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STT 231 – 001
PRACTICE EXERCISES ON STATISTICAL INFERENCES

SAMPLE DISTRIBUTION FOR ONE SAMPLE PROPORTION
QUESTIONS 1 – 2
Information on a packet of seeds claims that the germination rate is 84%. The packet contains 150 seeds. Let p̂
represent the proportion of seeds in the packet that will germinate.
1. The sampling distribution model for p̂ is
A. N(0.84, 0.0299)
B. N(0.16, 0.0299)
C. N(0.84, 0.0001)
D. N(0.0215, 0.92)
2. What’s the approximate probability that more than 90% of the 150 seeds in the packet will germinate?
A. 0.9778
B. 0.0224
C. 0.0299
D. 0.90
E. 0.16
3. When a truckload of apples arrives at a packing plant, a random sample of 180 is selected and examined for
bruises, discoloration, and other defects. The whole truckload will be rejected if more than 4% of the sample is
unsatisfactory. Suppose that in fact 7% of the apples on the truck do not meet the desired standard. What’s the
probability that the shipment will be accepted anyway?
A. 0.0571
B. 0.981
C. – 1.578
D. 0.943
E. 0.0409
QUESTIONS 4 – 5
It’s believed that 5% of children have a gene that may be linked to juvenile diabetes. Researchers hoping to track
30 of these children for several years test 742 newborns for the presence of this gene.
4. The sampling distribution of the proportion of children with gene linked to juvenile diabetes is best described as
A. N(0.04, 0.0072)
B. N(0.0072, 0.04)
C. N(0.008, 0.05)
D. N(0.05, 0.008)
5. What is the probability that they find enough subjects for their study?
A. 0.0404
B. 0.0072
C. 0.960
D. 0.1170
E. 0.884
6. Based on past experiences, a bank believes that 8% of the people who receive loans will not make payments on
time. The bank has recently approved 200 loans. What’s the probability that over 13% of these clients will not
make timely payments?
A. 0.9953
B. 0.0192
C. 0.0046
D. 0.92
E. 0.261
QUESTIONS 7 – 9
Suppose that 65% of the adult residents in North Dakota favor the death penalty. In a simple random sample of 100
North Dakota adult residents, let p̂ denote the proportion that favor the death penalty.
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7. What is the mean (expected) value of p̂ ?
(a) 50
(b) 60
(c) 65
(d) 0.50
(e) none of these
8. What is the standard deviation of p̂ ?
(a) 0.15
(b) 0.048
(c) 5
(e) none of these
(d) 0.09
9. What is the approximate probability that 57% or less of the sample residents favor death penalty, that is, what is
P( p̂ 0.57)
(a) 0.50
(b) 0.40
(c) 0.28
(d) 0.16
(e) 0.05
QUESTIONS 10 – 11
Information on a packet of seeds claims that the germination rate is 92%. The packet contains 160 seeds. Let p̂
represent the proportion of seeds in the packet that will germinate.
10. The sampling distribution model for p̂ is
A. N(0.92, 0.0215)
B. N(0.08, 0.0215)
C. N(147.2, 12.8)
D. N(0.0215, 0.92)
11. What’s the approximate probability that more than 95% of the 160 seeds in the packet will germinate?
A. 0.92
B. 0.081
C. 0.0215
D. 0.95
12. When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for
bruises, discoloration, and other defects. The whole truckload will be rejected if more than 5% of the sample is
unsatisfactory. Suppose that in fact 8% of the apples on the truck do not meet the desired standard. What’s the
probability that the shipment will be accepted anyway?
A. 0.0222
B. 0.9778
C. 0.088
D. 0.912
QUESTIONS 13 – 14
It’s believed that 4% of children have a gene that may be linked to juvenile diabetes. Researchers hoping to track
20 of these children for several years test 732 newborns for the presence of this gene.
13. The sampling distribution of the proportion of children with gene linked to juvenile diabetes is best described
as
A. N(0.04, 0.0072)
B. N(0.0072, 0.04)
C. N(0.027, 0.04)
D. N(0.027, 0.0072)
14. What is the probability that they find enough subjects for their study?
A. 0.0273
B. 0.0072
C. 0.960
D. 0.04
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
SAMPLING DISTRIBUTION FOR TWO SAMPLE PROPORTIONS
1. There has been debate among doctors over whether surgery can prolong life among men suffering from
prostrate cancer, a type of cancer that typically develops and spreads very slowly. In the summer of 2003, The
New England Journal of Medicine published results of some Scandinavian research. Men diagnosed with prostrate
cancer were randomly assigned to either undergo surgery or not. Among the 347 men who had surgery, 16
eventually died of prostrate cancer, compared with 31 of the 348 men who did not have surgery. What is the
standard error of the difference in the two proportions?
A. 0.0189
B. 0.00036
C. 0.0113
D. 0.037
E. 0.08908
2. Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed –
killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular
basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where no herbicides were
used, only 19 were found to have lymphoma. What’s the standard error of the difference in the two proportions?
A. 0.0354
B. 0.0013
C. 0.0172
D. 0.0309
E. 0.4258
3 - 4. A veterinarian wants to compare the rate of hip dysplasia in Boxers (breed of dog) and the rate hip dysplasia
in American Bulldogs (breed of dog). A random sample of 80 Boxers has 8 dogs with hip dysplasia. A random
sample of 140 American Bulldogs has 21 dogs with hip dysplasia.

3. From these two samples, the difference between 2 sample proportions, p1  p̂2 is
(a) 0.1
(b) 0.15
(c) -0.05
(d) 0.8
(e) -0.13
4. The standard error of the sampling distribution of p1  p2 is
(a) .085
(b) .071
(c) .045
(d) 0.037
(e) .101
5 - 7: At a small college somewhere on the East coast, 20% of the girls and 30% of the boys smoke at least once a
week. Independent random samples of 75 girls and 100 boys are to be selected and the proportions of smokers in
the samples are to be calculated.
5. What is the mean of the sampling distribution of the difference in the sample proportion of girl smokers
and the sample proportion of boy smokers?
KEY: –0.10
6. What is the standard deviation of the sampling distribution of the difference in the sample proportion of girl
smokers and the sample proportion of boy smokers?
KEY: 0.0651
7.
What is probability that the sample proportion of girl smokers is greater than the sample proportion of boy
smokers?
KEY: 0.0623
8 - 9: In the Youth Risk Behavior Survey (a study of public high school students), a random sample showed that
45 of 675 girls and 103 of 621 boys had been in a physical fight on school property one or more times during the
past 12 months.
8.
What is the difference in sample proportions of students who had been in a fight (boys – girls)?
KEY: 0.0992.
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9.
What is the standard error of the difference in sample proportions?
KEY: 0.0177
10 - 12: Suppose that 60% of all teenagers — both boys and girls — are classified as having good grades.
Independent random samples of 537 boys and 689 girls who go to school in the Washington, DC, area are to be
selected and surveyed and the proportions of teenagers with good grades are to be calculated.
10.
What is the mean of the sampling distribution of the difference between the two sample proportions
(boys – girls)?
KEY: 0
11.
What is the standard deviation of the sampling distribution of the difference between the two sample
proportions (boys – girls)?
KEY: 0.0282
What is the probability that the difference between the two sample proportions (boys – girls) is greater than 5
percentage points (0.05)?
KEY: 0.0381
12.
13 - 15: Suppose that 80% of all English majors and 85% of all engineering majors at a Minnesota college wear
winter boots when there is snow on the ground. Two independent random samples of 40 English majors and 60
engineering majors are to be selected during a day with snow on the ground and the proportions of students with
winter boots are to be calculated.
13.
What is the mean of the sampling distribution of the difference between the two sample proportions
(English majors – engineering majors)?
KEY: −0.05
14.
What is the standard deviation of the sampling distribution of the difference between the two sample
proportions (English majors – engineering majors)?
KEY: 0.0783
15. What is the probability that more English majors than engineering majors in the sample wear winter boots?
KEY: 0.2616
16 - 17: A survey at a large public university reveals that many students hold a (part-time) job while going to
college. Out of 127 female students surveyed, 95 have a job and out of 143 male students, 97 have a job.
16. What is the estimate for the difference between the proportions of female and male students who have a job
while going to college?
KEY: 0.0697
17. What is the standard error of the estimate?
KEY: 0.0549
18. Suppose that the mean of the sampling distribution for the difference in two sample means is 0. This tells
us that
A. the two sample means are both 0.
B. the two sample means are equal to each other.
C. the two population means are both 0.
D. the two population means are equal to each other.
KEY: D
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 SAMPLING DISTRIBUTION FOR ONE SAMPLE MEAN
 SAMPLING DISTRIBUTION FOR TWO SAMPLE MEANS
QUESTIONS 1 – 4
Statistics from Cornell’s Northeast Regional Climate Center indicate that Ithaca, NY, gets an average of 35.4
inches of rain each year, with a standard deviation of 4.2 inches.
