MATH 117 - Elements of Statistics
Fall 2014 Final Exam Review for Chapters 1 – 11, 15
From the textbook: Statistics: The Art and Science of Learning from Data (3rd ed.), by Alan Agresti and Christine Franklin
1.
UW Student Survey In a University of Wisconsin (UW)
study about alcohol abuse among students, 100 of the
40,858 members of the student body in Madison were
sampled and asked to complete a questionnaire. One
question asked was, “On how many days in the past
week did you consume at least one alcoholic drink?”
a. Identify the population and the sample.
b. For the 40,858 students at UW, one characteristic of
interest was the percentage who would respond
“zero” to this question. For the 100 students
sampled, suppose 29% gave this response. Does this
mean that 29% of the entire population of UW
students would make this response? Explain.
c. Is the numerical summary of 29% a sample statistic,
or a population parameter?
2.
Median versus mean sales price of new homes In
December 2010, the US Census Bureau reported that the
median US sales price of new homes was $241,500.
Would you expect the mean sales price to have been
higher or lower? Explain.
3.
Female Heights According to a recent report from the
US National Center for Health Statistics, females
between 25 and 34 years of age have a bell-shaped
distribution for height, with a mean of 65 inches and
standard deviation of 3.5 inches.
a. Give an interval within which about 95% of the
heights fall.
b. What is the height for a female who is 3 standard
deviations below the mean? Would this be a rather
unusual height? Why?
4.
High School Graduation Rates The distribution of high
school graduation rates in the United States in 2004 had
a minimum value of 78.3 (Texas), first quartile of 83.6,
median of 87.2, third quartile of 88.8, and maximum
value of 92.3 (Minnesota) (Statistical Abstract of the
United States, 2006).
a. Report the range and the interquartile range.
b. Would a box plot show any potential outliers?
Explain.
5.
Blood Pressure A World Health Organization study (the
MONICA project) of health in various countries reported
that in Canada, systolic blood pressure readings have a
mean 121 and a standard deviation of 16. A reading
above 140 is considered to be high blood pressure.
a. What is the z-score for a blood pressure reading of
140? How is this z-score interpreted?
b. The systolic blood pressure values have a bellshaped distribution. Report an interval within which
about 95% of the systolic blood pressure values fall.
6.
Life After Death for Males and Females In a recent General
Social Survey, respondents answered the question, “Do you
believe in a life after death?” The table shows the responses
cross-tabulated with gender.
Opinion About Life After Death by Gender
Gender
Opinion About Life After Death
Yes
No
Male
621
187
Female
834
145
a.
b.
7.
Construct a table of conditional proportions.
Summarize results. Is there much difference between
responses of males and females?
Women in Government and Economic Life The OECD
(Organization for Economic Cooperation and Development)
consists of advanced, industrialized countries that accept the
principles of representative democracy and a free market
economy. For the nations outside of Europe that are in the
OECD, the table shows UN data from 2007 on the percentage
of seats in parliament held by women and female economic
activity as a percentage of the male rate.
a. Treating women in parliament as the response variable,
prepare a scatterplot and find the correlation. Explain
how the correlation relates to the trend shown in the
scatterplot.
b. Use software or a calculator to find the regression
equation. Explain why the y-intercept is not meaningful.
c. Find the predicted value and residual for the United
States. Interpret the residual.
d. With UN data for all 23 OECD nations, the correlation
between these variables is 0.56. For women in
parliament, the mean is 26.5% and the standard
deviation is 9.8%. For female economic activity, the
mean is 76.8 and the standard deviation is 7.7. Find the
prediction equation, treating women in parliament as
the response variable.
Nation
Iceland
Australia
Canada
Japan
United States
New Zealand
Women in
Parliament (%)
33.3
28.3
24.3
10.7
15.0
32.2
Female Economic
Activity
87
79
83
65
81
81
2
8.
9.
Predict Crime Using Poverty A recent analyses of data for
the 50 US states on y = violent crime rate (measured as
number of violent crimes per 100,000 people in the state)
and x = poverty rate (percent of people in the state living
at or below the poverty level) yielded the regression
equation 𝑦̂ = 209.9 + 25.5𝑥.
a. Interpret the slope.
b. The state poverty rates ranged from 8.0 (for Hawaii)
to 24.7 (for Mississippi). Over this range, find the
range of predicted values for the violent crime rate.
c. Would the correlation between these variables be
positive or negative? Why?
Football Discipline A large southern university had
problems with 17 football players being disciplined for
team rule violations, arrest charges, and possible NCAA
violations. The online Atlanta Journal Constitution ran a
poll with the question, “Has the football coach lost control
over his players?” having possible responses, “Yes, he’s
been too lenient,” and “No, he can’t control everything
teenagers do.”
a. Was there potential for bias in this study? If so, what
types of bias?
b. The poll results after two days were
Yes
No
6012
487
93%
7%
Does this large sample size guarantee that the results are
unbiased? Explain.
10. Video games mindless? “Playing video games not so
mindless.” This was the headline of a CNN news report
about a study that concluded that young adults who
regularly play video games demonstrated better visual
skills than young adults who do not play regularly. Sixteen
young men volunteered to take a series of tests that
measured their visual skills; those who had played video
games in the previous six months performed better on the
test than those who hadn’t played.
a. What are the explanatory and response variables?
b. Was this an observational study or an experiment?
Explain.
c. Specify a potential lurking variable. Explain.
