Income ($1000s)
t Stat
F
61.7503
P-value
-11.8020
12.8441
-0.9189
2.1843
0.2780
7.8581 1.3768E-05
a.
yˆ = 11.802 + 2.1843x where x = Income ($1000s)
b.
Significant relationship: P-value = .000 <  = .05
c.
r2 = 0.86; A very good fit
13- 254
0.3798
Significance F
1.3768E-05
38. a.
Scatter diagram:
40.0
Average Rental Rate ($)
35.0
30.0
25.0
20.0
15.0
10.0
0.0
5.0
10.0
15.0
20.0
25.0
Vaacancy Rate (%)
b.
There appears to be a linear relationship between the two variables.
A portion of the Excel Regression tool output for this problem follows:
Regression STATISTICS
Multiple R
0.6589
R Square
0.4341
Adjusted R Square
0.3988
Standard Error
4.8847
Observations
18
ANOVA
SS
MS
F
Regression
df
1
292.9137
292.9137
12.2760
Residual
16
381.7712
23.8607
Total
17
674.6849
13- 255
Significance F
0.0029
Coefficients Standard Error
t Stat
P-value
Intercept
37.0747
3.5277
10.5097 1.36938E-08
Vacancy Rate (%)
-0.7792
0.2224
-3.5037
c.
yˆ = 37.0747 - 0.7792x where x = Vacancy Rate (%)
d.
Significant relationship: Significance F (or P-value) <  = .05
e.
39. a.
r2 = 0.43; Not a very good fit
s = 2.033
x 3
(x i  x )2  10
syˆ  s
2
1 (4  3)2
1 (x p  x )

 2.033 
 1.11
n (x  x )2
5
10
p
b.
y$  0.2  2.6x  0.2  2.6(4)  10.6
y$p  t /2 sy$
p
10.6  3.182(1.11) = 10.6  3.53
or 7.07 to 14.13
c.
sind  s 1
d.
y$p  t /2 sind
2
1 (4  3)2
1 (x p  x )

 2.033 1 
 2.32
n (x  x )2
5
10
10.6  3.182(2.32) = 10.6  7.38
or 3.22 to 17.98
40. a.
b.
s = 1.453
x  3.8
syˆ  s
p
(x i  x )2  30.8
2
1 (3  3.8)2
1 (x p  x )

 1.453 
 .068
n (x  x )2
5
30.8
y$  30.33  1.88x  30.33  1.88(3)  24.69
y$p  t /2 sy$
p
24.69  3.182(.68) = 24.69  2.16
13- 256
0.0029
or 22.53 to 26.85
c.
sind  s 1
d.
y$p  t /2 sind
2
1 (3  3.8)2
1 (x  x )

 1.453 1 
 1.61
n (x  x )2
5
30.8
24.69  3.182(1.61) = 24.69  5.12
or 19.57 to 29.81
41.
s = 1.33
x  5.2
syˆ  s
p
(x i  x )2  22.8
2
1 (3  5.2)2
1 (x p  x )

 1.33 
 0.85
n (x  x )2
5
22.8
y$  0.75  0.51x  0.75  0.51(3)  2.28
y$p  t /2 sy$
p
2.28  3.182 (.85) = 2.28  2.70
or -.40 to 4.98
sind  s 1
2
1 (3  5.2)2
1 (x p  x )
 1.33 1 
 1.58

n (x  x )2
5
22.8
y$p  t /2 sind
2.28  3.182 (1.58) = 2.28  5.03
or -2.27 to 7.31
42. a.
s = 145.89
x  3.2
syˆ  s
p
(x i  x )2  0.74
2
1  (3  3.2)2
1 (x p  x )

 145.89
 68.54
2
n (x  x )
6
0.74
yˆ = 1790.5 + 581.1x = 1790.5 + 581.1(3) = 3533.8
y$p  t /2 sy$
p
13- 257
3533.8  2.776(68.54) = 3533.8  190.27
or $3343.53 to $3724.07
b.
sind  s 1
2
1 (3  3.2)2
1 (x p  x )
 145.89 1 
 161.19

n (x  x )2
6
0.74
y$p  t /2 sind
3533.8  2.776(161.19) = 3533.8  447.46
or $3086.34 to $3981.26
43. a.
b.
yˆ  51.82  .1452x  51.82  .1452(200)  80.86
s = 3.5232
x  183.4667
(x i  x )2  11, 867.73
2
1 (200 183.4667)2
1 (x p  x )


 3.5232
 1.055
n (x  x )2
15
11, 867.73
syˆ  s
p
y$p  t /2 sy$
p
80.86  2.160(1.055) = 80.86  2.279
or 78.58 to 83.14
c.
sind  s 1
2
1  (200 183.4667)2
1 (x p  x )

 3.5232 1
 3.678
n (x  x )2
11,867.73
15
y$p  t /2 sind
80.86  2.160(3.678) = 80.86  7.944
or 72.92 to 88.80
44. a.
x  57
(x i  x )2  7648
s2 = 1.88
syˆ  s
p
s = 1.37
2
1  (52.5  57)2
1 (x p  x )

 1.37
 0.52
2
n (x  x )
7
7648
y$p  t /2 sy$
p
13.08  2.571 (.52) = 13.08  1.34
13- 258
or 11.74 to 14.42 or $11,740 to $14,420
b.
sind = 1.47
13.08  2.571 (1.47) = 13.08  3.78
or 9.30 to 16.86 or $9,300 to $16,860
c.
Yes, $20,400 is much larger than anticipated.
d.
Any deductions exceeding the $16,860 upper limit could suggest an audit.
45. a.
y$  1246.67  7.6(500)  $5046.67
b.
x  575 (x i  x )2  93, 750
s2 = MSE = 58,333.33 s = 241.52
sind  s 1
2
1 (500  575)2
1 (x p  x )
 241.52 1 
 267.50

n (x  x )2
6
93, 750
y$p  t /2 sind
5046.67  4.604 (267.50) = 5046.67  1231.57
or $3815.10 to $6278.24
c.
46.a.
Based on one month, $6000 is not out of line since $3815.10 to $6278.24 is the prediction interval.
However, a sequence of five to seven months with consistently high costs should cause concern.
Summations needed to compute the slope and y-intercept are:
x  227
i
b1 
y  2281.7
i
(x  x )( y  y)  6003.41
i
i
(x  x )2  1032.1
i
(xi  x )( yi  y) 6003.41

 5.816694
(x i  x )2
1032.1
b0  y  b1x  228.17  (5.816694)(27.7)  67.047576
y$  67.0476  5.8167 x
b.
SST = 39,065.14 SSE = 4145.141 SSR = 34,920.000
r2 = SSR/SST = 34,920.000/39,065.141 = 0.894
The estimated regression equation explained 89.4% of the variability in y; a very good fit.
c.
s2 = MSE = 4145.141/8 = 518.143
13- 259
s  518.143  22.76
2
1  (35  27.7)2
1 (x p  x )

 22.76
 8.86
n (x  x )2
10
1032.1
syˆ  s
p
y$  67.0476  5.8167x  67.0476  5.8167(35)  270.63
y$p  t /2 sy$
p
270.63  2.262 (8.86) = 270.63  20.04
or 250.59 to 290.67
d.
sind  s 1
2
1 (35  27.7)2
1 (x  x )