1. During what percentage of years does Ithaca get more than 30 inches of rain?
A. approximately 10.03%
C. approximately 90.07%
E. approximately 94.18%
B. approximately 10.95%
D. approximately 5.82
2. Less than how much rain falls in the driest 25% of all years?
A. 13.7 inches
B. 10.03 inches
C. 10.4 inches
D. 32.6 inches.
A Cornell University student is in Ithaca for 4 years. Let y represent the mean amount of rain for those 4 years.
3. The sampling distribution model of this mean, y , is best described as
A. N(35.4, 4.2)
B. N(4.2, 35.4)
C. N(35.4, 2.1)
D. N(2.1, 35.4)
4. What’s the probability that those 4 years average less than 30 inches of rain?
A. 0.995
B. 0
C. 0.11
D. 0.005
QUESTIONS 5 – 7
The weight of potato chips in a medium-size bag is stated to be 10 ounces. The amount that the packaging machine
puts in these bags is believed to have a Normal model with mean 10.2 ounces and standard deviation 0.12 ounces.
5. What fraction of all bags sold are underweight?
A. 0.9804
B. 0.0478
C. 0.863
D. 0.2
6. Some of the chips are sold in “bargain packs” of 3 bags. What’s the probability that none of the 3 is
underweight?
A. 0.863
B. 0.1434
C. 0.000109
D. 0.9522
E. None of these
7. What’s the probability that the mean weight of the 3 bags is below the stated amount?
A. 0.069
B. 0.0478
C. 0.9522
D. 0.0019
8. The average composite ACT score for Ohio students who took the test in 2003 was 21.4. Assume that the
standard deviation is 1.05. In a random sample of 25 students who took the exam in 2003, what is the probability
that the average composite ACT score is 22 or more? (HINT: Make sure to identify the sampling distribution you
use and check all necessary conditions before you proceed to solve the problem.)
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A. 0.21
B. 2.86
C. 0.0021
D. 0.9979
E. 0.79
QUESTIONS 9 – 12
Statistics from Cornell’s Northeast Regional Climate Center indicate that Ithaca, NY, gets an
inches of rain each year, with a standard deviation of 4.2 inches.
average of 35.4
9. During what percentage of years does Ithaca get more than 40 inches of rain?
A. approximately 13.7%
B. approximately 10.95%
C. approximately 31.2%
10. Less than how much rain falls in the driest 20% of all years?
A. 13.7 inches
B. 10.95 inches
C. 39.6 inches
D. 31.9 inches.
A Cornell University student is in Ithaca for 4 years. Let y represent the mean amount of rain for those 4 years.
11. The sampling distribution model of this mean, y is best described as
A. N(35.4, 4.2)
B. N(4.2, 35.4)
C. N(35.4, 2.1)
D. N(2.1, 35.4)
12. What’s the probability that those 4 years average less than 30 inches of rain?
A. 0.995
B. 0
C. 0.11
D. 0.005
13. Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and
standard deviation 16 days.
(a) What percentage of pregnancies should last between 270 and 280 days?
(b) At least how many days should the longest 25% of all pregnancies last? 276.8 days
(c) Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let y – bar represent
the mean length of their pregnancies. According to the Central Limit Theorem, what’s the distribution of this
sample mean, y – bar? Specify the model, mean, and standard deviation. Mean = 266 days; Standard deviation =
2.07 days
(d) What’s the probability that the mean duration of these patient’s pregnancies will be less than 260 days? 0.002
14. Carbon monoxide (CO) emissions for a certain kind of car vary with mean 2.9 g/mi and standard deviation 0.4
g/mi. A company has 80 of these cars in its fleet. Let y – bar represent the mean CO level for the company’s fleet.
(a) What’s the approximate model for the distribution of y – bar? Explain. N(2.9, 0.045)
(b) Estimate the probability that y – bar is between 3.0 and 3.1 g/mi.
(c) There is only a 5% chance that the fleet’s mean CO level is greater than what value? 2.97 g/mi
15. A fourth grade class of 28 students is given a standardized math test. The mean score of the 12 boys is 25
with a standard deviation of 3. The mean score of the 16 girls is 24 with a standard deviation of 4. What is
the standard error for the sampling distribution of xboys  xgirls ?
a. 1
b. 1.87
c. 1.75
d. 1.32
KEY: D
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16. A researcher wishes to determine if female students and male students differ in the size of their network of
friends. She selects random samples of 10 female and 15 male students. Descriptive statistics are given
below:
nfemale = 10 xfemale = 23 sfemale = 7.8
nmale = 15
xmale = 16
smale = 6.4
What is the standard error for the sampling distribution of xfemale  xmale ?
a. 1.10
b. 8.81
c. 2.97
d. 7.10
KEY: C
Questions 17 to 20: High school students can be categorized into two groups by the amount of activities they are
involved in. Let group 1 consist of all high school students who are very involved in sports and other activities and
group 2 consist of all high school students who aren’t. The distributions of GPAs in both groups are approximately
normal. The mean and standard deviation for group 1 are 2.9 and 0.4, respectively. The mean and standard
deviation for group 2 are 2.7 and 0.5, respectively. Independent random samples of 50 high school students are to
be selected from both groups (for a total of 100 students).
17. If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in
activities – not so involved in activities, what is the expected value for the difference in sample means?
a. 0
b. 0.2
c. 0.4
d. 0.6
KEY: B
18. If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in
activities – not so involved in activities, what is the standard deviation of the sampling distribution of the
difference in sample means?
a. 0.0082
b. 0.0905
c. 0.18
d. 0.45
KEY: B
19. What is the probability that the average GPA in the sample of students who are not so involved is higher
than the average GPA in the sample of students who are very involved?
a. < 0.0001
b. 0.0136
c. 0.0582
d. 0.9864
KEY: B
20. What is the probability that the average GPAs in the two samples differ by no more than 0.1?
a. 0.1341
b. 0.2266
c. 0.2415
d. 0.8659
KEY: A
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Questions 21 to 24: College students spend a lot of their money, not on things they would like to buy, but on text
books. College text books have increased in price significantly over the past few years. The amount students spend
on text books is approximately normally distributed with a standard deviation of $50. The average amount spent on
books each semester is $340 for undergraduate students and $250 for graduate students. Independent random
samples of 100 undergraduate students and 80 graduate students are to be selected and the average amount they
spent on text books last semester is to be compared (undergraduate – graduate).
21. What is the expected value for the difference in sample means?
KEY: $90
22. What is the standard deviation of the sampling distribution of the difference in sample means?
KEY: $7.50
23. What is the probability that the difference between the sample means is greater than $100?
KEY: 0.0912
24.
What is the probability that the undergraduate students in the sample spent more on books, on average, than
the graduate students in the sample?
KEY: < 0.0001
Questions 25 to 27: Many adults complain they do not get enough sleep at night. What happens when these adults
have children? Independent random samples of 150 adults who do not have children in the house and 130 adults
who do have children in the house are selected. The average amount of sleep the adults without children had was
7.1 hours with a standard deviation of 0.7 hours. The average amount of sleep the adults with children had was 6.5
hours with a standard deviation of 0.8 hours.
25.
What is the estimate for the difference in mean hours slept at night between adults with and adults without
children?
KEY: –0.6 hours (–36 minutes)
26.
What is the estimate for the standard deviation of the sampling distribution of the difference in sample
means?
KEY: 0.0905 hours (roughly 5 minutes)
27. What do we call the estimate in question 124?
KEY: standard error or standard error of x1  x2
Questions 28 to 30: Grade school students, especially the ones in higher grades, tend to spend a lot of time on the
computer already. The average amount of time 4th grade boys spend on the computer is 5 hours per week, with a
standard deviation of 1 hour. For 5th grade boys the average increases to 7 hours per week with a standard
deviation of 1.5 hours. Random sample of 20 4th grade students and 25 5th grade students are to be selected.
28. What is the expected value for the difference in sample means (5th − 4th grade)?
KEY: 2 hours
29. What is the standard deviation of the sampling distribution of the difference in sample means?
KEY: 0.374
8
30. What is the probability that the difference between the sample means is greater than 1 hour?
KEY: 0.9962
31. A randomly selected sample of 100 students had an average grade point average (GPA) of 3.2 with a
standard deviation of 0.2. The standard error of the sample mean is
A. 0.020
B. 0.200
C. 1.600
D. 2.000
KEY: B
32. A randomly selected sample of 30 students spent an average amount of $40.00 on a date, with a standard
deviation of $5.00. The standard error of the sample mean is
A. 0.063
B. 0.167
C. 0.913
D. 5.000
KEY: C
33. A random sample of 250 third graders scored an average of 3.2 on a standardized reading test. The standard
deviation was 0.95. What is the standard error of the sample mean?