11. Peyton Manning Completions As of the end of the 2010
NFL season, Indianapolis Colts quarterback Peyton
Manning, throughout his 13-year career, completed 65%
of all of his pass attempts. Suppose the probability each
pass attempted in the next season has probability 0.65 of
being completed.
a. Does this mean that if we watch Manning throw 100
times in the upcoming season, he would complete
exactly 65 passes? Explain.
b. Explain what this probability means in terms of
observing him over a longer period, say for 1000 passes
over the course of the next two seasons assuming
Manning is still at his typical playing level. Would it be
surprising if his completion percentage over a large
number of passes differed significantly from 65%?
12. Driver’s Exam Three 15-year-old friends with no particular
background in driver’s education decide to take the written
part of the Georgia Driver’s Exam. Each exam was graded as
a pass (P) or a failure (F).
a. How many outcomes are possible for the grades
received by the three friends together? Using a tree
diagram, list the sample space.
b. If the outcomes in the sample space in part a are
equally likely, find the probability that all three pass the
exam.
c. In practice, the outcomes in part a are not equally likely.
Suppose that statewide 70% of 15-year-olds pass the
exam. If these three friends were a random sample of
their age group, find the probability that all three pass.
d. In practice, explain why probabilities that apply to a
random sample are not likely to be valid for a sample of
three friends.
13. Grandparents Let X = the number of living grandparents
that a randomly selected adult American has. According to
recent General Social Surveys, its probability distribution is
approximately P(0) = 0.71, P(1) = 0.15, P(2) = 0.09, P(3) =
0.03, P(4) = 0.02.
a. Does this refer to a discrete or a continuous random
variable? Why?
b. Show that the probabilities satisfy the two conditions
for a probability distribution.
c. Find the mean of this probability distribution.
14. Z-score and Tail Probability
a. Find the z-score for the number that is less than only 1%
of the values of a normal distribution. Sketch a graph to
show where this value is.
th
b. Find the z-scores corresponding to the (i) 90 and (ii)
th
99 percentiles of a normal distribution.
15. Cloning Butterflies The wingspans of recently cloned
monarch butterflies follow a normal distribution with mean
9 inches and standard deviation 0.75 inches. What
proportion of the butterflies has a wingspan
a. less than 8 inches?
b. wider than 10 inches?
c. between 8 and 10 inches?
d. ten percent of the butterflies have a wingspan wider
than how many inches?
3
16. Exam Performance An exam consists of 50 multiplechoice questions. Based on how much you studied, for any
given question you think you have a probability of p = 0.70
of getting the correct answer. Consider the sampling
distribution of the sample proportion of the 50 questions
on which you get the correct answer.
a. Find the mean and standard deviation of the sampling
distribution of this proportion.
b. What do you expect for the shape of the sampling
distribution? Why?
c. If truly p = 0.70, would it be very surprising if you got
correct answers on only 60% of the questions? Justify
your answer by using the normal distribution to
approximate the probability of a sample proportion of
0.60 or less.
20. Grandpas Using Email When the GSS asked in 2004,
“About how many hours per week do you spend sending
and answering email?” the eight males in the sample of age
at least 75 responded:
0, 1, 2, 2, 7, 10, 14, 15
a. The TI-83+/84 screen shot shows results of a statistical
analysis for finding a 90% confidence interval. Identify
the results shown and explain how to interpret them.
b. Find and interpret a 90% confidence interval for the
population mean.
Explain why the population distribution may be skewed
right. If this is the case, is the interval you obtained in
part b useless, or is it still valid? Explain.
TInterval
(2.34, 10.41)
𝑥 = 6.38
𝑆𝑥 = 6.02
𝑛 = 8.00
17. Aunt Erma’s Restaurant In Example 5 about Aunt Erma’s
Restaurant, the daily sales follow a probability distribution
that has a mean of µ = $900 and a standard deviation of σ
21. US Popularity In 2007, a poll conducted for the BBC of
= $300. This past week the daily sales for the seven days
28,389 adults in 27 countries found that the United States
had a mean of $980 and a standard deviation of $276.
had fallen sharply in world esteem since 2001
a. Identify the mean and standard deviation of the
(www.globescan.com). The United States was rated third
population distribution.
most negatively (after Israel and Iran), with 30% of those
b. Identify the mean and standard deviation of the data
polled saying they had a positive image of the United States.
distribution. What does the standard deviation
a. In Canada, for a random sample of 1008 adults, 56%
describe?
said the United States is mainly a negative influence in
c. Identify the mean and the standard deviation of the
the world. True or false: The 99% confidence interval of
sampling distribution of the sample mean for samples
(0.52, 0.60) means that we can be 99% confident that
of seven daily sales. What does this standard
between 52% and 60% of the population of all Canadian
deviation describe?
adults have a negative image of the United States.
b. In Australia, for a random sample of 1004 people, 66%
18. Approval Rating for President Obama A July 2011 Gallup
said the United States is mainly a negative influence in
poll based on the responses of 1500 adults indicated that
the world. True or false: The 95% confidence interval of
46% of Americans approve of the job Barack Obama is
(0.63, 0.69) means that for a random sample of 100
doing as president. One way to summarize the findings of
people, we can be 95% confident that between 63 and
the poll is by saying, “It is estimated that 46% of American
69 people in the sample have a negative image of the
approve of the job Barack Obama is doing as president.
United States.
This estimate has a margin of error of plus or minus 3%.”
How could you explain the meaning of this to someone
22. Driving After Drinking In December 2004, a report based on
who has not taken a statistics course?
the National Survey on Drug Use and Health estimated that
20% of all Americans of ages 16 to 20 drove under the
19. Vegetarianism Time magazine (July 15, 2002) quoted a
influence of drugs or alcohol in the previous year (AP,
poll of 10,000 Americans in which only 4% said they were
December 30, 2004). A public health unit in Wellington, New
vegetarian.
