AP Statistics – Inference Review
7 questions
1. Survey Responses. Few people who receive questionnaires in the mail actually fill them out
and return them – often fewer than 10%! One researcher thinks he can improve the response
rate by including a coupon good for a free pint of ice cream along with each questionnaire.
The researcher believes that people will want the ice cream, and feel guilty if they don’t
return the questionnaire. To test this conjecture he mails questionnaires with ice cream
coupons to 150 randomly selected people. After two weeks 41 of the surveys have been
returned.
a. Create a 95% confidence interval to estimate the return rate.
b. Interpret your interval in context.
c. Encouraged by the response rate, this researcher now plans to replicate the study in hopes
of estimating the return rate for this strategy to within 5%. How many new
questionnaires must he mail out?
SOLUTION:
Part (a)
P: the proportion of those who return the questionnaire given coupon for a free pint of ice cream.
Each person receiving the questionnaire is independent, they are randomly chosen, 150 is less
than 10% of all who have received the questionnaire, and there were 41 who returned the
questionnaire and 109 who did not.
Since all of the assumptions and conditions have been met, we will proceed with a oneproportion z interval with 95% confidence.
 41  109 



41
150  150 

 1.96 
 .202, .345
150
150
Part (b)
We are 95% confident that the actual proportion of people who would return the questionnaire
when given a coupon for a free pint of ice cream is between 20.2 and 34.5 percent.
Part (c)
ME  z *
pˆ qˆ
n
z *2 pˆ qˆ
ME 2
2

1.96 .273.726
n
.052
n
n  305.21
He would need a sample size of at least 306 to have a margin of error of 5% with an interval at
95% confidence.
AP Statistics – Inference Review
2. Scoring. The table shows the average number of points scored in
home and away games by 8 randomly selected NFL teams during
the 2002 season. Assuming that the offensive performance of these
teams is representative of other teams during the season and others;
do these data provide evidence of a home field advantage when it
comes to scoring?
SOLUTION:
The data is paired based on the team, so we would want to do a
paired-t test.
7 questions
Team
Jets
Ravens
Texans
49ers
Giants
Bucs
Eagles
Packers
Home
24.3
19.8
13.5
25.2
18.9
26.5
24.1
24.6
Away
21.2
19.8
13.1
20.6
23.0
20.8
25.5
23.0
 d  The mean of the difference s of the home game and away game scores.
H o : d  0
H a : d  0
The data are paired based on team, each team is independent, they were randomly selected,
the 8 team scores for 2002 are less than 10% of all team scores for all the teams in the NFL,
and the distribution of the differences is unimodal and rough symmetric.
Since the assumptions and conditions have been met, we can proceed with a paired-t test.
t
1.2375  0
 1.088
3.218
8
p  value  0.156
Since the p-value is so large, we fail to reject the null hypothesis. There is not enough
evidence to suggest there is a home field advantage.
AP Statistics – Inference Review
7 questions
3. Car Insurance. An insurance company advertises that 90% of their accident claims are
settled within 30 days. A consumer group randomly selected 104 of last year’s claims from
the company’s files, and finds that only 89 of them were settled within 30 days. Is the
company guilty of false advertising?
SOLUTION:
p : the proportion of accident claims the insurance company settles within 30 days
H o : p  .90
H a : p  .90
The 104 files were randomly selected, we can assume they are independent of each other, they
are less than 10% of all of the files from the insurance company and there are 89 observed files
that were settled within 30 days and 15 that were not.
Since all the assumptions and conditions have been met, we can proceed with a one-proportion z
test.
z
.8558  .90
.90.10
 1.503
104
p  value  0.066
Since the p-value is larger than our significance level of 5%, we would fail to reject the null
hypothesis. We can conclude there is insufficient evidence to suggest that the company’s rate of
settling claims within 30 days is less than the advertised rate of 90%.
AP Statistics – Inference Review
7 questions
4. Grapes. An agronomist hopes that a new fertilizer she has developed will enable grape
growers to increase the yield of each grapevine by more than 5 pounds. To test this fertilizer
she applied it to 44 vines and used traditional growing strategies on 47 other vines. The
fertilized vines produced a mean of 58.4 pounds of grapes with a standard deviation of 3.7
pounds, while the unfertilized vines yielded an average of 52.1 pounds with standard
deviation of 3.4 pounds of grapes. Do these experimental results confirm the agronomist’s
expectations?
SOLUTION:
 F   T : The difference in the means of the fertilized vines and the tradition ally grown vine s
H O :  F  T  5
H A :  F  T  5
The two groups of vines are independent from each other and within each group are also independent,
they were randomly assigned to either the new fertilizer or a traditional growing strategy, the number of
grape vines is less than 10% of all grape vines, and we can assume the yield of grapes from the vines is
normally distributed.
Since all the conditions and assumptions have been met, we will proceed with a two-sample t-test.
t
58.4  52.1  5  1.741
3.7 2 3.4 2