A. 0.95
B. 0.202
C. 0.0038
D. 0.060
KEY: D
34. A randomly selected sample of 60 mathematics majors spent an average of $200.00 for textbooks one term,
while during the same term, a randomly selected sample of 40 literature majors spent an average of $180.00
for textbooks. The standard deviation for each sample was $20.00. The standard error for the difference
between the two sample means is
A. 0.057
B. 4.082
C. 5.744
D. 16.663
KEY: B
35. A random sample of 40 men drank an average of 20 cups of coffee per week during finals, while a sample
of 30 women drank an average of 15 cups of coffee per week. The sample standard deviations were 6 cups
for the men and 3 cups for the women. The standard error for the difference between the two sample
means is
A. 1.095
B. 1.200
C. 1.549
D. 2.400
KEY: A
36. Conscientiousness is a tendency to show self-discipline, act dutifully, and aim for achievement. The trait
shows a preference for planned rather than spontaneous behavior. A random sample of 650 students is
asked to fill out the Hogan Personality Inventory (HPI) to measure their level of conscientiousness. The
300 undergraduate students scored an average of 145 with a standard deviation of 16. The 350 graduate
students had a mean score of 153 with a standard deviation of 21. What is the standard error for the
difference between the two sample means?
A. 18.5
B. 2.11
9
C. 1.45
D. 2.05
KEY: C
37. Which of the following is not true about the standard error of a statistic?
A. The standard error measures, roughly, the average difference between the statistic and the population
parameter.
B. The standard error is the estimated standard deviation of the sampling distribution for the statistic.
C. The standard error can never be a negative number.
D. The standard error increases as the sample size(s) increases.
KEY: D
38. For a randomly selected sample of n = 36 men’s heights, it is reported that the standard error of the mean is
0.5 inches. Three of the following statements are true, while one is false. Which statement is false?
A. The standard error (0.5 inches) is an estimated value of the standard deviation of the sample mean.
B. If a new sample of n = 36 men’s heights is collected, the standard error of the mean might not equal 0.5
inches.
C. Over many different samples of n = 36 men’s heights, the average difference between the sample mean and
population mean will be roughly 0.5 inches.
D. In about 95% of all samples of n = 36 men’s heights, the sample mean will be within 0.5 inches of the
population mean.
KEY: D
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
 Confidence Intervals for One Sample Proportions, p
Confidence Intervals for the Difference in Two Sample Proportions
1. 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 2%?
A. 3382
D. 4145
B. 1692
E. None of the above
C. 2401
2. After conducting a survey, a researcher wishes to cut the standard error (and thus the margin of error) to
1
of
2
its original value. How will the necessary sample size change?
A. It will increase by a factor of 4.
C. It will decrease by a factor of 4.
E. None of the above.
B. It will decrease by factor of 9.
D. It will increase by a factor of 9.
QUESTIONS 3 – 6
In 1998 a San Diego reproductive clinic reported 49 births to 207 women under the age of 40 who had previously
been unable to conceive.
3. Find a 90% confidence interval for the success rate at this clinic.
A. (0.237, 0.763)
B. (0.285, 0.763)
E. (0.285, 0.188)
B. (0.188, 0.285)
D. (0.188, 0.237)
4. Would it be misleading for the clinic to advertise a 25% success rate? Explain.
A. Absolutely misleading because the sample size is not large enough.
B. It would not be misleading because the 10% condition fails.
C. It would not be misleading for the clinic to advertise a 25% success rate, since 25% is in the interval.
D. It would be misleading because pˆ  0.90
E. None of the above
5. The clinic wants to cut the stated margin of error in half. How many patients’ result must be used?
A. 103
C. 828
E. 79
B. 25
D. 158
6. Do you have any concerns about this sample? Explain.
A. A sample this size is too small to achieve the desired goal.
B. A sample this large may be more than 10% of the population of all potential patients.
C. There are no concerns about this sample.
D. No concerns since pˆ  0.237 is in the confidence interval.
E. None of the above
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7. Malcolm observes that 15 cars in the student parking lot have satellite radio, while 45 do not. He wants to
estimate the proportion of students’ cars with satellite radio. Find a 95% confidence interval for the proportion of
students’ cars with satellite radio.
A. (0.12, 0.36)
B. (0.24, 0.95)
C. (0.14, 0.36)
D. (0.19, 0.29)
E. Not enough information
QUESTIONS 8 – 10
A May 2002 Gallup poll, found that only 4% of a random sample of 812 adults approved of attempts to clone a
human.
8. Find the margin of error for this poll if we want 95% confidence in our estimate of the percent of American
adults who approve of cloning humans.
A. 1.960
B. 0.0135
C. 0.0167
D. 0.9865
E. None of these
9. If we only need to be 90% confident, will the margin of error be larger or smaller. What is this margin of error,
ME?
A. Larger, ME = 0.0167
B. Smaller, ME = 0.0135
C. Smaller, ME = 0.014
D. Smaller; 0.0113
10. In general, if all other aspects of the situation remain the same, would smaller samples produce smaller or
larger margins of error?
A. Larger margin of error B. Smaller margin of error
C. Margin of error remains the same.
11. The real estate industry claims that it is the best and most effective system to market residential real estate. A
survey of randomly selected home sellers in Illinois found that a 99% confidence interval for the proportion of
homes that are sold by a real estate agent is 70% to 80%. Explain what "99% confidence" means in this context.
A. About 99% of all random samples of home sellers in Illinois will produce a confidence interval that
contains the true proportion of homes sold by a real estate agent.
B. 99% of home sellers in Illinois will sell their home with a real estate agent between 70% and 80% of the time.
C. In 99% of the years, between 70% and 80% of homes in Illinois are sold by a real estate agent.
D. About 99% of all random samples of home sellers in Illinois will find that between 70% and 80% of homes are
sold by a real estate agent.
E. There is a 99% chance that the true proportion of home sellers in Illinois who sell their home with a real estate
agent is between 70% and 80%.
12. Laura observes that 15 cars in the student parking lot have satellite radio, while 45 do not. He wants to
estimate the proportion of students’ cars with satellite radio. Find a 95% confidence interval for the proportion of
students’ cars with satellite radio.
A. (14%, 36%)
B. (12%, 36%)
C. (24%, 95%)
D. (19%, 29%)
12
13. A political pollster wants to know what proportion of voters thinks James Burger should go away and stay
away, within 5 percentage points with 95% confidence. She does not have a good guess as to what this proportion
is. How large should her sample be?
A. 1448
B. 385
C. 519
D. 1227
E. 20
14. In a survey of 280 adults over 50, 75% said they were taking vitamin supplements. Find the margin of error for
this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin
supplements.
A. 0.133
B. 0.101
C. 0.067
D. 0.0507
E. 0.186
15. Betty wants to know what proportion of automobile customers order sunroofs on their Subarus. One day hour
she sees 92 Subarus with sunroofs and 34 without. Her 95% confidence interval for the proportion sold with
sunroofs is:
A. (.23, .51)
B. (.652, .805)
C. (41.7%, 50.3%)
D. (68%, 75%)
E. (5%, 95%)
16. Of 346 items tested, 12 are found to be defective. Construct a 98% confidence interval for the percentage of
all such items that are defective.
A. (1.18%, 5.76%)
D. (0.13%, 6.80%)
B. (0.93%, 6.00%)
E. (3.34%, 3.59%)
C. (1.85%, 5.09%)
17. Of 230 employees selected randomly from one company, 10.43% of them commute by carpooling.
Construct a 90% confidence interval for the percentage of all employees of the company who carpool.
A. (5.73%, 15.1%)
D. (5.23%, 15.6%)
B. (6.48%, 14.4%)
E. (7.11%, 13.7%)
C. (5.73%, 15.6%)
18. 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 2%?
A. 3382
B. 1692
C. 4145
D. 2401
E. Sufficient information is not provided
QUESTIONS 19 – 20
According to a September 2004 Gallup poll, about 73% of 18 – to – 29 – year – olds said that they were
registered to vote. A statistics professor asked her students whether or not they were registered to vote. In a
sample of 50 of her students (randomly sampled from her 700 students), 35 said they were registered to vote.
19. Find a 95% confidence interval for the true proportion of the professor’s students who were registered to
vote.
A. (0.573, 0.827)
E. (0.065, 0.573)
B. (0.827, 0.573)
C. (0.065, 0.70)
D. (0.065, 0.827)
13
20. If the professor only knew the information from the September 2004 Gallup poll and wanted to estimate the
percentage of her students who were registered to vote to within 4% with 95% confidence, how many students
should she sample?
A. approximately 200
B. approximately 474
C. approximately 13
D. approximately 800
E. approximately 140
21. After conducting a survey, a researcher wishes to cut the standard error (and thus the margin of error) to 1/3 of
its original value. How will the necessary sample size change?
A. It will increase by a factor of 3.
C. It will decrease by a factor of 3.
E. None of the above
B. It will decrease by factor of 9.
D. It will increase by a factor of 9.
QUESTIONS 22 – 24
A May 2002 Gallup poll, found that only 8% of a random sample of 1012 adults approved of attempts to clone a
human.
