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CHAPTER 12 SECTION 1: INFERENCE ABOUT A POPULATION
MULTIPLE CHOICE
1. Researchers determined that 60 Kleenex tissues is the average number of tissues used during a cold.
Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues
used during a cold: = 52 and s = 22. Suppose the test statistic does not fall in the rejection region at 
= 0.05. Which of the following conclusions is correct?
a. At  = 0.05, we do not reject H0.
b. At  = 0.05, we reject H0.
c. At  = 0.05, we accept H0.
d. Both a and c.
ANS: A
PTS: 1
REF: SECTION 12.1
2. A robust estimator is one that is:
a. unbiased and symmetrical about zero.
b. consistent and is also mound-shaped.
c. efficient and less spread out.
d. not sensitive to moderate nonnormality.
ANS: D
PTS: 1
REF: SECTION 12.1
3. For statistical inference about the mean of a single population when the population standard deviation
is unknown, the degrees for freedom for the t-distribution equal n  1 because we lose one degree of
freedom by using the:
a. sample mean as an estimate of the population mean.
b. sample standard deviation as an estimate of the population standard deviation.
c. sample proportion as an estimate of the population proportion.
d. sample size as an estimate of the population size.
ANS: A
PTS: 1
REF: SECTION 12.1
4. Researchers determined that 60 Kleenex tissues is the average number of tissues used during a cold.
Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues
used during a cold: = 52 and s = 22. Suppose the alternative we wanted to test was H1:  < 60. The
correct rejection region for  = 0.05 is:
a. reject H0 if t > 1.6604.
b. reject H0 if t < 1.6604.
c. reject H0 if t > 1.9842 or Z < 1.9842.
d. reject H0 if t < 1.9842.
ANS: B
PTS: 1
REF: SECTION 12.1
5. A random sample of size 15 taken from a normally distributed population revealed a sample mean of
75 and a sample variance of 25. The upper limit of a 95% confidence interval for the population mean
would equal:
a. 77.77
b. 72.23
c. 88.85
d. 77.27
ANS: A
PTS: 1
REF: SECTION 12.1
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6. A major department store chain is interested in estimating the average amount its credit card customers
spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly
sampled and analyzed with the following results: = $50.50 and s2 = 400. A 95% confidence interval
for the average amount the credit card customers spent on their first visit to the chain's new store in the
mall is:
a. $50.50  $9.09.
b. $50.50  $10.12.
c. $50.50  $11.08.
d. None of these choices.
ANS: C
PTS: 1
REF: SECTION 12.1
7. Researchers determine that 60 Kleenex tissues is the average number of tissues used during a cold.
Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues
used during a cold: = 52 and s = 22. Using the sample information provided, the value of the test
statistic is:
a. t = (52  60) / 22
b. t = (52  60) / (22 / 100)
c. t = (52  60) / (22 / 1002)
d. t = (52  60) / (22 / 10)
ANS: D
PTS: 1
REF: SECTION 12.1
8. For a 99% confidence interval of the population mean based on a sample of n = 25 with s = 0.05, the
critical value of t is:
a. 2.7969
b. 2.7874
c. 2.4922
d. 2.4851
ANS: A
PTS: 1
REF: SECTION 12.1
9. Based on sample data, the 90% confidence interval limits for the population mean are LCL = 170.86
and UCL = 195.42. If the 10% level of significance were used in testing the hypotheses H0:  = 201
vs. H1:   201, the null hypothesis:
a. would be rejected.
b. would be accepted.
c. would fail to be rejected.
d. would become H0:   201
ANS: A
PTS: 1
REF: SECTION 12.1
10. The degrees of freedom for the test statistic for  when  is unknown is:
a. 1
b. n
c. n  1
d. None of these choices.
ANS: C
PTS: 1
REF: SECTION 12.1
TRUE/FALSE
11. If a sample has 15 observations and a 95% confidence estimate for  is needed, the appropriate value
of t is 1.753.
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ANS: F
PTS: 1
REF: SECTION 12.1
12. If a sample has 18 observations and a 90% confidence estimate for  is needed, the appropriate value
of t is 1.740.
ANS: T
13. The statistic
degrees of freedom.
