Download Chapter 6 Inferences Based on a Single Sample: Tests of Hypothesis

Chapter 6
Inferences Based on a Single Sample:
Tests of Hypothesis
6.1
The null hypothesis is the "status quo" hypothesis, while the alternative hypothesis is the research hypothesis.
6.2
The test statistic is used to decide whether or not to reject the null hypothesis in favor of the alternative
hypothesis.
6.3
The "level of significance" of a test is α. This is the probability that the test statistic will fall in the
rejection region when the null hypothesis is true.
6.4
A Type I error is rejecting the null hypothesis when it is true.
A Type II error is accepting the null hypothesis when it is false.
α = the probability of committing a Type I error.
β = the probability of committing a Type II error.
6.5
The four possible results are:
1.
Rejecting the null hypothesis when it is true. This would be a Type I error.
2.
Accepting the null hypothesis when it is true. This would be a correct decision.
3.
Rejecting the null hypothesis when it is false. This would be a correct decision.
4.
Accepting the null hypothesis when it is false. This would be a Type II error.
6.6
We can compute a measure of reliability for rejecting the null hypothesis when it is true. This measure of
reliability is the probability of rejecting the null hypothesis when it is true which is α. However, it is
generally not possible to compute a measure of reliability for accepting the null hypothesis when it is false.
We would have to compute the probability of accepting the null hypothesis when it is false, β, for every
value of the parameter in the alternative hypothesis.
6.7
When you reject the null hypothesis in favor of the alternative hypothesis, this does not prove the
alternative hypothesis is correct. We are 100(1 − α)% confident that there is sufficient evidence to
conclude that the alternative hypothesis is correct.
If we were to repeatedly draw samples from the population and perform the test each time, approximately
100(1 − α)% of the tests performed would yield the correct decision.
6.8
a.
b.
311
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312
Chapter 6
c.
d.
e.
f.
6.9
g.
P(z > 1.96) = .025
P(z > 1.645) = .05
P(z > 2.575) = .005
P(z < −1.28) = .1003
P(z < −1.645 or z > 1.645) = .10
P(z < −2.575 or z > 2.575) = .01
a.
Let p = proportion of college presidents who believe that their online education courses are as good as
or superior to courses that utilize traditional face-to-face instruction. The null hypothesis would be:
H0: p = .60
6.10
b.
The rejection region requires α/2 = .01/2 = .005 in each tail if the z-distribution. From Table IV,
Appendix B, z.005 = 2.575. The rejection region for a two-tailed test is z < − 2.575 or z > 2.575.
a.
Let μ = average gain in green fees, lessons, or equipment expenditures for participating golf
facilities. The null and alternative hypotheses would be:
H0: μ = $2,400
Ha: μ > $2,400
b.
The α = .05 is the Type I error rate. This means that the probability of concluding that the average
gain in fees, lessons, or equipment expenditures for participation golf facilities exceeds $2,400 when
in fact, the average is $2,400 is .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
c.
6.11
313
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Let p = student loan default rate in this year. To see if the student loan default rate is less than .045, we
test:
H0: p = .045
Ha: p < .045
6.12
Let p = proportion of U.S. companies that have formal, written travel and entertainment policies for their
employees. The null hypothesis would be:
H0: p = .80
6.13
Let μ = mean caloric content of Virginia school lunches. To test the claim that after the testing period
ended, the average caloric content dropped, we test:
H0: μ = 863
Ha: μ < 863
6.14
Let μ = average Libor rate for 3-month loans. Since many Western banks think that the reported average
Libor rate (.054) is too high, they want to show that the average is less than .054. The appropriate
hypotheses would be:
H0: μ = .054
Ha: μ < .054
6.15
6.16
a.
Since the company must give proof the drug is safe, the null hypothesis would be the drug is unsafe.
The alternative hypothesis would be the drug is safe.
b.
A Type I error would be concluding the drug is safe when it is not safe. A Type II error would be
concluding the drug is not safe when it is. α is the probability of concluding the drug is safe when it
is not. β is the probability of concluding the drug is not safe when it is.
c.
In this problem, it would be more important for α to be small. We would want the probability of
concluding the drug is safe when it is not to be as small as possible.
a.
A Type I error would be concluding the proposed user is unauthorized when, in fact, the proposed
user is authorized.
A Type II error would be concluding the proposed user is authorized when, in fact, the proposed user
is unauthorized.
In this case, a more serious error would be a Type II error. One would not want to conclude that the
proposed user is authorized when he/she is not.
b.
The Type I error rate is 1%. This means that the probability of concluding the proposed user is
unauthorized when, in fact, the proposed user is authorized is .01.
The Type II error rate is .00025%. This means that the probability of concluding the proposed user is
authorized when, in fact, the proposed user is unauthorized is .0000025.
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Chapter 6
c.
The Type I error rate is .01%. This means that the probability of concluding the proposed user is
unauthorized when, in fact, the proposed user is authorized is .0001.
The Type II error rate is .005%. This means that the probability of concluding the proposed user is
authorized when, in fact, the proposed user is unauthorized is .00005.
6.17
a.
A Type I error is rejecting the null hypothesis when it is true. In a murder trial, we would be
concluding that the accused is guilty when, in fact, he/she is innocent.
A Type II error is accepting the null hypothesis when it is false. In this case, we would be
concluding that the accused is innocent when, in fact, he/she is guilty.
6.18
6.19
b.
Both errors are bad. However, if an innocent person is found guilty of murder and is put to death,
there is no way to correct the error. On the other hand, if a guilty person is set free, he/she could
murder again.
c.
In a jury trial, α is assumed to be smaller than β. The only way to convict the accused is for a
unanimous decision of guilt. Thus, the probability of convicting an innocent person is set to be
small.
d.
In order to get a unanimous vote to convict, there has to be overwhelming evidence of guilt. The
probability of getting a unanimous vote of guilt if the person is really innocent will be very small.
e.
If a jury is predjuced against a guilty verdict, the value of α will decrease. The probability of
convicting an innocent person will be even smaller if the jury if predjudiced against a guilty verdict.
f.
If a jury is predjudiced against a guilty verdict, the value of β will increase. The probability of
declaring a guilty person innocent will be larger if the jury is prejudiced against a guilty verdict.
a.
The null hypothesis is: Ho: There is no intrusion.
b.
The alternative hypothesis is: Ha: There is an intrusion.
c.
α = P(warning | no intrusion) =
1
 .001 .
1000
β = P(no warning | intrusion) =
500
 .5 .
1000
a.
The decision rule is to reject H0 if x > 270. Recall that
z=
x  μ0
σx
Therefore, reject H0 if x > 270
can be written reject H0 if
z>
z>
x  μ0
σx
270  255
63 / 81
z > 2.14
The decision rule in terms of z is to reject H0 if z > 2.14.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.20
b.
P(z > 2.14) = .5 − P(0 < z < 2.14)
= .5 − .4838
= .0162
a.
H0: μ = 100
Ha: μ > 100
The test statistic is z =
x  μ0
σx