22. Find the margin of error for this poll if we want 95% confidence in our estimate of the percent of American
adults who approve of cloning humans.
A. 1.960
B. 0.0085
C. 0.0167
D. 0.014
23. If we only need to be 90% confident, will the margin of error be larger or smaller. What is this margin of error,
ME?
A. Larger, ME = 0.0167
B. Smaller, ME = 0.0167
C. Smaller, ME = 0.014
24. In general, if all other aspects of the situation remain the same, would smaller samples produce smaller or
larger margins of error?
A. larger margin of error
B. larger margins of error C. margins of error remain the same.
QUESTIONS 25 – 26
Direct mail advertisers send solicitations [a.k.a. ‘junk mail’] to thousands of potential customers in the hope that
some will buy the company’s product. The response rate is usually quite low. Suppose a company wants to test the
response to a new flyer, and sends it to 1000 people randomly selected from their mailing list of over 200,000
people. They get orders from 123 of the recipients.
25. Create a 90% confidence interval for the percentage of people the company contacts who may buy something.
A. (0.123, 0.877)
B. (0.123, 0.0171)
C. (0.106, 0.14)
D. Cannot be calculated.
14
26. The company must decide whether to now do a mass mailing. The mailing won’t be cost-effective unless it
produces at least 5% return. What does your confidence interval suggest? Explain.
A. Our confidence interval suggests that the company should do the mass mailing. The entire interval is well
above the cutoff of 5%.
B. Our confidence interval suggests that the company should not go ahead with the mass mailing. The interval is
derived from estimates.
QUESTIONS 27 – 28
In 2004, ACT Inc. reported that 74% of 1644 randomly selected college freshmen returned to college the next year.
You are interested in estimating the national freshman-to-sophomore retention rate.
27. Construct a 98% confidence interval (CI) and interpret your interval.
A. CI = (0.715, 0.765); Between 71.5% and 76.5% of all college students return to college after freshman year.
B. CI = (0.765, 0.715); The proportion of all college students who return to college after freshman year is
between 0.715 and 0.765.
C. CI = (0.011, 0.02); We are 98% confident that between 1.1% and 2% of all college students return to
college after their freshman year.
D. CI = (0.74, 0.98); We are 98% confident that between 74% and 98% of all college students return to
college after their freshman year.
E. CI = (0.715, 0.765); We are 98% confident that between 71.5% and 76.5% of all college students
return to college after their freshman year.
28. Explain what “98%” confidence means in this context.
A. If we were to select repeated samples like this, we would expect about 98% of the confidence
intervals we created to contain the true proportion of all college students who return to college after
their freshman year.
B. 98% of the confidence intervals contain the true proportion of all college students who return to college
after their freshman year.
C. If we were to take several samples of the same size, we would expect just 2% of the confidence intervals to
contain the true proportion of all college students who return to college after their freshman year.
D. We are 98% confident that all college students will return to college after their freshman year.
E. None of the above
29. Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level
of confidence, and the margin of error. Which of the following statements is incorrect?
A. For a given sample size, reducing the margin of error will mean lower confidence.
B. For a given confidence level, a sample 9 times as large will make a margin of error one third as big.
C. For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample.
D. For a fixed margin of error, smaller samples will mean lower confidence.
E. For a given confidence level, halving the margin of error requires a sample twice as large.
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QUESTIONS 30 - 32
A Vermont study published in December 2001 by the American Academy of Pediatrics examined parental
influence on teenagers’ decision to smoke. A group of students who had never smoked were questioned about their
parents’ attitudes toward smoking. These students were questioned again two years later to see if they had started
smoking. The researchers found that, among the 284 students who indicated that their parents disapproved of kids
smoking, 54 had become established smokers. Among the 41 students who initially said their parents were lenient
about smoking, 11 became smokers.
30. Create a 95% confidence interval for the difference in the proportion of children who may smoke and have
approving parents and those who may smoke and have disapproving parents.
A. (- 0.065, 0.221)
C. (0.073, 0.221)
E. (0.221, 0.073)
B. (0.221, - 0.065)
D. (0.221, 1.960)
31. Interpret your interval in this context.
A. We are 95% confident that the proportions of teens whose parents disapprove of smoking who will
eventually smoke is between 6.5% less and 22.1% more than for teens with parents who are lenient
about smoking.
B. We are 95% confident that the proportions of teens whose parents disapprove of smoking who will
eventually smoke is between 6.5% and 22.1% more than for teens with parents who are lenient about
smoking.
C. The proportion of teens whose parents disapprove of smoking who will eventually smoke is between 6.5%
less and 22.1% more than for teens with parents who are lenient about smoking.
D. We are 95% confident that the proportions of teens whose parents disapprove of smoking who will
eventually smoke is between 7.3% and 22.1% higher than for the teens with parents who are lenient about
smoking.
E. We are 95% confident that the proportions of teens whose parents disapprove of smoking who will
eventually smoke is between 22.1% more and 19.6% less than for teens with parents who are lenient about
smoking.
32. Carefully explain what “95% confidence” means.
A. We expect 95% of random samples of this size to produce intervals that contain the true difference
between the proportions.
B. We are 95% confident that the true difference is (- 0.065, 0.221).
C. 95% of the teens whose parents disapprove of smoking and who will eventually smoke is contained in the
confidence interval 95% of the time.
D. We are 95% confident that the proportion of teens whose parents disapprove of smoking who will
eventually smoke is between 6.5% less and 22.1% more than for teens with parents who are lenient about
smoking.
E. We are confident that 5% of random samples of this size will produce intervals that contain the true
difference between the proportions.
QUESTIONS 33 – 35
There has been debate among doctors over whether surgery can prolong life among men suffering from prostrate
cancer, a type of cancer that typically develops and spreads very slowly. In the summer of 2003, The New England
Journal of Medicine published results of some Scandinavian research. Men diagnosed with prostrate cancer were
16
randomly assigned to either undergo surgery or not. Among the 347 men who had surgery, 16 eventually died of
prostrate cancer, compared with 31 of the 348 men who did not have surgery.
33. What is the standard error of the difference in the two proportions?
A. 0.0189
C. 0.0113
E. 0.08908
B. 0.00036
D. 0.037
34. Create a 95% confidence interval for the difference in rates of death for the two groups of men.
A. (0.043, 0.080)
C. (0.006, 0.043)
E. (0.046, 0.0891)
B. (0.006, 0.080)
D. (0.037, 0.043)
35. Based on your confidence intervals, is there evidence that surgery may be effective in preventing death from
prostrate cancer? Explain.
A. Since 0 is not contained in the interval, there is no evidence that surgery may be effective in preventing death
from prostrate cancer.
B. Since the lower bound of the interval is close to 0, there is no evidence that surgery may be effective in
preventing death from prostrate cancer.
C. Since 0 is not contained in the interval, there is evidence that surgery may be effective in preventing
death from prostrate cancer.
D. Since 0.080 – 0.006 = 0.074 is bigger than the significance level, there is evidence that surgery may be effective
in preventing death from prostrate cancer.
E. Not enough information provided to arrive at any conclusion.
QUESTIONS 36 – 38
Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed –
killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular
basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where no herbicides were
used, only 19 were found to have lymphoma.
36. What’s the standard error of the difference in the two proportions?
A. 0.0354
E. 0.4258
B. 0.0013
C. 0.0172
D. 0.0309
37. Construct a 95% confidence interval for this difference.
A. (0.495, 0.356)
C. (0.426, 0.461)
E. (0.0354, 0.426)
B. (0.356, 0.495)
D. (0.424, 0.572)
38. State an appropriate conclusion.
A. We are 95% confident that the proportion of pets with a malignant lymphoma is between 35.6% and 49.5%.
B. We are 95% confident that the true proportion of pets with a malignant lymphoma is in the interval (0.356,
0.495).
17
C. We are 95% confident that the proportion of pets with a malignant lymphoma in homes where herbicides
are used is between 35.6% and 49.5% higher than the proportion of pets with lymphoma in homes where no
pesticides are used.
D. We are 95% confident that the proportion of pets with a malignant lymphoma in homes where herbicides are
used is between 35.5% and 49.5% lower than the proportion of pets with lymphoma in homes where no pesticides
are used.
E. All of the above
Questions 39 and 40: Random samples from two age groups of brides (200 brides under 18 years and 100 brides
at least twenty years old) showed that 50% of brides in the under 18 group were divorced after 15 years, while
40% of brides in the 20 or older age group were divorced after 15 years. The difference between the two
proportions is 0.10, with a standard error of 0.0604.
39. What is a 95% confidence interval for the difference between the population proportions of brides who are
divorced within 15 years (brides under 18  brides at least 20)?