Zealand, plans a similar survey for young people of that age
a. What has to be assumed about this sample to
in New Zealand. They want a 95% confidence interval to
construct a confidence interval for the population
have a margin of error of 0.04.
proportion of vegetarians?
a. Find the necessary sample size if they expect results
b. Construct a 99% confidence interval for the
similar to those in the United States.
population proportion. Explain why the interval is so
b. Suppose that in determining the sample size, they use
narrow, even though the confidence level is high.
the safe approach that sets 𝑝̂ = 0.50 in the formula for
c. In interpreting this confidence interval, can you
n. Then, how many records need to be sampled?
conclude that fewer than 10% of Americans are
Compare this to the answer in part a. Explain why it is
vegetarians? Explain your reasoning.
better to make an educated guess about what to expect
for 𝑝̂ , when possible.
4
23. Mean property tax. A tax assessor wants to estimate the
mean property tax bill for all homeowners in Madison,
Wisconsin. A survey 10 years ago got a sample mean and
standard deviation of $1400 and $1000.
a. How many tax records should the tax assessor
randomly sample for 95% confidence interval for the
mean to have a margin of error equal to $100? What
assumption does your solution make?
b. In reality, suppose that they’d now get a standard
deviation equal to $1500. Using the sample size you
derived in part a, without doing any calculation,
explain whether the margin of error for a 95%
confidence interval would be less than $100, equal to
$100 or more than $100.
c. Refer to part b. Would the probability that the sample
mean falls with $100 of the population mean be less
than 0.95, equal to 0.95 or greater than 0.95? Explain.
24. H0 or Ha? For each of the following hypothesis explain
whether it is a null hypothesis or alternative hypothesis:
a. For females, the population mean on the political
ideology scale is equal to 4.0.
b. For males, the population proportion who support the
death penalty is larger than 0.50.
c. The diet has an effect; the population mean can
change in weight being less than 0.
d. For all subway sandwich stores worldwide, the
difference between sales this month and in the
corresponding month last year has been a mean of 0.
25. ESP A person who claims to possess extrasensory
perception (ESP) says she can guess more often than not
the outcome of a flip of a balanced coin. Out of 20 flips,
she guesses correctly 12 times. Would you conclude that
she truly has ESP? Answer by reporting all five steps of a
significance test of the hypothesis that each of her guesses
has probability 0.50 of being correct against the
alternative that corresponds to her having ESP.
26. Jurors and gender A jury list contains the names of all
individuals who may be called for jury duty. The
proportion of the available jurors on the list who are
women is 0.53. If 40 people are selected to serve as
candidates for being picked on the jury, show all steps of
significance test of the hypothesis that the selections are
random with respect to gender.
a. Set up notation and hypotheses, and specify
assumptions.
b. 5 of the 40 selected were women. Find the test
statistic.
c. Report the P-value, and interpret.
d. Explain how to make a decision using a significance
level or 0.01.
27. Type I and Type II errors Refer to the previous exercise.
a. Explain what type I and Type II errors mean in the
context of that exercise.
b. If you made an error with the decisions in part d, is it a
Type I or Type II error?
28. Tennis balls in control? When it is operating correctly a
machine for manufacturing tennis balls produces balls with a
mean weight of 57.6 grams. The last eight balls
manufactured had weights
57.3, 57.4, 57.2, 57.5, 57.4, 57.1, 57.3, 57.0
a.
Using a calculator or software, find the test statistic and
P-value for a test of whether the process is in control
against the alternative that the true mean of the
process now differs from 57.6.
b. For significance level of 0.05, explain what you would
conclude. Express your conclusion so it would be
understood by someone who never studied statistics.
c. If your decision in part b is in error, what type of error
have you made?
29. Wage claim false? Management claims that the mean
income for all senior-level assembly-line workers in a large
company equals $500 per week. An employee decides to
test the claim, believing that it is actually less than $500. For
a random sample of nine employees, the incomes are:
430, 450, 450, 440, 460, 420, 430, 450, 440.
Conduct a significance test of whether the population mean
income equals $500 per week against the alternative that is
less. Include all assumptions, the hypotheses, test statistics,
P-value, and interpret the results in context.
30. Legal trial errors Consider the analogy discussed in Section
9.4 between making a decision about null hypothesis in a
significance test and making a decision about the innocence
or guilt of a defendant in a criminal trial.
a. Explain the difference between Type I and Type II errors
in the trial setting.
b. In this context, explain intuitively why decreasing the
chance of Type I error increases the chance of Type II
error.
5
31. Gender and belief in afterlife. This table shows results from
the 2008 General Social Survey on gender and whether or
not on believes in an afterlife.
Belief in Afterlife
Yes
No
599
111
425
168
Gender
Female
Male
Total
710
593
a.
Denote the population proportion who believe in an
afterlife by p1 for females and by p2 for males.
Estimate p1, p2 and (p1 – p2 ).
b. Find the standard error for the estimate of (p1 – p2 ).
Interpret.
c. Construct a 95% confidence interval for (p1 – p2 ). Can
you conclude which of 𝑝1 and 𝑝2 is larger? Explain.
d. Suppose that, unknown to us, p1 = 0.81 and p2 = 0.72.
Does the confidence interval in part c contain the
parameter it is designed to estimate? Explain.
32. Belief depend on gender? Refer to the previous exercise.
a. Find the standard error of 𝑝̂1 − 𝑝̂2 for a test of
𝐻0 : 𝑝1 = 𝑝2 .
b. For two-sided test, find the test statistic and P-value,
and make a decision using significance level 0.05.
Interpret.
c. Suppose that actually p1 = 0.81 and p2 = 0.72. Was the
decision in part b in error?
d. State the assumption on which the methods in this
exercise are based.