ANS: F
PTS: 1
REF: SECTION 12.1
when the sampled population is normal is Student t-distributed with n
PTS: 1
REF: SECTION 12.1
14. In order to determine the p-value associated with hypothesis testing about the population mean , it is
necessary to know the value of the test statistic.
ANS: T
PTS: 1
REF: SECTION 12.1
15. In order to interpret the p-value associated with hypothesis testing about the population mean , it is
necessary to know the value of the test statistic.
ANS: F
PTS: 1
REF: SECTION 12.1
16. If the sampled population is nonnormal, the t-test of the population mean  is still valid, provided that
the condition is not extreme.
ANS: T
PTS: 1
REF: SECTION 12.1
17. A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 48.5
seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is
45.2 seconds to 51.8 seconds.
ANS: F
PTS: 1
REF: SECTION 12.1
18. In forming a 95% confidence interval for a population mean from a sample size of 20, the number of
degrees of freedom from the t-distribution equals 20.
ANS: F
PTS: 1
REF: SECTION 12.1
19. The t-distribution is used in a confidence interval for a mean when the actual standard error is not
known.
ANS: T
PTS: 1
REF: SECTION 12.1
20. The t-distribution allows the calculation of confidence intervals for means for small samples when the
population variance is not known, regardless of the shape of the distribution in the population.
ANS: F
PTS: 1
REF: SECTION 12.1
21. The t-distribution is used to develop a confidence interval estimate of the population mean when the
population standard deviation is unknown.
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copied, or distributed without the prior consent of the publisher.
ANS: T
PTS: 1
REF: SECTION 12.1
22. The t-distribution assumes that the population is normally distributed.
ANS: T
PTS: 1
REF: SECTION 12.1
23. In estimating the population mean with the population standard deviation unknown, if the sample size
is 16, there are 8 degrees of freedom.
ANS: F
PTS: 1
REF: SECTION 12.1
COMPLETION
24. When the population standard deviation is ____________________ and the population is normal, the
test statistic for testing hypotheses about  is the t-distribution with n  1 degrees of freedom.
ANS: known
PTS: 1
REF: SECTION 12.1
25. When the population standard deviation is unknown and the population is ____________________,
the test statistic for testing hypotheses about  is the t-distribution with n  1 degrees of freedom.
ANS: normal
PTS: 1
REF: SECTION 12.1
26. When the population standard deviation is unknown and the population is normal, the test statistic for
testing hypotheses about  is the ____________________-distribution with ____________________
degrees of freedom.
ANS:
t; n  1
t; n1
PTS: 1
REF: SECTION 12.1
27. The t-test for a population mean is ____________________, meaning that if the population is
nonnormal, the results of the test and confidence interval are still valid as long as the nonnormality is
not extreme.
ANS: robust
PTS: 1
REF: SECTION 12.1
28. When a population is small, we must adjust the test statistic and interval estimator using the
____________________ population correction factor.
ANS: finite
PTS: 1
REF: SECTION 12.1
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29. ____________________ populations allow us to use the confidence interval estimate of a mean to
produce a confidence interval estimate of the population total.
ANS: Finite
PTS: 1
REF: SECTION 12.1
30. The t-statistic has two variables: the sample ____________________ and the sample
____________________.
ANS:
mean; standard deviation
standard deviation; mean
PTS: 1
REF: SECTION 12.1
31. Because of the greater uncertainty, the t-statistic will display greater ____________________ than the
z-statistic.
ANS:
variability
variation
spread
PTS: 1
REF: SECTION 12.1
SHORT ANSWER
Students' Ages
A random sample of 10 college students was drawn from a large university. Their ages are 22, 17, 27,
20, 23, 19, 24, 18, 19, and 24 years.
32. {Students' Ages Narrative} Estimate the population mean with 90% confidence.
ANS:
. Thus, LCL = 19.446, and UCL = 23.154.
PTS: 1
REF: SECTION 12.1
33. {Students' Ages Narrative} Test to determine if we can infer at the 5% significance level that the
population mean is not equal to 20.
ANS:
H0:  = 20 vs. H1:   20
Rejection region: | t | > t0.025,9 = 2.262
Test statistic: t = 1.285
Conclusion: Don't reject H0. We can't infer at the 5% significance level that the population mean is not
equal to 20.