x  μ0
σ/ n

315
110  100
60 / 100
= 1.67
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region, (z = 1.67  1.645), H0 is
rejected. There is sufficient evidence to indicate the true population mean is greater than 100 at
α = .05.
b.
H0: μ = 100
Ha: μ ≠ 100
The test statistic is z =
x  μ0
σx

110  100
60 / 100
= 1.67
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic does not fall in the rejection region, (z = 1.67  1.96),
H0 is not rejected. There is insufficient evidence to indicate μ does not equal 100 at α = .05.
6.21
c.
In part a, we rejected H0 and concluded the mean was greater than 100. In part b, we did not reject
H0. There was insufficient evidence to conclude the mean was different from 100. Because the
alternative hypothesis in part a is more specific than the one in b, it is easier to reject H0.
a.
H0: μ = .36
Ha: μ < .36
The test statistic is z =
x  μ0
σx

.323  .36
.034 / 64
= −1.61
The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.10 = 1.28. The rejection region is z < −1.28.
Since the observed value of the test statistic falls in the rejection region (z = −1.61 < −1.28), H0 is
rejected. There is sufficient evidence to indicate the mean is less than .36 at α = .10.
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Chapter 6
b.
H0: μ = .36
Ha: μ ≠ .36
The test statistic is z = −1.61 (see part a).
The rejection region requires α/2 = .10/2 = .05 in the each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = −1.61 
−1.645), H0 is not rejected. There is insufficient evidence to indicate the mean is different from .36
at α = .10.
6.22
a.
Let μ = mean Mach rating score for all purchasing managers. To determine if the mean Mach rating
score is different from 85, we test:
Ho: µ = 85
Ha: µ ≠ 85
6.23
6.24
b.
The rejection requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < -1.645 or z > 1.645.
c.
The test statistic is z =
d.
Since the observed value of the test statistic falls in the rejection region (z = 12.80 > 1.645), H0 is
rejected. There is sufficient evidence to indicate that the true mean Mach rating score of all
purchasing managers is not 85 at α = .10.
a.
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z < −2.33.
b.
The test statistic is z 
c.
Since the observed value of the test statistics does not fall in the rejection region (z = −.40 </ −2.33),
H0 is not rejected. There is insufficient evidence to indicate the true mean number of latex gloves
used per week by all hospital employees is less than 20 at α = .01.
a.
To determine whether the mean July, 2006 dealer price of the Toyota Prius differs
from $25,000, we test:
x − μo
σx
x  μo
σx
≈

99.6 − 85
12.6 / 122
19.3  20
11.9
46
= 12.80
 .40
H0: μ = 25,000
Ha: μ ≠ 25,000
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Inferences Based on a Single Sample: Tests of Hypothesis
b.
The sample mean is x 
x
i
n

317
4,076,271
 25, 476.69
160
The sample variance is:
s2 

xi2
 x 

2
i
n
n 1

104,788,653,115 
160  1
4,076,2712
160
 5, 904, 057.862
The sample standard deviation is: s  s 2  5, 904, 057.862  2, 429.8267
6.25
x  μo
The test statistic is z 
d.
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From
Table IV, Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
e.
Since the observed value of the test statistic falls in the rejection region (z = 2.48 > 1.96),
Ho is rejected. There is sufficient evidence to indicate the mean July, 2006 dealer price of
the Toyota Prius differs from $25,000 at α = .05.
σx

