Practice Test 13
AP Statistics
Name:
Directions: Work on these sheets. Answer completely, but be concise. Tables are attached.
Part 1: Multiple Choice. Circle the letter corresponding to the best answer.
1. We wish to test if a new feed increases the mean weight gain compared to an old feed. At the
conclusion of the experiment it was found that the new feed gave a 10 kg bigger gain than the old
feed. A two-sample t test with the proper one-sided alternative was done and the resulting P-value
was 0.082. This means that
(a) there is an 8.2% chance the null hypothesis is true.
(b) there was only an 8.2% chance of observing an increase greater than 10 kg (assuming the null
hypothesis was true).
(c) there was only an 8.2% chance of observing an increase greater than 10 kg (assuming the null
hypothesis was false).
(d) there is an 8.2% chance the alternative hypothesis is true.
(e) there is only an 8.2% chance of getting a 10 kg increase.
2. A study was conducted to investigate the effectiveness of a new drug for treating Stage 4 AIDS
patients. A group of AIDS patients was randomly divided into two groups. One group received the
new drug; the other group received a placebo. The difference in mean subsequent survival (those
with drugs – those without drugs) was found to be 1.04 years, and a 95% confidence interval was
found to be 1.04 ± 2.37 years. Based upon this information, we can conclude that
(a) the drug was effective since those taking the drug lived, on average, 1.04 years longer.
(b) the drug was ineffective since those taking the drug lived, on average, 1.04 years less.
(c) there is no evidence the drug was effective since the 95% confidence interval covers zero.
(d) there is evidence the drug was effective since the 95% confidence interval does not cover zero.
(e) we can make no conclusions since we do not know the sample size or the actual mean survival
of each group.
3. Thirty-five people from a random sample of 125 workers from Company A admitted to using sick
leave when they weren’t really ill. Seventeen employees from a random sample of 68 workers
from Company B admitted that they had used sick leave when they weren’t ill. A 95% confidence
interval for the difference in the proportions of workers at the two companies who would admit to
using sick leave when they weren’t ill is
(0.28)(0.72) (0.25)(0.75)
(a) 0.03 

125
68
(0.28)(0.72) (0.25)(0.75)
(b) 0.03  1.96

125
68
(0.28)(0.72) (0.25)(0.75)
(c) 0.03  1.645

125
68
1
 1
(d) 0.03  1.96 
 0.2690.731
 125 68 
1
 1
(e) 0.03  1.645 
 0.2690.731
 125 68 
Chapter 13
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Practice Test 13
4. An SRS of size 100 is taken from a population having proportion 0.8 successes. An independent
SRS of size 400 is taken from a population having proportion 0.5 successes. The sampling
distribution of the difference in sample proportions has what mean?
(a)
(b)
(c)
(d)
(e)
0.3
0.15
The smaller of 0.8 and 0.5
The mean cannot be determined without the sampling results.
None of the above. The answer is _____________________________.
The next two questions refer to this scenario.
Different varieties of fruits and vegetables have different amounts of nutrients. These differences are
important when these products are used to make baby food. We wish to compare the carbohydrate
content of two varieties of peaches. The data were analyzed with SAS, and the following output was
obtained:
VARIETY
1
2
VARIANCES
UNEQUAL
EQUAL
N
5
7
MEAN STD DEV STD ERROR
33.6
3.781
1.691
25.0 10.392
3.927
T
2.0110
1.7490
MIN
29.000
2.000
MAX
38.000
33.000
DF PROB > |T|
8.0
0.0791
10.0
0.1109
5. We wish to test if the two varieties are significantly different in their mean carbohydrate content.
The null and alternative hypotheses are
(a) H 0 : 1   2 ; H a : 1   2
(b) H 0 : 1   2 ; H a : 1   2
(c) H 0 : 1   2 ; H a : 1   2
(d) H 0 : x1  x2 ; H a : x1  x2
(e) H 0 : x1  x2 ; H a : x1  x2
6. The test statistic and P-value are
(a) 1.7490; 0.0318
(b) 1.7490; 0.0554
(c) 2.0110; 0.1582
(d) 2.0110; 0.0791
(e) 2.0110; 0.0396
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Practice Test 13
7. A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random
samples of nests: one of small nests and the other of large nests. She weighs one egg from each
nest. The data are summarized below.
sample size
sample mean (g)
sample variance
Small nests
60
37.2
24.7
Large nests
159
35.6
39.0
A 95% confidence interval for the difference between the average mass of eggs in small and
large nests is:
24.7 2 39.02
(a)  37.2  35.6   2.000

