> _0bjbjAbbc...8fz.$UF"h(kkkTTTTTTTV|YvT-:)kk:):)TT...:)T.:)T..PXSJ)vgQ(TT0$UQDYb*YPSSYSk".$&akkkTTJ.kkk$U:):):):)Ykkkkkkkkk(: CHAPTER 9: FUNDAMENTALS OF HYPOTHESIS TESTING: ONE-SAMPLE TESTS
Which of the following would be an appropriate null hypothesis?
The mean of a population is equal to 55.
The mean of a sample is equal to 55.
The mean of a population is greater than 55.
Only (a) and (c) are true.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: form of hypothesis
Which of the following would be an appropriate null hypothesis?
The population proportion is less than 0.65.
The sample proportion is less than 0.65.
The population proportion is no less than 0.65.
The sample proportion is no less than 0.65.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: form of hypothesis
Which of the following would be an appropriate alternative hypothesis?
The mean of a population is equal to 55.
The mean of a sample is equal to 55.
The mean of a population is greater than 55.
The mean of a sample is greater than 55.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: form of hypothesis
Which of the following would be an appropriate alternative hypothesis?
The population proportion is less than 0.65.
The sample proportion is less than 0.65.
The population proportion is no less than 0.65.
The sample proportion is no less than 0.65.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: form of hypothesis
A Type II error is committed when
we reject a null hypothesis that is true.
we don't reject a null hypothesis that is true.
we reject a null hypothesis that is false.
we don't reject a null hypothesis that is false.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: type II error
A Type I error is committed when
we reject a null hypothesis that is true.
we don't reject a null hypothesis that is true.
we reject a null hypothesis that is false.
we don't reject a null hypothesis that is false.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: type I error
The power of a test is measured by its capability of
rejecting a null hypothesis that is true.
not rejecting a null hypothesis that is true.
rejecting a null hypothesis that is false.
not rejecting a null hypothesis that is false.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: power
If we are performing a two-tailed test of whether EMBED Equation.2 = 100, the probability of detecting a shift of the mean to 105 will be ________ the probability of detecting a shift of the mean to 110.
less than
greater than
equal to
not comparable to
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: power
True or False: For a given level of significance, if the sample size is increased, the power of the test will increase.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: power, level of significance, sample size
True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type I error will increase.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: level of significance, sample size, type I error
True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type II error will increase.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: level of significance, sample size, type II error
True or False: For a given sample size, the probability of committing a Type II error will increase when the probability of committing a Type I error is reduced.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sample size, type I error, type II error
If an economist wishes to determine whether there is evidence that average family income in a community exceeds $25,000
either a one-tailed or two-tailed test could be used with equivalent results.
a one-tailed test should be utilized.
a two-tailed test should be utilized.
None of the above.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test
If an economist wishes to determine whether there is evidence that average family income in a community equals $25,000
either a one-tailed or two-tailed test could be used with equivalent results.
a one-tailed test should be utilized.
a two-tailed test should be utilized.
None of the above.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: two-tailed test
If the p-value is less than EMBED Equation.2 in a two-tailed test,
the null hypothesis should not be rejected.
the null hypothesis should be rejected.
a one-tailed test should be used.
no conclusion should be reached.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: p-value, level of significance
If a test of hypothesis has a Type I error probability ( EMBED Equation.2 ) of 0.01, we mean
if the null hypothesis is true, we don't reject it 1% of the time.
if the null hypothesis is true, we reject it 1% of the time.
if the null hypothesis is false, we don't reject it 1% of the time.
if the null hypothesis is false, we reject it 1% of the time.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: type I error, level of significance
If the Type I error ( EMBED Equation.2 ) for a given test is to be decreased, then for a fixed sample size n
the Type II error ( EMBED Equation.2 ) will also decrease.
the Type II error ( EMBED Equation.2 ) will increase.
the power of the test will increase.
a one-tailed test must be utilized.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: type I error, type II error, sample size
For a given sample size n, if the level of significance ( EMBED Equation.2 ) is decreased, the power of the test
will increase.
will decrease.
will remain the same.
cannot be determined.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: level of significance, power, sample size
For a given level of significance ( EMBED Equation.2 ), if the sample size n is increased, the probability of a Type II error ( EMBED Equation.2 )
will decrease.
will increase.
will remain the same.
cannot be determined.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: level of significance, beta-risk, sample size
If a researcher rejects a true null hypothesis, she has made a _______error.
