AP Statistics Chapters 18-22 Final Exam Review
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Determine whether the Normal model may be used to describe the distribution of the sample proportions. If the Normal
model may be used, list the conditions and explain why each is satisfied. If Normal model may not be used, explain
which condition is not satisfied.
1) In a large statistics class, the professor has each student toss a coin 54 times and calculate the
1)
proportion of tosses that come up tails. The students then report their results, and the professor
plots a histogram of these several proportions. May a Normal model be used here?
A) A Normal model should not be used because the population distribution is not Normal.
B) A Normal model may not be used because the success/failure condition is not satisfied: np =
27 = nq = 27 which are not less than 10
C) A Normal model may not be used because the 10% condition is not satisfied: the sample size,
54, is more than 10% of the population of all coin flips.
D) A Normal model may be used:
Coin flips are independent of one another - no need to check the 10% condition. The
success/failure condition is satisfied because np = 27 = nq = 27 which are both greater than 10
E) A Normal model may be used:
The 10% condition is satisfied: the sample size, 54, is more than 10% of the population of all
coin flips.
The success/failure condition is satisfied because np = 27 = nq = 27 which are both greater
than 10
Find the mean of the sample proportion.
2) Based on past experience, a bank believes that 6% of the people who receive loans will not make
payments on time. The bank has recently approved 300 loans. What is the mean of the proportion
of clients in this group who may not make timely payments?
A) μ = 3%
B) μ = 4.24%
C) μ = 18%
D) μ = 6%
E) μ = 1.37%
Find the standard deviation of the sample proportion.
3) Assume that 15% of students at a university wear contact lenses. We randomly pick 200 students.
What is the standard deviation of the proportion of students in this group who may wear contact
lenses?
A) σ = 6.38%
B) σ = 5.48%
C) σ = 2.74%
D) σ = 2.52%
E) σ = 5.05%
2)
3)
In a large class, the professor has each person toss a coin several times and calculate the proportion of his or her tosses
that come up heads. The students then report their results, and the professor plots a histogram of these proportions. Use
the 68-95-99.7 Rule to provide the appropriate response.
4) If each student tosses the coin 200 times, about 95% of the sample proportions should be between
4)
what two numbers?
A) 0.495 and 0.505
B) 0.429 and 0.571
C) 0.025 and 0.975
D) 0.071 and 0.106
E) 0.2375 and 0.7375
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Find the specified probability, from a table of Normal probabilities. Assume that the necessary conditions and
assumptions are met.
5) When a truckload of oranges arrives at a packing plant, a random sample of 125 is selected and
5)
examined. The whole truckload will be rejected if more than 8% of the sample is unsatisfactory.
Suppose that in fact 9% of the oranges on the truck do not meet the desired standard. What's the
probability that the shipment will be rejected?
A) 0.3483
B) 0.6517
C) 0.6966
D) 0.2197
E) 0.7803
Answer the question.
6) A national study reported that 74% of high school graduates pursue a college education
immediately after graduation. A private high school advertises that 155 of their 196 graduates last
year went on to college. Does this school have an unusually high proportion of students going to
college?
A) This school can boast an unusually high proportion of students going to college. Their
proportion is 1.30 standard deviations above the mean.
B) This school can boast an unusually high proportion of students going to college. Their
proportion is 2.61 standard deviations above the mean.
C) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 0.97 standard deviations above the mean.
D) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 1.62 standard deviations above the mean.
E) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 1.30 standard deviations above the mean.
Describe the indicated sampling distribution.
7) The heights of people in a certain population are normally distributed with a mean of 67 inches and
a standard deviation of 3.9 inches. Describe the sampling distribution of the mean for samples of
size 41. In particular, state whether the distribution of the sample mean is normal or
approximately normal and give its mean and standard deviation.
A) Normal, mean = 67 inches, standard deviation = 3.9 inches
B) Normal, mean = 67 inches, standard deviation = 0.1 inches
C) Normal, mean = 67 inches, standard deviation = 0.61 inches
D) Approximately normal, mean = 67 inches, standard deviation = 0.61 inches
E) Approximately normal, mean = 67 inches, standard deviation = 0.1 inches
6)
7)
Find the specified probability, from a table of Normal probabilities. Assume that the necessary conditions and
assumptions are met.
8) The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine
8)
puts in the boxes is believed to have a Normal model with mean 16.15 ounces and standard
deviation 0.3 ounces. What is the probability that the mean weight of a 16-box case of crackers is
above 16 ounces?
A) 0.9544
B) 0.9772
C) 0.9316
D) 0.0228
E) 0.0456
9) The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine
puts in the boxes is believed to have a Normal model with mean 16.15 ounces and standard
deviation 0.3 ounces. What is the probability that the mean weight of a 10-box case of crackers is
below 16 ounces?