PTS: 1
REF: SECTION 12.1
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copied, or distributed without the prior consent of the publisher.
34. {Students' Ages Narrative} What is the required condition of the techniques used in the previous
questions? What graphical device can you use to check to see if that required condition is satisfied?
ANS:
The condition is that ages in the population are normally distributed. A histogram of the data can be
used to check if the normality assumption is satisfied.
PTS: 1
REF: SECTION 12.1
Grocery Receipts
A simple random sample of 100 grocery receipts was drawn from a normal population. The mean and
standard deviation of the sample were $120 and $25, respectively.
35. {Grocery Receipts Narrative} Test the hypothesis H0:  = 125 vs. H1:   125 at the 10% significance
level.
ANS:
Rejection region: | t | > t0.05,99 = 1.66
Test statistic: t = 2.0
Conclusion: Reject H0. We can infer that the population mean is not equal to 125. According to our
data, the mean is lower than that.
PTS: 1
REF: SECTION 12.1
36. {Grocery Receipts Narrative} Estimate the population mean with 90% confidence.
ANS:
. Thus, LCL =115.85, and UCL = 124.15.
PTS: 1
REF: SECTION 12.1
37. {Grocery Receipts Narrative} Explain how to use the confidence interval to test the hypotheses at  =
0.10.
ANS:
Since the hypothesized value 0 = 125 does not lie in the 90% confidence interval, we reject H0 at  =
0.10.
PTS: 1
REF: SECTION 12.1
Hourly Wages
A random sample of 15 hourly wages for restaurant servers (including tips) was drawn from a normal
population. The sample mean and sample standard deviation were = $14.9 and s = $6.75.
38. {Hourly Wages Narrative} Can we infer at the 5% significance level that the mean wage for restaurant
servers (including tips) is greater than 12?
ANS:
H0:  = 12, H1:  > 12
Rejection region: t > t0.05,14 = 1.761
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copied, or distributed without the prior consent of the publisher.
Test statistic: t = 1.664
Conclusion: Don't reject H0. We can't infer at the 5% significance level that the mean wage for
restaurant servers (including tips) is greater than 12.
PTS: 1
REF: SECTION 12.1
39. {Hourly Wages Narrative} Can we infer at the 5% significance level that the population mean is
greater than 12, assuming that you know the population standard deviation is equal to 6.75?
ANS:
Rejection region: z > z0.05 = 1.645
Test statistic: z = 1.664
Conclusion: Reject H0. (Note this is an unrealistic situation however.)
PTS: 1
REF: SECTION 12.1
Oil Change
The manager of a service station is in the process of analyzing the number of times car owners change
the oil in their cars. She believes that the average motorist changes his or her car's oil less frequently
than recommended by the owner's manual (two times per year). In a preliminary survey she asked 14
car owners how many times they changed their cars' oil in the last 12 months. The results are 1, 1, 2, 0,
3, 3, 0, 1, 0, 1, 2, 3, 3, and 1.
40. {Oil Change Narrative} Does this data provide sufficient evidence at the 10% significance level to
indicate that the manager is correct?
ANS:
H0:  = 2, H1:  < 2
Rejection region: t < t0.10,13 = 1.35
Test statistic: t = 1.613
Conclusion: Reject H0. There is sufficient evidence that the manager is correct.
PTS: 1
REF: SECTION 12.1
41. {Oil Change Narrative} What condition is required in order to analyze this data using a t-test?
ANS:
The number of times car owners change the oil in their cars is normally distributed.
PTS: 1
REF: SECTION 12.1
42. The air pumps at service stations come equipped with a gauge to regulate the air pressure of tires. A
mechanic believes that the gauges are in error by at least 3 pounds per square inch. To test his belief he
takes a random example of 50 air pump gauges and determines the difference between the true
pressure (as measured by an accurate measuring device) and the pressure shown on the air pump
gauge. The mean and the standard deviation of the sample are = 3.4 and s = 1.2. Can the mechanic
infer that he is correct at the 5% significance level? Assume tire pressures have a normal distribution.
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copied, or distributed without the prior consent of the publisher.