ANSWER:
Type I
TYPE: FI DIFFICULTY: Easy
KEYWORDS: type I error
If a researcher accepts a true null hypothesis, she has made a _______decision.
ANSWER:
correct
TYPE: FI DIFFICULTY: Easy
KEYWORDS: decision
If a researcher rejects a false null hypothesis, she has made a _______decision.
ANSWER:
correct
TYPE: FI DIFFICULTY: Easy
KEYWORDS: decision
If a researcher accepts a false null hypothesis, she has made a _______error.
ANSWER:
Type II
TYPE: FI DIFFICULTY: Easy
KEYWORDS: type II error
It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if
a two-tailed test for EMBED Equation.2 is used.
a one-tailed test for EMBED Equation.2 is used.
Both of the previous statements are true.
None of the previous statements is true.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: confidence interval, two-tailed test
The power of a statistical test is
the probability of not rejecting H0 when it is false.
the probability of rejecting H0 when it is true.
the probability of not rejecting H0 when it is true.
the probability of rejecting H0 when it is false.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: power
The symbol for the power of a statistical test is
EMBED Equation.2 .
1 EMBED Equation.2 .
EMBED Equation.2 .
1 EMBED Equation.2 .
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: power
Suppose we wish to test H0: EMBED Equation.2 47 versus H1: EMBED Equation.2 > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52?
We have made a Type I error.
We have made a Type II error.
We have made a correct decision
None of the above are correct.
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, conclusion
How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 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: EMBED Equation.2 = 52, s = 22. Give the null and alternative hypotheses to determine if the number of tissues used during a cold is less than 60.
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean, t test
How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 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: EMBED Equation.2 = 52, s = 22. Using the sample information provided, calculate the value of the test statistic.
EMBED Equation.DSMT4
EMBED Equation.DSMT4
EMBED Equation.DSMT4
EMBED Equation.DSMT4
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test
How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 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: EMBED Equation.2 = 52, s = 22. Suppose the alternative we wanted to test was EMBED Equation.2 . State the correct rejection region for EMBED Equation.2 = 0.05.
Reject H0 if t > 1.6604.
Reject H0 if t < 1.6604.
Reject H0 if t > 1.9842 or Z < 1.9842.
Reject H0 if t < 1.9842.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, rejection region
How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 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: EMBED Equation.2 = 52, s = 22. Suppose the test statistic does fall in the rejection region at EMBED Equation.2 = 0.05. Which of the following decision is correct?
At EMBED Equation.2 = 0.05, we do not reject H0.
At EMBED Equation.2 = 0.05, we reject H0.
At EMBED Equation.2 = 0.05, we accept H0.
At EMBED Equation.2 = 0.10, we do not reject H0.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, decision
How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 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: EMBED Equation.2 = 52, s = 22. Suppose the test statistic does fall in the rejection region at EMBED Equation.2 = 0.05. Which of the following conclusion is correct?
At EMBED Equation.2 = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues.
At EMBED Equation.2 = 0.05, there is sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues.
At EMBED Equation.2 = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues.
At EMBED Equation.2 = 0.10, there is sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, conclusion
We have created a 95% confidence interval for EMBED Equation.2 with the result (10, 15). What decision will we make if we test EMBED Equation.2 at EMBED Equation.2 = 0.05?
Reject H0 in favor of H1.
Accept H0 in favor of H1.
Fail to reject H0 in favor of H1.
We cannot tell what our decision will be from the information given.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: two-tailed test, confidence interval, mean, t test, decision
We have created a 95% confidence interval for EMBED Equation.2 with the result (10, 15). What decision will we make if we test EMBED Equation.2 at EMBED Equation.2 = 0.10?
Reject H0 in favor of H1.
Accept H0 in favor of H1.
Fail to reject H0 in favor of H1.
We cannot tell what our decision will be from the information given.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
EXPLANATION: The 90% confidence interval is narrower than (10, 15), which still does not contain 16.
