The value obtained for the test statistic, z, in a one–mean z–test is given. Also given is whether the test is two tailed, left
tailed, or right tailed. Determine the P–value.
232)
A two–tailed test:
z = – 0.17
232)
A)
0.1350
B)
0.4325
C)
0.8650
D)
0.5675
Determine the critical value(s) for a one–mean z–test.
233)
A two–tailed test with =0.05.
233)
A)
±1.96
B)
±2.575
C)
±1.645
D)
±1.764
For the given hypothesis test, determine the probability of a Type II error or the power, as specified.
234)
A hypothesis test is to be performed to determine whether the mean hematocrit (percentage by
volume of the blood occupied by red blood cells) for women differs from the mean hematocrit for
men which is known to be 47%. Preliminary data analyses indicate that it is reasonable to apply a
z–test. The hypotheses are
H0: µ= 47%
Ha: µ 47%.
Assume that the population standard deviation is 2.8%. The sample size is 10. The significance
level is 0.01. Find the probability of a Type II error if in fact the mean hematocrit for women, µ , is
43%.
234)
A)
0.9738
B)
0.0262
C)
0.01
D)
0.99
Find the critical value(s) for the specified Wilcoxon signed–rank test.
235)
Sample size = 9, right–tailed test, significance level = 0.05
235)
A)
6
B)
39
C)
8
D)
37
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
236)
In 2000, the average duration of long–distance telephone calls originating in one town was
9.4 minutes. Five years later, in 2005, a long–distance telephone company wants to perform a
hypothesis test to determine whether the average duration of long–distance phone calls has
changed from the 2000 mean of 9.4 minutes. The hypotheses are:
H0: µ= 9.4 minutes
Ha: µ 9.4 minutes
where µ is the mean duration, in 2005, of long–distance telephone calls originating in the town.
Explain the meaning of a correct decision.
236)
A)
A correct decision would occur if, in fact, µ
9.4 minutes and the results of the sampling do
not lead to rejection of the null hypothesis that µ= 9.4 minutes.
B)
A correct decision would occur if, in fact, µ= 9.4 minutes, and the results of the sampling do
not lead to rejection of that fact; or if, in fact, µ 9.4 minutes and the results of the sampling
lead to that conclusion.
C)
A correct decision would occur if, in fact, µ= 9.4 minutes, and the results of the sampling lead
to rejection of the null hypothesis; or if, in fact, µ
9.4 minutes and the results of the sampling
lead to that conclusion.
D)
A correct decision would occur if, in fact, µ= 9.4 minutes, and the results of the sampling do
not lead to rejection of that fact; or if, in fact, µ 9.4 minutes and the results of the sampling
do not lead to rejection of the null hypothesis.
Use a table of t–values to estimate the P–value for the specified one–mean t–test.
237)
Right–tailed test, n = 19, t = 2.318
237)
A)
0.05 < P < 0.10
B)
0.01 < P < 0.025
C)
P > 0.10
D)
0.025 < P < 0.05
Explanation:
Find the critical value(s) for the specified Wilcoxon signed–rank test.
238)
Sample size = 10, two–tailed test, significance level = 0.05
238)
A)
11, 47
B)
8, 47
C)
11, 44
D)
8, 48
Explanation:
Explanation:
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
239)
The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 parts per
million (ppm). A consumer health group measured the level of the chemical in a random sample
of tomatoes obtained from one producer to determine whether the mean level of the chemical in
these tomatoes exceeds the recommended limit.
The hypotheses are
H0: µ= 0.4 ppm
Ha: µ> 0.4 ppm
where µ is the mean level of the chemical in tomatoes from this producer. Suppose that the results
of the sampling lead to nonrejection of the null hypothesis. Classify that conclusion as a Type I
error, a Type II error, or a correct decision, if in fact the mean level of the chemical in these tomatoes
is greater than 0.4 ppm.
239)
A)
Correct decision
B)
Type I error
C)
Type II error
A sample mean, sample standard deviation, and sample size are given. Use the one–mean t–test to perform the required
hypothesis test about the mean, µ, of the population from which the sample was drawn. Use the P–value approach. Also,
assess the strength of the evidence against the null hypothesis.