A. (0.018, 0.218)
B. (0.123, 0.023)
C. (0.040, 0.160)
D. None of the above
KEY: A
40. What is a 99% confidence interval for the difference between the population proportions of brides who are
divorced within 15 years (brides under 18  brides at least 20)?
A. (0.123, 0.023)
B. (0.056, 0.256)
C. (0.040, 0.160)
D. None of the above
KEY: B
41. In a recent poll of 500 13-year-olds, many indicated to enjoy their relationships with their parents. Suppose
that 200 of the 13-year olds were boys and 300 of them were girls. We wish to estimate the difference in
proportions of 13-year old boys and girls who say that their parents are very involved in their lives. In the
sample, 93 boys and 172 girls said that their parents are very involved in their lives. What is a 96%
confidence interval for the difference in proportions (proportion of boys minus proportion of girls)?
A. (0.2015, 0.0151)
B. (0.1973, 0.0194)
C. (0.1978, 0.0289)
D. None of the above
KEY: A
42. In the Youth Risk Behavior Survey (a study of public high school students), a random sample showed that
45 of 675 girls and 103 of 621 boys had been in a physical fight on school property one or more times
during the past 12 months. What is an approximate 95% confidence interval for the difference in
proportions of students who had been in a fight (proportion of boys minus proportion of girls)?
KEY: (6.38%, 13.5%)
43.
A study was conducted to learn about the proportions of U.S. residents who do not have medical insurance.
Independent random samples of 942 U.S. women and 754 U.S. men were obtained. Of those sampled, 181
women and 181 men said that they do not have medical insurance. Calculate a 90% confidence interval for
the difference in proportions (proportion of men minus proportion of women).
KEY: (0.0147, 0.0911)
18
Questions 44 to 46: Many people now-a-days use e-mail to communicate with family and friends and as a primary
source of communication at their job. Is there a difference between the use of e-mail for men with a college
education and those without? A survey in a large city reveals that out of 230 males with a college degree, 212 use
e-mail more than twice a day. Out of the 150 without a college education, 122 use e-mail more than twice a day.
44.
What is the estimate for the difference between the proportions of men with and without a college degree
who use e-mail often?
KEY: 0.1084
45. What is the standard error of the estimate?
KEY: 0.0364
46. Calculate a 98% confidence interval for the difference in proportions.
KEY: (0.0237, 0.1931)
Questions 47 to 49: A survey at a large public university reveals that many students hold a (part-time) job while
going to college. Out of 127 female students surveyed, 95 have a job and out of 143 male students, 97 have a job.
47.
What is the estimate for the difference between the proportions of female and male students who have a job
while going to college?
KEY: 0.0697
48. What is the standard error of the estimate?
KEY: 0.0549
49. Calculate a 95% confidence interval for the difference in proportions.
KEY: (0.0378, 0.1772)
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
 Confidence Intervals for One Population Mean
Confidence Intervals for the Difference in Two Population Means
1. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310 mg, with
a standard deviation of 36 mg. Find a 95% confidence interval for the mean sodium content of this brand of hot
dog. Explain clearly what your interval means.
A. CI = (321.5, 298.5): We are 95% confident that the interval (321.5, 298.5) contains the true mean sodium
content in this type of “reduced sodium” hot dogs.
B. CI = (298.5, 321.5): The interval 298.5mg to 321.5 mg contains the true mean sodium content in this type of
“reduced sodium” hot dogs.
C. CI = (298.5, 321.5): If we take several samples of size 40, and construct confidence intervals, we will find out
that the interval 298.5 mg to 321.5mg contains the true mean sodium content in this type of “reduced sodium” hot
dogs.
D. CI = (298.5, 321.5): We are 95% confident that the interval 298.5mg to 321.5mg contains the true mean
sodium content in this type of “reduced sodium” hot dogs.
E. None of the above
2. Which of the following helps determine the standard error for a confidence interval for a mean?
A. The sample size(s).
B. The sample estimate.
C. The true value of the population parameter.
D. The confidence level.
KEY: A
3. The distinction between a sampling distribution and a confidence interval is:
A. A confidence interval gives possible values for a sample statistic when the population parameter is assumed
known, while a sampling distribution gives possible values for a population parameter when only a single
value of a sample statistic is known.
B. A sampling distribution gives possible values for a sample statistic when the population parameter is
assumed known, while a confidence interval gives possible values for a population parameter when only a
single value of a sample statistic is known.
C. Sampling distributions exist only for situations involving means, while confidence intervals can be
computed for situations involving means and proportions.
D. Confidence intervals exist only for situations involving means, while sampling distributions can be
computed for situations involving means and proportions.
KEY: B
4. What is the primary purpose of a 95% confidence interval for a mean?
A. to estimate a sample mean
B. to test a hypothesis about a sample mean
C. to estimate a population mean
D. to provide an interval that covers 95% of the individual values in the population
KEY: C
5. The confidence level for a confidence interval for a mean is
A. the probability the procedure provides an interval that covers the sample mean.
B. the probability of making a Type 1 error if the interval is used to test a null hypothesis about the population
mean.
C. the probability that individuals in the population have values that fall into the interval.
20
D. the probability the procedure provides an interval that covers the population mean.
KEY: D
6. Which of the following will not result in paired data?
A. The same measurement is taken twice on each person, under different conditions or at different times.
B. Similar individuals are paired before giving the treatments in an experiment. Each member of a pair then
receives a different treatment. The same response variable is measured for all individuals.
C. Two different variables are measured for each individual. There is interest in the amount of difference
between the two variables.
D. One random sample is taken, and a variable is recorded for each individual, but then units are categorized
as belonging to one population or another.
KEY: D
7. Which of the following statements is most correct about a confidence interval for a mean?
A. It provides a range of values, any of which is a good guess at the possible value of the sample mean.
B. It provides a range of values, any of which is a good guess at the possible value of the population mean.
C. It provides a good guess for the range of values the sample mean is likely to have in repeated samples.
D. It provides a good guess for the range of values the population mean is likely to have in repeated samples.
KEY: B
8. The weights of a sample of n = 8 college men will be used to create a 95% confidence interval for the mean
weight of all college men. What is the correct t* multiplier involved in calculating the interval?
A. 1.89
B. 2.00
C. 2.31
D. 2.36
KEY: D
9. The heights of a sample of n = 18 female college students will be used to create a 98% confidence interval
for the mean height of all female college students. What is the correct t* multiplier for this interval?
A. 2.11
B. 2.55
C. 2.57
D. 2.90
KEY: C
10. The WISC scores (similar to IQ test scores) of a sample of n = 20 5th graders will be used to create a 99%
confidence interval for the mean WISC score of all 5th graders. What is the correct t* multiplier for this interval?
A. 2.86
B. 3.55
C. 2.54
D. 2.85
KEY: A
11. The heights of a random sample of 100 women are recorded. The sample mean is 65.3 inches and the sample
standard deviation is 3 inches. Which of the following is an approximate 95% confidence interval for the
population mean?
A. 65.3 (2)(0.03)
B. 65.3  (2)(0.3)
C. 65.3  (2)(3)
D. 65.3  (2)(30)
KEY: B
21
12. A random sample of 100 students had a mean grade point average (GPA) of 3.2 with a standard deviation of
0.2. The standard error of the sample mean in this case is 0.02. Calculate an approximate 95% confidence
interval for the mean GPA for all students.
A. (2.8, 3.6)
B. (3.16, 3.24)
C. (3.18, 3.22)
D. None of the above
KEY: B
13. A random sample of 30 students spent an average amount of $40.00 on a date, with a standard deviation of
$5.00. The standard error of the sample mean is 0.913. Calculate an approximate 95% confidence interval for
the average amount spent by all students on a date.
A. (40.00, 90.00)
B. (79.10, 80.90)
C. (78.20, 81.80)
D. None of the above
KEY: C
14. The cholesterol levels of a random sample of 100 men are measured. The sample mean is 188 and the sample
standard deviation is 40. Which of the following provides a 95% confidence interval for the population mean?
A. 188  (1.98)(0.4)
B. 188  (1.98)(4)
C. 188  (1.98)(40)
D. 188  (1.98)(4000)
KEY: B
15. A random sample of 30 airline flights during a storm had a mean delay of 40 minutes. The standard deviation
was 5 minutes, and the standard error of the mean is 0.9129. Calculate a 90% confidence interval for the
average delay for all flights during a storm.
A. (38.2, 41.8)
B. (38.4, 41.6)
C. (31.5, 48.5)
D. None of the above
KEY: B
16. A randomly selected sample of n = 51 men in Brazil had an average lifespan of 59 years. The standard
deviation was 10 years, and the standard error of the mean is 1.400. Calculate a 90% confidence interval for
the average lifespan for all men in Brazil.
A. (42.2, 75.8)
B. (56.6, 61.4)
C. (56.2, 61.8)
D. None of the above
KEY: B
17. When constructing a confidence interval for the difference in two population means, it is appropriate to use the
pooled standard error only when
A. the population standard deviations can be assumed to be equal.
B. the sample standard deviations are exactly equal.
C. the population means can be assumed to be equal.
D. the sample means are exactly equal.
KEY: A
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18. Which one of the following ways of collecting data would result in two independent samples?