 22.76 1
 24.42
n (x  x )2
10
1032.1
y$p  t /2 sind
270.63  2.262 (24.42) = 270.63  55.24
or 215.39 to 325.87
47. a.
Using Excel's Regression tool, the estimated regression equation is:
yˆ = 7.0222 + 1.5873x or yˆ = 7.02 + 1.59x
b.
The residuals calculated using yˆ = 7.02 + 1.59x are 3.48, -2.47, -4.83, -1.60, and 5.22
c.
6
Residuals
4
2
0
-2 0
5
10
15
20
25
-4
-6
x
With only 5 observations it is difficult to determine if the assumptions are satisfied; however, the plot
does suggest curvature in the residuals which would indicate that the error team assumptions are not
satisfied. The scatter diagram for these DATA also indicates that the underlying relationship between x
and y may be curvilinear.
13- 260
d.
x = 14 s = 4.8765
 xi  x 
2
xi
xi x
(x i  x )2
6
11
15
18
20
-8
-3
1
4
6
64
9
1
16
36
126
∑ x  x 
i
hi
sy  yˆ
y i  yˆi
Standardized
Residual
.5079
.0714
.0079
.1270
.2857
.7079
.2714
.2079
.3270
.4857
2.6356
4.1625
4.3401
4.0005
3.4972
3.48
-2.47
-4.83
-1.60
5.22
1.32
-.59
-1.11
-.40
1.49
2
i
i
e.
Standardized Residuals
2
1.5
1
0.5
0
-0.5 0
5
10
15
20
25
-1
-1.5
x
The shape of the standardized residual plot is the same shape as the residual plot. The conclusions
reached in part (c) are also appropriate here.
48. a.
Using Excel's Regression tool, the estimated regression equation is:
yˆ = 2.322 + 0.6366x or yˆ = 2.32 + 0.64x
b.
13- 261
Residuals
-2
-3
-4
x
The assumption that the variance is the same for all values of x is questionable. The variance appears
to increase for larger values of x.
49. a.
Using Excel's Regression tool, the estimated regression equation is:
yˆ = 29.3991 + 1.5475x or yˆ = 29.40 + 1.55x
b.
Significant relationship: Significance F (or P-value) <  = .05
c.
13- 262
10
Residuals
5
0
0
5
10
15
20
25
-5
-10
-15
x
d.
The residual plot here leads us to question the assumption of a linear relationship between x and y.
Even though the relationship is significant at the  = .05 level, it would be extremely dangerous to
extrapolate beyond the range of the DATA. (e.g. x > 20).
50. a. From the solution to Exercise 9 we know that y$ = 80 + 4x
Residuals
8
-4
-6
-8
x
b.
51. a.
The assumptions concerning the error terms appear reasonable.
Let x = return on investment (ROE) and y = price/earnings (P/E) ratio.
yˆ  32.13  3.22x
b.
13- 263
Standardized Residuals
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
10
20
30
40
50
60
x
c.
There is an unusual trend in the residuals. The assumptions concerning the error term appear
questionable.
52.
No. Regression or correlation analysis can never prove that two variables are casually related.
53.
The estimate of a mean value is an estimate of the average of all y values associated with the same x.
The estimate of an individual y value is an estimate of only one of the y values associated with a
particular x.
54.
To determine whether or not there is a significant relationship between x and y. However, if we reject
B1 = 0, it does not imply a good fit.
55. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8624
R Square
0.7438
Adjusted R Square
0.7118
Standard Error
1.4193
Observations
10
ANOVA
MS
F
Regression
df
1
SS
46.7838
46.7838
23.2233
Residual
8
16.1162
2.0145
Total
9
62.9
Coefficients
Standard Error
t Stat
0.0013
P-value
Intercept
9.2649
1.0991
8.4293
2.99E-05
Shares
0.7105
0.1474
4.8191
0.0013
b.
Significance F
Since the p-value corresponding to F = 23.223 = .001 <  = .05, the relationship is significant.
13- 264
c.
r 2 = .744; a good fit. The least squares line explained 74.4% of the variability in Price.
d.
yˆ  9.26 .711(6)  13.53
56. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9586
R Square
0.9189
Adjusted R Square
0.9116
Standard Error
11.0428
Observations
13
ANOVA
SS
MS
Regression
df
1
15208.3982
15208.4 124.7162
Residual
11
1341.3849
121.9441
Total
12
16549.7831
Coefficients
Standard Error
t Stat
F
Significance F
2.42673E-07
P-value
Intercept
-3.8338
5.9031
-0.6495
Common Shares Outstanding (millions)
0.2957
0.0265
11.1676 2.43E-07
0.5294
b.
yˆ  3.83  .296(150)  40.57 ; approximately 40.6 million shares of options grants outstanding.
c.
r 2 = .919; a very good fit. The least squares line explained 91.9% of the variability in Options.
57. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.6852
R Square
0.4695
Adjusted R Square
0.4032
Standard Error
2.6641
Observations
10
ANOVA
MS
F
Regression
df
1
SS
50.2554
50.2554
7.0807
Residual
8
56.7806
7.0976
Total
9
107.036
13- 265
Significance F
0.0288
Coefficients
Standard Error
t Stat
P-value
Intercept
0.2747
0.9004
0.3051
0.7681
S&P 500
0.9498
0.3569
2.6609
0.0288
b.
Since the p-value = 0.029 is less than  = .05, the relationship is significant.
c.
r2 = .470. The least squares line does not provide a very good fit.
d.
Woolworth has higher risk with a market beta of 1.25.
58. a.
100
High Temperature
90
80
70
60
50
40
35
45
55
65
75
85
Low Temperature
b.
It appears that there is a positive linear relationship between the two variables.
c.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8837
R Square
0.7810
Adjusted R Square
0.7688
Standard Error
5.2846
Observations
20
ANOVA
SS
MS
F
Regression
df
1
1792.2734
1792.273
64.1783
Residual
18
502.6766
27.9265
Total
19
2294.95
13- 266
Significance F
2.40264E-07
Coefficients
Standard Error
t Stat
P-value
Intercept
23.8987
6.4812
3.6874
0.0017
Low
0.8980
0.1121
8.0111
2.4E-07
d.
Since the p-value corresponding to F = 64.18 = .000 <  = .05, the relationship is significant.
e.
r 2 = .781; a good fit. The least squares line explained 78.1% of the variability in high temperature.
f.
rxy  .781  .88
59.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9253
R Square
0.8562
Adjusted R Square
0.8382
Standard Error
4.2496
Observations
10
ANOVA
SS
MS
F
Regression
df
1
860.0509486
860.0509
47.6238
Residual
8
144.4740514
18.0593
Total
9
1004.525
Coefficients
Standard Error
t Stat
Significance F
P-value
Intercept
10.5280
3.7449
2.8113
0.0228
Weekly Usage
0.9534
0.1382
6.9010
0.0001
a.
yˆ = 10.528 + .9534x
b.
Since the p-value corresponding to F = 47.62 = .0001 <  = .05, we reject H0: 1 = 0.
c.
Using the PredInt macro, the 95% prediction interval is 28.74 to 49.52 or $2874 to $4952
d.
Yes, since the expected expense is $3913.
13- 267
0.0001
60. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8597
R Square
Adjusted R
Square
0.7391
Standard Error
1.4891
0.6739
Observations
6
ANOVA
MS
F
Regression
df
1
SS
25.1304
25.1304
11.3333
Residual
4
8.8696
2.2174
Total
5
34
Coefficients
Standard Error
t Stat
Significance F
0.0281
P-value
Intercept
22.1739
1.6527
13.4164
0.0002
Line Speed
-0.1478
0.0439
-3.3665
0.0281
b.
Since the p-value corresponding to F = 11.33 = .0281 <  = .05, the relationship is significant.
c.
r 2 = .739; a good fit. The least squares line explained 73.9% of the variability in the number of
defects.
d.
Using the PredInt macro, the 95% confidence interval is 12.294 to 17.271.
The scatter diagram follows:
10
8
Days Absent
61. a.
6
4
2
0
0
5
10
Distance to Work
A negative linear relationship appears to be reasonable.
13- 268
15
20
b.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8431
R Square
0.7109
Adjusted R Square
0.6747
Standard Error
1.2894
Observations
10
ANOVA
MS
F
Regression
df
1
SS
32.6993
32.6993
19.6677
Residual
8
13.3007
1.6626
Total
9
46
Coefficients
Standard Error
t Stat
Significance F
0.0022
P-value
Intercept
8.0978
0.8088
10.0119
8.41E-06
Distance to Work
-0.3442
0.0776
-4.4348
0.0022
c.
Since the p-value corresponding to F = 419.67 is .0022 <  = .05. We reject H0 : 1 = 0.
d.
r2 = .711. The estimated regression equation explained 71.1% of the variability in y; this is a
reasonably good fit.
e.
Using the PredInt macro, the 95% confidence interval is 5.195 to 7.559 or approximately 5.2 to 7.6
days.
62. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9341
R Square
0.8725
Adjusted R Square
0.8566
Standard Error
75.4983
Observations
10
ANOVA
df
Regression
Residual
Total
Intercept
Age
SS
1
8
9
312050
45600
357650
Coefficients
Standard Error
220
58.4808
131.6667
17.7951
13- 269
MS
312050
5700
F
Significance F
54.7456
7.62662E-05
t Stat
P-value
3.7619
0.0055
7.3990 7.63E-05
b.
Since the p-value corresponding to F = 54.75 is .000 <  = .05, we reject H0: 1 = 0.
c.
r2 = .873. The least squares line provided a very good fit.
d.
Using the PredInt macro, the 95% prediction interval is 559.5 to 933.9 or $559.50 to $933.90
63. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9369
R Square
0.8777
Adjusted R Square
0.8624
Standard Error
7.5231
Observations
10
ANOVA
SS
MS
F
Regression
df
1
3249.720752
3249.721
57.4182
Residual
8
452.7792483
56.5974
Total
9
3702.5
Coefficients
Standard Error
t Stat
Significance F
P-value
Intercept
5.8470
7.9717
0.7335
0.4842
Hours Studying
0.8295
0.1095
7.5775
6.44E-05
b.
Since the p-value corresponding to F = 57.42 is .000 <  = .05, we reject H0: 1 = 0.
c.
84.65 points
d.
Using the PredInt macro, the 95% prediction interval is 65.35 to 103.96
64. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.4659
R Square
0.2171
Adjusted R Square
0.1736
Standard Error
0.2088
Observations
20
13- 270
6.43959E-05
ANOVA
MS
F
Regression
df
1
SS
0.2175
0.2175
4.9901
Residual
18
0.7845
0.0436
Total
19
1.002
Coefficients
Intercept
Adjusted
Gross Income
Standard Error
t Stat