ANS:
H0:  = 3, H1:  > 3
Rejection region: t > t0.05,49 = 1.676
Test statistics: t = 2.357
Conclusion: Reject H0. Yes, we can infer that the mechanic is correct at the 5% significance level.
PTS: 1
REF: SECTION 12.1
43. A life insurance representative believes that the mean age of people who buy their first life insurance
plan is less than 35. To test his belief he takes a random sample of 15 customers who have just
purchased their first life insurance. Their ages are 42, 43,28, 34, 30, 36, 25, 29, 32, 33, 27, 30, 22, 37,
and 40. There is not enough evidence to say the data are nonnormal. Can we conclude at the 1%
significance level that the insurance representative is correct?
ANS:
H0:  = 35, H1:  < 35
Rejection region: t < t.01,14 = 2.624
Test statistics: t = 1.547
Conclusion: Cannot reject H0. We can't conclude at the 1% significance level that the average age of
people who buy their first life insurance plan is less than 35, according to this data.
PTS: 1
REF: SECTION 12.1
Domino's Pizza
Domino's Pizza in Big Rapids, Michigan, advertises that they deliver your pizza within 15 minutes of
placing an order or it is free. A sample of 25 customers is selected at random. The average delivery
time in the sample was 13 minutes with a sample standard deviation of 4 minutes.
44. {Domino's Pizza Narrative} We want to know whether Domino's can make this claim or not. Test to
determine if we can infer at the 5% significance level that the population mean delivery time is less
than 15 minutes.
ANS:
H0:  = 15, H1:  < 15
Rejection region: t < t0.05,24 = 1.711
Test statistic: t = 2.50
Conclusion: Reject the null hypothesis. There is enough evidence to say the mean delivery time is less
than 15 minutes.
PTS: 1
REF: SECTION 12.1
45. {Domino's Pizza Narrative} What is the required condition of the technique used in the previous
question?
ANS:
Delivery times are normally distributed.
PTS: 1
REF: SECTION 12.1
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copied, or distributed without the prior consent of the publisher.
Coffee Consumption
A researcher at the University of Michigan medical school believes that coffee consumption may
increase heart rate. Suppose it is known that heart rate (in beats per minute) is normally distributed
with an average of 70 bpm for adults. A random sample of 25 adults was selected and it was found that
their average heartbeat was 73 bpm after coffee consumption, with a standard deviation of 7 bpm.
46. {Coffee Consumption Narrative} Formulate the null and alternative hypotheses.
ANS:
H0:  = 70, H1:  > 70
PTS: 1
REF: SECTION 12.1
47. {Coffee Consumption Narrative} Test the hypotheses in the previous question at the 10% significance
level to determine if we can infer that coffee consumption increases heart rate.
ANS:
Rejection region: t > t0.10,24 = 1.318
Test statistic: t = 2.143
Conclusion: Reject H0. Yes, we can infer that coffee consumption increases heart rate at the 10% level,
according to this data.
PTS: 1
REF: SECTION 12.1
48. Employees in a large company are entitled to 15-minute coffee breaks. A random sample of the
duration of coffee breaks for 10 employees was taken with the times shown as: 12, 16, 14, 18, 21, 17,
19, 15, 18, and 16. Assuming that the times are normally distributed, is there enough evidence at the
5% significance level to indicate that on average employees are taking longer coffee breaks than they
are entitled to?
ANS:
H0:  = 15, H1:  > 15
Rejection region: t > t0.05,9 = 1.833
Test statistics: t = 1.953
Conclusion: Reject H0. Yes, there is enough evidence at the 5% significance level to indicate that on
average employees are taking longer coffee breaks than they are entitled to.
PTS: 1
REF: SECTION 12.1
49. During a water shortage, a water company randomly sampled residential water meters in order to
monitor daily water consumption. On a particular day, a sample of 100 meters showed a sample mean
of 250 gallons and a sample standard deviation of 50 gallons. Provide a 90% confidence interval
estimate of the mean water consumption for the population.
ANS:
. Thus, LCL = 241.7, and UCL = 258.3. We estimate that the mean
water consumption for the population lies between 241.7 gallons and 258.3 gallons.
PTS: 1
REF: SECTION 12.1
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copied, or distributed without the prior consent of the publisher.