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.
126)
A graphical display of the decision criterion follows.
Determine the rejection region.
126)
A)
–2.33
z
2.33
B)
z –2.33 or z
2.33
C)
z 0.01
D)
z = – 2.33 or z = 2.33
Solve the problem.
127)
Silicon Stuff is a discount home electronics reseller with stores located nationwide. Silicon Stuff
managers know that a salesperson in the computer department will sell on average 3.2 computers
each day, with a standard deviation of 1.1 computers per day. The store manager at the Booleville
store decides to investigate the performance of his 3 computer salespersons. He assumes the test
hypotheses:
Ho: µ= 3.2 computers per day
Ha: µ> 3.2 computers per day
and he decides to test 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 Booleville store.
True Sample Mean, µ P Power
3.9
127)
A)
Curve IV
B)
Curve III
C)
Curve II
D)
Curve I
Find the critical value(s) for the specified Wilcoxon signed–rank test.
128)
Sample size = 24, left–tailed test, significance level = 0.05
128)
A)
92
B)
81
C)
219
D)
208
129)
Sample size = 17, left–tailed test, significance level = 0.01
129)
A)
23
B)
28
C)
130
D)
135
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.
130)
P = 0.075
130)
A)
Moderate
B)
Very strong
C)
Strong
D)
Weak or none
Find the critical value(s) for the specified Wilcoxon signed–rank test.
131)
Sample size = 21, right–tailed test, significance level = 0.01
131)
A)
188
B)
43
C)
49
D)
182
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.
132)
x= 24.4, s = 9.2, n = 25, H0: µ= 26, Ha: µ< 26, = 0.05
132)
A)
Test statistic: t = – 4.35. Critical value: t = – 1.711. Reject H0. There is sufficient evidence to
conclude that the mean is less than 26.
B)
Test statistic: t = – 4.35. Critical value: t = – 1.708. Reject H0. There is sufficient evidence to
conclude that the mean is less than 26.
C)
Test statistic: t = – 0.87. Critical value: t = – 1.711. Do not reject H0. There is not sufficient
evidence to conclude that the mean is less than 26.
D)
Test statistic: t = – 0.87. Critical value: t = – 1.708. Do not reject H0. There is not sufficient
evidence to conclude that the mean is less than 26.
The significance level and P–value of a hypothesis test are given. Decide whether the null hypothesis should be rejected.
133)
= 0.01, P–value = 0.02
133)
A)
Reject the null hypothesis.
B)
Do not reject the null hypothesis.
Classify the hypothesis test as two–tailed, left–tailed, or right–tailed.
134)
In 1990, the average 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 average duration of long–distance phone calls has changed from the 1990 mean of
7.2 minutes.
134)
A)
Two–tailed
B)
Left–tailed
C)
Right–tailed
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
135)
In 2000, the mean math SAT score for students at one school was 471. Five years later, in 2005, a
teacher performed a hypothesis test to determine whether the average math SAT score of students
at the school had changed from the 2000 mean of 471. The hypotheses were:
H0: µ=471
Ha: µ471
where µ is the mean math SAT score, in 2005, for students at the school
Explain the meaning of a Type I error.
135)
A)
A Type I error would occur if, in fact, µ=471, but the results of the sampling lead to the
conclusion that µ471
B)
A Type I error would occur if, in fact, µ471, and the results of the sampling lead to that
conclusion.
C)
A Type I error would occur if, in fact, µ471, but the results of the sampling fail to lead to
that conclusion.
D)
A Type I error would occur if, in fact, µ=471, but the results of the sampling do not lead to
rejection of that fact
136)
In the past, the mean running time for a certain type of flashlight battery has been 9.7 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 increased as a result. The
hypotheses are:
H0: µ=9.7 hours
Ha: µ>9.7 hours
where µ is the mean running time of the new batteries . Explain the meaning of a Type I error.
136)
A)
A Type I error would occur if, in fact, µ>9.7 hours, but the results of the sampling fail to lead
to that conclusion.
B)
A Type I error would occur if, in fact, µ>9.7 hours, but the results of the sampling lead to the
conclusion that µ<9.7 hours.