33. Heavier horseshoe crabs more likely to mate? A study of a
sample of horseshoe crabs on a Florida island (J. Brockman,
Ethology, vol. 102, 1996, pp. 1-21) investigated the factors
that were associated with whether or not female crabs had a
male crab mate. Basic statistics, including five-number
summary on weight (kg) for the 111 female crabs who had a
male crab nearby and for the 62 female crabs who did not
have a male nearby, are given in the table. Assume that
these horseshoe crabs have the properties of random
sample of all such crabs.
Summary Statistics for Weights of Horseshoe Crabs
Mate
No Mate
a.
b.
c.
d.
#
111
62
Mean
2.6
2.1
Std. Dev.
0.6
0.4
Min
1.5
1.2
Q1
2.2
1.8
Med
2.6
2.1
Q3
3.0
2.4
Max
5.2
3.2
Sketch box plots for the weight distributions of the two
groups. Interpret by comparing the groups with respect
to the shape, center, and variability.
Estimate the difference between the mean weights of
the female crabs who have mates and those who don't.
Find the standard error for the estimate in part b.
Construct a 90% confidence interval for the difference
between the population mean weights, and interpret.
34. Sex roles A study of the effect of the gender of the tester on
16
the sex-role differentiation scores in Manhattan gave a
random sample of preschool children the Occupational
Preference Test. Children were asked to give three choices of
what they wanted to be when they grew up. Each occupation
was rated on a scale from 1 (traditionally feminine) to 5
(traditionally masculine), and a child’s score was the mean of
the three selections. When the tester was male, the 50 girls
had 𝑥 = 2.9 and s = 1.4, whereas when the tester was
female, the 90 girls had 𝑥 = 3.2 and s = 1.2. Show all steps of
a test of the hypothesis that the population mean is the same
for the female and male testers, against the alternative that
they differ. Report the P-value and interpret.
35. Internet book prices Anna’s project for her introductory
statistics course was to compare the selling prices of
textbooks at two Internet bookstores. She first took a
random sample of 10 textbooks used that term in courses at
her college, based on the list of texts compiled by the college
bookstore. The prices of those textbooks at two Internet
sites were
Site A: $115, $79, $43, $140, $99, $30, $80, $99, $119, $69
Site B: $110, $79, $40, $129, $99, $30, $69, $99, $109, $66
a.
Are these independent samples or dependent samples?
Justify your answer.
b. Find the mean for each sample. Find the mean of the
difference scores. Compare, and interpret.
c. Using software or a calculator, construct a 90%
confidence interval comparing the population mean
prices of all the textbooks used that term at her college.
Interpret.
36. Comparing book prices 2 For the data in the previous
exercise, use software or a calculator to perform a
significance test comparing the population mean prices.
Show all steps of the test, and indicate whether you would
conclude that the mean price is lower at one or the two
Internet bookstores.
6
37. Down syndrome diagnostic test The table shown, from
Example 8 in Chapter 5, cross-tabulates whether a fetus
has Down syndrome by whether or not the triple blood
diagnostic test for Down syndrome is positive (that is,
indicates that the fetus has Down syndrome).
a.
Tabulate the conditional distributions for the blood
test result given the true Down syndrome status.
b.
For the Down cases, what percentage was diagnosed
as positive by the diagnostic test? For the unaffected
cases, what percentage got a negative test result?
Does the diagnostic test appear to be a good one?
c.
Construct the conditional distribution on Down
syndrome status, for those who have a positive test
result. (Hint: You condition on the first column total
and find proportions in that column.) Of those cases,
what percentage truly have Down syndrome? Is the
result surprising? Explain why this probability is
small.
Down Syndrome Status
D (Down)
0
D (unaffected)
Total
Blood Test Result
Positive Negative Total
48
6
54
1307
3921
5228
1355
3927
5282
38. Down and chi-squared For the data in the previous
exercise, 𝑥 2 = 114.4. Show all steps of the chi-squared
test of independence.
39. Gender gap? Exercise 11.1 showed a 2 x 3 table relating
gender and political party identification, shown again
here. The chi-squared statistic for these data equals 8.294.
Conduct all five steps of chi-squared test.
Sex
Dem
F
M
422
299
Political Party
Indep Repub
381
365
273
232
40. Tanning experiment Suppose the tanning experiment
described in Examples 1 and 2 used only four participants,
two for each treatment.
a. Show the six possible ways the four ranks could be
allocated, two to each treatment, with no ties.
b. For each possible sample, find the mean rank for each
treatment and the difference between the mean ranks.
c. Presuming H0 is true of identical treatment effects,
construct the sampling distribution of the difference
between the sample mean ranks for the two
treatments.
41. Test for tanning experiment Refer to the previous exercise.
For the actual experiment, suppose the participants using
the tanning studio got ranks 1 and 2 and the participants
using the tanning lotion got ranks 3 and 4.
a. Find and interpret the P-value for the alternative
hypothesis that the tanning studio tends to give better
tans than the tanning lotion.
b. Find and interpret the P-value for the alternative
hypothesis that the treatments have different effects.
c. Explain why it is a waste of time to conduct this
experiment if you plan to use a 0.05 significance level to
make a decision.
7
Answers
1.
UW Student Survey
a. The population is the entire UW student body of 40,858. The sample is the 100 students who were asked to complete the
questionnaire.
b. This value would not necessarily equal the value of the entire population of UW students. It is quite possible that the sample of 100 is not exactly
c.
representative of the whole student body. This percentage is only an estimate of the percentage of all students who would respond this way. It is
unlikely that any single sample of 100 would have a percentage that was exactly the percentage of the entire population.
The numerical summery is the sample statistics because it only summarizes for a sample, not for a population.
2.