240)
x= 24.4, s = 9.2, n = 25, H0: µ= 26, Ha: µ< 26, = 0.05
240)
A)
Test statistic: t = – 0.87. P–value = 0.1966. Do not reject H0. There is not sufficient evidence to
conclude that the mean is less than 26. The evidence against the null hypothesis is weak or
none.
B)
Test statistic: t = – 0.87. P–value = 0.8078. Do not reject H0. There is not sufficient evidence to
conclude that the mean is less than 26. The evidence against the null hypothesis is strong.
C)
Test statistic: t = – 0.87. P–value = 0.8034. Do not reject H0. There is not sufficient evidence to
conclude that the mean is less than 26. The evidence against the null hypothesis is weak or
none.
D)
Test statistic: t = – 0.87. P–value = 0.1922. Do not reject H0. There is not sufficient evidence to
conclude that the mean is less than 26. The evidence against the null hypothesis is weak or
none.
Provide an appropriate response.
241)
The P–value for a one–mean t–test is estimated using a t–table as 0.05 < P < 0.10. Based on this
information, for what significance levels can the null hypothesis be rejected?
241)
A)
We can reject H0 at any significance level smaller than 0.05.
B)
We can reject H0 at any significance level smaller than 0.10.
C)
We can reject H0 at any significance level 0.10 or larger.
D)
We can reject H0 at any significance level 0.05 or larger.
The significance level and P–value of a hypothesis test are given. Decide whether the null hypothesis should be rejected.
242)
= 0.05, P–value = 0.017
242)
A)
Reject the null hypothesis.
B)
Do not reject the null hypothesis.
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
243)
A psychologist has designed a test to measure stress levels in adults. She has determined that
nationwide the mean score on her test is 27. A hypothesis test is to be conducted to determine
whether the mean score for trial lawyers exceeds the national mean score. The hypotheses are
H0: µ=27
Ha: µ>27
where µ is the mean score for all trial lawyers. Suppose that the results of the sampling lead to
nonrejection of the null hypothesis. Classify that conclusion as a Type I error, a Type II error, or a
correct decision, if in fact the mean score for all trial lawyers is equal to 27.
243)
A)
Type II error
B)
Correct decision
C)
Type I error
Two graphical displays are given for a set of data. A hypothesis test is to be performed for the mean of the population
from which the data were obtained. Use the graphs to determine whether a z–test, a t–test, a Wilcoxon signed–rank test, or
none of these is appropriate.
104
244)
A normal probability plot and a stem–and–leaf diagram of the data are given below. is
unknown. Which procedure is appropriate?
244)
A)
t–test
B)
z–test
C)
Wilcoxon signed–rank test
D)
None of these tests is appropriate.
Determine the critical value(s) for a one–mean z–test.
245)
A left–tailed test with = 0.04.
245)
A)
–2.05
B)
–1.75
C)
2.05
D)
1.75
B
D
Provide an appropriate response.
246)
Suppose that you wish to perform a hypothesis test for a population mean. Suppose that the
population standard deviation is unknown and the sample size is small. If the distribution of the
variable under consideration is J–shaped, should you use a t–test, a Wilcoxon signed–rank test, or
neither?
246)
A)
Wilcoxon signed–rank test
B)
t–test
C)
Neither
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
247)
The recommended dietary allowance (RDA) of vitamin C for women is 75 milligrams per day. A
hypothesis test is to be performed to decide whether adult women are, on average, getting less than
the RDA of 75 milligrams per day. The hypotheses are
H0: µ= 75 mg
Ha: µ< 75 mg
where µ is the mean vitamin C intake (per day) of all adult females. Explain the meaning of a Type
II error.
247)
A)
A Type II error would occur if, in fact, µ= 75 mg, and the results of the sampling do not lead
to rejection of that fact.
B)
A Type II error would occur if, in fact, µ= 75 mg, but the results of the sampling lead to the
conclusion that µ< 75 mg
C)
A Type II error would occur if, in fact, µ< 75 mg, but the results of the sampling fail to lead to
that conclusion.
D)
A Type II error would occur if, in fact, µ< 75 mg, and the results of the sampling lead to
rejection of the null hypothesis that µ= 75 mg.