25, 476.69  25, 000
c.
2, 429.8267
160
 2.48
Let μ = mean lacunarity measurement for all grassland pixels. To determine if the area sampled is
grassland, we test:
H0: μ = 220
Ha: μ ≠ 220
The test statistic is z 
x  μo
σx

225  220
20 / 100
 2.50 .
The rejection region requires α / 2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.575. The rejection region is z < −2.575 or z > 2.575.
Since the observed value of the test statistic does not fall in the rejection region (z = 2.50  2.575),
H0 is not rejected. There is insufficient evidence to conclude that the area sampled is not grassland at
α = .01.
6.26
a.
A Type I error is rejecting H0 when H0 is true. In this case, we would conclude that the mean number
of carats per diamond is different from .6 when, in fact, it is equal to .6.
A Type II error is accepting H0 when H0 is false. In this case, we would conclude that the mean
number of carats per diamond is equal to .6 when, in fact, it is different from .6.
b.
From Exercise 5.18, the random sample of 30 diamonds yielded x = .691 and s = .262.
Let μ =
mean number of carats per diamond. To determine if the mean number of carats per diamond is
different from .6, we test:
H0: μ = .6
Ha: μ ≠ .6
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Chapter 6
x  μ0
The test statistic is z 
σx

.691  .6
.262
30
 1.90
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z > 1.96 or z < −1.96.
Since the observed value of the test statistic does not fall in the rejection region
(z = 1.90  1.96), H0 is not rejected. There is insufficient evidence to indicate the mean number of
carats per diamond is different from .6 carats at α = .05.
c.
When α is changed, H0, Ha, and the test statistic remain the same.
The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645 or z < −1.645.
Since the observed value of the test statistic falls in the rejection region
(z = 1.90 > 1.645), H0 is rejected. There is sufficient evidence to indicate the mean number of carats
per diamond is different from .6 carats at α = .10.
6.27
d.
When the value of α changes, the decision can also change. Thus, it is very important to include the
level of α used in all decisions.
a.
To determine if the process is not operating satisfactorily, we test:
H0: μ = .250
Ha: μ ≠ .250
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tees
Variable
Tees
N
40
Mean
0.25248
Median
0.25300
TrMean
0.25256
Variable
Tees
Minimum
0.24700
Maximum
0.25600
Q1
0.25100
Q3
0.25400
The test statistic is z =
x  μ0
σx

StDev
0.00223
SE Mean
0.00035
.25248  .250
= 7.03
.00223/ 40
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.575. The rejection region is z < −2.575 or z > 2.575.
Since the observed value of the test statistic falls in the rejection region (z = 7.03 > 2.575), H0 is
rejected. There is sufficient information to indicate the process is performing in an unsatisfactory
manner at α = .01.
c.
α is the probability of a Type I error. A Type I error, in this case, is to say the process is
unsatisfactory when, in fact, it is satisfactory. The risk, then, is to the producer since he will be
spending time and money to repair a process that is not in error.
β is the probability of a Type II error. A Type II error, in this case, is to say the process is
satisfactory when it, in fact, is not. This is the consumer's risk since he could unknowingly purchase
a defective product.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.28
319
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: GASTURBINE
Variable
GASTURBINE
N
67
N*
0
Mean
11066
Variable
GASTURBINE
Maximum
16243
SE Mean
195
StDev
1595
Minimum
8714
Q1
9918
Median
10656
Q3
11842
To determine if the mean heat rate of gas turbines augmented with high pressure inlet fogging exceeds
10,000 kJ/kWh, we test:
H0: μ = 10,000
H0: μ > 10,000
The test statistic is z 
x  μo
σx

11, 066  10, 000
1, 595
67
 5.47
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B,
z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistics falls in the rejection region (z = 5.47 > 1.645),
H0 is rejected. There is sufficient evidence to indicate the true mean heat rate of gas turbines augmented
with high pressure inlet fogging exceeds 10,000 kJ/kWh at α = .05.
6.29
To determine if the mean point-spread error is different from 0, we test:
H0: μ = 0
Ha: μ ≠ 0
The test statistic is z 
x  μ0
σx

1.6  0
13.3 / 240
= −1.86
The rejection region requires α/2 = .01/2 = .005 in each tail of the z distribution. From Table IV, Appendix
B, z.005 = 2.575. The rejection region is z > 2.575 or z < −2.575.
Since the observed value of the test statistic does not fall in the rejection region (z = −1.86  2.575), H0 is
not rejected. There is insufficient evidence to indicate that the true mean point-spread error is different
from 0 at α = .01.
6.30
a.
Let μ = average full-service fee (in thousands of dollars) of U.S. funeral homes in 2009. To
determine if the average full-service fee exceeds $6,500, we test:
H0: μ = 6.50
Ha: μ > 6.50
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Chapter 6
b.
Using MINTAB, the output is:
Descriptive Statistics: FUNERAL
Variable
Fee
Variable
Fee
N
36
Mean
6.819
Minimum
5.200
Median
6.600
Maximum
11.600
StDev
1.265
Q1
6.025
SE Mean
0.211
Q3
7.400
H0: μ = 6.50
Ha: μ > 6.50
The test statistic is z 
x  μ0
σx