P-value
-0.4710
0.5842
-0.8061
0.4307
3.86778E-05
1.73143E-05
2.2339
0.0384
b.
Since the p-value = 0.0384 is less than  = .05, the relationship is significant.
c.
r2 = .217. The least squares line does not provide a very good fit.
d.
Using the PredInt macro, the 95% confidence interval is .7729 to .9927.
13- 271
Significance F
0.0384
Chapter 13
Multiple Regression
Learning Objectives
1.
Understand how multiple regression analysis can be used to develop relationships involving one
dependent variable and several independent variables.
2.
Be able to interpret the coefficients in a multiple regression analysis.
3.
Know the assumptions necessary to conduct statistical tests involving the hypothesized regression
model.
4.
Understand the role of Excel in performing multiple regression analysis.
5.
Be able to interpret and use Excel's Regression tool output to develop the estimated regression
equation.
6.
Be able to determine how good a fit is provided by the estimated regression equation.
7.
Be able to test for the significance of the regression equation.
8.
Understand how multicollinearity affects multiple regression analysis.
9.
Know how residual analysis can be used to make a judgement as to the appropriateness of the model,
identify outliers, and determine which observations are influential.
13- 272
Solutions:
1.
a.
b1 = .5906 is an estimate of the change in y corresponding to a 1 unit change in x1 when x2 is held
constant.
b2 = .4980 is an estimate of the change in y corresponding to a 1 unit change in x2 when x1 is held
constant.
2.
a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8124
R Square
0.6600
Adjusted R Square
0.6175
Standard Error
25.4009
Observations
10
ANOVA
SS
MS
F
Regression
df
1
10021.24739
10021.25
15.5318
Residual
8
5161.652607
645.2066
Total
9
15182.9
Coefficients
Standard Error
P-value
45.0594
25.4181
1.7727
0.1142
X1
1.9436
0.4932
3.9410
0.0043
An estimate of y when x1 = 45 is
yˆ = 45.0594 + 1.9436(45) = 132.52
b.
t Stat
Intercept
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.4707
R Square
0.2215
Adjusted R Square
0.1242
Standard Error
38.4374
Observations
10
14 - 273
Significance F
0.0043
ANOVA
SS
MS
F
Regression
df
1
3363.4142
3363.414
2.2765
Residual
8
11819.4858
1477.436
Total
9
15182.9
Coefficients
Standard Error
t Stat
Significance F
0.1698
P-value
Intercept
85.2171
38.3520
2.2220
0.0570
X2
4.3215
2.8642
1.5088
0.1698
An estimate of y when x2 = 15 is
yˆ = 85.2171 + 4.3215(15) = 150.04
c.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9620
R Square
0.9255
Adjusted R Square
0.9042
Standard Error
12.7096
Observations
10
ANOVA
SS
MS
F
Regression
df
2
14052.15497
7026.077
43.4957
Residual
7
1130.745026
161.535
Total
9
15182.9
Coefficients
Intercept
Standard Error
t Stat
Significance F
0.0001
P-value
-18.3683
17.97150328
-1.0221
0.3408
X1
2.0102
0.2471
8.1345
8.19E-05
X2
4.7378
0.9484
4.9954
0.0016
An estimate of y when x1 = 45 and x2 = 15 is
yˆ = -18.3683 + 2.0102(45) + 4.7378(15) = 143.16
3.
a.
b1 = 3.8 is an estimate of the change in y corresponding to a 1 unit change in x1 when x2, x3, and x4
are held constant.
b2 = -2.3 is an estimate of the change in y corresponding to a 1 unit change in x2 when x1, x3, and x4
are held constant.
13 - 274
b3 = 7.6 is an estimate of the change in y corresponding to a 1 unit change in x3 when x1, x2, and x4
are held constant.
b4 = 2.7 is an estimate of the change in y corresponding to a 1 unit change in x4 when x1, x2, and x3
are held constant.
4.
5.
a.
yˆ = 25 + 10(15) + 8(10) = 255; sales estimate: $255,000
b.
Sales can be expected to increase by $10 for every dollar increase in inventory investment when
advertising expenditure is held constant. Sales can be expected to increase by $8 for every dollar
increase in advertising expenditure when inventory investment is held constant.
a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.8078
R Square
0.6526
Adjusted R Square
0.5946
Standard Error
1.2152
Observations
8
ANOVA
df
SS
16.6401
8.8599
25.5
MS
16.6401
1.4767
Coefficients Standard Error
88.6377
1.5824
t Stat
56.0159
P-value
2.174E-09
0.4778
3.3569
0.0153
SS
23.4354
2.0646
25.5
MS
11.7177
0.4129
Regression
Residual
Total
Intercept
Television
Advertising ($1000s)
b.
1
6
7
1.6039
F
Significance F
11.2688
0.0153
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9587
R Square
0.9190
Adjusted R Square
0.8866
Standard Error
0.6426
Observations
8
ANOVA
df
Regression
Residual
Total
2
5
7
13 - 275
F
Significance F
28.3778
0.0019
Intercept
Television
Advertising ($1000s)
Newspaper
Advertising ($1000s)
6.
Coefficients Standard Error
83.2301
1.5739
t Stat
P-value
52.8825 4.57E-08
2.2902
0.3041
7.5319
0.0007
1.3010
0.3207
4.0567
0.0098
c.
No, it is 1.6039 in part (a) and 2.2902 above. In this exercise it represents the marginal change in
revenue due to an increase in television advertising with newspaper advertising held constant.
d.
Revenue = 83.2301 + 2.2902(3.5) + 1.3010(1.8) = $93.59 or $93,590
a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.5579
R Square
0.3112
Adjusted R Square
0.2620
Standard Error
7.0000
Observations
16
ANOVA
SS
MS
F
Regression
df
1
309.9516
309.9516
6.3255
Residual
14
686.0028
49.0002
Total
15
995.9544
Coefficients
b.
Standard Error
t Stat
P-value
Intercept
49.7800
19.1062
2.6054
0.0208
Curb Weight (lb.)
0.0151
0.0060
2.5151
0.0247
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9383
R Square
0.8804
Adjusted R Square
0.8620
Standard Error
3.0274
Observations
16
ANOVA
13 - 276
Significance F
0.0247
SS
MS
F
Regression
df
2
876.8049822
438.4025
47.8327
Residual
13
119.1493928
9.1653
Total
15
995.954375
Coefficients
c.
Standard Error
t Stat
Significance F
1.01401E-06
P-value
Intercept
80.4873
9.1393
8.8067
Curb Weight (lb.)
-0.0031
0.0035
-0.8968
7.69E-07
0.3861
Horsepower
0.1047
0.0133
7.8643
2.7E-06
yˆ = 80.4873 - 0.0031(2910) + 0.1047(296) = 102
Note to instructor: The Excel output shows that Curb Weight is not very significant (p-value = .3861)
given the effect of Horsepower. In Section 15.5, students will learn how to test for the significance
of the individual parameters.
7.
a.
The Excel output is shown below:
Regression STATISTICS
0.9121
Multiple R
R Square
0.8318
Adjusted R Square
0.7838
Standard Error
51.1363
Observations
10
ANOVA
df
Regression
Residual
Total
Intercept
Capacity
Comfort
8.
2
7
9
SS
MS
90548.0554 45274.03
18304.4446 2614.921
108852.5
Coefficients Standard Error
356.1208
197.1740
-0.0987
0.0459
122.8672
21.7998
t Stat
1.8061
-2.1524
5.6362
F
Significance F
17.3137
0.0019
P-value
0.1139
0.0684
0.0008
b.
b1 = -.0987 is an estimate of the change in the price with respect to a 1 cubic inch change in capacity
with the comfort rating held constant. b2 = 122.8672 is an estimate of the change in the price with
respect to a 1 unit change in the comfort rating with the capacity held constant.
c.
yˆ = 356.1208 - .0987(4500) + 122.8672(4) = $403
a.
The Excel output is shown below:
13 - 277
Regression STATISTICS
Multiple R
0.7629
R Square
0.5820
Adjusted R Square
0.5329
Standard Error
16.9770
Observations
20
ANOVA
df
9.
SS
MS
Regression
2
6823.2072
3411.604 11.8368
F
Residual
17
4899.7428
288.2202
Total
19
11722.95
Coefficients
Standard Error
Intercept
247.3579
110.4462
2.2396
0.0388
Safety Rating
-32.8445
13.9504
-2.3544
0.0308
Annual Expense Ratio (%)
34.5887
14.1294
2.4480
0.0255
b.
yˆ  247.3579  32.8445(7.5)  34.5887(2)  70.2
a.
The Excel output is shown below:
t Stat
Significance F
0.0006
P-value
Regression STATISTICS
Multiple R
0.6182
R Square
0.3821
Adjusted R Square
0.2998
Standard Error
12.4169
Observations
18
ANOVA
df
Regression
Residual
Total
Intercept
Average
Class Size
Combined
SAT Score
b.
2
15
17
SS
1430.4194
2312.6917
3743.111111
Coefficients Standard Error
26.7067
51.6689
MS
715.2097
154.1794
F
Significance F
4.6388
0.0270
t Stat
P-value
0.5169 0.6128
-1.4298
0.9931
-1.4397
0.1705
0.0757
0.0391
1.9392
0.0715
yˆ = 26.7067 - 1.4298(20) + 0.0757(1000) = 73.8 or 73.8%
13 - 278
Note to instructor: the Excel output shows that Average Class Size is not very significant (p-value =
.1705) given the effect of Combined SAT Score. In Section 15.5, students will learn how to test for
the significance of the individual parameters.
10. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9616
R Square
0.9246
Adjusted R Square
Standard Error
0.9188
226.6709
Observations
15
ANOVA
df
SS
MS
F
Regression
1
8192067.3605
8192067.3605
Residual
13
667935.9155
51379.6858
Total
14
8860003.2760
Coefficients
Standard Error
t Stat
Significance F
159.4418
1.13179E-08
P-value
Intercept
33.3352
83.0767
0.4013
Cars
7.9840
0.6323
12.6270 1.13179E-08
0.6947
b.
An increase of 1000 cars in service will result in an increase in revenue of $7.984 million.
c.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9703
R Square
0.9416
Adjusted R Square
Standard Error
0.9318
207.7292
Observations
15
ANOVA
df
SS
MS
Regression
2
8342186.4020
4171093.2010
Residual
12
517816.8740
43151.4062
Total
14
8860003.2760
Coefficients Standard Error
Intercept
105.9727
85.5166
13 - 279
t Stat
1.2392
F
96.6618
P-value
0.2390
Significance F
3.98523E-08
11. a.
Cars
8.9427
0.7746
11.5451 7.42955E-08
Locations
-0.1914
0.1026
-1.8652
SSE = SST - SSR = 6,724.125 - 6,216.375 = 507.75
SSR 6, 216.375