A. A sample of college students is measured for stress both before and after taking an important exam.
B. A group of college students are paired based on gender, age, major, and grades. One student in a pair
receives stress management training and the other one does not. After a few weeks all students are
measured for stress right before an important exam.
C. A sample of English majors is measured for stress before taking an important exam. A sample of
engineering majors is also measured for stress before taking their important exam.
D. A sample of college students is measured for stress before taking an important exam. Their exam grade is
also recorded.
KEY: C
19. The amount of time single men and women spend on house work is measured for 15 single women and 20
single men. What are the appropriate degrees of freedom for the multiplier t* for a pooled confidence interval
for the difference in mean time spent on housework between single men and women?
A. 14
B. 19
C. 33
D. 35
KEY: C
20. The head circumference is measured for a sample of 15 girls and a separate sample of 15 boys. What is the
correct combination of degrees of freedom and the value of the multiplier t* for a 90% confidence interval for
the difference in mean head circumference between girls and boys?
A. df = 13, t* = 1.77
B. df = 28, t* = 1.70
C. df = 28, t* = 2.05
D. df = 30, t* = 1.70
KEY: B
21. The amount of time single men and women spend on house work is measured for 15 single women and 25
single men. For the women the mean was 7 hours/week with a standard deviation of 1.5. For the men the mean
was 4.5 hours/week with a standard deviation of 1.1. What is the value of the pooled standard deviation for the
difference in mean time spent on housework between single men and women?
A. 1.30
B. 0.45
C. 1.59
D. 1.26
KEY: D
22. The head circumference is measured for a sample of 17 girls and a separate sample of 12 boys. The mean for
the girls was 49 cm with a standard deviation of 1.25 cm. The mean for the boys was 50 cm with a standard
deviation of 0.95 cm. What is the value of the pooled standard error for the difference in mean head
circumference between girls and boys?
A. 0.41
B. 1.29
C. 0.43
D. 1.14
KEY: C
23
23. A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, with a
standard deviation of $22.50. A random sample of 40 English majors spent an average of $180.00 for
textbooks that term, with a standard deviation of $18.30. What is the value of the pooled standard error for the
difference in mean amount spent?
A. 20.93
B. 4.27
C. 16.81
D. 4.10
KEY: B
24. Random samples of 200 men and 200 women were collected and their resting pulse rates were measured, to
estimate how much mean resting pulse rates differ for men and women in the population. An analyst
mistakenly paired the observations and constructed an approximate 95% confidence interval for the mean
difference to be (5  20.2) beats per minute. If the data had been analyzed correctly, finding an approximate
95% confidence interval for the difference in population means, which of the following parts of the interval
would be different?
A. The sample statistic of 5 beats per minute.
B. The multiplier of 2.
C. The standard error of 0.2.
D. None of the parts would be different; it is an equivalent analysis.
KEY: C
25. A researcher asked random samples of 50 kindergarten teachers and 50 12th grade teachers how much money
they spent out-of-pocket on school supplies in the previous school year, to see if teachers at one grade level
spend more than the other. A 95% confidence interval for K  12 is $30 to $50. Based on this result, it is
reasonable to conclude that
A. 95% of all kindergarten and 12th grade teachers spend between $30 and $50 on average.
B. 95% of all kindergarten teachers spend between $30 and $50 more then 95% of all 12th grade teachers.
C. kindergarten teachers spend more on average than do 12th grade teachers.
D. 12th grade teachers spend more on average than do kindergarten teachers.
KEY: C
26. A 95% confidence interval for the difference between the mean handspans of men and the mean handspans of
women is determined to be 2.7 centimeters to 3.3 centimeters. Which of the following statements is the best
interpretation of this interval?
A. It is likely that the difference in the population mean handspans of men and women is covered by the
interval 2.7 centimeters to 3.3 centimeters.
B. It is likely that the difference in the sample mean handspans of men and women is covered by the interval
2.7 centimeters to 3.3 centimeters.
C. It is likely that if new samples of the same size were to be taken, the difference in sample means would be
contained in the interval 2.7 centimeters to 3.3 centimeters.
D. It is likely that for 95% of married couples, the husband's handspan is between 2.7 and 3.3 centimeters
longer than the wife's hand span.
KEY: A
27. A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, while a
random sample of 40 literature majors spent an average of $180.00 for textbooks that term. The standard
deviation for each sample was $20.00. The standard error for the difference between the two sample means is
4.082. Calculate an approximate 95% confidence interval for the difference in average amounts spent on
textbooks (math majors – literature majors).
A. (7.8, 32.2)
B. (11.8, 28.2)
24
C. (15.9, 24.1)
D. None of the above
KEY: B
28. A random sample of 40 men drank an average of 20 cups of coffee per week during finals, while a sample of
30 women drank an average of 15 cups of coffee per week. The sample standard deviations were 6 cups for
the men and 3 cups for the women. The standard error for the difference between the two sample means is
1.095. Calculate an approximate 95% confidence interval for the difference in average cups of coffee drunk
(men –women).
A. (2.81, 7.19)
B. (3.02, 6.98)
C. (3.91, 6.10)
D. None of the above
KEY: A
29. A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, with a
standard deviation of $22.50. A random sample of 40 English majors spent an average of $180.00 for
textbooks that term, with a standard deviation of $18.30. Calculate a 90% pooled confidence interval for the
difference in average amounts spent on textbooks (math majors – English majors).
A. (12.91, 27.09)
B. (11.50, 28.50)
C. (13.19, 26.81)
D. None of the above
KEY: A
30. An experiment is conducted with two groups of 25 college students. The treatment group scored an average of
3.67 with a standard deviation of 0.86. The placebo group scored an average of 3.29 with a standard deviation
of 0.97. Calculate a 95% pooled confidence interval for the difference in means between the treatment and the
placebo.
A. (−0.44, 0.82)
B. (−0.14, 0.90)
C. (−0.25, 1.01)
D. None of the above
KEY: B
31. Reaction time is measured in a driving simulator for a random sample of 16 year-old boys (n = 12, x = 4.5,
s = 1.1) and a random sample of 24 year-old young men (n = 9, x = 2.9, s = 1.3). Calculate a 99% pooled
confidence interval for the difference in mean reaction time (16 year-olds – 24 year-olds).
A. (0.48, 2.72)
B. (0.51, 2.70)
C. (0.07, 3.13)
D. None of the above
KEY: D
Questions 32 and 33: For a randomly selected sample of 20 new mothers in the year 2000, the mean age was 24.6
years. For a randomly selected sample of 10 new mothers in 1970, the mean age was 21.4 years. The difference
between the mean ages is 3.2 years, and the standard error of the difference is 1.366. Assume that the ages of new
mothers are normally distributed, do not assume the population variances are equal, and use the conservative “by
hand” estimate for the degrees of freedom.
25
32. Calculate a 90% confidence interval for the difference in population mean ages of new mothers in the two
years (year 2000 – year 1970).
A. (0.84, 5.56)
B. (0.78, 5.67)
C. (0.70, 5.70)
D. None of the above
KEY: C
33. Calculate a 99% confidence interval for the difference in population mean ages of new mothers in the two
years (year 2000 – year 1970).
A. (−1.24, 7.64)
B. (−1.13, 7.53)
C. (0.47, 5.93)
D. None of the above
KEY: A
34. Random samples of 5 Japanese women and 10 Japanese men showed an average life span of 83 years for the
women and 77 years for the men. The standard deviation was 2 years for the women and 1 year for the men.
Calculate a 95% confidence interval for the difference in average life spans (women  men). Assume that the
life spans are normally distributed, but do not assume the population variances are equal, and use the
conservative “by-hand” estimate for the degrees of freedom.
KEY: (3.36, 8.64)
35. After surveying students at Dartmouth College, a campus organization calculated that a 95% confidence
interval for the mean cost of food for one term (of three in the Dartmouth trimester calendar) is ($780, $920). Now
the organization is trying to write its report, and considering the following interpretations. Which interpretation is
most appropriate?
A. 95% of all students pay between $780 and $920 for food.
B. 95% of the sampled students paid between $780 and $920.
C. We are 95% sure that students in this sample averaged between $780 and $920 for food.
D. 95% of all samples of students will have average food costs between $780 and $920.
E. We are 95% sure that the average amount all students pay is between $780 and $920.
MIXED REVIEW QUESTIONS
1-2. From a random sample of 25 three year old girls, the mean height was found to be 38.2 inches and standard
deviation 3.5 inches.
1. The t* multiplier necessary to construct the 98% confidence interval (CI) for the population mean height is
(a) 2.58
(b) 2.49
(c) 2.33
(d) 1.28
(e) none of these
2. The 98% confidence interval for the population mean height is
(a) 38.2  1.74
(b) 3.8  0.36
(c) 38.2  1.63
(d) 25  1.63
3-5. Of 100 people vaccinated against a disease, 17 had side effects. Regard this as a random sample.