C
C
Two graphical displays are given for a set of data. A hypothesis test is to be performed for the mean of the population
from which the data were obtained. Use the graphs to determine whether a z–test, a t–test, a Wilcoxon signed–rank test, or
none of these is appropriate.
248)
A normal probability plot and a stem–and–leaf diagram of the data are given below. is
unknown. Which procedure is appropriate?
248)
A)
t–test
B)
z–test
C)
Wilcoxon signed–rank test
D)
None of these tests is appropriate.
The value obtained for the test statistic, z, in a one–mean z–test is given. Also given is whether the test is two tailed, left
tailed, or right tailed. Determine the P–value.
249)
A left–tailed test:
z = – 0.58
249)
A)
0.5620
B)
0.4380
C)
0.7190
D)
0.2810
D
D
The P–value for a hypothesis test is given. Determine whether the strength of the evidence against the null hypothesis is
weak/none, moderate, strong, or very strong.
250)
P = 0.002
250)
A)
Moderate
B)
Very strong
C)
Weak or none
D)
Strong
Provide an appropriate response.
251)
A hypothesis test for a population mean is to be performed. True or false: The probability of a Type
I error depends on the true mean.
251)
A)
True
B)
False
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
252)
A psychologist has designed a test to measure stress levels in adults. She has determined that
nationwide the mean score on her test is 24. A hypothesis test is to be conducted to determine
whether the mean score for trial lawyers exceeds the national mean score. The hypotheses are
H0: µ=24
Ha: µ>24
where µ is the mean score for all trial lawyers. Explain the meaning of a Type I error.
252)
A)
A Type I error would occur if, in fact, µ=24, but the results of the sampling lead to the
conclusion that µ>24.
B)
A Type I error would occur if, in fact, µ>24, but the results of the sampling lead to rejection
of the null hypothesis that µ=24.
C)
A Type I error would occur if, in fact, µ=24, but the results of the sampling fail to lead to
rejection of that fact.
D)
A Type I error would occur if, in fact, µ>24, but the results of the sampling fail to lead to that
conclusion.
The graph portrays the decision criterion for a one–mean z–test. The curve in the graph is the normal curve for the test
statistic under the assumption that the null hypothesis is true. Use the graph to solve the problem.
253)
A graphical display of the decision criterion follows.
Determine the nonrejection region.
253)
A)
z –1.96
B)
–1.96
z
1.96
C)
z –1.96
D)
z 0.025
Find the critical value(s) for the specified Wilcoxon signed–rank test.
254)
Sample size = 19, two–tailed test, significance level = 0.10
254)
A)
54, 134
B)
52, 136
C)
46, 144
D)
54, 136
A sample mean, sample size, and population standard deviation are given. Use the one–mean z–test to perform the
required hypothesis test at the given significance level. Use the critical –value approach.
255)
x=7.1, n = 18 , =1.5, H0: µ= 10; Ha: µ< 10, = 0.01
255)
A)
z= – 8.20; critical value = 1.96; do not reject H0
B)
z= – 8.20; critical value = – 2.33; do not reject H0
C)
z= – 8.20; critical value = – 2.33; reject H0
D)
z= – 8.20; critical value = – 1.96; reject H0
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
256)
A health insurer has determined that the “reasonable and customary” fee for a certain medical
procedure is $1200. They suspect that the mean fee charged by one particular clinic for this
procedure is higher than $1200. The insurer wants to perform a hypothesis test to determine
whether their suspicion is correct. The hypotheses are:
H0: µ= $1200
Ha: µ> $1200
where µ is the mean amount charged by the clinic for this procedure. Explain the meaning of a
correct decision.
256)
A)
A correct decision would occur if, in fact, µ= $1200, and the results of the sampling lead to
rejection of the null hypothesis; or if, in fact, µ> $1200 and the results of the sampling lead to
that conclusion.
B)
A correct decision would occur if, in fact, µ= $1200, and the results of the sampling do not
lead to rejection of that fact; or if, in fact, µ> $1200 and the results of the sampling do not lead
to rejection of the null hypothesis that µ= $1200.
C)
A correct decision would occur if, in fact, µ> $1200 and the results of the sampling do not
lead to rejection of the null hypothesis that µ= $1200.