 .924
SST 6, 724.125
b.
R2 
c.
R2  1 (1 R2 ) n 1  1 (1.924) 10 1  .902
a
n  p 1
10  2 1
d.
The estimated regression equation provided an excellent fit.
12. a.
b.
c.
13. a.
R2 
SSR 14, 052.2

 .926
SST 15,182.9
R2  1 (1 R2 ) n 1  1 (1.926) 10 1  .905
a
n  p 1
10  2 1
Yes; after adjusting for the number of independent variables in the model, we see that 90.5% of the
variability in y has been accounted for.
R2 
SSR
SST

1760
 .975
1805
b.
R2  1 (1 R2 ) n 1  1 (1.975) 30 1  .971
a
n  p 1
30  4 1
c.
The estimated regression equation provided an excellent fit.
14. a.
b.
R2 
SSR 12, 000

 .75
SST 16, 000
R2  1 (1 R2 )
a
c.
15. a.
n 1  1.25 9  .68
n  p 1
7
The adjusted coefficient of determination shows that 68% of the variability has been explained by the
two independent variables; thus, we conclude that the model does not explain a large amount of
variability.
R2 
SSR 23.435

 .919
SST
25.5
R2  1 (1 R2 )
a
b.
0.0868
n 1  1 (1.919) 8 1  .887
n  p 1
8  2 1
Multiple regression analysis is preferred since both R2 and R2a show an increased percentage of the
variability of y explained when both independent variables are used.
13 - 280
16.
Note: the Excel output is shown with the solution to Exercise 6.
a.
No; R Square = .3112
b.
Multiple regression analysis is preferred since both R Square and Adjusted R Square show an
increased percentage of the variability of y explained when both independent variables are used.
17. a.
b.
18.
R Square = .3821
Adjusted R Square = .2998
The fit is not very good
Note: The Excel output is shown with the solution to Exercise 10.
a.
R Square = .9416
b.
The fit is very good.
19. a.
Adjusted R Square = .9318
MSR = SSR/p = 6,216.375/2 = 3,108.188
MSE  SSE  507.75  72.536
n  p 1 10  2 1
b.
F = MSR/MSE = 3,108.188/72.536 = 42.85
F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator)
Since F = 42.85 > F.05 = 4.74 the overall model is significant.
c.
t = .5906/.0813 = 7.26
t.025 = 2.365 (7 degrees of freedom)
Since t = 2.365 > t.025 = 2.365,  is significant.
d.
t = .4980/.0567 = 8.78
Since t = 8.78 > t.025 = 2.365,  is significant.
20.
A portion of the Excel output is shown below.
Regression STATISTICS
Multiple R
0.9620
R Square
0.9255
Adjusted R Square
0.9042
Standard Error
12.7096
Observations
10
ANOVA
df
SS
13 - 281
MS
F
Significance F
Regression
2
14052.15497
7026.077
Residual
7
1130.745026
161.535
Total
9
15182.9
Coefficients
Intercept
Standard Error
t Stat
43.4957
0.0001
P-value
-18.36826758
17.97150328
-1.0221
0.3408
X1
2.0102
0.2471
8.1345
8.19E-05
X2
4.7378
0.9484
4.9954
0.0016
a.
Since the p-value corresponding to F = 43.4957 is .0001 <  = .05, we reject H0:  =  = 0; there is
a significant relationship.
b.
Since the p-value corresponding to t = 8.1345 is .000 <  = .05, we reject H0:  = 0;  is
significant.
c.
Since the p-value corresponding to t = 4.9954 is .0016 <  = .05, we reject H0:  = 0;  is
significant.
21. a.
b.
22. a.
In the two independent variable case the coefficient of x1 represents the expected change in y
corresponding to a one unit increase in x1 when x2 is held constant. In the single independent variable
case the coefficient of x1 represents the expected change in y corresponding to a one unit increase in
x1.
Yes. If x1 and x2 are correlated, one would expect a change in the coefficient of x1 when x2 is
dropped from the model.
SSE = SST - SSR = 16000 - 12000 = 4000
s2 
SSE