6.819  6.50
1.265
36
 1.51
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region
(z = 1.51  1.645), H0 is not rejected. There is insufficient evidence to indicate the true mean fullservice fee of U.S. funeral homes in 2009 exceeds $6,500 at α = .05.
6.31
c.
No. Since the sample size (n = 36) is greater than 30, the Central Limit Theorem applies. The
distribution of x is approximately normal regardless of the population distribution.
a.
No. Since the hypothesized value of μM (60,000) falls in the 95% confidence interval, it is a likely
candidate for the true mean. Thus, we would not reject H0. There is no evidence that the mean
salary for males differs from $60,000.
b.
To determine if the true mean salary of males with post-graduate degrees differs from $60,000, we
test:
H0: μ = 60,000
Ha: μ ≠ 60,000
The test statistic is z 
x  μ0
σx

61, 340  60, 000
 0.61
2,185
The rejection region requires α/2 = .05/2 = .025 in each tail of the z -distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z > 1.96 or z < −1.96.
Since the observed value of the test statistic does not fall in the rejection region
(z = 0.61  1.96), H0 is not rejected. There is insufficient evidence to indicate the true mean salary
of males with post-graduate degrees differs from $60,000 at α = .05.
c.
Parts a and b must agree. In both cases, a two-sided test / confidence interval is used. The z -score
used in both parts is the same, as are x and s x .
d.
No. Since the hypothesized value of μF (33,000) falls in the 95% confidence interval, it is a likely
candidate for the true mean. Thus, we would not reject H0. There is no evidence that the mean
salary for females differs from $33,000.
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Inferences Based on a Single Sample: Tests of Hypothesis
e.
321
To determine if the true mean salary of females with post-graduate degrees differs from
$33,000, we test:
H0: μ = 33,000
Ha: μ ≠ 33,000
x  μ0
The test statistic is z 
σx

32, 227  33, 000
 0.83
932
The rejection region requires α/2 = .05/2 = .025 in each tail of the z -distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z > 1.96 or z < −1.96.
Since the observed value of the test statistic does not fall in the rejection region
(z = −0.83  −1.96), H0 is not rejected. There is insufficient evidence to indicate the true mean
salary of females with post-graduate degrees differs from $33,000 at α = .05.
6.32
f.
Parts d and e must agree. In both cases, a two-sided test / confidence interval is used. The z-score
used in both parts is the same, as are x and s x .
a.
To determine if the sample data refute the manufacturer's claim, we test:
H0: μ = 10
Ha: μ < 10
b.
A Type I error is concluding the mean number of solder joints inspected per second is less than 10
when, in fact, it is 10 or more.
A Type II error is concluding the mean number of solder joints inspected per second is at least 10
when, in fact, it is less than 10.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: PCB
Variable
PCB
N
48
Mean
9.292
Median
9.000
TrMean
9.432
Variable
PCB
Minimum
0.000
Maximum
13.000
Q1
9.000
Q3
10.000
StDev
2.103
SE Mean
0.304
H0: μ = 10
Ha: μ < 10
The test statistic is z =
x  μ0
σx

9.292  10
2.103 / 48
= −2.33
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < −1.645.
Since the observed value of the test statistic falls in the rejection region (z = −2.33 < −1.645), H0 is
rejected. There is sufficient evidence to indicate the mean number of inspections per second is less
than 10 at α = .05.
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322
6.33
Chapter 6
a.
To determine if CEOs at all California small firms generally agree with the statement, we test:
H0: μ = 3.5
Ha: μ > 3.5
The test statistic is z 
x  μo
σx

3.85  3.5
1.5
137
 2.73
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistics falls in the rejection region (z = 2.73 > 1.645), H0 is
rejected. There is sufficient evidence to indicate CEOs at all California small firms generally agree
with the statement (true mean scale score exceeds 3.5) at α = .05.
6.34
6.35
6.36
b.
Although the sample mean of 3.85 is far enough away from 3.5 to statistically conclude the
population mean score is greater than 3.5, a score of 3.85 may not be practically different from 3.5 to
make any difference.
c.
No. Since the sample size (n = 137) is greater than 30, the Central Limit Theorem applies. The
distribution of x is approximately normal regardless of the population distribution.
We will reject H0 if the p-value < α.
a.
.06  .05, do not reject H0.
b.
.10  .05, do not reject H0.
c.
.01 < .05, reject H0.
d.
.001 < .05, reject H0.
e.
.251  .05, do not reject H0.
f.
.042 < .05, reject H0.
a.
Since the p-value = .10 is greater than α = .05, H0 is not rejected.
b.
Since the p-value = .05 is less than α = .10, H0 is rejected.
c.
Since the p-value = .001 is less than α = .01, H0 is rejected.
d.
Since the p-value = .05 is greater than α = .025, H0 is not rejected.
e.
Since the p-value = .45 is greater than α = .10, H0 is not rejected.
z=
x  μ0
σx

49.4  50
4.1 / 100
= −1.46
p-value = P(z ≥ −1.46) = .5 + .4279 = .9279
There is no evidence to reject H0 for α ≤ .10.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.37
323
p-value = P(z ≥ 2.17) = .5 − P(0 < z < 2.17) = .5 − .4850 = .0150 (using Table IV, Appendix B)
The probability of observing a test statistic of 2.17 or anything more unusual if the true mean is 100 is
.0150. Since this probability is so small, there is evidence that the true mean is greater than 100.
6.38
First, find the value of the test statistic:
z=
x  μ0
σx