C)
A Type I error would occur if, in fact, µ=9.7 hours, but the results of the sampling lead to the
conclusion that µ>9.7 hours.
D)
A Type I error would occur if, in fact, µ=9.7 hours, but the results of the sampling do not lead
to rejection of that fact.
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.
137)
A two–tailed test:
z = 1.31
137)
A)
0.8098
B)
0.9049
C)
0.0951
D)
0.1902
Provide an appropriate response.
138)
A one–sample z–test for a population mean is to be performed. Let z 0 denote the observed value
of the test statistic, z. True or false: For a left–tailed test, the P–value is the area under the standard
normal curve to the left of z0.
138)
A)
True
B)
False
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.
139)
A normal probability plot and a stem–and–leaf diagram of the data are given below. is
unknown. Which procedure is appropriate?
139)
A)
Wilcoxon signed–rank test
B)
z–test
C)
t–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.
140)
A right–tailed test:
z = 0.75
140)
A)
0.4532
B)
0.7734
C)
0.5468
D)
0.2266
D
A
141)
A right–tailed test:
z = 2.38
141)
A)
0.0087
B)
0.9826
C)
0.9913
D)
0.0174
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.
142)
x= 137, s = 14.2, n = 20, H0: µ= 132, Ha: µ
132, = 0.1
142)
A)
Test statistic: t = 0.35. Critical values: t = ±1.645. Do not reject H0. There is not sufficient
evidence to conclude that the mean is different from 132.
B)
Test statistic: t = 1.57. Critical values: t = ±1.645. Do not reject H0. There is not sufficient
evidence to conclude that the mean is different from 132.
C)
Test statistic: t = 0.35. Critical values: t = ±1.729. Do not reject H0. There is not sufficient
evidence to conclude that the mean is different from 132.
D)
Test statistic: t = 1.57. Critical values: t = ±1.729. Do not reject H0. There is not sufficient
evidence to conclude that the mean is different from 132.
Classify the hypothesis test as two–tailed, left–tailed, or right–tailed.
143)
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.
143)
A)
Two–tailed
B)
Left–tailed
C)
Right–tailed
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
144)
In the past, the mean running time for a certain type of flashlight battery has been 9.0 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 increased as a result. The
hypotheses are:
H0: µ=9.0 hours
Ha: µ>9.0 hours
where µ is the mean running time of the new batteries
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 running time
has not increased.
144)
A)
Correct decision
B)
Type I error
C)
Type II error
Use a table of t–values to estimate the P–value for the specified one–mean t–test.
145)
Right–tailed test, n = 23, t = 1.571
145)
A)
P > 0.10
B)
0.01 < P < 0.025
C)
0.025 < P < 0.05
D)
0.05 < P < 0.10
Explanation:
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
146)
In the past, the mean running time for a certain type of flashlight battery has been 8.8 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 increased as a result. The
hypotheses are:
H0: µ=8.8 hours
Ha: µ>8.8 hours
where µ is the mean running time of the new batteries . Explain the meaning of a Type II error.
146)
A)
A Type II error would occur if, in fact, µ=8.8 hours, but the results of the sampling do not
lead to rejection of that fact.
B)
A Type II error would occur if, in fact, µ>8.8 hours, but the results of the sampling lead to the
conclusion that µ<8.8 hours.
C)
A Type II error would occur if, in fact, µ>8.8 hours, but the results of the sampling fail to
lead to that conclusion.
D)
A Type II error would occur if, in fact, µ=8.8 hours, but the results of the sampling lead to the
conclusion that µ>8.8 hours.
Explanation:
Explanation:
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
147)
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. 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 vitamin C intake of adult females is 75 mg.
147)
A)
Correct decision
B)
Type II error
C)
Type I error
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.
148)
A left–tailed test:
z = – 2.65
148)
A)
0.0080
B)
0.9920
C)
0.0040
D)
0.9960
For the given hypothesis test, determine the probability of a Type II error or the power, as specified.