3. What is the sample proportion, i.e. a sample statistic for the population proportion p of people who have side
effects following this vaccination?
(a) 17
(b) 50
(c) 100
(d) .25
(e) none of these
26
4. What is the standard error for the sample proportion based on this sample?
(a) 0.030
(b) 0.038
(c) 0.042
(d) 0.050
(e) none of these
5. Estimate with a 90% confidence interval the proportion p of people who have side effects following this
vaccination
(a) .1  .03
(b) .17  .04
(c) .17  .06
(d) .90  .06
(e) none of these
6-9. In a study aiming to estimate the mean  mileage per gallon for certain brand of cars, a sample of n=81 cars
produced x = 24.3, and s=8.
6. The margin of error in the 90% CI estimate of  is
(a) 0.89
(b) 0.95
(c) 1.48
(d) 2.00
(e) 24.3
7. The left end point of the 90% CI is
(a) 22.82
(b) 25.78
(c) 29.62
(d) 20.82
(e) 21.91
8. The 90% CI estimate is __________________________ the 95% CI estimate.
(a) wider than
(b) narrower than
(c) the same width as
9. If 100 samples are taken, and a 90% confidence interval is constructed based on each sample, about how many
of the resulting intervals are expected to cover the population mean ?
(a) 100
(b) 90
(c) 0
(d) 50
(e) 10
10. A nationwide poll of 200 first year college students showed that 132 of them identified “being well-off
financially” as an important personal goal. Consider this to be a random sample. Find a 99% confidence interval
estimate for p, the proportion of all first year college students nationwide who identified “being well-off
financially” as an important personal goal.
(a) .132  .09 (b) 0.66  .07
(c) .20  .07
(d) 0.66  .09
(e) .20  .09
11-13. From a random sample of first year college students, a researcher finds .57  .05 to be a 95% CI for the
population proportion p of all first year college students nationwide who identified “being well-off financially” as
an important personal goal.
11. The sample proportion for this sample was
(a) 0.95
(b) .62
(c) .52
(d) .57
(e) .05
12. The standard error was
(a) 0.05
(b) 0.026
(c) 0.57
(d) 0.126
(e) 0.301
13. The sample size n was
(a) 377
(b) 950
(c) 50
(d) 570
(e) 542
14-17. An air pollution index for 16 randomly selected days during the summer months was recorded for a major
western city. Assuming that the conditions for the t-interval are met, use the descriptive statistics from Minitab
shown below to answer the questions concerning estimation of the mean index  for the entire summer.
Descriptive Statistics
Variable
N
Index
16
Mean
75.2
Median
64.00
StDev
10.2
14. What is the t* multiplier for construction of the 90% interval estimate of ?
(a) 1.75 (b) 2.95
(c) 1.65
(d) 1.34
(e) 2.13
27
15. What is the lower confidence limit of the 90% interval estimate of ?
(a) 64.5 (b) 68.7
(c) 68.9
(d) 70.7
(e) 72.0
16. Each of 100 researchers takes a sample of 16 days in summer, records the pollution index for each day, and
constructs a 95% confidence interval for the mean pollution index. About how many of the intervals are expected
to contain the value of the population mean?
(a) 16
(b) 15
(c) 100
(d) 95
(e) 5
17. Each of 100 researchers takes a sample of various numbers of days in summer, records the pollution index for
each day, and constructs a 95% confidence interval for the mean pollution index. The widest confidence intervals
are based on samples with the largest numbers of days.
(a) true
(b) false
Formulas:
Confidence Intervals for one proportion or difference between two proportions:
pˆ (1  pˆ )
One sample: pˆ  z *
n
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )
Two independent samples: pˆ 1  pˆ 2  z *

n1
n2
Confidence Intervals for one mean or difference between two means:
s
One sample: x  t *
, t* has n-1 degrees of freedom
n
Two independent samples (not assuming equal
variances):
x1  x2  t *
s12 s22

where t * is based on the smaller of n1  1 and n2  1 (conservat ive approach t o d.f.).
n1 n2
Test Statistics for tests about one proportion or difference between two proportions:
One sample: z 
pˆ  p0
p0 (1  p0 )
n
Two independent samples:
pˆ1  pˆ 2
x x
z
where pˆ  1 2 , the estimate of the common proportion under H 0 .
n1  n2
1 1
pˆ (1  pˆ )  
 n1 n2 
Test Statistics for tests about one mean or difference between two means:
One sample: t 
x  0
s/ n
with n-1 degrees of freedom
28
Two independent samples (not assuming equal variances):
t
x1  x 2
s12 s 22

n1 n2
, where with the conservative approach, the degrees of freedom for t is the smaller of n1  1 and n2  1 .
Answers:
1. b The appropriate t* multiplier is found from Table with 24 degrees of freedom.
2. a 38.2  2.49×3.5/5=38.2  1.74.
3. e The sample proportion p̂ =17/100=0.17.
0.17(1  0.17)
 0.038 .
100
c The z* multiplier for 90% confidence is 1.645, the CI is (sample statistic) z*×(standard error), or 0.17 
1.645×0.038.
c From table A.2, the t* multiplier for 80 degrees of freedom is 1.66. The margin of error is
1.66  8 / 81  1.48
a The left end point is (sample statistic)-(margin of error), or 24.3  1.48  22.82 .
b The z* multiplier for 90% confidence is 1.645, which is smaller than 1.96, the z* multiplier for 95%
confidence. Therefore for 90% confidence, the margin of error is smaller, and the interval is narrower.
b The coverage rate is 90%, so 90 out of 100 are expected to cover the population mean.
4. b The standard error is
5.
6.
7.
8.
9.
10. d The sample proportion p̂ =132/200=0.66, the z* multiplier is 2.575. The CI is
0.66(1  0.66)
200
11. d Compare the interval to the formula according to which the interval was calculated to see that p̂ =0.57.
12. b The margin of error is 0.05, z*=1.96 for 95%, so the standard error is 0.05/1.96=0.026.
.57(1  .57)
13. a The standard error is 0.026=
. Solve for n to get n=376.63, or n=377.
n
14. a t*=1.75 for 15 degrees of freedom
15. d 75.2  1.75  10.2 / 16
16. d The reliability of the method is 95% out of 100 repeated applications.
17. b The large the sample size, the smaller the margin of error, the narrower the interval.
0.66  2.575
29

HYPOTHESES TESTING FOR PROPORTIONS
 HYPOTHESES TESTING FOR MEANS
1. A clean standard requires that vehicle exhaust not exceed specified limits for various pollutants. Many states
require that cars be tested annually to be sure they meet these standards. Suppose state regulators double check a
random sample of cars that a suspect repair shop has certified as okay. They will revoke the shop’s license if they
find significant evidence that the shop is certifying vehicle that do not meet standards.
The appropriate hypotheses are:
A. Null hypothesis: The regulators decide that the shop is meeting standards.
Alternative hypothesis: The regulators decide that the shop is not meeting standards.
B. Null hypothesis: The shop’s emission standards are higher than for other shops.
Alternative hypothesis: The shop’s emission standards are not higher than for other shops.
C. Null hypothesis: The shop is meeting the emissions standards.
Alternative hypothesis: The shop is not meeting the emission standards.
D. Null hypothesis: The repair shop’s license is revoked.
Alternative hypothesis: The repair shop’s license is not revoked.
QUESTIONS 2 – 4
The National center for Education Statistics monitors many aspects of elementary and secondary education
nationwide. Their 1996 numbers are often used as a baseline to assess changes. In 1996 31% of students reported
that their mothers had graduated from college. In 2000, responses from 8368 students found that this figure had
grown to 32%. Is this evidence that there has been an increase in education level among mothers?
2. The appropriate hypotheses are:
A. Ho: p = 0.31; HA: p > 0.31
C. Ho: p = 0.32; HA: p  0.32
B. Ho: p>0.31; HA: p<0.31
D. Ho: p<0.31; HA: p>0.32
3. Let all necessary conditions be satisfied. Find the test statistic (z), and the P-value.
A. z = 1.978, P-value = 0.0047
C. z = 0.0051, P-value = -1.978
B. z = 1.978, P-value = 0.024
D. z = 0.048, P-value = 1.978
4. State your conclusion.
A. With a P-value of 1.978, we fail to reject the null hypothesis.
B. No conclusion can be drawn because the P-value is too big to be considered in this context.
C. With a P-value of 0.024, we reject the null hypothesis. There is evidence to suggest that the percentage
of students whose mothers are college graduates has changed since 1996. In fact, the evidence suggests
that the percentage has increased.
D. The confidence level was not given, so no conclusion can be arrived at.
5. National data in the 1960s showed that about 44% of the adult population had never smoked cigarettes. In 1995
a national health survey interviewed a random sample of 881 adults and found that 52% had never been smokers.