10.7  10
3.1 / 50
= 1.60
p-value = P(z ≤ −1.60 or z ≥ 1.60) = 2P(z ≥ 1.60) = 2(.5 − .4452) = 2(.0548) = .1096
(using Table IV, Appendix B)
There is no evidence to reject H0 for α ≤ .10.
6.39
p-value = P(z ≥ 2.17) + P(z ≤ −2.17) = (.5 − .4850)2 = .0300 (using Table IV, Appendix B)
6.40
a.
The p-value reported by SAS is for a two-tailed test. Thus, P(z ≤ 1.63) + P(z ≥ 1.63) = .1032. For
this one-tailed test, the p-value = P(z ≤ 1.63) = .1032/2 = .0516.
Since the p-value = .0516 > α = .05, H0 is not rejected. There is insufficient evidence to indicate μ <
75 at α = .05.
b.
For this one-tailed test, the p-value = P(z ≤ 1.63). Since P( z  1.63) = .1032/2 = .0516, P(z ≤ 1.63)
= 1  .0516 = .9484.
Since the p-value = .9484 > α = .10, H0 is not rejected. There is insufficient evidence to indicate μ <
75 at α = .10.
c.
For this one-tailed test, the p-value = P(z ≥ 1.63) = .1032/2 = .0516.
Since the p-value = .0516 < α = .10, H0 is rejected. There is sufficient evidence to indicate μ > 75 at
α = .10.
d.
For this two-tailed test, the p-value = .1032.
Since the p-value = .1032 > α = .01, H0 is not rejected. There is insufficient evidence to indicate μ ≠
75 at α = .01.
6.41
The smallest value of α for which the null hypothesis would be rejected is just greater than .06.
6.42
The p-value is p = 0.014. The probability of observing a test statistic of t = 2.48 or anything more unusual
if μ = 25,000 is 0.014. Since p = 0.014 is so small, we would reject H0. There is sufficient evidence to
indicate the mean prices for hybrid Toyota Prius cars is different than $25,000 for any value of α > .014.
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324
6.43
Chapter 6
a.
Using MINITAB and the data sampled from Exercise 5.18, we get the following:
One-Sample Z: carats-samp
Test of mu = 0.6 vs mu not = 0.6
The assumed sigma = 0.262
Variable
carats-samp
Variable
carats-samp
N
30
(
Mean
0.6910
StDev
0.2620
95.0% CI
0.5972, 0.7848)
Z
1.90
SE Mean
0.0478
P
0.057
From the printout, the p-value is p = .057.
b.
Since the p-value is greater than α (.057 > .05), H0 is not rejected. There is insufficient evidence to
indicate the mean number of carats per diamond is different from .6 at α = .05.
6.44
From the printout, the p-value = .000. Since the p-value = .000 < α = .01, H0 is rejected. There is
sufficient evidence to indicate that the true population mean weight of plastic golf tees is different from
.250 at α = .01.
6.45
a.
The p-value from the printout is .0693.
b.
From the printout, the set of hypotheses used are:
H0: μ = 6.5
Ha: μ > 6.5
Since the p-value is not small (.0693), H0 is not rejected. There is insufficient evidence to indicate the
true mean full-service fee of U.S. funeral homes in 2009 exceeds $6,500 at α = .05. This agrees with
the decision in Exercise 6.30.
6.46
x  μ0

10.2  0
a.
z=
b.
For this two-sided test, the p-value = P(z ≥ 2.30) + P(z ≤ −2.30) = (.5 − .4893)
+ (.5 − .4893) = .0214. Since this value is so small, there is evidence to reject H0. There is sufficient
evidence to indicate the mean level of feminization is different from 0% for any value of α > .0214.
c.
z=
σx
x - μ0
σx

31.3 / 50
15.0  0
25.1 / 50
= 2.30
= 4.23
For this two-sided test, the p-value = P(z ≥ 4.23) + P(z ≤ −4.23) ≈ (.5 − .5) + (.5 − .5) = 0. Since this
value is so small, there is evidence to reject H0. There is sufficient evidence to indicate the mean
level of feminization is different from 0% for any value of α > 0.0.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.47
a.
325
To determine if the true mean forecast error for buy-side analysts is positive, we test:
Ho: μ = 0
Ha: μ > 0
The test statistic is z 
x  μo
σx

.85  0
1.93 / 3, 526
 26.15 .
The observed p-value of the test is p = P(z > 26.15) ≈ 0 (Using Table IV, Appendix B)
Since the p-value is less than α (p ≈ 0 < .01), H0 is rejected. There is sufficient evidence to indicate
that the true mean forecast error for buy-side analysts is positive at α = .01. This means that the
buy-side analysts are overestimating earnings.
b.
To determine if the true mean forecast error for sell-side analysts is negative; we test:
H0: μ = 0
Ha: μ < 0
The test statistic is z =
x − μo
σx
≈
−.05 − 0
.85 / 58,562
= −14.24 .
The observed p-value of the test is p = P(z < −14.24) ≈ 0 (using Table IV, Appendix B)
Since the p-value is less than α (p ≈ 0 < .01), H0 is rejected. There is sufficient evidence to indicate
that the true mean forecast error for sell-side analysts is negative at α = .01. This means that the sellside analysts are not underestimating earnings.
6.48
a.
z
x  μo
σx