44
47
p  value  0.043
Since the p-value is low, we would reject the null hypothesis. We have sufficient evidence to suggest that
the difference in the mean amount of grapes produced on the vines with the new fertilizer is more than 5
pounds that those vines used with traditional growing techniques.
AP Statistics – Inference Review
7 questions
5. Cigarettes. Some of the cigarettes sold in the US claim to be “low tar”. How much less tar
would a smoker get by smoking low tar brands instead of regular cigarettes? Samples of 15
brands of each type were randomly selected from the 1206 varieties (no kidding) that are
marketed. Their tar contents (mg/cig) are listed in the table below. Find a 95% confidence
interval for the difference between regular and low tar brands.
Type
Regular
Low
Tar
18
9
10
5
14
10
15
4
Milligrams of tar per cigarette
15 12 17 11 14 17 12
8
9
9
3
7
12 6
14
10
15
8
15
11
12
8
SOLUTION:
 R : the mean amount of tar per cigarette in regular cigarettes
 L : the mean amount of tar per cigarette in low tar cigarettes
The sample of low tar and regular cigarettes are independent groups, the cigarettes within each
group are independent, the samples were randomly selected, the 15 cigarettes in each group are
less than 10% of all cigarettes, and each groups’ distribution is roughly symmetric and unimodal.
Since the assumptions and conditions have been met, we can proceed with a 2-sample t-interval.
14.067  7.933  t 27.74 *
2.313 2 2.548 2

 4.312, 7.954
15
15
We are 95% confident that the mean amount of tar in regular cigarettes is between 4.312 and
7.954 milligrams more than low tar cigarettes.
AP Statistics – Inference Review
7 questions
6. Taxes. Waiters are expected to keep track of their income from tips and report it on their
income tax forms. The Internal Revenue Service suspects that one waiter (we’ll call him
“Jared”) has been under-reporting his income, so they are auditing his tax return. An IRS
agent goes through the restaurant’s files and obtains a random sample of 80 credit card
receipts from people Jared served. The average tip size shown on these receipts was $9.68
with a standard deviation of $2.72.
a. Create and interpret a 90% confidence interval for the mean size of all of Jared’s credit
card tips.
b. On his tax return Jared claimed that his tips averaged $8.73. Based on their confidence
interval, does the IRS have a case against him? Explain.
SOLUTION:
Part (a)
 : the average tip size for all of Jared' s receipts
The 80 receipts are a random sample of Jared’s receipts, the receipts are independent of each
other, the 80 receipts are less than 10% of all of Jared’s receipts, and we can assume the
distribution of receipts is nearly normal based on the CLT with a sample size of 80.
Since the assumptions and conditions for inference have been met, we can proceed with a onesample t-interval.
 2.72 
9.68  t 79 * 
  $9.17, $10.19
 80 
Part (b)
The estimation from the interval has Jared’s mean tips should be between $9.17 and $10.19 with
90% confidence. Since $8.73 is below this interval, the IRS agent does have sufficient evidence
to show Jared is claiming lower tips on his tax returns than what he should.
AP Statistics – Inference Review
7 questions
7. Shopping. A survey of 430 randomly selected chosen adults found that 21% of 222 men and
18% of 208 women had purchased books online.
a. Is there evidence that men are more likely than women to make online purchases of
books? Test an appropriate hypothesis and state your conclusion in context.
b. If your conclusion were in fact proves to be wrong, did you make a Type I error or Type
II error?
SOLUTION:
Part (a)
p M : the proportion of men who purchase books online
pW : the proportion of women who purchase books online
H O : p M  pW  0
H A : p M  pW  0
The sample of 222 men and 208 women were randomly selected, then men are independent from
the women, and the individuals within each group are independent. The 222 men and 208
women are both less than 10% of all men and women, and there were 46 men and 37 women
who purchased books online and 176 men and 171 women who haven’t purchased books online.
Since the assumptions and conditions have been met, we can proceed with a two-proportion ztest.
pˆ pooled 
z
.22222  .18208  .201
222  208
.22  .18
.201.799  .201.799
222
p  value  .2207
 .7699
208
Since the p-value is so large, we would fail to reject the null hypothesis. We can conclude there
is insufficient evidence to suggest the proportion of men who purchase books online is more than
the proportion of women.
Part (b)
If the conclusion was false, we would have made a Type II error.