60
159
24.7 2 39.02
(b)  37.2  35.6   2.009

59
158
24.7 39.0
(c) 37.2  35.6 

59
158
(e) None of these
24.7 2  39.0 2
(d) 37.2  35.6 
59  158
The next two questions refer to this situation.
All of us nonsmokers can rejoice—the mosaic tobacco virus that affects and injures tobacco plants is
spreading! Meanwhile, a tobacco company is investigating if a new treatment is effective in reducing
the damage caused by the virus. Eleven plants were randomly chosen. On each plant, one leaf was
randomly selected, and one half of the leaf (randomly chosen) was coated with the treatment, while the
other half was left untouched (control). After two weeks, the amount of damage to each half of the leaf
was assessed. The output from SAS follows:
VARIABLE
1ST:
CONTROL
2ND:
TRT
1ST-2ND:
DIFF
N_USED
MEAN
MEDIAN
11
15.7273
13
11
13.3636
11
2.36364
SD
MIN
MAX
9.1224
5
36
12
10.0725
2
32
3
3.32484
-6
6
8. What is the best reason for performing a paired experiment rather than a two–independent sample
experiment in this case?
(a) It is easier to do since we need fewer experimental units and each unit receives more than one
treatment.
(b) It allows us to remove variation in the results caused by other factors since we can compare
both treatments within the same experimental unit.
(c) The computer program is more accurate since we work only with the differences.
(d) It requires fewer assumptions since we are interested only in the difference between treatments.
(e) It allows us to do more experiments since we use each experimental unit twice.
9. What are the rejection region (  = 0.05) and P-value for the paired t test?
(a) Reject if t*> 1.812; P-value = 0.040
(b) Reject if t*> 1.812; P-value = 0.020
(c) Reject if t*> 2.358; P-value = 0.040
(d) Reject if t*> 2.358; P-value = 0.020
(e) Reject if t*> 1.645; P-value = 0.020
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Practice Test 13
10. Popular wisdom is that eating presweetened cereal tends to increase the number of dental caries
(cavities) in children. A sample of children was (with parental consent) entered into a study and
followed for several years. Each child was classified as a sweetened-cereal lover or a nonsweetened
cereal lover. At the end of the study, the amount of tooth damage was measured. Here are the
summary data:
Group
n
Mean
Std. Dev
Sugar bombed
No sugar
10
15
6.41
5.20
5.0
15.0
An approximate 95% confidence interval for the difference in the mean tooth damage is
(a) 6.41  5.20  2.26
5 15