Median versus mean sales price of new homes
We would expect the mean sales price to have been higher due to the distribution being skewed to the right. A few very expensive
homes will greatly affect the mean, but not the median sales price.
3. Female Heights
a. According to the Empirical Rule, 95% of the scores in a bell-shaped distribution fall within two standard deviation of the mean.
𝑥̅ − 2𝑆 = 65 − 2(3.5) = 58
𝑥̅ + 2𝑆 = 65 + 2(3.5) = 72
Thus, 95% of the heights likely fall between 58 and 72 inches.
b. The height for a woman who is three standard deviations below the mean is 54.5.
𝑥̅ − 3𝑆 = 65 − 3(3.5) = 54.5
This is on the cusp of what would be considered an outlier according to the z-score criterion. Scores that are beyond three standard
deviations from the mean are considered to be potential outliers. So, yes, this height is bordering on unusual.
4.
a.
High School Graduation Rates
The range is the difference between the lowest and the highest scores: 92.3 − 78.3 = 14.
th
th
The interquartile range (IQR) is the difference between scores at the 25 and 75 percentiles: IQR=Q3−Q1= 88.8 − 83.6 = 5.2.
b. 1.5(IQR)= 7.8 from Q1 or Q3; this criterion suggests that potential outliers would be scores less than 75.8 and greater than 96.6.
There are no scores beyond these values, so it would not indicate any potential outliers.
5. Blood Pressure
𝑥−𝑥̅
140−121
a. 𝑧 =
=
= 1.19
𝑠
16
A 𝑧 −score of 1.19 indicates that a person with a blood pressure of 140, the cutoff having a high blood pressure, falls 1.19
standard deviations above the mean.
b. About 95% of all the values in a bell-shaped distribution fall within two standard deviation of the mean−in this case 32. Subtracting
two times the standard deviation from the mean and adding two times the standard deviation to the mean tells us about 95% of
systolic blood pressures fall between 89 and 153.
6.
Life After Death for Males and Females
Using 2008 data:
a.
Opinion about life after death
Gender
Yes
No
Total
Male
0.77
0.23
808
Female
0.85
0.15
979
Overall, both male and women are more likely to believe in life after death than not, but women are somewhat more likely to do
so.
b.
8
7.
Women in Government and Economic Life
a.
The correlation between women in parliament and female economic activity is 0.745. This correlation is supported by the positive
linear trend evident in the scatterplot, but note this is largely driven by the point (for Japan) having female economic activity very low
(65).
b. The regression equation is 𝑦̂ = −48.91 + 0.1986𝑥. Since the 𝑦 –intercept correspond to an 𝑥 value of 0, the 𝑦 –intercept is not
meaningful in this case. (Female economic activity=0 is outside the range of the observed data).
c. The predicted value for U.S. is −48.91 + 0.1986(81) = 25.5 with 15.0 − 25.5 = −10.5 as the corresponding residual. The
regression equation underestimates the percentage of women in parliament by 10.5% for the U.S.
9.8
d. 𝑏 = 0.56 ( ) = 0.7127 and 𝑎 = 26.5 − 0.7127(76.8) = −28.24. Thus the prediction equation is 𝑦̂ = −28.24 + 0.7127𝑥.
7.7
8.
a.
Predict Crime Using Poverty
The slope of 25.5 indicates that for each percentage increase in the poverty, the predicted violent crime rate increases by 25.5 crimes
per 100, 000people statewide.
b. The range predicted values runs from 413.9 to 839.8 crimes per 100, 000 people.
𝑦̂ = 209.9 + 25.5(8.0) = 413.9
𝑦̂ = 209.9 + 25.5(24.7) = 839.75 (rounds to 839.8)
c.
9.
a.
The correlation would be positive we know this because the slope is positive and the slope and the correlation have always the same
sign.
Football Discipline
Because this is a volunteer sample, there is the potential for sampling bias, both because of the sample is not selected randomly
(those who have responded might have been those felt the most strongly) and because of the undercover age (anyone without
Internet access would not have been able to participate). There also is potential for response biases because of the statements are
leading.
b. If the sample is biased due to undercoverage and lack of random sampling, it does not matter how big the sample. It is almost always
better to have a small random sample than a large volunteer sample.
9
10. Video games mindless?
a. The explanatory is the history playing video games, and the response variable is visual skills.
b. This was an observation study because the men were not randomly assigned to treatment (played video games versus hadn’t played);
those who already were in these groups were observed.
c. One possible lurking variable is reaction time. Excellent reaction times might make it easier and therefore more fun, to play video
games, leading young people to be more likely to play. Excellent reaction times also might lead young men to perform better on tasks
measuring visual skills. These young men might have performed better on tasks measuring visual skills regardless of whether they
played video games.
11. Peyton Manning Completions
a. No. What it means that in 100 passes we expect to see about 65 completions, but actually the number may vary somewhat.
b. If Manning is still at his typical paying level, it would be quite surprising if his completion percentage over a large number of passes
differed significantly from 0.6. The more passes he throws, the closer the observed percentage should be to 0.65.
12. Driver’s Exam
a. There are 2 × 2 × 2 = 8 possible outcomes. Construct the tree yourself. An alternative way of presenting the outcomes:
let P=”pass” and F=”failure”, the outcomes are: PPP, PPF, PFP, FPP, PFF, FPF, FFP, FFF.
b.
1
If the eight outcomes are equally likely, all three pass the exam is = 0.125. This also be calculated by multiplying the probability that
8
15.
the first would pass(0.5), by the probability that the second would pass(0.5), by the probability that the third would pass(0.5),
0.5 × 0.5 × 0.5 = 0.125.
If the three friends were a random sample of their age group, the probability that all would pass 0.7 × 0.7 × 0.7 = 0.343.