A) 0.9429
B) 0.1142
C) 0.0571
D) 0.8858
E) 0.2273
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9)
Solve the problem.
10) At a shoe factory, the time taken to polish a finished shoe has a mean of 3.7 minutes and a standard
deviation of 0.48 minutes. If 44 shoes are polished, there is a 5% chance that the mean time to
polish the shoes is below what value?
A) 3.89 minutes
B) 3.51 minutes
C) 3.58 minutes
D) 3.82 minutes
E) 3.53 minutes
Find the margin of error for the given confidence interval.
11) A survey found that 69% of a random sample of 1024 American adults approved of cloning
endangered animals. Find the margin of error for this survey if we want 90% confidence in our
estimate of the percent of American adults who approve of cloning endangered animals.
A) 4.27%
B) 2.83%
C) 2.38%
D) 24.35%
E) 5.09%
10)
11)
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion.
12) Of 92 adults selected randomly from one town, 60 have health insurance. Construct a 90%
12)
confidence interval for the percentage of all adults in the town who have health insurance.
A) (52.4%, 78.0%)
B) (53.6%, 76.8%)
C) (57.0%, 73.4%)
D) (57.7%, 72.7%)
E) (55.5%, 74.9%)
Solve the problem.
13) A pollster wishes to estimate the true proportion of U.S. voters who oppose capital punishment.
How many voters should be surveyed in order to be 95% confident that the true proportion is
estimated to within 3%?
A) 1068
B) 1842
C) 752
D) 1503
E) Not enough information is given.
14) A university's administrator proposes to do an analysis of the proportion of graduates who have
not found employment in their major field one year after graduation. In previous years, the
percentage averaged 13%. He wants the margin of error to be within 4% at a 99% confidence level.
What sample size will suffice?
A) 563
B) 191.4
C) 469
D) 272
E) 19
What confindence level did the pollsters use?
15) A newspaper reports that the governor's approval rating stands at 60%. The article adds that the
poll is based on a random sample of 2,547 adults and has a margin of error of 2.5%.
A) 98%
B) 90%
C) 99%
D) 95%
E) Not enough information is given.
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13)
14)
15)
Provide an appropriate response.
16) In a poll of 861 voters in a certain city, 72% said that they backed a bill which would limit growth
and development in their city. The margin of error in the poll was reported as 3 percentage points
(with a 95% degree of confidence). Make a statement about the adequacy of the sample size for the
given margin of error.
A) The sample size is too small to achieve the stated margin of error.
B) The sample size is too large to achieve the stated margin of error.
C) For the given sample size, the margin of error should be smaller than stated
D) The reported margin of error is consistent with the sample size.
E) There is not enough information to determine whether the margin of error is consistent with
the sample size.
16)
17) We have calculated a confidence interval based on a sample of size n = 100. Now we want to get a
better estimate with a margin of error that is only one-fourth as large. How large does our new
sample need to be?
A) 50
B) 25
C) 400
D) 1600
E) 200
17)
18) A relief fund is set up to collect donations for the families affected by recent storms. A random
sample of 400 people shows that 28% of those 200 who were contacted by telephone actually made
contributions compared to only 18% of the 200 who received first class mail requests. Which
formula calculates the 95% confidence interval for the difference in the proportions of people who
make donations if contacted by telephone or first class mail?
(0.23)(0.77)
A) (0.28 - 0.18) ± 1.96
400
18)
B) (0.28 - 0.18) ± 1.96
(0.28)(0.72) (0.18)(0.82)
+
200
200
C) (0.28 - 0.18) ± 1.96
(0.23)(0.77) (0.23)(0.77)
+
200
200
D) (0.28 - 0.18) ± 1.96
(0.28)(0.72) (0.18)(0.82)
+
400
400
E) (0.28 - 0.18) ± 1.96
(0.23)(0.77)
200
Write the null and alternative hypotheses you would use to test the following situation.
19) 3% of trucks of a certain model have needed new engines after being driven between 0 and 100
miles. The manufacturer hopes that the redesign of one of the engine's components has solved this
problem.
A) H0 : p = 0.03
HA: p < 0.03
B) H0 : p < 0.03
HA: p = 0.03
C) H0 : p < 0.03
HA: p > 0.03
D) H0 : p = 0.03
HA: p > 0.03
E) H0 : p > 0.03
HA: p = 0.03
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19)
20) The federal guideline for smog is 12% pollutants per 10,000 volume of air. A metropolitan city is
trying to bring its smog level into federal guidelines. The city comes up with a new policy where
city employees are to use of city transportation to and from work. A local environmental group
does not think the city is doing enough and no real change will occur. An independent agency,
hired by the city, runs a test on the air. What are the null and alternative hypotheses?