n - p -1
MSR 
SSR
p
b.
4000
 571.43
7

12000
 6000
2
F = MSR/MSE = 6000/571.43 = 10.50
F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator)
Since F = 10.50 > F.05 = 4.74, we reject H0. There is a significant relationship among the variables.
23. a.
F = 28.38
F.01 = 13.27 (2 degrees of freedom, numerator and 1 denominator)
Since F > F.01 = 13.27, reject H0.
b.
Alternatively, the p-value of .002 leads to the same conclusion.
t = 7.53
13 - 282
t.025 = 2.571
Since t > t.025 = 2.571,  is significant and x1 should not be dropped from the model.
c.
t = 4.06
t.025 = 2.571
Since t > t.025 = 2.571,  is significant and x2 should not be dropped from the model.
24.
Note: The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9383
R Square
0.8804
Adjusted R Square
0.8620
Standard Error
3.0274
Observations
16
ANOVA
SS
MS
F
Regression
df
2
876.8049822
438.4025
47.8327
Residual
13
119.1493928
9.1653
Total
15
995.954375
Coefficients
a.
Standard Error
t Stat
80.4873
9.1393
8.8067
Curb Weight (lb.)
-0.0031
0.0035
-0.8968
0.3861
Horsepower
0.1047
0.0133
7.8643
2.7E-06
F = 47.8327
F.05 = 3.81 (2 degrees of freedom numerator and 13 denominator)
Since F = 47.8327 > F.05 = 3.81, we reject H0:  =  = 0.
Alternatively, since the p-value = .000 <  = .05 we can reject H0.
b.
P-value
Intercept
For Curb Weight:
H0:  = 0 Ha:   0
Since the p-value = 0.3861 >  = 0.05, we cannot reject H0
For Horsepower:
13 - 283
7.69E-07
Significance F
1.01401E-06
H0:  = 0
H a:    0
Since the p-value = 0.000 <  = 0.05, we can reject H0
25. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.6867
R Square
0.4715
Adjusted R Square
0.3902
Standard Error
5.4561
Observations
16
ANOVA
MS
F
Regression
df
2
345.2765787
SS
172.6383
5.7992
Residual
13
387.0034213
29.7695
Total
15
732.28
Coefficients Standard Error
t Stat
Significance F
0.0158
P-value
Intercept
6.0382
4.5893
1.3157
0.2110
Gross Profit Margin (%)
0.6916
0.2133
3.2421
0.0064
Sales Growth (%)
0.2648
0.1871
1.4154
0.1805
b.
Since the p-value = 0.0158 <  = 0.05, there is a significant relationship among the variables.
c.
For Gross Profit Margin (%): Since the p-value = 0.0064 <  = 0.05, Profit% is significant.
For Gross Profit Margin (%): Since the p-value = 0.1805 >  = 0.05, Sales% is not significant.
26.
Note: The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9703
R Square
0.9416
Adjusted R Square
Standard Error
Observations
0.9318
207.7292
15
ANOVA
13 - 284
df
SS
MS
F
Regression
2
8342186.4020
4171093.2010
Residual
12
517816.8740
43151.4062
Total
14
8860003.2760
Coefficients
Intercept
Standard Error
t Stat
96.6618
Significance F
3.98523E-08
P-value
105.9727
85.5166
1.2392
Cars
8.9427
0.7746
11.5451 7.42955E-08
0.2390
Locations
-0.1914
0.1026
-1.8652
0.0868
a.
Since the p-value corresponding to F = 96.6618 is 0.000 <  = .05, there is a significant relationship
among the variables.
b.
For Cars: Since the p-value = 0.000 <  = 0.05, Cars is significant
c.
For Location: Since the p-value = 0.0868 >  = 0.05, Location is not significant
27. a.
b.
yˆ = 29.1270 + .5906(180) + .4980(310) = 289.8150
The point estimate for an individual value is yˆ = 289.8150, the same as the point estimate of the
mean value.
28. a.
Using the PredInt macro, the 95% confidence interval is 132.16 to 154.16.
b.
Using the PredInt macro, the 95% prediction interval is 111.13 to 175.18.
29. a.
yˆ = 83.2 + 2.29(3.5) + 1.30(1.8) = 93.555 or $93,555
Note: In Exercise 5b, the Excel output also shows that b0 = 83.2301, b1 = 2.2902,
and b2 = 1.3010; hence, yˆ = 83.2301 + 2.2902x1 + 1.3010x2. Using this estimated regression
equation, we obtain
yˆ = 83.2301 + 2.2902(3.5) + 1.3010(1.8) = 93.588 or $93,588
The difference, $93,588 - $93,555 = $33, is simply due to the fact that additional significant digits
are used in the computations.
b.
Using the PredInt macro, the confidence interval estimate: 92.840 to 94.335 or $92,840 to $94,335
c.
Using the PredInt macro, the prediction interval estimate: 91.774 to 95.401 or $91,774 to $95,401
13 - 285
30. a.
Since Curb Weight is not statistically significant (see Exercise 24), we will use an estimated
regression equation which uses only Horsepower to predict the speed at 1/4 mile. The Excel output is
shown below:
Regression STATISTICS
Multiple R
0.9343
R Square
0.8730
Adjusted R Square
0.8639
Standard Error
3.0062
Observations
16
ANOVA
df
SS
MS
Regression
1
869.4340
869.434
Residual
14
126.5204
9.0372
Total
15
995.9544
Coefficients
Standard Error
t Stat
F
Significance F
96.2064
1.18632E-07
P-value
Intercept
72.6500
2.6555
27.3586 1.49E-13
Horsepower
0.0968
0.0099
9.8085 1.19E-07
Using the PredInt macro, the point estimate is a speed of 101.29 miles per hour.
b.
Using the PredInt macro, the 95% confidence interval is 99.490 to 103.089 miles per hour.
c.
Using the PredInt macro, the 95% prediction interval is 94.596 to 107.984 miles per hour.
31. a.
Using the PredInt macro, the 95% confidence interval is 58.37% to 75.03%.
b.
Using the PredInt macro, the 95% prediction interval is 35.24% to 90.59%.
32. a.
E(y) =  +  x1 +  x2 where
x2 = 0 if level 1 and 1 if level 2
b.
E(y) =  +  x1 + (0) =  +  x1
c.
E(y) =  +  x1 + (1) =  +  x1 + 
d.
 = E(y | level 2) - E(y | level 1)
 is the change in E(y) for a 1 unit change in x1 holding x2 constant.
13 - 286
33. a.
b.
two
E(y) =  +  x1 +  x2 +  x3 where
x2
0
1
0
c.
x3
0
0
1
Level
1
2
3
E(y | level 1) =  +  x1 + (0) + (0) = +  x1
E(y | level 2) =  +  x1 + (1) + (0) =  +  x1 + 
E(y | level 3) =  +  x1 + (0) + (0) =  +  x1 + 
 = E(y | level 2) - E(y | level 1)
 = E(y | level 3) - E(y | level 1)
 is the change in E(y) for a 1 unit change in x1 holding x2 and x3 constant.
34. a.
$15,300
b.
Estimate of sales = 10.1 - 4.2(2) + 6.8(8) + 15.3(0) = 56.1 or $56,100
c.
Estimate of sales = 10.1 - 4.2(1) + 6.8(3) + 15.3(1) = 41.6 or $41,600
35. a.
Let Type = 0 if a mechanical repair
Type = 1 if an electrical repair
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.2952
R Square
0.0871
Adjusted R Square
-0.0270
Standard Error
1.0934
Observations
10
ANOVA
df
Regression
Residual
Total
Intercept
Type
b.
SS
1
8
9
0.9127
9.5633
10.476
Coefficients Standard Error
3.45
0.5467
0.6167
0.7058
MS
0.9127
1.1954
F
Significance F
0.7635
0.4077
t Stat
P-value
6.3109
0.0002
0.8738
0.4077
The estimated regression equation did not provide a good fit. In fact, the p-value of .4077 shows that
the relationship is not significant for any reasonable value of .
13 - 287
c.
Person = 0 if Bob Jones performed the service and Person = 1 if Dave Newton performed the
service. The Excel output is shown below:
Regression STATISTICS
Multiple R
0.7816
R Square
0.6109
Adjusted R Square
0.5623
Standard Error
0.7138
Observations
10
ANOVA
df
Regression
Residual
Total
Intercept
Person
d.
36. a.
SS
1
8
9
MS
6.4
4.076
10.476
Coefficients Standard Error
4.62
0.3192
-1.6
0.4514
6.4
0.5095
F
Significance F
12.5613
0.0076
t Stat
P-value
14.4729 5.08E-07
-3.5442
0.0076
We see that 61.1% of the variability in repair time has been explained by the repair person that
performed the service; an acceptable, but not good, fit.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9488
R Square
0.900199692
Adjusted R Square
0.850299539
Standard Error
0.4174
Observations
10
ANOVA
df
Regression
Residual
Total
Intercept
Months Since Last Service
Type
Person
b.
SS
3
6
9
9.4305
1.0455
10.476
Coefficients Standard Error
1.8602
0.7286
0.2914
0.0836
1.1024
0.3033
-0.6091
0.3879
MS
3.1435
0.1743
F
Significance F
18.0400
0.0021
t Stat
P-value
2.5529
0.0433
3.4862
0.0130
3.6342
0.0109
-1.5701
0.1674
Since the p-value corresponding to F = 18.04 is .0021 <  = .05, the overall model is statistically
significant.
13 - 288
c.
37. a.
b.
The p-value corresponding to t = -1.57 is .1674 >  = .05; thus, the addition of Person is not
statistically significant. Person is highly correlated with Months (the SAMPLE correlation coefficient
is
-.691); thus, once the effect of Months has been accounted for, Person will not add much to the
model.
Let Position = 0 if a guard
Position = 1 if an offensive tackle.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.6895
R Square
0.4755
Adjusted R Square
0.4005
Standard Error
0.6936
Observations
25
ANOVA
MS
F
Regression
df
3
SS
9.1562
3.0521
6.3451
Residual
21
10.1014
0.4810
Total
24
19.2576
Coefficients
Standard Error
t Stat
Significance F
0.0031
P-value
Intercept
11.2233
4.5226
2.4816
0.0216
Position
0.7324
0.2893
2.5311
0.0194
Weight
0.0222
0.0104
2.1352
0.0447
Speed
-2.2775
0.9290
-2.4517
0.0231
c.
Since the p-value corresponding to F = 6.3451 is .0031 <  = .05, there is a significant relationship
between rating and the independent variables.
d.
The value of Adjusted R Square is .4005; the estimated regression equation did not provide a very
good fit.
e.
Since the p-value for Position is .0194 <  = .05, position is a significant factor in the player’s rating.
f.
yˆ  11.2233 .7324(1) .0222(300)  2.2775(5.1)  7.0
13 - 289
38. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9346
R Square
0.8735
Adjusted R Square
0.8498
Standard Error
5.7566
Observations
20
ANOVA
df
Regression
Residual
Total
Intercept
Age
Pressure
Smoker
3
16
19
SS
MS
3660.7396 1220.247
530.2104 33.1382
4190.95
Coefficients Standard Error
-91.7595
15.2228
1.0767
0.1660
0.2518
0.0452
8.7399
3.0008
F
Significance F
36.8230
2.06404E-07
t Stat
P-value
-6.0278 1.76E-05
6.4878 7.49E-06
5.5680 4.24E-05
2.9125
0.0102
b.
Since the p-value corresponding to t = 2.9125 is .0102 <  = .05, smoking is a significant factor.
c.
