Download STT 201, sections 7-9 - Michigan State University`s Statistics

Final Examination Practice Exercises Sheet II
STT 200 – STATISTICAL METHODS
Lecture 2, Lecture 3, and Lecture 4
SPRING 2016
PART I:
 Confidence Intervals For One Sample Proportions
 Confidence Intervals For The Difference Between 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
C. 2401
E. Not enough information provided
2. After conducting a survey, a researcher wishes to cut the standard error (and thus the margin of error) to
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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
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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
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.
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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. Sh e 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%)
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%)
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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
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)
B. (0.827, 0.573)
C. 0.065, 0.70)
D. (0.065, 0.827)
E. (0.065, 0.573)
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.
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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.
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.
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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.
QUESTIONS 30 - 31
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.
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30. Create a 95% confidence interval for the difference in the proportion of children who may smoke and have
disapproving parents and those who may smoke and have approving (lenient) 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)
30. 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.
31. 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 32 – 34
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.
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32. What is the standard error of the difference in the two proportions?
A. 0.0189
C. 0.0113
B. 0.00036
D. 0.037
E. 0.08908
33. 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)
34. 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 35 – 37
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.
35. 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
36. 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)
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37. 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).
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 38 and 39: 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.
38. 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
39. 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
40. 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
41. 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%)
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42.
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)
Questions 43 to 45: 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.
43.
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
44. What is the standard error of the estimate?
KEY: 0.0364
45. Calculate a 98% confidence interval for the difference in proportions.
KEY: (0.0237, 0.1931)
Questions 46 to 48: 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.
46.
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
47. What is the standard error of the estimate?
KEY: 0.0549
48. Calculate a 95% confidence interval for the difference in proportions.
KEY: (0.0378, 0.1772)
Also:
CHAPTER 19: Nos. 7 – 10; 13 – 17; 27 – 28; 33, 35 -36; 41 – 42. (See class examples or recitation solutions)
CHAPTER 22: Nos. 8 – 11; 13, 15 – 16. (See class examples or recitation solutions)
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PART 11

 Confidence Intervals for One Population Mean
Confidence Intervals for the Difference Between 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
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4. 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
5. 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
6. 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
7. 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
8. 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
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9. 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
C. 1.45
D. 2.05
KEY: C
10. 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
11. 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
12. 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
13. 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.
D. the probability the procedure provides an interval that covers the population mean.
KEY: D
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14. 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
15. 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
16. 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
17. 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
18. 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
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19. 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
20. 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
21. 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
22. 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
23. 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
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24. 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
CONFIDENCE INTERVALS FOR THE DIFFERENCE IN POPULATION MEANS
25. 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
26. 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
27. 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
28. 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 = 14, t* = 1.76
B. df = 28, t* = 1.70
C. df = 28, t* = 2.05
D. df = 30, t* = 1.70
KEY: A
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29. 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
30. 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
31. 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
32. 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
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33. 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
34. 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
35. 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)
C. (15.9, 24.1)
D. None of the above
KEY: B
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36. 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
37. 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
38. 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
39. 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
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Questions 40 and 41: 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.
40. 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
41. 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
42.
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)
SOME TEXTBOOK EXERCISES: INTRO STATS 3RD EDITION
CHAPTER 19: Nos. 5 – 6; (See class examples and recitation solutions)
CHAPTER 23: Nos. 1 – 2; 7 – 10; 13 – 14; 21 – 22; (See class examples and recitation solutions)
CHAPTER 24: Nos. 11 – 14; (See class examples and recitation solutions)
20
Formulas:
Confidence Intervals for one proportion or difference between two proportions:
One sample: pˆ  z *
pˆ (1  pˆ )
n
Two independent samples: pˆ 1  pˆ 2  z *
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
Confidence Intervals for one mean or difference between two means:
One sample: x  t *
s
, 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
.
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