149)
At one school, the average amount of time that tenth–graders spend watching television each week
is 21.6 hours. The principal introduces a campaign to encourage the students to watch less
television. One year later, the principal wants to perform a hypothesis test to determine whether
the average amount of time spent watching television per week has decreased. Preliminary data
analyses indicate that it is reasonable to apply a z–test. The hypotheses are
H0: µ= 21.6 hours
Ha: µ< 21.6 hours.
Assume that = 7.5 hours, n = 50, and the significance level is 0.05. Find the power of the test if in
fact µ= 21 hours.
149)
A)
0.1401
B)
0.8599
C)
0.2802
D)
0.95
Provide an appropriate response.
150)
A hypothesis test for a population mean is to be performed at the 1% level of significance. True or
false: If the null hypothesis is true, the probability that the test statistic will fall in the rejection
region is 0.01.
150)
A)
True
B)
False
For the given hypothesis test, determine the probability of a Type II error or the power, as specified.
151)
The average diastolic blood pressure of a population of migraine sufferers is 90 mmHg. During a
clinical trial each person in a random sample from the population receives a drug to relieve the
migraines. The researcher is interested in investigating whether the drug also affects blood pressure
and wants to perform a hypothesis test to determine whether the average diastolic blood pressure
for people receiving the drug (µ) differs from 90 mmHg. Preliminary data analyses indicate that it
is reasonable to apply a z–test. The hypotheses are
H0: µ= 90 mmHg
Ha: µ 90 mmHg.
Assume that = 4.8 mmHg, n = 65, and the significance level is 0.05. Find the power of the test if in
fact µ= 91.5 mmHg.
151)
A)
0.4176
B)
0.2912
C)
0.9500
D)
0.7088
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.
152)
x= 251, n = 15, = 19, H0: µ = 250 , Ha: µ > 250 , = 0.05
152)
A)
z = 0.20; P–value = 0.4207; reject H0
B)
z = 0.20; P–value = 0.5793; reject H0
C)
z = 0.20; P–value = 0.4207; do not reject H0
D)
z = 0.20; P–value = 0.5793; do not reject H0
Classify the hypothesis test as two–tailed, left–tailed, or right–tailed.
153)
The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.5 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.
153)
A)
Two–tailed
B)
Left–tailed
C)
Right–tailed
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.
154)
A graphical display of the decision criterion follows.
Determine the critical value(s).
154)
A)
z –2.05 or z
2.05
B)
= 0.04
C)
z = ±2.05
D)
z = 2.05
Provide an appropriate response.
155)
A one–sample z–test for a population mean is to be performed. Let z 0 denote the observed value
of the test statistic, z. Assume that a two–tailed test is being performed. True or false: If z0 is
negative, the P–value is twice the area under the standard normal curve to the right of z0.
155)
A)
True
B)
False
71
For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.
156)
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. Explain the meaning of
a Type I error.
156)
A)
A Type I error would occur if, in fact, µ= 16.1 ounces, but the results of the sampling lead to
the conclusion that µ< 16.1 ounces.
B)
A Type I error would occur if, in fact, µ< 16.1 ounces, but the results of the sampling lead to
the conclusion that µ> 16.1 ounces.
C)
A Type I error would occur if, in fact, µ= 16.1 ounces, but the results of the sampling do not
lead to rejection of that fact.
D)
A Type I error would occur if, in fact, µ< 16.1 ounces, but the results of the sampling fail to
lead to that conclusion.
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.
157)
x= 78, n = 28, = 11, H0: µ = 72 , Ha: µ > 72 , = 0.01
157)
A)
z = 0.55; P–value = 0.2912; do not reject H0
B)
z = 0.55; P–value = 0.2912; reject H0
C)
z = 2.89; P–value = 0.0019; reject H0
D)
z = 2.89; P–value = 0.0039; reject H0
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.
158)
P = 0.057
158)
A)
Very strong
B)
Strong
C)
Moderate
D)
Weak or none
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.
159)
x=7, s =1.8, n = 18, H0: µ= 10, Ha: µ< 10, = 0.01
159)
A)
Test statistic: t = – 7.07. Critical value: t = – 2.33. Do not reject H0. There is not sufficient
evidence to support the claim that the mean is less than 10 minutes.