KEYWORDS: two-tailed test, confidence interval, mean, t test, decision
We have created a 95% confidence interval for EMBED Equation.2 with the result (10, 15). What decision will we make if we test EMBED Equation.2 at EMBED Equation.2 = 0.025?
Reject H0 in favor of H1.
Accept H0 in favor of H1.
Fail to reject H0 in favor of H1.
We cannot tell what our decision will be from the information given.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
EXPLANATION: The 97.5% confidence interval is wider than (10, 15), which could have contained 16 or not have contained 16.
KEYWORDS: two-tailed test, confidence interval, mean, t test, decision
Suppose we want to test EMBED Equation.2 . Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0 in favor of H1?
EMBED Equation.2 = 28, s = 6
EMBED Equation.2 = 27, s = 4
EMBED Equation.2 = 32, s = 2
EMBED Equation.2 = 26, s = 9
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, rejection region
Which of the following statements is not true about the level of significance in a hypothesis test?
The larger the level of significance, the more likely you are to reject the null hypothesis.
The level of significance is the maximum risk we are willing to accept in making a Type I error.
The significance level is also called the EMBED Equation.2 level.
The significance level is another name for Type II error.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: level of significance
If, as a result of a hypothesis test, we reject the null hypothesis when it is false, then we have committed
a Type II error.
a Type I error.
no error.
an acceptance error.
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: decision, type I error, type II error
The value that separates a rejection region from a non-rejection region is called the _______.
ANSWER:
critical value
TYPE: FI DIFFICULTY: Easy
KEYWORDS: critical value, rejection region
A is a numerical quantity computed from the data of a sample and is used in reaching a decision on whether or not to reject the null hypothesis.
significance level
critical value
test statistic
parameter
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: test statistic
The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. The appropriate hypotheses to test are:
EMBED Equation.2 .
EMBED Equation.2 .
EMBED Equation.2 .
EMBED Equation.2 .
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean
The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to be 99% confident in her decision, what rejection region should she use?
Reject H0 if t < 2.34.
Reject H0 if t < 2.55.
Reject H0 if t > 2.34.
Reject H0 if t > 2.58.
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, t test, rejection region
The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what decision should she make?
Reject H0.
Accept H0.
Fail to reject H0.
We cannot tell what her decision should be from the information given.
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, decision
The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what conclusion can she make?
There is not sufficient evidence that the mean age of her customers is over 30.
There is sufficient evidence that the mean age of her customers is over 30.
There is not sufficient evidence that the mean age of her customers is not over 30.
There is sufficient evidence that the mean age of her customers is not over 30.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, conclusion
The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. What is the p-value associated with the test statistic?
0.3577
0.1423
0.0780
0.02
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, p-value
A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to:
4.12
2.33
1.86
0.07
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, test statistic
A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose that the test statistic is 2.20. Can we conclude that H0 should be rejected at the (a) EMBED Equation.2 = 0.10, (b) EMBED Equation.2 = 0.05, and (c) EMBED Equation.2 = 0.01 level of Type I error?
(a) yes; (b) yes; (c) yes
(a) no; (b) no; (c) no
(a) no; (b) no; (c) yes
(a) yes; (b) yes; (c) no
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, proportion, decision
A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose you reject the null hypothesis. What conclusion can you draw?
There is not sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90.
There is sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90.
There is not sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90.
There is sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, proportion, conclusion
A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. State the test of interest to the rental chain.
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, form of hypothesis, form of hypothesis
A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The value of the test statistic in this problem is approximately equal to:
2.80
2.60
1.94
1.30
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, Z test, test statistic
A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The p-value associated with the test statistic in this problem is approximately equal to:
0.0100
0.0051
0.0026
0.0013
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, Z test, p-value
A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The decision on the hypothesis test using a 3% level of significance is:
to reject H0 in favor of H1.
to accept H0 in favor of H1.
to fail to reject H0 in favor of H1.
we cannot tell what the decision should be from the information given.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, Z test, decision
A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The rental chain's conclusion from the hypothesis test using a 3% level of significance is:
to open a new store.
not to open a new store.
to delay opening a new store until additional evidence is collected.
we cannot tell what the decision should be from the information given.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, Z test, conclusion
An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675, which test would you use?
Z-test of a population mean
Z-test of a population proportion
t-test of a population mean
t-test of a population proportion
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: two-tailed test, mean, Z test
An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675 and decide not to reject the null hypothesis, what can you conclude?