A) H0 : p = 0.12
20)
HA: p < 0.12
B) H0 : p = 0.12
HA: p ≠ 0.12
C) H0 : p = 0.12
HA: p > 0.12
D) H0 : p < 0.12
HA: p > 0.12
E) H0 : p ≠ 0.12
HA: p = 0.12
Provide an appropriate response.
21) A state university wants to increase its retention rate of 4% for graduating students from the
previous year. After implementing several new programs during the last two years, the university
reevaluates its retention rate and comes up with a P-value of 0.075. What is reasonable to conclude
about the new programs?
A) We can say there is a 7.5% chance of seeing the new programs having no effect on retention in
the results we observed from natural sampling variation. There is no evidence the new
programs are more effective, but we cannot conclude the new programs have no effect on
retention.
B) There is a 92.5% chance of the new programs having no effect on retention.
C) We can say there is a 7.5% chance of seeing the new programs having an effect on retention in
the results we observed from natural sampling variation. We conclude the new programs are
more effective.
D) There's only a 7.5% chance of seeing the new programs having no effect on retention in the
results we observed from natural sampling variation. We conclude the new programs are
more effective.
E) There is a 7.5% chance of the new programs having no effect on retention.
22) A state university wants to increase its retention rate of 4% for graduating students from the
previous year. After implementing several new programs during the last two years, the university
reevaluated its retention rate using a random sample of 352 students and found the retention rate at
5%. Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions
and conditions are satisfied before you proceed.
A) H0 : p = 0.04; HA: p > 0.04; z = 0.96; P-value = 0.1685. This data does not show that more than
4% of students are retained; the university should not continue with the new programs.
B) H0 : p = 0.04; HA: p < 0.04; z = 1.07; P-value = 0.8577. This data shows that more than 4% of
students are retained; therefore, the university should continue with the new programs.
C) H0 : p = 0.04; HA: p ≠ 0.04; z = 1.07; P-value = 0.2846. This data does not show that more than
4% of students are retained; the university should not continue with the new programs.
D) H0 : p = 0.04; HA: p > 0.04; z = -1.07; P-value = 0.1423. This data does not show that more
than 4% of students are retained; the university should not continue with the new programs.
E) H0 : p = 0.04; HA: p < 0.04; z = -1.07; P-value = 0.8577. This data shows that more than 4% of
students are retained; the university should continue with the new programs.
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21)
22)
23) A researcher investigating whether joggers are less likely to get colds than people who do not jog
found a P-value of 3%. This means that:
A) Joggers get 3% fewer colds than non-joggers.
B) There's a 3% chance that joggers don't get fewer colds.
C) There's a 3% chance that joggers get fewer colds.
D) 3% of joggers get colds.
E) None of these.
Explain what the P-value means in the given context.
24) A new manager, hired at a large warehouse, was told to reduce the 26% employee sick leave. The
manager introduced a new incentive program for employees with perfect attendance. The manager
decides to test the new program to see if it's better and receives a P-value of 0.06. What is
reasonable to conclude about the new strategy using α = 0.05?
A) We can say there is a 6% chance of seeing the new program having no effect on employee
attendance in the results we observed from natural sampling variation. There is no evidence
the new program is more effective, but we cannot conclude the program has no effect on
employee attendance.
B) There's only a 6% chance of seeing the new program having no effect on employee attendance
in the results we observed from natural sampling variation. We conclude the new program is
more effective.
C) We can say there is a 6% chance of seeing the new program having an effect on employee
attendance in the results we observed from natural sampling variation. We conclude the new
program is more effective.
D) There is a 94% chance of the new program having no effect on employee attendance.
E) There is a 6% chance of the new program having no effect on employee attendance.
Provide an appropriate response.
25) A state university wants to increase its retention rate of 4% for graduating students from the
previous year. After implementing several new programs during the last two years, the university
reevaluated its retention rate. Identify the Type I error in this context.
A) The product of the university's sample size and sample proportion was less than 10.
B) The university stops all new programs, but in fact retention is on the rise and the programs
help.
C) The university sampled all students at the university.
D) The university concludes that retention is on the rise, but in fact the new programs do not
help retention.
E) The university concludes that retention is on the rise since the retention rate can only increase.
26) A psychologist claims that more than 1.4% of the population suffers from professional problems
due to extreme shyness. Identify the Type II error in this context.
A) The error of rejecting the claim that the true proportion is at most 1.4% when it really is at
most 1.4%.
B) The error of accepting the claim that the true proportion is more than 1.4% when it really is
more than 1.4%.
C) The error of rejecting the claim that the true proportion is more than 1.4% when it really is
more than 1.4%.
D) The error of failing to reject the claim that the true proportion is at most 1.4% when it is
actually more than 1.4%.
E) The error of failing to accept the claim that the true proportion is at most 1.4% when it is
actually more than 1.4%.