B)
Test statistic: t = – 7.07. Critical value: t = – 2.552. Reject H0. There is sufficient evidence to
support the claim that the mean is less than 10 minutes.
C)
Test statistic: t = – 7.07. Critical value: t = – 2.33. Reject H0. There is sufficient evidence to
support the claim that the mean is less than 10 minutes.
D)
Test statistic: t = – 7.07. Critical value: t = – 2.567. Reject H0. There is sufficient evidence to
support the claim that the mean is less than 10 minutes.
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.
160)
After training intensively for six months, John hopes that his mean time to run 100 meters has
decreased from last year’s mean time of 11.9 seconds. He performs a hypothesis test to determine
whether his mean time has decreased. Preliminary data analyses indicate that it is reasonable to
apply a z–test. The hypotheses are
H0: µ=11.9 seconds
Ha: µ<11.9 seconds.
Assume that =0.33 seconds, n =26, and the significance level is 0.10. Express the decision
criterion for the hypothesis test in terms of x.
160)
A)
Reject H0 if x<11.48 seconds.
B)
Reject H0 if x>11.82 seconds.
C)
Reject H0 if x>12.01 seconds or x<11.79 seconds.
D)
Reject H0 if x<11.82 seconds.
Determine the critical value(s) for a one–mean z–test.
161)
A right–tailed test with = 0.03.
161)
A)
2.17
B)
1.88
C)
–1.88
D)
–2.17, 2.17
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.
162)
x= 84.5, s = 11.2, n = 16, H0: µ= 80, Ha: µ< 80, = 0.01.
162)
A)
Test statistic: t = 1.61. P–value = 0.9463. Do not reject the null hypothesis. There is not
sufficient evidence to conclude that the mean is less than 80. The evidence against the null
hypothesis is weak or none.
B)
Test statistic: t = 1.61. P–value = 0.9356. Do not reject the null hypothesis. There is not
sufficient evidence to conclude that the mean is less than 80. The evidence against the null
hypothesis is weak or none.
C)
Test statistic: t = 1.61. P–value = 0.0644. Do not reject the null hypothesis. There is not
sufficient evidence to conclude that the mean is less than 80. The evidence against the null
hypothesis is moderate.
D)
Test statistic: t = 1.61. P–value = 0.9356. Reject the null hypothesis. There is sufficient
evidence to conclude that the mean is less than 80. The evidence against the null hypothesis is
very strong.
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.
163)
A normal probability plot and a histogram plot of the data are given below. is unknown. Which
procedure is appropriate?
163)
74
A)
t–test
B)
Wilcoxon signed–rank test
C)
z–test
D)
None of these tests is appropriate.
A hypothesis test is to be performed. Determine the null and alternative hypotheses.
164)
The mean credit card debt among households in one state is $8400. A hypothesis test is to be
performed to decide whether the mean credit card debt for households in the formerly affluent
town of Rich–No–More differs from the mean credit card debt for the state
164)
A)
H0: µ $8400
Ha: µ= $8400
B)
H0: µ< $8400
Ha: µ> $8400
C)
H0: µ= $8400
Ha: µ $84000
D)
H0: µ= $8400
Ha: µ> $8400
C
Classify the hypothesis test as two–tailed, left–tailed, or right–tailed.
165)
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.
165)
A)
Two–tailed
B)
Left–tailed
C)
Right–tailed
B
D
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.
166)
In 1990, the average duration of long–distance telephone calls originating in one town was
7.8 minutes. 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 1990 mean of
7.8 minutes. Preliminary data analyses indicate that it is reasonable to apply a z–test. The
hypotheses are
H0: µ=7.8 minutes
Ha: µ7.8 minutes.
Assume that =5.2 minutes, n =77, and the significance level is 0.05. Express the decision criterion
for the hypothesis test in terms of x.
166)
A)
Reject H0 if x>8.96 minutes.
B)
Reject H0 if x>8.96 minutes or x<6.64 minutes.
C)
Reject H0 if x<6.64 minutes.
D)
Reject H0 if 6.64 minutes <x<8.96 minutes.