There is not enough evidence to conclude that the daily average revenue was $675.
There is not enough evidence to conclude that the daily average revenue was not $675.
There is enough evidence to conclude that the daily average revenue was $675.
There is enough evidence to conclude that the daily average revenue was not $675.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: two-tailed test, mean, Z test, conclusion
A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you wanted to test whether the auditor should conclude that there is evidence that the average balance is different from $75, which test would you use?
Z-test of a population mean
Z-test of a population proportion
t-test of a population mean
t-test of a population proportion
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: two-tailed test, mean, t test
A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you were to conduct a test to determine whether the average balance is different from $75 and decided to reject the null hypothesis, what conclusion could you draw?
There is not evidence that the average balance is $75.
There is not evidence that the average balance is not $75.
There is evidence that the average balance is $75.
There is evidence that the average balance is not $75.
ANSWER:
d
TYPE: MC DIFFICULTY: moderate
KEYWORDS: two-tailed test, mean, t test, conclusion
The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30, which test would you use?
Z-test of a population mean
Z-test of a population proportion
t-test of population mean
t-test of a population proportion
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: two-tailed test, proportion
The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you draw?
There is sufficient evidence that the proportion is 0.30.
There is not sufficient evidence that the proportion is 0.30.
There is sufficient evidence that the proportion is 0.30.
There is not sufficient evidence that the proportion is not 0.30.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: two-tailed test, proportion, conclusion
TABLE 9-1
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 Monarch butterflies:
n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82;
Null Hypothesis: EMBED Equation.2 ; ( = 0.10; df = 45; T Test Statistic = 2.09;
One-Tailed Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
Referring to Table 9-1, the parameter the biologist is interested in is:
the mean number of butterflies in Pismo Beach State Park.
the mean number of parasites on these 46 butterflies.
the mean number of parasites on Monarch butterflies in Pismo Beach State Park.
the proportion of butterflies with parasites.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: mean, t test, parameter
Referring to Table 9-1, state the alternative hypothesis for this study.
ANSWER:
EMBED Equation.2
TYPE: PR DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, form of hypothesis
Referring to Table 9-1, what critical value should the biologist use to determine the rejection region?
1.6794
1.3011
1.3006
0.6800
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, critical value
True or False: Referring to Table 9-1, the null hypothesis would be rejected.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, decision
True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 4% probability of committing a Type I error is allowed.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, decision
True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, decision
Referring to Table 9-1, the lowest level of significance at which the null hypothesis can be rejected is ______.
ANSWER:
0.021
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, p-value
True or False: Referring to Table 9-1, the evidence proves beyond a doubt that the mean number of parasites on butterflies in Pismo Beach State Park is over 20.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, conclusion
True of False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per butterfly is over 20 using a level of significance of 0.10.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, conclusion
True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per butterfly is over 20 with no more than a 5% probability of incorrectly rejecting the true null hypothesis.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, decision
True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per butterfly is over 20 with no more than a 1% probability of incorrectly rejecting the true null hypothesis.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, t test, decision
True or False: Referring to Table 9-1, the value of EMBED Equation.2 is 0.90.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, beta-risk
True or False: Referring to Table 9-1, if these data were used to perform a two-tailed test, the p-value would be 0.042.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, p-value
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.0355
TYPE: FI DIFFICULTY: Difficult
EXPLANATION: The power is computed for the Z test because the population standard deviation is given.
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.9645
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.4071
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.5929
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.9091
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.1 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.0909
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.0151
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.9849
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.2749
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.7251
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-1, the power of the test is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.8344
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-1, the probability of committing a Type II error is _____ if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.05 level of significance and assuming that the population standard deviation is 25.92.