The probabilities that apply to a random sample are not likely to be valid for a sample of three friends because the three friends are
likely to be similar on many characteristics that might affect the performance on such a test ( e.g., IQ) . In addition it is possible that
they studied together.
Grandparents
This refers to a discrete random variable because there can only be whole numbers of grandparents. One can’t have 1.78
grandparents.
The probabilities satisfy the two conditions for a probability distribution because they each fall between 0 and 1, and the sum of the
probabilities of all possible values is 1.
The mean of this probability is 0(0.71) + 1(0.15) + 2(0.09) + 3(0.03) + 4(0.02) ≈ 0.5.
Z-score and Tail Probability
The z- score that is less than only 1% of the values would be greater than 99% of the values. If we look up 0.99 on Table A, we see the
z-score is 2.33.
(i) The z-score that is above 0.90 is 1.28.
(ii) The z-score that is above 0.99 is 2.33.
Cloning Butterflies
a.
𝑧=
c.
d.
13.
a.
b.
c.
14.
a.
b.
b. 𝑧 =
8−9
= −1.33; according to Table A, 0.092 of the butterflies have a wingspan less than 8 inches.
0.75
10−9
0.75
= 1.33; according to Table A, 0.092 of the butterflies have a wingspan wider than 10 inches.
c. From parts a and b, 1 − 2(0.092) = 0.816 of the butterflies have wingspans between 8 and 10 inches.
th
d. From Table A, the 90 percentile is 1.28. 1.28 × 0.75 + 9 = 9.96 inches. Thus 10% of the butterflies have wingspan wider than 9.96
inches.
16. Exam Performance
a.
𝑝(1−𝑝)
Mean= 𝑝 = 0.70, standard error= √
𝑛
=√
0.7(1−0.7)
50
= 0.0648.
b. Since 𝑛 = 50, by the Central Limit Theorem, we would expect the shape of the sampling distribution to be approximately normal
with mean=0.70 and standard deviation≈ 0.0648.
c.
The z-score for 0.60 is
answers correct.
0.60−0.70
0.0648
= −1.54 giving a cumulative probability of 0.06. It would not be surprising to only get 60% of the
1
0
17. Aunt Erma’s Restaurant
a. The population distribution has a mean $900 and a standard deviation of $300.
b. The population distribution has a mean $980 and a standard deviation of $276. The standard deviation of the data distribution
describes the spread of the daily sales values for this past week.
c.
The mean of the sampling distribution of the sample mean = 𝜇 =$900; standard error=
𝜎
√𝑛
=
300
√7
≈ 113.4 dollars. The standard
error describes the spread of the sample means based on sample of seven days sales.
18. Approval Rating for President Obama
We could tell to someone who hadn’t taken a statistics course that we do not know the exact percentage of the population who
approve of the job Barack Obama is doing as president, but we are quite sure that it is within 3% of 46%, that is, between 43% and
49%.
19. Vegetarianism
a. We must assume that the data were obtained randomly.
b.
c.
𝑠𝑒 = √
𝑝̂(1−𝑝̂)
𝑛
0.04(1−0.04)
=√
10,000
≈ 0.002.
The confidence interval is 𝑝̂ ± 𝑧(𝑠𝑒)
Lower limit: 0.04 − 2.58(0.002) = 0.035.
Upper limit: 0.04 + 2.58(0.002) = 0.045.
The interval is so narrow, even though the confidence level is high, mainly because of the very large sample size. The very large
sample size contributes to a small standard error by providing a very large denominator for the standard error calculation.
We can conclude that fewer than 10% of Americans are vegetarians because 10% falls above the highest believable value in the
confidence interval.
20. Grandpas Using Email
a. The first result is a 90% confidence interval for the mean hours spent per week sending and answering e-mail for males at least age
75. The sample mean, 𝑥̅ , is listed as 6.38 hours. Thus, the estimated mean spent per week sending and answering e-mail for males at
least age 75 is 6.38 hours. The sample standard deviation is 6.02. This quantity estimates the population standard deviation which
tells us how far we can expect a typical observation to vary from the mean. These estimates are based on a sample of size 8.
b. The confidence interval is 2.34 to 10.41. We can be 90% confident that the populations mean numbers of hours spent per week
sending and answering e-mail for males at least age 75 is 2.34 to 10.41 hours.
c. Since there are likely to be a lot of men over the age of 75 who do not use email but also some who use e-mail regularly, this
distribution is likely skewed right. Since the t-distribution is robust to violations of normal assumption, the interval is still valid.
21. US Popularity
a. True
b. False
22. Driving After Drinking
a.
b.
𝑛=
𝑛=
[𝑝̂(1−𝑝̂)]𝑧 2
𝑚2
[𝑝̂(1−𝑝̂)]𝑧 2
𝑚2
=
=
[0.2(1−0.2)](1.96)2
(0.04)2
[0.5(1−0.5)](1.96)2
(0.04)2
= 385.
= 601.
This is larger than the answer in a. If we can make an educated guess about what to expect for the proportion, we can use a smaller
sample size, saving possibly unnecessary time and money.
23. Mean property tax
a.
𝑛=
𝜎2𝑧 2
𝑚2
=
(1000)2 (1.96)2
(100)2
= 385.
The solution makes the assumption that the standard deviation will be similar now.
b. The margin of error would be more than $100 because the standard error would be larger than predicted.
c. With a larger margin of error, the 95% confidence interval is wider; thus, the probability the sample mean is within $100(which is less
than the margin of error from b) of the population mean is less than 0.95.