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23)
24)
25)
26)
27) A state university wants to increase its retention rate of 4% for graduating students from the
previous year. After implementing several new programs during the last two years, the university
reevaluated its retention rate using a random sample of 352 students and retained 18 students.
Should the university continue its new programs? Test an appropriate hypothesis using α = 0.10
and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before
you proceed.
A) z = 1.07; P-value = 0.1423. The university should not continue with the new programs. There
is a 14.23% chance of having 18 or more of 352 students in a random sample be retained if in
fact 4% are retained. The P-value of 0.1423 is greater than the alpha level of 0.10.
B) z = -1.07; P-value = 0.1423. The change is statistically significant. A 98% confidence interval is
(2.7%, 7.5%). This is clearly lower than 4%. The chance of observing 18 or more retained
students of 352 is only 14.23% if the dropout rate is really 4%.
C) z = 1.07; P-value = 0.8577. The change is statistically significant. A 90% confidence interval is
(3.4%, 6.8%). This is clearly higher than 4%. The chance of observing 18 or more retained
students of 352 is only 85.77% if the dropout rate is really 4%.
D) z = -1.07; P-value = 0.8577. The university should continue with the new programs. There is
an 85.77% chance of having 18 or more of 352 students in a random sample be retained if in
fact 4% are retained.
E) z = 1.07; P-value = 0.2846. The change is statistically significant. A 95% confidence interval is
(3.1%, 67.2%). This is clearly lower than 4%. The chance of observing 18 or more retained
students of 352 is only 28.46% if the dropout rate is really 4%.
27)
28) We test the hypothesis that p = 35% versus p < 35%. We don't know it but actually p = 26%. With
which sample size and significance level will our test have the greatest power?
A) α = 0.01, n = 250
B) The power will be the same as long as the true proportion p remains 26%
C) α = 0.03, n = 250
D) α = 0.01, n = 400
E) α = 0.03, n = 400
28)
Construct the indicated confidence interval for the difference in proportions. Assume that the samples are independent
and that they have been randomly selected.
29) A survey of randomly selected college students found that 50 of the 95 freshmen and 52 of the 107
29)
sophomores surveyed had purchased used textbooks in the past year. Construct a 98% confidence
interval for the difference in the proportions of college freshmen and sophomores who purchased
used textbooks.
A) (0.388, 0.664)
B) (-0.124, 0.204)
C) (-0.098, 0.664)
D) (0.362, 0.664)
E) (0.362, 0.690)
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Interpret the given confidence interval.
30) Suppose the proportion of women who follow a regular exercise program is pw and the proportion
30)
of men who follow a regular exercise program is pm . A study found a 90% confidence interval for
pw - pm is (-0.025, 0.113). Give an interpretation of this confidence interval.
A) We are 90% confident that the proportion of women who follow a regular exercise program is
between 2.5% less and 11.3% more than the proportion of men who follow a regular exercise
program.
B) We are 90% confident that the proportion of men who follow a regular exercise program is
between 2.5% less and 11.3% more than the proportion of women who follow a regular
exercise program.
C) We know that 90% of men exercise between 2.5% less and 11.3% more than women.
D) We know that 90% of all random samples done on the population will show that the
proportion of women who follow a regular exercise program is between 2.5% less and 11.3%
more than the proportion of men who follow a regular exercise program.
E) We know that 90% of women exercise between 2.5% less and 11.3% more than men.
A two-sample z-test for two population proportions is to be performed using the P-value approach. The null hypothesis
is H : p = p and the alternative is H : p ≠ p . Use the given sample data to find the P-value for the hypothesis
0
1
2
a
1
2
test. Give an interpretation of the p-value.
31) A state university found it retained 25 students out of 352 in 2003 and 36 students out of 334 in
31)
2004.
A) P-value = 0. 0455; There is about a 4.55% chance that the two proportions are equal.
B) P-value = 0.9545; If there is no difference in the proportions, there is about a 95.45% chance of
seeing the observed difference or larger by natural sampling variation.
C) P-value = 0.0455; If there is no difference in the proportions, there is about a 4.55% chance of
seeing the observed difference or larger by natural sampling variation.
D) P-value = 0.091; If there is no difference in the proportions, there is about a 9.1% chance of
seeing the observed difference or larger by natural sampling variation.
E) P-value = 0. 091; There is about a 9.1% chance that the two proportions are equal.
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Answer Key
Testname: FINAL EXAM REVIEW SECOND TRIMESTER
1) D
2) D
3) D
4) B
5) B
6) D
7) C
8) B
9) C
10) C
11) C
12) C
13) A
14) C
15) C
16) D
17) D
18) B
19) A
20) B
21) A
22) A
23) E
24) A
25) D
26) D
27) A
28) D
29) B
30) A
31) D
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