ANSWER:
0.1656
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, type II error
True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be 0.043. The null hypothesis should be rejected if the chosen level of significance is 0.05.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: mean, t test, p-value, level of significance, decision
True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be 0.034. The null hypothesis should be rejected if the chosen level of significance is 0.01.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: p-value, level of significance, decision
True or False: Suppose, in testing a hypothesis about a proportion, the Z test statistic is computed to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tailed test is used.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: proportion, Z test, test statistic, critical value, decision
True or False: In testing a hypothesis, statements for the null and alternative hypotheses as well as the selection of the level of significance should precede the collection and examination of the data.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: ethical issues
True or False: The test statistic measures how close the computed sample statistic has come to the hypothesized population parameter.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: test statistic
True or False: The statement of the null hypothesis always contains an equality.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: form of null hypoehsis
True or False: The larger is the p-value, the more likely one is to reject the null hypothesis.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: p-value
True or False: The smaller is the p-value, the stronger is the evidence against the null hypothesis.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: p-value
True or False: A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: confidence interval, two-tailed test, decision
True or False: A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 18 versus the alternative hypothesis that the mean of the population differs from 18, the null hypothesis could be rejected at a level of significance of 0.05.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: confidence interval, two-tailed test, decision
True or False: A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.10.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: confidence interval, two-tailed test, decision
True or False: A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.02.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
EXPLANATION: We are not sure if 20 will be in the wider confidence interval.
KEYWORDS: confidence interval, two-tailed test, decision
True or False: A sample is used to obtain a 95% confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be accepted at a level of significance of 0.02.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
EXPLANATION: We are not sure if 20 will be in the wider confidence interval.
KEYWORDS: confidence interval, two-tailed test, decision
TABLE 9-2
A student claims that he can correctly identify whether a person is a business major or an agriculture major by the way the person dresses. Suppose in actuality that if someone is a business major, he can correctly identify that person as a business major 87% of the time. When a person is an agriculture major, the student will incorrectly identify that person as a business major 16% of the time. Presented with one person and asked to identify the major of this person (who is either a business or agriculture major), he considers this to be a hypothesis test with the null hypothesis being that the person is a business major and the alternative that the person is an agriculture major.
Referring to Table 9-2, what would be a Type I error?
Saying that the person is a business major when in fact the person is a business major.
Saying that the person is a business major when in fact the person is an agriculture major.
Saying that the person is an agriculture major when in fact the person is a business major.
Saying that the person is an agriculture major when in fact the person is an agriculture major.
ANSWER:
c
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: form of hypothesis, form of hypothesis
Referring to Table 9-2, what would be a Type II error?
Saying that the person is a business major when in fact the person is a business major.
Saying that the person is a business major when in fact the person is an agriculture major.
Saying that the person is an agriculture major when in fact the person is a business major.
Saying that the person is an agriculture major when in fact the person is an agriculture major.
ANSWER:
b
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: type II error
Referring to Table 9-2, what is the actual level of significance of the test?
0.13
0.16
0.84
0.87
ANSWER:
a
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: level of significance
Referring to Table 9-2, what is the actual confidence coefficient?
0.13
0.16
0.84
0.87
ANSWER:
d
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: confidence coefficient
Referring to Table 9-2, what is the value of EMBED Equation.2 ?
0.13
0.16
0.84
0.87
ANSWER:
a
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: level of significance
Referring to Table 9-2, what is the value of EMBED Equation.2 ?
0.13
0.16
0.84
0.87
ANSWER:
b
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: beta-risk
Referring to Table 9-2, what is the power of the test?
0.13
0.16
0.84
0.87
ANSWER:
c
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: power
TABLE 9-3
An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W.
Referring to Table 9-3, the population of interest is
the power consumption in the 20 microwave ovens.
the power consumption in all such microwave ovens.
the mean power consumption in the 20 microwave ovens.
the mean power consumption in all such microwave ovens.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test
Referring to Table 9-3, the parameter of interest is
the mean power consumption of the 20 microwave ovens.
the mean power consumption of all such microwave ovens.
250
257.3
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, parameter
Referring to Table 9-3, the appropriate hypotheses to determine if the manufacturer's claim appears reasonable are:
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis
Referring to Table 9-3, for a test with a level of significance of 0.05, the critical value would be ________.
ANSWER:
Z = 1.645
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, critical value
Referring to Table 9-3, the value of the test statistic is ________.
ANSWER:
2.18
TYPE: FI DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, test statistic
Referring to Table 9-3, the p-value of the test is ________.