10 15
(b) 6.41  5.20  2.26
25 225

10 15
(c) 6.41  5.20  1.96
25 225

10 15
(d) 6.41  5.20  2.26
25 225

100 225
(e) 6.41  5.20  1.96
25 225

100 225
11. The power takeoff driveline on tractors used in agriculture is a potentially serious hazard to operators
of farm equipment. The driveline is covered by a shield in new tractors, but for a variety of reasons,
the shield is often missing on older tractors. Two types of shields are the bolt-on and the flip-up. It
was believed that the bolt-on shield was perceived as a nuisance by the operators and deliberately
removed, but the flip-up shield is easily lifted for inspection and maintenance and may be left in
place. In a study initiated by the U.S. National Safety Council, a sample of older tractors with both
types of shields was taken to see what proportion were removed. Of 183 tractors designed to have
bolt-on shields, 35 had been removed. Of the 136 tractors with flip-up shields, 15 were removed. We
wish to test the hypothesis H0: pb = pf vs. Ha: pb  pf where pb and pf are the proportion of tractors with
the bolt-on and flip-up shields removed, respectively. Which of the following conditions for
performing the appropriate significance test is satisfied in this case?
(a) Both population distributions are Normally distributed.
(b) Two independent simple random samples were chosen.
(c) Both sample sizes are at least 30.
(d) np and n(1 – p) are both large enough to use Normal calculations.
(e) The sample size is at least 10 times the population size.
12. A study was carried out to investigate the effectiveness of a treatment. 1000 subjects participated in
the study, with 500 being randomly assigned to the “treatment group” and the other 500 to the
“control (or placebo) group.” A statistically significant difference was reported between the
responses of the two groups (P < 0.005). Thus,
(a) there is a large difference between the effects of the treatment and the placebo.
(b) there is strong evidence that the treatment is very effective.
(c) there is strong evidence that there is some difference in effect between the treatment and the
placebo.
(d) there is little evidence that the treatment has any effect.
(e) there is evidence of a strong treatment effect.
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Practice Test 13
13. A study was conducted to estimate the effectiveness of doing assignments in an introductory
statistics course. Students in one section, taught by Instructor A, received no assignments. Students
in another section, taught by Instructor B, received assignments. The final grade of each student
was recorded. A 95% confidence interval for the difference in the mean grades (Section A 
Section B) was computed to be  3.5 ± 1.8. This means that
(a) there is evidence that doing assignments improves the average grade since the difference in the
population means is less than zero.
(b) there is little evidence that doing assignments improves the average grade since the 95%
confidence interval does not cover 0.
(c) there is evidence that doing assignments improves the average grade since the 95% confidence
interval does not cover 0.
(d) there is evidence that doing assignments does not improve the average grade since the 95%
confidence interval does not cover 0.
(e) there is little evidence that doing assignments does not improve the average grade since the
95% confidence interval does cover 0.
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Practice Test 13
Part 2: Free Response
Answer completely, but be concise. Communicate your thinking clearly and completely.
14. Nitrites are often added to meat products as preservatives. In a study of the effect of these chemicals
on bacteria, the rate of uptake of a radio-labeled amino acid was measured for a number of cultures of
bacteria, some growing in a medium to which nitrites had been added. Here are the summary
statistics from this study.
x
Group
n
s
Nitrite
Control
30
30
7880
8112
1115
1250
(a) Carry out a test of the research hypothesis that nitrites decrease amino acid uptake and report your
results.
(b) Construct and interpret a 95% confidence interval for the difference in population means.
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Practice Test 13
15. The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella
germanica. A study investigated the persistence of this pesticide on various types of surfaces.
Researchers applied a 0.5% emulsion of diazinon to glass and plasterboard. After 14 days, they
placed 18 cockroaches on each surface and recorded the number that died within 48 hours. On glass,
9 cockroaches died, while on plasterboard, 13 died.
(a) Chemical analysis of the residues of diazinon suggests that it may persist longer on
plasterboard than on glass because it binds to the paper covering on the plasterboard. The
researchers therefore expected the mortality rate to be greater on plasterboard than on glass.
Conduct a significance test to assess the evidence that this is true.
(b) Construct and interpret a 95% confidence interval for the difference in the two population
proportions.
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Practice Test 13
16. Researchers studying the learning of speech often compare measurements made on the recorded
speech of adults and children. One variable of interest is called the voice onset time (VOT). Here
are the results for 6-year-old children and adults asked to pronounce the word “bees.” The VOT is
measured in milliseconds and can be either positive or negative.
(a) The researchers were investigating whether VOT distinguishes adults from children. Construct
a 95% confidence interval for the difference in mean VOTs when pronouncing the word
“bees.”
(b) Based on your work in part (b), how would you answer the researchers’ question? Explain.
(c) The researchers in the study looked at VOTs for adults and children pronouncing several
different words. Explain why they should not perform a separate two-sample t test for each
word and conclude that the words with a significant difference (say, P < 0.05) distinguish
children from adults. (The researchers did not make this mistake.)
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Practice Test 13
17. An association of Christmas tree growers in Indiana sponsored a sample survey of 500 randomly
selected Indiana households to help improve the marketing of Christmas trees. One question the
researchers asked was “Did you have a Christmas tree this year?”
Respondents who had a tree during the holiday season were asked whether the tree was natural
or artificial. Respondents were also asked if they lived in an urban area or in a rural area. Of the
421 households displaying a Christmas tree, 160 lived in rural areas and 261 were urban residents.
The tree growers want to know if there is a difference in preference for natural trees versus
artificial trees between urban and rural households. Here are the data:
Population
1 (rural)
2 (urban)
n
160
261
X(natural)
64
89
(a) Construct and interpret a 95% confidence interval for the difference in the proportion of rural
and urban Indiana residents who had a natural Christmas tree this year.
(b) Do these data provide evidence of a significant difference in the proportion of rural and urban
Indiana residents who had a natural Christmas tree this year? State hypotheses, then use your
results from part (a) to help answer the question.
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Practice Test 13