D)
A correct decision would occur if, in fact, µ= $1200, and the results of the sampling do not
lead to rejection of that fact; or if, in fact, µ> $1200 and the results of the sampling lead to that
conclusion.
A hypothesis test is to be performed. Determine the null and alternative hypotheses.
257)
The manufacturer of a refrigerator system for beer kegs produces refrigerators that are supposed to
maintain a mean temperature, µ, of 49°F, ideal for a certain type of German pilsner. The owner of
the brewery does not agree with the refrigerator manufacturer, and wants to conduct a hypothesis
test to determine whether the true mean temperature differs from this value.
257)
A)
H0: µ49°F
Ha: µ=49°F
B)
H0: µ49°F
Ha: µ>49°F
C)
H0: µ49°F
Ha: µ<49°F
D)
H0: µ=49°F
Ha: µ49°F
A sample mean, sample size, and population standard deviation are given. Use the one–mean z–test to perform the
required hypothesis test at the given significance level. Use the P–value approach.
258)
x= 6.7, n = 20, = 0.6, H0: µ = 7.1 , Ha: µ < 7.1 , = 0.05
258)
A)
z = – 2.98; P–value = 0.0014; reject H0
B)
z = – 2.98; P–value = 0.9986; do not reject H0
C)
z = – 2.98; P–value = 0.0014; do not reject H0
D)
z = – 0.67; P–value = 0.2514; do not reject H0
A z–test is to be performed for a population mean. Express the decision criterion for the hypothesis test in terms of x.
That is, determine for what values of x the null hypothesis would be rejected.
259)
A hypothesis test is to be performed to determine whether the mean waiting time during peak
hours for customers in a supermarket has increased from the previous mean waiting time of
8.9 minutes. Preliminary data analyses indicate that it is reasonable to apply a z–test. The
hypotheses are
H0: µ=8.9 minutes
Ha: µ>8.9 minutes.
The population standard deviation is 3.3 minutes. The sample size is 34. The significance level is
0.05. Express the decision criterion for the hypothesis test in terms of x.
259)
A)
Reject H0 if x<9.83 minutes.
B)
Reject H0 if x>9.83 minutes.
C)
Reject H0 if x>10.01 minutes.
D)
Reject H0 if x> 1.645 minutes.
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
260)
The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 parts per
million (ppm). A consumer health group measured the level of the chemical in a random sample
of tomatoes obtained from one producer to determine whether the mean level of the chemical in
these tomatoes exceeds the recommended limit.
The hypotheses are
H0: µ= 0.4 ppm
Ha: µ> 0.4 ppm
where µ is the mean level of the chemical in tomatoes from this producer. Explain the meaning of a
Type I error.
260)
A)
A Type I error would occur if, in fact, µ= 0.4 ppm, but the results of the sampling lead to the
conclusion that µ> 0.4 ppm
B)
A Type I error would occur if, in fact, µ> 0.4 ppm, and the results of the sampling lead to
rejection of the null hypothesis that µ= 0.4 ppm.
C)
A Type I error would occur if, in fact, µ> 0.4 ppm, but the results of the sampling fail to lead
to that conclusion.
D)
A Type I error would occur if, in fact, µ= 0.4 ppm, but the results of the sampling fail to lead
to rejection of that fact.
A sample mean, sample standard deviation, and sample size are given. Use the one–mean t–test to perform the required
hypothesis test about the mean, µ, of the population from which the sample was drawn. Use the critical–value approach.
261)
x=40.9, s =6.5, n = 15, H0: µ= 32.6, Ha: µ
32.6, = 0.05.
261)
A)
Test statistic: t =4.95. Critical values: t = ±2.145. Do not reject H0: µ= 32.6. There is not
sufficient evidence to support the claim that the mean is different from 32.6.
B)
Test statistic: t =4.95. Critical values: t = ±1.96. Reject H0: µ= 32.6. There is sufficient evidence
to support the claim that the mean is different from 32.6.
C)
Test statistic: t =4.95. Critical values: t = ±2.145. Reject H0: µ= 32.6. There is sufficient
evidence to support the claim that the mean is different from 32.6.