ANSWER:
0.0148 using Excel or 0.0146 using Table E.2
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, p-value
True or False: Referring to Table 9-3, for this test to be valid, it is necessary that the power consumption for microwave ovens has a normal distribution.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, assumptions
True or False: Referring to Table 9-3, the null hypothesis will be rejected at 5% level of significance.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, decision
True or False: Referring to Table 9-3, the null hypothesis will be rejected at 1% level of significance.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, decision
True or False: Referring to Table 9-3, the consumer group can conclude that there is enough evidence to prove that the manufacturers claim is not true when allowing for a 5% probability of committing a type I error.
ANSWER:
True
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, conclusion
Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.05 level of significance?
ANSWER:
0.7025
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.05 level of significance?
ANSWER:
0.2975
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 248 W using a 0.05 level of significance?
ANSWER:
0.0125
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 248 W using a 0.05 level of significance?
ANSWER:
0.9875
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.10 level of significance?
ANSWER:
0.8146
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.10 level of significance?
ANSWER:
0.1854
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 248 W using a 0.10 level of significance?
ANSWER:
0.0302
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 248 W using a 0.10 level of significance?
ANSWER:
0.9698
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
TABLE 9-4
A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision.
Referring to Table 9-4, the appropriate hypotheses are:
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis
Referring to Table 9-4, for a test with a level of significance of 0.10, the critical value would be ________.
ANSWER:
-1.28
TYPE: FI DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, critical value
Referring to Table 9-4, the value of the test statistic is ________.
ANSWER:
-1.50
TYPE: FI DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, Z test, test statistic
Referring to Table 9-4, the p-value of the test is ________.
ANSWER:
0.0668
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, p-value
True or False: Referring to Table 9-4, the null hypothesis will be rejected with a level of significance of 0.10.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, decision
True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, decision
True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the company would market the new anesthetic.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, Z test, conclusion
Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?
ANSWER:
0.6388
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance?
ANSWER:
0.3612
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance?
ANSWER:
0.0160
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance?
ANSWER:
0.9840
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance?
ANSWER:
0.7638
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance?
ANSWER:
0.2362
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance?
ANSWER:
0.0374
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, power
Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance?
ANSWER:
0.9626
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: one-tailed test, mean, Z test, type II error
TABLE 9-5
A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202.
True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of EMBED Equation.2 = 0.05.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, decision
True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of EMBED Equation.2 = 0.01.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, decision
True or False: Referring to Table 9-5, the bank can conclude that the average age is greater than 45 at a significance level of EMBED Equation.2 = 0.01.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, conclude
Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what would be that test's p-value?
0.0202
0.0404
0.9596
0.9798
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, p-value
TABLE 9-6
The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test.
True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, decision
True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, conclusion
True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, decision
True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
EXPLANATION: the engineer can conclude that there is insufficient evidence to show that the mean amount of force needed is not 650 but cannot conclude that there is evidence to show that the force needed is 650.
KEYWORDS: one-tailed test, mean, conclusion
True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, mean, decision
Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. What would be the p-value of this one-tailed test?
0.040
0.160
0.840
0.960
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, p-value
Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p-value of this one-tailed test?
0.040
0.160
0.840
0.960
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, p-value
True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, decision
TABLE 9-7
A major home improvement store conducted its biggest brand recognition campaign in the companys history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who like the ads a lot. A study of 1,189 adults who viewed the ads reported that 230 indicated that they like the ads a lot. The percentage of a typical television advertisement receiving the like the ads a lot score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of like the ads a lot for the companys ads is less than 0.22) at a 0.01 level of significance.
Referring to Table 9-7, the parameter the company officials is interested in is:
the mean number of viewers who like the ads a lot.
the total number of viewers who like the ads a lot.
the mean number of company officials who like the ads a lot.
the proportion of viewers who like the ads a lot.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: proportion, t test, parameter
Referring to Table 9-7, state the null hypothesis for this study.
ANSWER:
EMBED Equation.3
TYPE: PR DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, form of hypothesis
Referring to Table 9-7, state the alternative hypothesis for this study.
ANSWER:
EMBED Equation.3
TYPE: PR DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, form of hypothesis
Referring to Table 9-7, what critical value should the company officials use to determine the rejection region?