52.3  51
7.1
50
 1.29
The p-value is p  P ( z  1.29)P( z  1.29)  (.5  .4015)  (.5  .4015)  .1970 .
(Using Table IV, Appendix B.)
b.
The p-value is p  P ( z  1.29) (.5  .4015)  .0985 . (Using Table IV, Appendix B.)
c.
z=
x − μo
σx
≈
52.3 − 51
10.4
50
= 0.88
The p-value is p  P ( z  0.88)P ( z  0.88)  (.5  .3106)  (.5  .3106)  .3788 .
(Using Table IV, Appendix B.)
d.
In part a, in order to reject H0, α would have to be greater than .1970. In part b, in order to reject H0,
α would have to be greater than .0985. In part c, in order to reject H0, α would have to be greater
than .3788.
e.
For a two-tailed test, α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58.
z
x  μo
σx
 2.58 
52.3  51
s
50
 2.58
s
50
 52.3  51  .3649 s  1.3  s  3.56
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326
Chapter 6
For a one-tailed test, α = .01. From Table IV, Appendix B, z.01 = 2.33.
z
6.49
6.50
6.51
x  μo
σx
 2.33 
52.3  51
s
50
 2.33
s
50
 52.3  51  .3295s  1.3  s  3.95
a.
We should use the t-distribution in testing a hypothesis about a population mean if the sample size is
small, the population being sampled from is normal, and the variance of the population is unknown.
b.
Both distributions are mound-shaped and symmetric. The t-distribution is flatter than the zdistribution.
a.
P(t > 1.440) = .10
(Using Table V, Appendix B, with df = 6)
b.
P(t < −1.782) = .05
(Using Table V, Appendix B, with df = 12)
c.
P(t < −2.060) + P(t > 2.060) = .025 + .025 = .05
(Using Table V, Appendix B, with df = 25)
d.
The probability of a Type I error is computed above for each of the parts.
a.
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 =
14 − 1 = 13. From Table V, Appendix B, t.025 = 2.160. The rejection region is t < −2.160 or t > 2.160.
b.
The rejection region requires α = .01 in the upper tail of the t-distribution with df = n − 1 = 24 − 1 =
23. From Table V, Appendix B, t.01 = 2.500. The rejection region is t > 2.500.
c.
The rejection region requires α = .10 in the upper tail of the t-distribution with df = n − 1 = 9 − 1 = 8.
From Table V, Appendix B, t.10 = 1.397. The rejection region is t > 1.397.
d.
The rejection region requires α = .01 in the lower tail of the t-distribution with df = n − 1 = 12 − 1 =
11. From Table V, Appendix B, t.01 = 2.718. The rejection region is t < −2.718.
e.
The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = n − 1 = 20 −
1 = 19. From Table V, Appendix B, t.05 = 1.729. The rejection region is t < −1.729 or t > 1.729.
f.
The rejection region requires α = .05 in the lower tail of the t-distribution with df = n − 1 = 4 − 1 = 3.
From Table V, Appendix B, t.05 = 2.353. The rejection region is t < −2.353.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.52
a.
327
H0: μ = 6
Ha: μ < 6
The test statistic is t 
x  μ0
s/ n

4.8  6
1.3 / 5
= −2.064
The necessary assumption is that the population is normal.
The rejection region requires α = .05 in the lower tail of the t-distribution with df = n − 1 = 5 − 1 = 4.
From Table V, Appendix B, t.05 = 2.132. The rejection region is t < −2.132.
Since the observed value of the test statistic does not fall in the rejection region (t = −2.064
</ −2.132), H0 is not rejected. There is insufficient evidence to indicate the mean is less than 6 at
α = .05.
b.
H0: μ = 6
H a: μ ≠ 6
The test statistic is t = −2.064 (from a).
The assumption is the same as in a.
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 = 5 −
1 = 4. From Table V, Appendix B, t.025 = 2.776. The rejection region is t < −2.776 or t > 2.776.
Since the observed value of the test statistic does not fall in the rejection region (t = −2.064
</ −2.776), H0 is not rejected. There is insufficient evidence to indicate the mean is different from 6
at α = .05.
c.
For part a, the p-value = P(t ≤ −2.064).
From Table V, with df = 4, .05 < P(t ≤ −2.064) < .10 or .05 < p-value < .10.
For part b, the p-value = P(t ≤ −2.064) + P(t ≥ 2.064).
From Table V, with df = 4, 2(.05) < p-value < 2(.10) or .10 < p-value < .20.
6.53
a.
We must assume that a random sample was drawn from a normal population.
b.
The hypotheses are:
H0: μ = 1000
Ha: μ > 1000
The test statistic is t = 1.89.
The p-value is .038.
There is evidence to reject H0 for α > .038. There is evidence to indicate the mean is greater than
1000 for α > .038.
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328
Chapter 6
c.
The hypotheses are:
H0: μ = 1000
Ha: μ ≠ 1000
The test statistic is t = 1.89.
The p-value is 2(.038) = .076.
There is no evidence to reject H0 for α = .05. There is insufficient evidence to indicate the mean is
different than 1000 for α = .05.
There is evidence to reject H0 for α > .076. There is evidence to indicate the mean is different than
1000 for α > .076.
6.54
a.
To determine if the true mean breaking strength of the new bonding adhesive is less
than 5.70 Mpa, we test:
H0: μ = 5.70
Ha: μ < 5.70
6.55
b.
The rejection region requires α = .01 in the lower tail of the t-distribution with
df = n – 1 = 10 – 1 = 9. From Table V, Appendix B, t.01 = 2.821. The rejection region is
t < −2.821.
c.
The test statistic is t 
d.
Since the observed value of the test statistic falls in the rejection region (t = −4.33 < −2.821), H0 is
rejected. There is sufficient evidence to indicate the true mean breaking strength of the new bonding
adhesive is less than 5.70 Mpa at α = .01.
e.
We must assume that the sample was random and selected from a normal population.
a.
To determine if the mean surface roughness of coated interior pipe differs from 2 micrometers, we
test:
x  μo
s
n