D)
Test statistic: t =4.95. Critical values: t = ±1.96. Do not reject H0: µ= 32.6. There is not
sufficient evidence to support the claim that the mean is different from 32.6.
Provide an appropriate response.
262)
Suppose that you wish to perform a hypothesis test for a population mean. Suppose that the
population standard deviation is unknown and the sample size is small. If the distribution of the
variable under consideration is symmetric multimodal, should you use a t–test, a Wilcoxon
signed–rank test, or neither?
262)
A)
Wilcoxon signed–rank test
B)
t–test
C)
Neither
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
263)
In 2000, the average duration of long–distance telephone calls originating in one town was
9.4 minutes. Five years later, in 2005, a long–distance telephone company performs a hypothesis
test to determine whether the average duration of long–distance phone calls has changed from the
2000 mean of 9.4 minutes. The hypotheses are:
H0: µ= 9.4 minutes
Ha: µ 9.4 minutes
where µ is the mean duration, in 2005, of long–distance telephone calls originating in the town.
Suppose that the results of the sampling lead to nonrejection of the null hypothesis. Classify that
conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean duration of
long–distance phone calls has changed from the 2000 mean of 9.4 minutes.
263)
A)
Type I error
B)
Type II error
C)
Correct decision
B
Use a table of t–values to estimate the P–value for the specified one–mean t–test.
264)
Left–tailed test, n = 12, t = – 3.612
264)
A)
0.01 < P < 0.025
B)
P < 0.005
C)
P > 0.005
D)
0.005 < P < 0.01
B
A
For the given hypothesis test, determine the probability of a Type II error or the power, as specified.
265)
A health insurer has determined that the “reasonable and customary” fee for a certain medical
procedure is $1200. They suspect that the average fee charged by one particular clinic for this
procedure is higher than $1200. The insurer wants to perform a hypothesis test to determine
whether their suspicion is correct. Preliminary data analyses indicate that it is reasonable to apply a
z–test. The hypotheses are
H0: µ= $1200
Ha: µ> $1200.
Assume that = $180, n = 38, and the significance level is 0.10. Find the power of the test if in fact
µ= $1290.
265)
A)
0.9641
B)
0.0718
C)
0.90
D)
0.0359
Solve the problem.
266)
The mercury level in adult pacific pink salmon is known to be 1.3 ppm with a standard deviation of
0.4 ppm. A sample of 5 fish was collected from a particular river basin. The sample was analyzed
and the average mercury concentration was determined. For this sample, assume the test
hypotheses:
Ho: µ= 1.3 ppm
Ha: µ< 1.3 ppm
at the 5% level of significance. Complete the table below and use the resulting power to decide
which curve below describes the power curve for the salmon sample.
True Sample Mean, µ P Power
1.1
266)
A)
Curve III
B)
Curve IV
C)
Curve II
D)
Curve I
D)
Provide an appropriate response.
267)
A two–tailed hypothesis test for a population mean is to be performed at the 1% level of
significance. The population standard deviation is known. True or false: The critical values are the
two z–scores which divide the area under the standard normal curve into a middle 0.98 area and
two outside areas of 0.01.
267)
A)
True
B)
False
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
268)
A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. A
consumer advocacy group wants to perform a hypothesis test to determine whether the mean
amount is actually less than this. The hypotheses are:
H0: µ= 16.1 ounces
Ha: µ< 16.1 ounces
where µ is the mean amount of juice in the manufacturer’s 16 ounce bottles.
Suppose that the results of the sampling lead to rejection of the null hypothesis. Classify that
conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean amount of
juice, µ, is less than 16.1 ounces.
268)
A)
Type II error
B)
Correct decision
C)
Type I error
Provide an appropriate response.
269)
Suppose that you wish to perform a hypothesis test for a population mean. Suppose that the
population standard deviation is unknown, the sample size is small, and the variable under
consideration has a symmetric distribution but is not normally distributed. True or false: If a
Wilcoxon signed–rank test is used, the probability of rejecting a false null hypothesis is higher than
if a t–test is used.
269)
A)
True
B)
False
The graph portrays the decision criterion for a one–mean z–test. The curve in the graph is the normal curve for the test
statistic under the assumption that the null hypothesis is true. Use the graph to solve the problem.