ANSWER:
-2.3263
TYPE: PR DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, critical value
Referring to Table 9-7, the null hypothesis will be rejected if the test statistics is
greater than 2.3263
less than 2.3263
greater than -2.3263
less than -2.3263
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, rejection region
True or False: Referring to Table 9-7, the null hypothesis would be rejected.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, decision
Referring to Table 9-7, the lowest level of significance at which the null hypothesis can be rejected is ______.
ANSWER:
0.0135
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, proportion, t test, p-value
Referring to Table 9-7, the largest level of significance at which the null hypothesis will not be rejected is ______.
ANSWER:
0.0135
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: one-tailed test, proportion, t test, p-value
True of False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, conclusion
True or False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.05.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: one-tailed test, proportion, t test, decision
True or False: Referring to Table 9-7, the value of EMBED Equation.2 is 0.90.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, mean, t test, beta-risk
Referring to Table 9-7, what will be the p-value if these data were used to perform a two-tailed test?
ANSWER:
0.027
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: one-tailed test, proportion, t test, p-value
PAGE 280 Fundamentals of Hypothesis Testing: One-Sample Tests
Fundamentals of Hypothesis Testing: One-Sample Tests PAGE 281
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currentpoint translate 64 62 translate
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16 th 904 8 moveto 234 0 rlineto stroke
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/sh {moveto show} def
384 /Times-Italic f1
(H) 9 322 sh
(X) 875 322 sh
(H) 2888 322 sh
(X) 3717 322 sh
224 /Times-Roman f1
(0) 297 418 sh
(1) 3157 418 sh
384 ns
(60) 1525 322 sh
(60) 4368 322 sh
384 /Times-Roman f1
(:) 519 322 sh
( ) 699 322 sh
( ) 1962 322 sh
( ) 2720 322 sh
(:) 3361 322 sh
( ) 3541 322 sh
(.) 4752 322 sh
384 /Symbol f1
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384 /Times-Roman f1
(and) 2107 322 sh
end MTsave restore
dqMATHei
H0
: 2X60 and H1
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16 th 904 8 moveto 234 0 rlineto stroke
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384 /Times-Italic f1
(H) 9 322 sh
(X) 875 322 sh
(H) 2901 322 sh
(X) 3730 322 sh
224 /Times-Roman f1
(0) 297 418 sh
(1) 3170 418 sh
384 ns
(52) 1538 322 sh
(52) 4393 322 sh
384 /Times-Roman f1
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end MTsave restore
dqMATHer
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384 /Times-Italic f1
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end MTsave restore
dMATH5
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end MTsave restore
dMATH5
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end MTsave restore
d4MATH(
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(a) -30 261 sh
end MTsave restore
dMATH
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end MTsave restore
dMATH5
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end MTsave restore
dMATH
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
aFMicrosoft Equation Editor 2.0DNQEEquation.2
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/fs 0 def /cf 0 def
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/f1 {ff dup /cf exch def sf} def
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
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(a) -30 261 sh
end MTsave restore
dMATH
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end MTsave restore
dMATH
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end MTsave restore
dMATH
mFMicrosoft Equation Editor 2.0DNQEEquation.2
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/fs 0 def /cf 0 def
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/f1 {ff dup /cf exch def sf} def
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384 /Times-Italic f1
(H) 9 261 sh
(H) 3079 261 sh
224 /Times-Roman f1
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(16) 4328 261 sh
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( ) 1729 261 sh
( ) 2911 261 sh
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(m) 700 261 sh
(m) 3733 261 sh
384 /Symbol f1
(=) 1026 261 sh
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end MTsave restore
dkMATH_
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end MTsave restore
dMATH
aFMicrosoft Equation Editor 2.0DNQEEquation.2
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(H) 9 261 sh
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end MTsave restore
dkMATH_
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end MTsave restore
dMATH5
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end MTsave restore
dMATH5
2XmFMicrosoft Equation Editor 2.0DNQEEquation.2CompObj}YObjInfoDFEquation Native +_963915380SIFJJ
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16 th 46 8 moveto 234 0 rlineto stroke