5.07  5.70
.46
10
 4.33 .
H0: μ = 2
Ha: μ ≠ 2
b.
From the printout, the test statistic is t = −1.02.
c.
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with
df = n – 1 = 20 – 1 = 19. From Table V, Appendix B, t.025 = 2.093. The rejection region is
t < −2.093 or t > 2.093.
d.
Since the observed value of the test statistic does not fall in the rejection region
(t = −1.02 </ −2.093), H0 is not rejected. There is insufficient evidence to indicate the true mean
surface roughness of coated interior pipe differs from 2 micrometers at α = .05.
e.
The p-value is p = .322. Since the p-value is not less than α = .05, H0 is not rejected. There is
insufficient evidence to indicate the true mean surface roughness of coated interior pipe differs from
2 micrometers at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
f.
6.56
329
From Exercise 5.30, we found the 95% confidence interval for the mean surface roughness of coated
interior pipe to be (1.636, 2.126). Since the hypothesized value of μ (μ = 2) falls in the confidence
interval, it is a likely value. We cannot reject it. The confidence interval and the test of hypothesis
lead to the same conclusion because the critical values for the 2 techniques are the same.
Some preliminary calculations are:
x
s2 =
 x  736
n

7
x2
= 105.14
 x 

n
n 1
2

(736) 2
7
= 218.4762
7 1
78696 
s=
218.4762 = 14.7809
a.
To determine if the mean consumption rate of salad dressings in the Southeastern U.S. is different
than the mean national consumption rate, we test:
H0: μ = 100
Ha: μ ≠ 100
b.
Since the sample size is so small, we must assume that the population being sampled is normal. In
addition, we must assume that the sample is random.
c.
The test statistic is t =
x  μ0
s/ n

105.14  100
14.7809 / 7
= .92
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution. From Table V,
Appendix B, with df = n − 1 = 7 − 1 = 6, t.025 = 2.447. The rejection region is
t > 2.447 or t < −2.447.
Since the value of the test statistic does not fall in the rejection region (t = .92 >/ 2.447), H0 is not
rejected. There is insufficient evidence to indicate the mean consumption rate of salad dressings in
the Southeastern U.S. is different than the mean national consumption rate at α = .05.
6.57
d.
The observed significance level is p-value = P(t ≥ .92) + P(t ≤ −.92). Since we did not reject H0 in
part c, we know that the p-value must be greater than .05. Using Table V, Appendix B, with
df = n − 1 = 7 − 1 = 6, p-value = P(t ≥ .92) + P(t ≤ −.92) > .1 + .1 = .2 Thus, with this table, we only
know that the p-value is greater than .2.
a.
To determine if the mean number of suicide bombings for all Al Qaeda attacks against the U.S.
differs from 2.5, we test:
H0: μ = 2.5
Ha: μ ≠ 2.5
From the printout, the test statistic is t = −2.46.
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330
Chapter 6
The rejection requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = n − 1 = 21 − 1 =
20. From Table V, Appendix B, t.05 = 1.725. The rejection region is t < −1.725 or t > 1.725.
Since the observed value of the test statistic falls in the rejection region (t = −2.46 < −1.725), H0 is
rejected. There is sufficient evidence to indicate that the true mean number of suicide bombings for
all Al Qaeda attacks against the US differs from 2.5 at α = .10.
In addition, we could make our decision based on the p-value given in the printout. The p-value for
the test is p = .0229. Since the p-value is less than α = .10, we reject H0.
b.
From Exercise 5.33, we found the 90% confidence interval for the mean number of suicide bombings
for all Al Qaeda attacks against the US to be (1.41, 2.31). Since the hypothesized value of
μ ( μ = 2.5) does not fall in the confidence interval, it is an unlikely value. We reject it. Therefore,
we conclude that the true mean differs from 2.5 at α = .10.
c.
The confidence interval and the test of hypothesis lead to the same conclusion because the critical
values for 2 techniques are the same.
d.
Since the sample size is small, we must assume that the population being sampled is normal. In
addition, we must assume that the sample is random.
e.
Using MINITAB, a histogram of the data is:
Histogram of bombings
12
10
Frequency
8
6
4
2
0
1
2
3
bombings
4
5
Based on the histogram, the data do not look like they are from a normal distribution. Thus, the
validity of the inference in part a is questionable.
6.58
a.
To determine whether the mean amount charged to credit and debit cards in 2007 by all banks with
recent acquisitions differs from $200 billion, we test:
H0: μ = 200
Ha: μ ≠ 200
b.
From the printout, the test statistic is t = 2.708 and the p-value is .054.
Since the p-value (.054) is not less than .05, there is no evidence to reject H0. There is insufficient
evidence to indicate that the mean amount charged to credit and debit cards in 2007 by all banks with
recent acquisitions differs from $200 billion at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.59
331
c.
Since the p-value (.054) is less than α = .10, there is evidence to reject H0. There is sufficient
evidence to indicate that the mean amount charged to credit and debit cards in 2007 by all banks with
recent acquisitions differs from $200 billion at α = .10.
d.
This sample may impact the validity of our conclusions in parts b and c in the sense that this sample
was not randomly selected since the banks sampled represent the top ranked banks in the US with
recent acquisitions. These top banks may not be representative of all banks with recent acquisitions.
Using MINITAB, the descriptive statistics are:
One-Sample T: Skid
Test of mu = 425 vs < 425
Variable
Skid
N
20
Mean
358.450
StDev
117.817
SE Mean
26.345
95%
Upper
Bound
T
404.004 -2.53
P
0.010
To determine if the mean skidding distance is less than 425 meters, we test:
H0: μ = 425
Ha: μ < 425
The test statistics is t 
x  μo
s
n