270)
A graphical display of the decision criterion follows.
Determine the significance level.
270)
A)
= 2.575
B)
= 0.01
C)
z 2.575
D)
= 0.005
A hypothesis test is to be performed. Determine the null and alternative hypotheses.
271)
In the past, the mean running time for a certain type of flashlight battery has been 8.0 hours. The
manufacturer has introduced a change in the production method and wants to perform a
hypothesis test to determine whether the mean running time has changed as a result.
271)
A)
H0: µ8.0 hours
Ha: µ=8.0 hours
B)
H0: µ=8.0 hours
Ha: µ8.0 hours
C)
H0: µ8.0 hours
Ha: µ=8.0 hours
D)
H0: µ=8.0 hours
Ha: µ>8.0 hours
The graph portrays the decision criterion for a one–mean z–test. The curve in the graph is the normal curve for the test
statistic under the assumption that the null hypothesis is true. Use the graph to solve the problem.
272)
A graphical display of the decision criterion follows.
Identify the hypothesis test as two–tailed, left–tailed, or right–tailed.
272)
A)
Left–tailed
B)
Two–tailed
C)
Right–tailed
D)
There is not enough information to answer the question.
A hypothesis test is to be performed. Determine the null and alternative hypotheses.
273)
In 1990, the mean duration of long–distance telephone calls originating in one town was
7.2 minutes. A long–distance telephone company wants to perform a hypothesis test to determine
whether the mean duration of long–distance phone calls has changed from the 1990 mean of
7.2 minutes.
273)
A)
H0: µ< 7.2 minutes
Ha: µ> 7.2 minutes
B)
H0: µ= 7.2 minutes
Ha: µ 7.2 minutes
C)
H0: µ= 7.2 minutes
Ha: µ 7.2 minutes
D)
H0: µ 7.2 minutes
Ha: µ= 7.2 minutes
Explanation:
Explanation:
The graph portrays the decision criterion for a one–mean z–test. The curve in the graph is the normal curve for the test
statistic under the assumption that the null hypothesis is true. Use the graph to solve the problem.
274)
A graphical display of the decision criterion follows.
Determine the nonrejection region.
274)
A)
z = – 2.33 or z = 2.33
B)
–2.33
z
2.33
C)
z 0.01
D)
z –2.33 or z
2.33
Provide an appropriate response.
275)
True or false: If it is important to reject a false null hypothesis, the probability should be small.
275)
A)
True
B)
False
A sample mean, sample size, and population standard deviation are given. Use the one–mean z–test to perform the
required hypothesis test at the given significance level. Use the P–value approach.
276)
x= 94, n = 27, = 14, H0: µ = 89 , Ha: µ
89, = 0.10
276)
A)
z = 1.86; P–value = 0.0314; reject H0
B)
z = 0.36; P–value = 0.7188; do not reject H0
C)
z = 0.36; P–value = 0.3594; do not reject H0
D)
z = 1.86; P–value = 0.0628; reject H0
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
277)
At one school, in 2005, the average amount of time that tenth–graders spent watching television
each week was 21.6 hours. The principal introduced a campaign to encourage the students to
watch less television. One year later, in 2006, the principal performed a hypothesis test to determine
whether the average amount of time spent watching television per week had decreased. The
hypotheses were:
H0: µ= 21 hours
Ha: µ< 21 hours
where µ is the mean amount of time, in 2006, that tenth–graders spend watching television each
week.
Suppose that the results of the sampling lead to nonrejection of the null hypothesis. Classify that
conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean amount of
time, µ, spent watching television had not decreased.
277)
A)
Type II error
B)
Type I error
C)
Correct decision
Provide an appropriate response.
278)
Suppose that you wish to perform a hypothesis test for a population mean. Suppose that the
population standard deviation is unknown, the sample size is small, and the population is
normally distributed. True or false: If a Wilcoxon signed–rank test is used the probability of making
a Type II error will be smaller than if a t–test is used.
278)
A)
True
B)
False
The P–value for a hypothesis test is given. Determine whether the strength of the evidence against the null hypothesis is
weak/none, moderate, strong, or very strong.
279)
P = 0.05
279)
A)
Very strong
B)
Moderate
C)
Weak or none
D)
Strong