1
1
24. H0 or Ha?
a. null hypothesis
b. alt alternative hypothesis
c. alternative hypothesis
d. null hypothesis
25. ESP
1. Assumptions: The data are categorical (correct vs. incorrect guesses) and are obtained randomly. The expected successes and failures
are less than 15 under null hypothesis 𝐻0 : 𝑛𝑝 = (20)(0.5) = 10 < 15 and 𝑛(1 − 𝑝) = (20)(0.5) = 10 < 15, so this test is
approximate.
2. Hypotheses: 𝐻0 : 𝑝 = 0.5; 𝐻𝑎 : 𝑝 > 0.5
3.
Test statistic: 𝑧 =
0.6−0.5
√0.5(1−0.5)/20
= 0.89
4.
5.
P-value:0.19
Conclusion: If the null hypothesis were true, the probability would be 0.19 of getting a test statistic at least as extreme as the value
observed. There is no strong evidence that the population proportion correct guesses is higher than 0.50.
26. Jurors and gender
a.
The proportion of jurors who are women
Hypotheses: 𝐻0 : 𝑝 = 0.53; 𝐻𝑎 : 𝑝 ≠ 0.53
b.
Test statistic: 𝑧 =
0.125−0.53
√0.53(1−0.53)/40
= −5.1
c.
P-value is 0.000. If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as
the value observed.
d. This P-value is more extreme than the significance level of 0.01. We can reject the null hypothesis; we have strong evidence that
women are not being selected in numbers proportionate to their representation in their jury pool.
27. Type I and Type II errors
a. In the previous exercises, a Type I error would have occurred if we had rejected the null hypothesis, concluding that the women were
passed over jury duty, when they really were not. A Type II error would have occurred if we had failed to reject the null hypothesis,
but women were picked disproportionate to their representation in their jury pool.
b. If we made an error, it was a Type I error.
28. Tennis balls in control?
a. Software indicates a test statistic of −5.5 and a P-value of 0.001.
b. For a significance level of 0.05, we conclude that the process is not in control. The machine is producing tennis balls that weight less
than they are supposed to.
c. If we rejected the null hypothesis when in fact it is true, we have made a Type I error and conclude that the process is not in control
when it actually is.
d.
29. Wage claim false?
1.
2.
3.
Assumptions: The data are quantitative. The data seem to have been produced using randomization. We also assume an
approximately normal population distribution.
Hypotheses: 𝐻0 : 𝜇 = 500; 𝐻𝑎 : 𝜇 < 500
The sample mean is 441.11, and the standard deviation is 12.69.
Test statistic: 𝑡 =
4.
5.
0441.11−500
12.69
√9
= −13.9
P-value:0.000
Conclusion: If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the
value observed. There is extremely strong evidence that the population mean is less than 500; we can conclude that the mean income
is less than $500 per week.
1
2
30. Legal trial errors
a. A Type I error in a trail setting would occur if we convicted a defendant who was not guilty. A Type II error would occur if we failed to
convict a guilty defendant.
b. To decrease the chance of a Type I error, we would decrease the significance level. In doing this, it is more difficult to reject the null
hypothesis (i.e., find someone guilty) . Thus, there will be more guilty people who are not found guilty, a Type II error.
31. Gender and belief in afterlife.
a.
The sample proportions who report that they believe in afterlife for females is
599
710
= 0.8437, for males
425
593
= 0.7167, and for the
difference between females and males is 0.8437 − 0.7167 = 0.127.
b. The standard error for the estimate of (𝑝1 − 𝑝2 ) is
𝑠𝑒 = √
𝑝̂1 (1−𝑝̂1 )
𝑛1
+
𝑝̂2 (1−𝑝̂2 )
𝑛2
=√
0.8437(1−0.8437)
710
+
0.7167(1−0.7167)
593
= 0.0230
This is the standard deviation for the difference between males and females for the sample of these sizes.
The confidence interval is 𝑝̂1 − 𝑝̂2 ± 𝑧(𝑠𝑒); lower endpoint: 0.127−1.96(0.0230) = 0.0819; upper end point:
0.127+1.96(0.0230) = 0.172. The confidence interval is (0.0819, 0.172).
Because 0 is less than all plausible values given in the confidence interval, we can conclude that the proportion of females who report
that they believe in afterlife is higher than the proportion of males.
d. The difference between these population proportions, 0.09, is in the confidence interval. The confidence interval in part c) contains
the parameter it is designated to estimate.
32. Belief depend on gender
c.
a.
𝑠𝑒 = √𝑝̂ (1 − 𝑝̂ ) (
b. 𝑧 =
(𝑝̂1 −𝑝̂2 )−0
𝑠𝑒
=
1
𝑛1
+
0.127
0.0228
1
𝑛2
) = √0.7859(1 − 0.7859) (
1
710
+
1
593
) = 0.0228.
= 5.57; P-value≈ 0.
If the null hypothesis were true, the probability would be approximately 0 of getting a test statistic at least as extreme as the value
observed. Therefore, we reject the null hypothesis, and conclude that the population proportions believing in afterlife are different
for females and males.
c. If the population difference were 0.81−0.72=0.09, our decision would have been correct.
d. The assumptions on which the methods in this exercises are based are independent random samples for the two groups and that we
had at least 5 successes and 5 failures.
33. Heavier horseshoe crabs more likely to mate?
a. Construct yourself two boxplots of weight for female carbs who have mates and who do not have mates on the same scale. Then
conclude that the female crabs have a higher median and a bigger spread if they had a mate than if they did not have a mate. The
distribution for female crabs with a mate is right-skewed, whereas the distribution for female crabs without a mate is symmetrical.
b. The estimated difference between the mean weights of female crabs who have mates and do not have mates is 2.6−2.1=0.5.
c.