(a) Create a 95% confidence interval for the proportion of adults (in 1995) who had never been smokers.
(b) Does this provide evidence of a change in behavior among Americans? Using your confidence interval, test
an appropriate hypothesis and state your conclusion.
30
ANSWERS
(a) (.1531, .5789) (b) Null Hypo: p = 0.44; Alter Hypo: p not equal to 0.44. Since 44% is not in the 95% CI,
we will reject the Null Hypo. There is strong evidence that, in 1995, the percentage of adults who have never
smoked was not 44%.
6. A company hopes to improve customer satisfaction, setting as a goal no more than 5% negative comments. A
random survey of 350 customers found only 10 with complaints.
(a) Create a 95% confidence interval for the true level of dissatisfaction among customers.
(b) Does this provide evidence that the company has reached its goal? Using your confidence interval, test an
appropriate hypothesis and state your conclusion.
ANSWERS
(a) (0.011, 0.046) (b) Null Hypo: p = 0.05; Alter. Hypo: p < 0.05. Since 5% is not in the 95% CI, we will reject
the Null Hypo. There is strong evidence that less that 5% of customers have complaints.
7. A company with a fleet of 150 cars found that the emissions systems of 7 out of 22 they tested failed to meet
pollution control guidelines. Is this strong evidence that more than 20% of the fleet might be out of
compliance? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and
conditions are satisfied before you proceed.
ANSWER
Null Hypo: p = 0.20; Alter. Hypo: p > 0.20. Two conditions are not satisfied (verify!!!!!!!) We cannot proceed
with a hypothesis test.
8. In a rural area only about 30% of the wells that are drilled find adequate water at a depth of 100 feet or less. A
local man claims to be able to find water by “dowsing” – using a forked stick to indicate where the well should be
drilled. You check with 80 of his customers and find that 27 have wells less than 100 feet deep. What do you
conclude about this claim? (We consider a P – value of around 5% to represent strong evidence)
(a)
(b)
(c)
(d)
(e)
Write appropriate hypotheses.
Check the necessary assumptions.
Perform the mechanics of the test. What is the P – value?
Explain carefully what the P – value means in this context.
What’s your conclusion?
9. In the 1980s it was generally believed that congenital abnormalities affected about 5% of the nation’s children.
Some people believe that the increase in the number of chemicals in the environment has led to an increase in the
incidence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an
abnormality. Is this strong evidence that the risk has increased? (We consider a P – value of around 5% to
represent strong evidence)
(a)
(b)
(c)
(d)
(e)
(f)
Write appropriate hypotheses.
Check the necessary assumptions.
Perform the mechanics of the test. What is the P – value?
Explain carefully what the P – value means in this context.
What’s your conclusion?
Do environmental chemicals cause congenital abnormalities?
31
QUESTIONS 10 – 12
The National Center for Education Statistics monitors many aspects of elementary and secondary education
nationwide. Their 1996 numbers are often used as a baseline to assess changes. In 1996, 34% of students had not
been absent from school even once during the previous month. In the 2000 survey, responses from 8302 students
showed that this figure had slipped to 33%. Officials would of course be concerned if student attendance were
declining. Do these figures give evidence of a decrease in student attendance?
10. Write appropriate hypotheses.
A.
B.
C.
D.
E.
Ho : p = 0.33;
Ho : p > 0.34;
Ho : p = 0.34;
Ho : p < 0.33;
Ho : p = 0.34;
HA: p < 0.33
HA: p < 0.34
HA: p < 0.34
HA: p = 0.34
HA: p = 0.33
11. Perform the test by finding the Test statistics and the P – value.
A.
B.
C.
D.
E.
Zo = 0.0052; P – value = 0.027
Zo = - 1. 923; P – value = 0.027
Zo = 0.048; P – value = 0.0052
Zo = 0.01; P – value = 0.048
Zo = 1.923; P – value = 0.027
12. State your conclusion.
A. With a P – value of 0.027, we fail to reject the null hypothesis. There is no evidence to suggest that the
percentage of students with perfect attendance in the previous month has decreased in 2000.
B. With a P – value of 0.0052, we reject the null hypothesis. There is evidence to suggest that the percentage of
students with perfect attendance in the previous month has decreased in 2000.
C. With a P – value of 0.048, essentially 0.05, equal to the significance level, we retain the null hypothesis.
There is no evidence to suggest that the percentage of students with perfect attendance in the previous month
has decreased in 2000.
D. With a P – value of 0.027, we reject the null hypothesis. There is evidence to suggest that the
percentage of students with perfect attendance in the previous month has decreased in 2000.
E. With a P – value of 0.027, we retain the null hypothesis. There is evidence to suggest that the percentage of
students with perfect attendance in the previous month has decreased in 2000.
17. The seller of a loaded die claims that it will favor the outcome 6. We don’t believe that claim, and roll the die
200 times to test an appropriate hypothesis. Our P – value turns out to be 0.03. Which conclusion is appropriate?
A. There’s a 3% chance that the die is fair.
B. There’s a 97% chance that the die is fair.
C. There’s a 3% chance that a loaded die could randomly produce the results we observed, so it’s reasonable
to conclude that the die is fair.
D. There’s a 3% chance that a fair die could randomly produce the results we observed, so it’s
reasonable to conclude that the die is loaded.
E. None of the above.
32
QUESTIONS 18 – 21
In 1960, census results indicated that the age at which American men first married had a mean of 23.3 years. It is
widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of
first marriage has increased during the past 40 years.
18. Write appropriate hypotheses.
A.
B.
C.
D.
E.
H o :   23.3;
H A :   23.3
H o :   23.3;
H A :   23.3
H o :   23.3;
H A :   23.3
H o :   23.3;
H A :   23.3
None of the above
19. Describe the approximate sampling distribution model for the mean age in such samples.
 

A. N  23.3, 
40 


 
B. t 39  23.3,

40 


s 
C. N  23.3,

40 


s 
D. t 39  23.3,

40 

E. None of the above
20. The men in our sample married at an average age of 24.2 years, with a standard deviation of 5.3 years. What is
the P – value for this result? Explain in context what this P – value means.
A. P – value = 0.8544; If the mean age at first marriage is still 23.3 years, there is a 0.8544 chance of getting a
sample mean of 24.2 years or older simply from natural sampling variation.
B. P – value = 1.07; If the mean age at first marriage is still 23.3 years, then, the P – value is what we observe
from data due to natural sampling variation.
C. P – value = 0.1456; If the mean age at first marriage is still 23.3 years, there is a 0.1456 chance of getting a
sample mean of 24.2 years or older simply from natural sampling variation.
D. P – value = 0.1456; If the mean age at first marriage is still 23.3 years, there is a 0.1456 chance of getting a
sample mean of 24.2 years or younger simply from natural sampling variation.
E. P – value = 1.07; If the mean age at first marriage is still 23.3 years, then, the P – value is what we observe in a
sample of size n = 40, with mean 24.2 years or older.
21. What is your conclusion?
A. We fail to reject the null hypothesis. There is no evidence to suggest that the mean age at first marriage has
changed from 23.3 years, the mean in 1960, to 24.2 years.
B. We fail to reject the null hypothesis. There is strong evidence to suggest that the mean age at first marriage
has changed from 23.3 years, the mean in 1960, to 24.2 years.
C. We reject the null hypothesis. There is strong evidence to suggest that the mean age at first marriage has
changed from 23.3 years, the mean in 1960, to 24.2 years.
D. We reject the null hypothesis. There is no strong evidence to suggest that the mean age at first marriage has
changed from 23.3 years, the mean in 1960, to 24.2 years.
33
E. We cannot draw any conclusion in this case because the P – value = 1.07 is exceedingly big.
TYPE I AND TYPE II ERRORS ?????????????????????????????????????????????
1. Production managers on an assembly line must monitor the output to be sure that the level of
defective products remains small. They periodically inspect a random sample of the items produced. If
they find a significant increase in the proportion of items that must be rejected, they will halt the
assembly process until the problem can be identified and repaired.
(a) In this context, what is a Type I error?
(b) In this context, what is a Type II error?
(c) Which type of error would the factory owner consider more serious?
(d) Which type of error might customers consider more serious?
[Solution to problem 1 is posted on the class website]
2. Consider again the task of the quality control inspectors in Exercise 1.
(a) In this context, what is meant by the power of the test the inspectors conduct?
(b) They are currently testing 5 items each hour. Someone has proposed that they test 10 each hour
instead. What are the advantages and disadvantages of such a change?
(c) Their test currently uses a 5% level of significance. What are the advantages and disadvantages of
changing to an alpha level of 1%?
(d) Suppose that, as a day passes, one of the machines on the assembly line produces more and more
items that are defective. How will this affect the power of the test?
[Solution to problem 2 is posted on the class website]
34