Find the critical value(s) for the specified Wilcoxon signed–rank test.
167)
Sample size = 22, two–tailed test, significance level = 0.10
167)
A)
72, 178
B)
66, 187
C)
75, 187
D)
75, 178
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.
168)
A two–tailed test:
z = 2.83
168)
A)
0.0046
B)
0.9954
C)
0.9977
D)
0.0023
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.
169)
x= 3.7, n = 32, = 1.8, H0: µ = 4.2 , Ha: µ < 4.2 , = 0.05
169)
A)
z = – 1.57; P–value = 0.0582; reject H0
B)
z = – 1.57; P–value = 0.1164; reject H0
C)
z = – 1.57; P–value = 0.0582; do not reject H0.
D)
z = – 1.57; P–value = 0.1164; do not reject H0
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.
170)
x=3.1 , s =0.52, n = 9, H0: µ= 2.85, Ha: µ> 2.85, = 0.01
170)
A)
Test statistic: t =1.44. Critical value: t =2.896. Reject H0. There is sufficient evidence to
support the claim that the mean is greater than 2.85.
B)
Test statistic: t =1.44. Critical value: t =2.33. Do not reject H0. There is not sufficient evidence
to support the claim that the mean is greater than 2.85.
C)
Test statistic: t =1.44. Critical value: t =2.896. Do not reject H0. There is not sufficient
evidence to support the claim that the mean is greater than 2.85.
D)
Test statistic: t =1.44. Critical value: t =2.821. Do not reject H0. There is not sufficient
evidence to support the claim that the mean is greater than 2.85.
Use a table of t–values to estimate the P–value for the specified one–mean t–test.
171)
Left–tailed test, n = 4, t = – 1.527
171)
A)
P < 0.10
B)
0.05 < P < 0.10
C)
0.025 < P < 0.05
D)
P > 0.10
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.
172)
x= 21, n = 18, = 7, H0: µ = 24 , Ha: µ < 24 , = 0.05
172)
A)
z = – 0.43; P–value = 0.0336; reject H0
B)
z = – 1.82; P–value = 0.0688; do not reject H0
C)
z = – 1.82; P–value = 0.0344; reject H0
D)
z = – 0.43; P–value = 0.3336; do not reject H0
A hypothesis test is to be performed. Determine the null and alternative hypotheses.
173)
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.
173)
A)
H0: µ= $1200
Ha: µ< $1200
B)
H0: µ= $1200
Ha: µ> $1200
C)
H0: µ> $1200
Ha: µ= $1200
D)
H0: µ= $1200
Ha: µ $1200
Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.
174)
In the past, the mean running time for a certain type of flashlight battery has been 9.2 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 increased as a result. The
hypotheses are:
H0: µ=9.2 hours
Ha: µ>9.2 hours
where µ is the mean running time of the new batteries
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 running time
has increased.
174)
A)
Type I error
B)
Correct decision
C)
Type II error
78
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.
175)
A normal probability plot and a stem–and–leaf diagram of the data are given below. is known.
Which procedure is appropriate?
175)
A)
Wilcoxon signed–rank test
B)
z–test
C)
t–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.
176)
A right–tailed test:
z = – 1.89
176)
A)
0.9412
B)
0.0294
C)
0.9706
D)
0.0588
C
B
Classify the hypothesis test as two–tailed, left–tailed, or right–tailed.
177)
At one school, the average amount of time that tenth–graders spend watching television each week
is 21.6 hours. The principal introduces a campaign to encourage the students to watch less
television. One year later, the principal wants to perform a hypothesis test to determine whether
the average amount of time spent watching television per week has decreased from the previous
mean of 21.6 hours.
177)
A)
Two–tailed
B)
Left–tailed
C)
Right–tailed
Provide an appropriate response.
178)
A one–sample z–test for a population mean is to be performed. True or false: The larger the
P–value, the stronger the evidence against the null hypothesis?
178)
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.
179)
A graphical display of the decision criterion follows.
Determine the significance level.
179)
A)
z –2.05 or z
2.05
B)
z = ±2.05
C)
= 0.02
D)
= 0.04