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} def
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findfont
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/fs 0 def /cf 0 def
/sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def
/f1 {ff dup /cf exch def sf} def
/ns {cf sf} def
/sh {moveto show} def
384 /Times-Italic f1
(X) 17 322 sh
end MTsave restore
dMATH5
2XmCompObjYObjInfoLNEquation Native +_963915383QFJJFMicrosoft Equation Editor 2.0DNQEEquation.2
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16 th 46 8 moveto 234 0 rlineto stroke
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} def
/ff {
dup FontDirectory exch known not {
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/fs 0 def /cf 0 def
/sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def
/f1 {ff dup /cf exch def sf} def
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384 /Times-Italic f1
(X) 17 322 sh
end MTsave restore
dMATH5
2XmFMicrosoft Equation Editor 2.0DNQEEquation.2
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} def
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/f1 {ff dup /cf exch def sf} def
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/f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def
384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
a+FMicrosoft Equation Editor 2.0DNQEEquation.2
aCompObjYObjInfo\^Equation Native (_963927155aFJJOle
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/fs 0 def /cf 0 def
/sf {exch dup /fs exch def dup neg matrix scale makefont setfont} def
/f1 {ff dup /cf exch def sf} def
/ns {cf sf} def
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384 /Times-Italic f1
(H) 9 261 sh
(H) 3096 261 sh
224 /Times-Roman f1
(0) 297 357 sh
(1) 3365 357 sh
384 ns
(30) 1309 261 sh
(30) 4360 261 sh
384 /Times-Roman f1
(:) 519 261 sh
( ) 1746 261 sh
( ) 2928 261 sh
(:) 3569 261 sh
/f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def
384 /Symbol f2
(m) 700 261 sh
(m) 3750 261 sh
384 /Symbol f1
(\263) 1019 261 sh
(<) 4071 261 sh
384 /Times-Roman f1
(versus) 1902 261 sh
end MTsave restore
dkMATH_
H0
:m30 versus H1
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/fs 0 def /cf 0 def
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384 /Times-Italic f1
(H) 9 261 sh
(H) 3132 261 sh
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(30) 1333 261 sh
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( ) 1773 261 sh
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384 /Times-Roman f1
(versus) 1932 261 sh
end MTsave restore
dkMATH_
H0
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(X) 716 322 sh
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end MTsave restore
dqMATHeg
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384 /Times-Italic f1
(H) 9 322 sh
(X) 716 322 sh
(H) 3150 322 sh
(X) 3820 322 sh
224 /Times-Roman f1
(0) 297 418 sh
(1) 3419 418 sh
384 ns
(30) 1363 322 sh
(30) 4468 322 sh
384 /Times-Roman f1
(:) 519 322 sh
( ) 1800 322 sh
( ) 2982 322 sh
(:) 3623 322 sh
384 /Symbol f1
(\243) 1073 322 sh
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384 /Times-Roman f1
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end MTsave restore
dqMATHeY
H0
:2X30 versus H1
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
aFMicrosoft Equation Editor 2.0DNQEEquation.2CompObjYObjInfoEquation Native (_963936677gFJJ
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
aFMicrosoft Equation Editor 2.0CompObjYObjInfoEquation Native (_963936719FJJDNQEEquation.2
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
aFMicrosoft Equation Editor 2.0DNQEEquation.2
aTw4
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/f1 {ff dup /cf exch def sf} def
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384 /Times-Italic f1
(H) 9 264 sh
(H) 3636 264 sh
224 /Times-Roman f1
(0) 299 360 sh
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(5) 1337 264 sh
(000) 1645 264 sh
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384 /Times-Roman f1
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( ) 2277 264 sh
( ) 3465 264 sh
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/f2 {ff matrix dup 2 .22 put makefont dup /cf exch def sf} def
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(m) 714 264 sh
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384 /Times-Roman f1
(versus) 2436 264 sh
end MTsave restore
d}MATHqC
H0
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384 /Symbol f1
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end MTsave restore
d@MATH4
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384 /Symbol f2
(b) -5 284 sh
end MTsave restore
dMATH
bFMicrosoft Equation Editor 2.0DNQEEquation.2
bCompObjYObjInfoEquation Native (_964013776FJJOle
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384 /Symbol f2
(a) -30 261 sh
end MTsave restore
dMATH
a۠FMicrosoft Equation Editor 2.0DNQEEquation.2
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384 /Symbol f2
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end MTsave restore
dMATH
bmFMicrosoft Equation Editor 2.0DNQEEquation.2CompObjYObjInfoEquation Native (_970067709FJJ
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