𝑠𝑒 = √
𝑠12
𝑛1
+
𝑠22
𝑛2
0.36
=√
111
+
0.16
62
= 0.076
d. (𝑥̅1 − 𝑥̅2 ) ± 𝑡0.05 (𝑠𝑒)
Because 𝑛1 and 𝑛2 are large, we will approximate t with the normal distribution using z=1.645.
0.5 − 1.645(0.076) = 0.375
0.5 + 1.645(0.076) = 0.625
(0.375, 0.625)
We can be 90% confidence the difference between the population mean weights of female crabs with and without a mate is between
0.375 and 0.625. Because 0 does not fall in this interval, we can conclude that female crabs with a met weight more than do female
crabs without a mate.
1
3
34. Sex roles
1. Assumptions: The data are quantitative (Child’s score); the samples are independent and we will assume that they were obtained
randomly; we will assume that the populations scores distribution are approximately normal for each group.
2. Hypotheses: 𝐻0 : 𝜇1 = 𝜇2; 𝐻𝑎 : 𝜇1 ≠ 𝜇2 where group 1 represents with the group with male tester group 2 represents the group with
female tester.
3.
𝑠𝑒 = √
1.4 2
50
+
1.22
90
= 0.2349, 𝑡 =
2.9−3.2
0.2349
= −1.28
4.
5.
P-value:0.205
If the null hypothesis were true, the probability would be 0.205 of getting a test statistic at least as extreme as the value observed.
Since the P-value is quite large, there is not much evidence of a difference in the population mean of the children’s scores when the
tester is male than female.
35. Internet book prices
a. The samples are dependent because they are prices of the same ten books at two different internet sites
b. Let group 1 be the prices from Site A and group 2 from Site B. Then, 𝑥̅1 = $87.30, 𝑥̅2 = $83.00, 𝑥̅𝑑 = $4.30. Sample mean price for
the books from Site A is higher than the sample mean price for the book from Site B. Thus, the sample mean of the difference
between the prices from the two sites is positive.
c. A 90% confidence interval for 𝜇1 − 𝜇2 is given by (1.53, 7.03). Since 0 is less than the values in the confidence interval, we can
conclude that the prices for textbooks used at her college are more expensive at Site A than at Site B.
36. Comparing book prices 2
1. Assumptions: the differences in prices are a random sample from a population that is approximately normal.
2. 𝐻0 : 𝜇𝑑 = 0; 𝐻𝑎 : 𝜇𝑑 ≠ 0
3.
𝑡=
𝑑̅ −0
𝑠𝑑/√𝑛
=
4.3
4.7152/√10
= 2.88
4.
5.
P-value:0.02
If the null hypothesis is true, the probability obtaining a difference in sample means extreme as the value observed is 0.02. We
would reject the null hypothesis and conclude there is a significant difference in the prices of textbooks used at her college between
the two sites for 𝛼 =0.05, or 0.10, but not for 𝛼 =0.01.
37. Down syndrome diagnostic test
a.
BLOOD TEST RESULT
STATUS
Positive Negative Total
D(Down)
0.89
0.11
54
𝑐
𝐷 (Unafected) 0.25
0.75
5228
b. For the Down cases, 89% were correctly diagnosed. For the unaffected cases, 75% get a negative result. The test seems fairly good,
but there are a good number of false positives and false negatives.
c.
BLOOD TEST RESULT
STATUS
Positive Negative
D(Down)
0.035
0.002
𝐷𝑐 (Unafected) 0.965
0.998
Total
1355
3927
Of the positive cases, only 0.035 truly have Down syndrome. This result is not surprising because there are so few cases overall. The
fairly large numbers of false positives will overwhelm the much smaller number of actual cases.
38. Down and chi-squared
1. The assumptions are there are two categorical variables (Down syndrome status and blood test result), that randomization was used
to obtain the data and that the expected count was at least five in all cells.
2. 𝐻0 : Down syndrome status and blood test result are independent
𝐻𝑎 : Down syndrome status and blood test result are dependent
3. 𝜒 2 = 114.4, df=1
4. P-value:0.000
1
4
5.
If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the value
observed. There is very strong evidence of an association between test result and actual status.
39. Gender gap?
1.
The assumptions are there are two categorical variables (party identification and gender), that randomization was used to obtain the
data and that the expected count was at least five in all cells.
2. 𝐻0 : Party identification and gender are independent
𝐻𝑎 : Party identification and gender are dependent
3. 𝜒 2 = 8.294, df=2
4. P-value:0.016
5.
If the null hypothesis were true, the probability would be 0.016 of getting a test statistic at least as extreme as the value observed.
There is very strong evidence that party identification depends on gender.
40. Tanning experiment
a.
Treatments
Ranks
Lotion
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
Studio
(3, 4)
(2, 4)
(2, 3)
(1, 4)
(1, 3)
(1, 2)
b.
Lotion mean rank
Studio mean rank
Difference of mean
ranks
1.5
3.5
-2.0
2.0
3.0
-1.0
2.5
2.5
0.0
2.5
2.5
0.0
3.0
2.0
1.0
3.5
1.5
2.0
c.
Difference between mean ranks
-2.0
-1.0
0.0
1.0
2.0
probability
1/6
1/6
2/6
1/6
1/6
41. Test for tanning experiment
a. The P-value is 1/6 =0.17; if the treatments had identical effects, the probability would be 0.17 of getting sample like we observed, or
even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the studio does not lead to better
results than the lotion.
b. The P-value is 2/6 =0.33; if the treatments had identical effects, the probability would be 0.33 of getting sample like we observed, or
even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the treatments do not lead to different
results.
c. It is a waste of time to conduct this experiment if we plan to use a 0.05 significance level because the smallest possible P-value is 0.17.