Using the PredInt macro, the point estimate is 34.27; the 95% prediction interval is 21.35 to 47.18.
Thus, the probability of a stroke (.2135 to .4718 at the 95% confidence level) appears to be quite
high. The physician would probably recommend that Art quit smoking and begin some type of
treatment designed to reduce his blood pressure.
39. a.
Job satisfaction can be expected to decrease by 8.69 units with a one unit increase in length of
service if the wage rate does not change. A dollar increase in the wage rate is associated with a 13.5
point increase in the job satisfaction score when the length of service does not change.
b.
40. a.
b.
41. a.
yˆ = 14.4 - 8.69(4) + 13.5(6.5) = 67.39
The expected increase in final college grade point average corresponding to a one point increase in
high school grade point average is .0235 when SAT mathematics score does not change. Similarly,
the expected increase in final college grade point average corresponding to a one point increase in
the SAT mathematics score is .00486 when the high school grade point average does not change.
yˆ = -1.41 + .0235(84) + .00486(540) = 3.19
The regression equation is
Regression STATISTICS
Multiple R
0.9681
R Square
0.9373
Adjusted R Square
0.9194
Standard Error
0.1298
Observations
10
13 - 290
ANOVA
df
SS
MS
2
1.7621
0.8810
Residual
7
0.1179
0.0168
Total
9
1.88
Coefficients Standard Error
b.
F
Regression
t Stat
Significance F
52.3053
6.17838E-05
P-value
Intercept
X1
-1.4053
0.0235
0.4848
0.0087
-2.8987
2.7078
0.0230
0.0303
X2
0.0049
0.0011
4.5125
0.0028
F.05 = 4.74 (2 degrees of freedom numerator and 7 degrees of freedom denominator)
F = 52.44 > F.05; significant relationship.
c.
R2 
SSR
 .937
SST
R2  1 (1.937)
a
9
 .919
7
good fit
d.
t.025 = 2.365 (7 DF)
for B: t = 2.71 > 2.365; reject H0 : B = 0
for B: t = 4.51 > 2.365; reject H0 : B = 0
42. a.
The regression equation is
Regression STATISTICS
Multiple R
0.9493
R Square
0.9012
Adjusted R Square
0.8616
Standard Error
3.773
Observations
8
ANOVA
df
SS
MS
Regression
2
648.83
324.415
Residual
5
71.17
14.234
Total
7
720
13 - 291
F
22.7916
Significance F
0.0031
Coefficients
b.
Standard Error
t Stat
P-value
Intercept
14.4
8.191
1.7580
0.1391
X1
-8.69
1.555
-5.5884
0.0025
X2
13.517
2.085
6.4830
0.0013
F.05 = 5.79 (5 degrees of freedom)
F = 22.79 > F.05; significant relationship.
c.
R2 
SSR
 .901
SST
R2  1 (1.901)
a
7
 .861
5
good fit
d.
t.025 = 2.571 (5 degrees of freedom)
for : t = -5.59 < -2.571; reject H0 :  = 0
for : t = 6.48 > 2.571; reject H0 :  = 0
43. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.5423
R Square
0.2941
Adjusted R Square
0.2689
Standard Error
19.4957
Observations
30
ANOVA
SS
MS
F
Regression
df
1
4433.856352
4433.856
11.6656
Residual
28
10642.25117
380.0804
Total
29
15076.10752
Coefficients
b.
Standard Error
t Stat
Significance F
P-value
Intercept
12.7928
6.6242
1.9312
0.0636
Book Value Per Share
2.2649
0.6631
3.4155
0.0020
The value of R Square is .2941; the estimated regression equation does not provide a good fit.
13 - 292
0.0020
c.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.7528
R Square
0.5667
Adjusted R Square
0.5346
Standard Error
15.5538
Observations
30
ANOVA
MS
F
Regression
df
2
8544.237582
SS
4272.119
17.6591
Residual
27
6531.869938
241.9211
Total
29
15076.10752
Coefficients
Standard Error
t Stat
Significance F
1.24768E-05
P-value
Intercept
5.8766
5.5448
1.0598
0.2986
Book Value Per Share
2.5356
0.5331
4.7562
5.87E-05
Return on Equity Per
Share (%)
0.4841
0.1174
4.1220
0.0003
Since the p-value corresponding to the F test is 0.000, the relationship is significant.
44. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9747
R Square
0.9500
Adjusted R Square
0.9319
Standard Error
2.1272
Observations
16
ANOVA
MS
F
Regression
df
4
946.1809495
SS
236.5452
52.2768
Residual
11
49.7734
4.5249
Total
15
995.954375
13 - 293
Significance F
4.33829E-07
Coefficients
Standard Error
t Stat
P-value
Intercept
97.5702
11.7926
8.2738
4.74E-06
Price ($1000s)
0.0693
0.0380
1.8210
0.0959
Curb Weight (lb.)
-0.0008
0.0026
-0.3145
0.7590
Horsepower
0.0590
0.0154
3.8235
0.0028
Zero to 60 (Seconds)
-2.4836
0.9601
-2.5869
0.0253
b.
Since the p-value corresponding to the F test is 0.000, the relationship is significant.
c.
Since the p-values corresponding to the t test for both Horsepower (p-value = .0028) and Zero to 60
(p-value = .0253) are less than .05, both of these independent variables are significant.
d.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9648
R Square
0.9309
Adjusted R Square
0.9203
Standard Error
2.3011
Observations
16
ANOVA
SS
MS
F
Regression
df
2
927.1181
463.559
87.5449
Residual
13
68.8363
5.2951
Total
15
995.9544
t Stat
P-value
Intercept
Coefficients
103.1028
Standard Error
9.4478
10.9129
6.47E-08
Horsepower
0.0558
0.0145
3.8436
0.0020
Zero to 60 (Seconds)
-3.1876
0.9658
-3.3006
0.0057
13 - 294
Significance F
2.86588E-08
e.
The standardized residual plot is shown below:
Standard Residuals
3
2
1
0
-1
80
90
100
110
120
-2
Predicted y
There is an unusual trend in the plot and one observation appears to be an outlier.
f.
The Excel output is shown below:
Household Exposures
45. a.
The Excel output indicates that observation 2 is an outlier
700
600
500
400
300
200
100
0
0
20
40
60
Times Ad Aired
b.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9829
R Square
0.9660
Adjusted R Square
Standard Error
Observations
0.9618
31.70350482
10
13 - 295
80
100
ANOVA
MS
F
Regression
df
1
228519.8983
SS
228519.9
227.3576
Residual
8
8040.897745
1005.112
Total
9
236560.796
Coefficients Standard Error
t Stat
Significance F
3.70081E-07
P-value
Intercept
53.2448
16.5334
3.2204
0.0122
Times Ad Aired
6.7427
0.4472
15.0784
3.7E-07
Since the p-value is 0.000, the relationship is significant.
c.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.9975
R Square
0.9949
Adjusted R Square
0.9935
Standard Error
13.0801
Observations
10
ANOVA
MS
F
Regression
df
2
235363.1688
SS
117681.6
687.836
Residual
7
1197.62722
171.0896
Total
9
236560.796
Coefficients Standard Error
t Stat
9.23264E-09
P-value
Intercept
73.0634
7.5067
9.7331
2.56E-05
Times Ad Aired
5.0368
0.3268
15.4131
1.17E-06
101.1129
15.9877
6.3244
0.0004
BigAds
Significance F
d.
The p-value corresponding to the t test for BigAds is 0.0004; thus, the dummy variable is significant.
e.
The dummy variable enables us to fit two different lines to the DATA; this approach is referred to
as piecewise linear approximation.
13 - 296
46. a.
The Excel output is shown below:
Regression STATISTICS
Multiple R
0.6059
R Square
0.3671
Adjusted R Square
0.3445
Standard Error
5.4213
Observations
30
ANOVA
SS
MS
F
Regression
df
1
477.2478
477.2478
16.2385
Residual
28
822.9189
29.3900
Total
29
1300.1667
Coefficients Standard Error
t Stat
Significance F
0.0004
P-value
Intercept
38.7718
4.3481
8.9170
1.13E-09
Suggested Retail Price ($)
0.0008
0.0002
4.0297
0.0004
Since the p-value corresponding to F = 16.24 is .0004 <  = .05, there is a significant relationship
between the resale value (1%) and the suggested price.
b.
R-Square = .3671; not a very good fit.
c.
Let Type1 = 0 and Type2 = 0 if a small pickup; Type1 = 1 and Type2 = 0 if a full-size pickup; and
Type1 = 0 and Type2 = 1 if a sport utility.
The Excel output using Type1, Type2, and Price is shown below:
Regression STATISTICS
Multiple R
0.7940
R Square
0.6305
Adjusted R Square
0.5879
Standard Error
4.2985
Observations
30
ANOVA
MS
F
Regression
df
3
819.7710938
SS
273.257
14.7892
Residual
26
480.3955729
18.4768
Total
29
1300.166667
13 - 297
Significance F
8.11183E-06
Coefficients Standard Error
d.
t Stat
P-value
Intercept
42.5539
3.5618
11.9472
4.62E-12
Type1
9.0903
2.2476
4.0444
0.0004
Type2
Suggested Retail
Price ($)
7.9172
2.1634
3.6596
0.0011
0.0003
0.0002
1.8972
0.0690
Since the p-value corresponding to F = 14.7892 is .000 <  = .05, there is a significant relationship
between the resale value and the independent variables. Note that individually, Suggested retail Price
is not significant at the .05 level of significance. If we rerun the regression using just Type1 and
Type2 the value of Adjusted R-Square decreases to .5482, a drop of approximately .04. Thus, it
appears that for these DATA, the type of vehicle is the strongest predictor of the resale value.
13 - 298
Chapter 14
Statistical Methods for Quality Control
Learning Objectives
1.
Learn about the importance of quality control and how statistical methods can assist in the quality
control process.
2.
Learn about acceptance sampling procedures.
3.
Know the difference between consumer’s risk and producer’s risk.
4.
Be able to use the binomial probability distribution to develop acceptance sampling plans.
5.
Know what is meant by multiple sampling plans.
6.
Be able to construct quality control charts and understand how they are used for statistical process control.
7.
Know the definitions of the following terms:
producer's risk
assignable causes
consumer's risk
common causes
acceptance sampling
control charts
acceptable criterion
upper control limit
operating characteristic curve
lower control limit
13 - 299
Solutions:
1.
a.
b.
For n = 4
UCL =  + 3( /
n ) = 12.5 + 3(.8 /
4 ) = 13.7
LCL =  - 3( /
n ) = 12.5 - 3(.8 /
4 ) = 11.3
For n = 8
UCL =  + 3(.8 / 8 ) = 13.35
LCL =  - 3(.8 / 8 ) = 11.65
For n = 16
UCL =  + 3(.8 / 16 ) = 13.10
LCL =  - 3(.8 / 16 ) = 11.90
2.
c.
UCL and LCL become closer together as n increases. If the process is in control, the larger
SAMPLEs should have less variance and should fall closer to 12.5.
a.

677.5
 5.42
25(5)
UCL =  + 3( /
b.
LCL =  - 3( /
3.
n ) = 5.42 + 3(.5 / 5 ) = 6.09
n ) = 5.42 - 3(.5 / 5 ) = 4.75
135
 0.0540
25(100)
a.
p 
b.
 p 
p(1 p)

n
0.0540(0.9460)
 0.0226
100
UCL = p + 3  p = 0.0540 + 3(0.0226) = 0.1218
c.
LCL = p - 3  p = 0.0540 -3(0.0226) = -0.0138
Use LCL = 0
4.
R Chart:
UCL = RD4 = 1.6(1.864) = 2.98
LCL = RD3 = 1.6(0.136) = 0.22
x Chart:
UCL = x  A2 R = 28.5 + 0.373(1.6) = 29.10
LCL = x  A2 R = 28.5 - 0.373(1.6) = 27.90
5.
a.
UCL =  + 3( /
LCL =  - 3( /
n ) = 128.5 + 3(.4 / 6 ) = 128.99
n ) = 128.5 - 3(.4 / 6 ) = 128.01
14 - 300
6.
b.
x  xi / n 
772.4
 128.73
6
in control
c.
x  xi / n 
774.3
 129.05
6
out of control
20.12  19.90
 20.01
2
Process Mean =
UCL =  + 3( /
n ) = 20.01 + 3( /
5 ) = 20.12
Solve for :

(20.12  20.01) 5
 0.082
3
7.
SAMPLE
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Observations
31
26
25
17
38
41
21
32
41
29
26
23
17
43
18
30
28
40
18
22
42
18
30
25
29
42
17
26
34
17
31
19
24
35
25
42
36
29
29
34
28
35
34
21
35
36
29
28
33
30
40
25
32
17
29
31
32
31
28
26
xi
33.67
26.33
29.67
21.00
34.00
39.67
22.33
28.67
36.00
25.33
32.33
22.33
24.33
31.67
24.00
34.33
32.00
33.33
25.00
27.33
R = 11.4 and x  29.17
R Chart:
UCL = RD4 = 11.4(2.575) = 29.35
LCL = RD3 = 11.4(0) = 0
x Chart:
UCL = x  A2 R = 29.17 + 1.023(11.4) = 40.8
LCL = x  A2 R = 29.17 - 1.023(11.4) = 17.5
13 - 301
Ri
14
17
9
8
9
6
12
6
8
13
14
6
15
26
11
12
8
11
11
12
R Chart:
30
UCL = 29.3
20
R = 11.4
10
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
LCL = 0
SAMPLE Number
x Chart:
UCL = 40.8
40
=
30
x = 29.17
20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
SAMPLE Number
8.
141
 0.0470
20(150)
a.
p 
b.
 p 
p(1 p)

n
0.0470(0.9530)
 0.0173
150
UCL = p + 3  p = 0.0470 + 3(0.0173) = 0.0989
LCL = p - 3  p = 0.0470 -3(0.0173) = -0.0049
13 - 302
LCL = 17.5
c.
Use LCL = 0
12
p 
 0.08
150
Process should be considered in control.
d.
p = .047, n = 150
UCL = np + 3
np(1 p) = 150(0.047) + 3 150(0.047)(0.953) = 14.826
LCL = np - 3 np(1 p) = 150(0.047) - 3 150(0.047)(0.953) = -0.726
Thus, the process is out of control if more than 14 defective packages
are found in a SAMPLE of 150.
9.
e.
Process should be considered to be in control since 12 defective packages were found.
f.
The np chart may be preferred because a decision can be made by simply counting the number of
defective packages.
a.
Total defectives: 165
p
b.
165
 0.0413
20(200)
 p 
0.0413(0.9587)
 0.0141
200
p(1 p)

n
UCL = p + 3  p = 0.0413 + 3(0.0141) = 0.0836
LCL = p - 3  p = 0.0413 + 3(0.0141) = -0.0010
Use LCL = 0
20
 0.10
200
c.
p 
d.
p = .0413, n = 200
Out of control
UCL = np + 3
np(1 p) = 200(0.0413) + 3 200(0.0413)(0.9587) = 16.702
LCL = np - 3 np(1 p) = 200(0.0413) - 3 200(0.0413)(0.9587) = 0.1821
e.
10.
The process is out of control since 20 defective pistons were found.
f ( x) 
n!
p x (1  p)nx
x!(n  x)!
When p = .02, the probability of accepting the lot is
13 - 303
f (0) 
25!
0!(25  0)!
(0.02)0 (1 0.02)25  0.6035
When p = .06, the probability of accepting the lot is
f (0) 
11. a.
25!
0!(25  0)!
(0.06) 0 (1  0.06)25  0.2129
Using binomial probabilities with n = 20 and p0 = .02.
P (Accept lot) = f (0) = .6676
Producer’s risk:  = 1 - .6676 = .3324
b.
P (Accept lot) = f (0) = .2901
Producer’s risk:  = 1 - .2901 = .7099
12.
At p0 = .02, the n = 20 and c = 1 plan provides
P (Accept lot) = f (0) + f (1) = .6676 + .2725 = .9401
Producer’s risk:  = 1 - .9401 = .0599
At p0 = .06, the n = 20 and c = 1 plan provides
P (Accept lot) = f (0) + f (1) = .2901 + .3703 = .6604
Producer’s risk:  = 1 - .6604 = .3396
For a given SAMPLE size, the producer’s risk decreases as the acceptance number c is increased.
13. a.
Using binomial probabilities with n = 20 and p0 = .03.
P(Accept lot) = f (0) + f (1)
= .5438 + .3364 = .8802
Producer’s risk:  = 1 - .8802 = .1198
b.
With n = 20 and p1 = .15.
P(Accept lot) = f (0) + f (1)
= .0388 + .1368 = .1756
Consumer’s risk:  = .1756
c.
The consumer’s risk is acceptable; however, the producer’s risk associated with the n = 20, c = 1 plan is
a little larger than desired.
13 - 304
14.
c
0
1
2
P (Accept)
p0 = .05
.5987
.9138
.9884
Producer’s
Risk 
.4013
.0862
.0116
P (accept)
p1 = .30
.0282
.1493
.3828
Consumer’s
Risk 
.0282
.1493
.3828
(n = 15)
0
1
2
3
.4633
.8291
.9639
.9946
.5367
.1709
.0361
.0054
.0047
.0352
.1268
.2968
.0047
.0352
.1268
.2968
(n = 20)
0
1
2
3
.3585
.7359
.9246
.9842
.6415
.2641
.0754
.0158
.0008
.0076
.0354
.1070
.0008
.0076
.0354
.1070
(n = 10)
The plan with n = 15, c = 2 is close with  = .0361 and  = .1268. However, the plan with n = 20,
c = 3 is necessary to meet both requirements.
15. a.
P (Accept) shown for p values below:
c
0
1
2
p = .01
.8179
.9831
.9990
p = .05
.3585
.7359
.9246
p = .08
.1887
.5169
.7880
p = .10
.1216
.3918
.6770
p = .15
.0388
.1756
.4049
The operating characteristic curves would show the P (Accept) versus p for each value of c.
b.
P (Accept)
c
0
1
2
16. a.

x
20
At p0 = .01
.8179
.9831
.9990

1908
Producer’s Risk
.1821
.0169
.0010
At p1 = .08
.1887
.5169
.7880
Consumer’s Risk
.1887
.5169
.7880
 95.4
20
b.
UCL =  + 3( /
LCL =  - 3( /
c.
n ) = 95.4 + 3(.50 / 5 ) = 96.07
n ) = 95.4 - 3(.50 / 5 ) = 94.73
No; all were in control
13 - 305
17. a.
For n = 10
UCL =  + 3( /
LCL =  - 3( /
n ) = 350 + 3(15 / 10 ) = 364.23
n ) = 350 - 3(15 / 10 ) = 335.77
For n = 20
UCL = 350 + 3(15 /
20 ) = 360.06
LCL = 350 - 3(15 /
20 ) = 339.94
For n = 30
UCL = 350 + 3(15 /
30 ) = 358.22
LCL = 350 - 3(15 /
30 ) = 343.78
b.
Both control limits come closer to the process mean as the SAMPLE size is increased.
c.
The process will be declared out of control and adjusted when the process is in control.
d.
The process will be judged in control and allowed to continue when the process is out of control.
e.
All have z = 3 where each tail area = 1 - .9986 = .0014
P (Type I) = 2 (.0014) = .0028
f.
18.
The Type II error probability is reduced as the SAMPLE size is increased.
R Chart:
UCL = RD4 = 2(2.115) = 4.23
LCL = RD3 = 2(0) = 0
x Chart:
UCL = x  A2 R = 5.42 + 0.577(2) = 6.57
LCL = x  A2 R = 5.42 - 0.577(2) = 4.27
Estimate of Standard Deviation:
$ 
R

d2
19.
R = 0.665
2
 0.86
2.326
x = 95.398
x Chart:
UCL = x  A2 R = 95.398 + 0.577(0.665) = 95.782
LCL = x  A2 R = 95.398 - 0.577(0.665) = 95.014
R Chart:
13 - 306
UCL = RD4 = 0.665(2.115) = 1.406
LCL = RD3 = 0.665(0) = 0
The R chart indicated the process variability is in control. All SAMPLE ranges are within the control
limits. However, the process mean is out of control. SAMPLE 11 ( x = 95.80) and SAMPLE 17
( x =94.82) fall outside the control limits.
20.
R = .053
x = 3.082
x Chart:
UCL = x  A2 R = 3.082 + 0.577(0.053) = 3.112
LCL = x  A2 R = 3.082 - 0.577(0.053) = 3.051
R Chart:
UCL = RD4 = 0.053(2.115) = 0.1121
LCL = RD3 = 0.053(0) = 0
All DATA points are within the control limits for both charts.
21. a.
.0 8
UCL
.0 6
.0 4
.0 2
LCL
0
Warning: Process should be checked. All points are within control limits; however, all points are also
greater than the process proportion defective.
13 - 307
b.
25
UCL
24
23
LCL
22
Warning: Process should be checked. All points are within control limits yet the trend in points show a
movement or shift toward UCL out-of-control point.
22. a.
p = .04
 p 
0.04(0.96)
p(1 p)

 0.0139
n
200
UCL = p + 3  p = 0.04 + 3(0.0139) = 0.0817
LCL = p - 3  p = 0.04 - 3(0.0139) = -0.0017
Use LCL = 0
b.
13 - 308
out of control
UCL
(.082)
.04
LCL
(0)
For month 1 p = 10/200 = 0.05. Other monthly values are .075, .03, .065, .04, and .085. Only the last
month with p = 0.085 is an out-of-control situation.
23. a.
Use binomial probabilities with n = 10.
At p0 = .05,
P(Accept lot) = f (0) + f (1) + f (2)
= .5987 + .3151 + .0746 = .9884
Producer’s Risk:  = 1 - .9884 = .0116
At p1 = .20,
P(Accept lot) = f (0) + f (1) + f (2)
= .1074 + .2684 + .3020 = .6778
Consumer’s risk:  = .6778
b.
The consumer’s risk is unacceptably high. Too many bad lots would be accepted.
c.
Reducing c would help, but increasing the SAMPLE size appears to be the best solution.
24. a.
P (Accept) are shown below: (Using n = 15)
f (0)
f (1)
p = .01
.8601
.1303
.9904
p = .02
.7386
.2261
.9647
p = .03
.6333
.2938
.9271
p = .04
.5421
.3388
.8809
p = .05
.4633
.3658
.8291
 = 1 - P (Accept)
.0096
.0353
.0729
.1191
.1709
Using p0 = .03 since  is close to .075. Thus, .03 is the fraction defective where the producer will
tolerate a .075 probability of rejecting a good lot (only .03 defective).
b.
f (0)
p = .25
.0134
13 - 309
f (1)
.0668
.0802
 =
25. a.
P (Accept) when n = 25 and c = 0. Use the binomial probability function with
n!
f ( x) 
p x (1  p)nx
x!(n  x)!
or
25! 0
f (0) 
p (1 p)25  (1 p)25
0!25!
If
p = .01
p = .03
p = .10
p = .20
f (0)
.7778
.4670
.0718
.0038
b.
1.0
P (Accept)
.8
.6
.4
.2
.00
.02
.04
.06
.08
.10
.12
Percent Defective
c.
1 - f (0) = 1 - .778 = .222
13 - 310
.14
.16
.18
.20