358.45  425
117.817
20
 2.53 .
The rejection region requires α = .10 in the lower tail of the t-distribution with df = n – 1 = 20 – 1 = 19.
From Table V, Appendix B, t.10 = 1.328. The rejection region is t < −1.328.
Since the observed value of the test statistic falls in the rejection region (t = −2.53 < −1.328), H0 is rejected.
There is sufficient evidence to indicate the true mean skidding distance is less than 425 meters at α = .10.
There is sufficient evidence to refute the claim.
6.60
a.
To determine if the mean repellency percentage of the new mosquito repellent is less than 95, we
test:
H0: μ = 95
Ha: μ < 95
The test statistic is t =
x  μ0
s/ n

83  95
15 / 5
= −1.79
The rejection region requires α = .10 in the lower tail of the t distribution. From Table V,
Appendix B, with df = n − 1 = 5 − 1 = 4, t.10 = 1.533. The rejection region is t < −1.533.
Since the observed value of the test statistic falls in the rejection region (t = −1.79 < −1.533), H0 is
rejected. There is sufficient evidence to indicate that the true mean repellency percentage of the new
mosquito repellent is less than 95 at α = .10.
b.
We must assume that the population of percent repellencies is normally distributed.
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332
6.61
Chapter 6
To determine if the true mean crack intensity of the Mississippi highway exceeds the AASHTO
recommended maximum, we test:
H0: μ = .100
Ha: μ > .100
The test statistic is t =
x - μ0
=
s/ n
.210  .100
.011 / 8
= 2.97
The rejection region requires α = .01 in the upper tail of the t-distribution with df = n − 1 = 8 − 1 = 7.
From Table V, Appendix B, t.01 = 2.998. The rejection region is t > 2.998.
Since the observed value of the test statistic does not fall in the rejection region (t = 2.97  2.998), H0 is
not rejected. There is insufficient evidence to indicate that the true mean crack intensity of the Mississippi
highway exceeds the AASHTO recommended maximum at α = .01.
6.62
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Plants
Variable
Plants
N
20
Mean
4.050
Median
3.500
TrMean
3.833
Variable
Plants
Minimum
1.000
Maximum
11.000
Q1
2.250
Q3
5.000
StDev
2.665
SE Mean
0.596
Let μ = mean number of active nuclear power plants operating in all states. To determine if the mean
number of active nuclear power plants operating in all states exceeds 3, we test:
H0: μ = 3
Ha: μ > 3
The test statistic is t 
x  μo
s
n

4.05  3
2.665
20
 1.762
The rejection region requires α = .10 in the upper tail of the t-distribution with
df = n – 1 = 20 – 1 = 19. From Table V, Appendix B, t.10 = 1.328. The rejection region is
t > 1.328.
Since the observed value of the test statistic falls in the rejection region (t = 1.762 > 1.328), H0 is
rejected. There is sufficient evidence to indicate the mean number of active nuclear power plants
operating in all states exceeds 3 at α = .10.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
We will look at the 4 methods for determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the number of power plants is:
Histogram of PLANTS
Normal
9
Mean
StDev
N
8
4.05
2.665
20
7
6
Frequency
b.
333
5
4
3
2
1
0
-2
0
2
4
6
PLANTS
8
10
12
From the histogram, the data appear to be skewed to the right. This indicates that the data may not be
normal.
Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal.
x  s  4.05  2.665  (1.385, 6.715) 13 of the 20 values fall in this interval. The proportion is
.65. This is close to the .68 we would expect if the data were normal.
x  2s  4.05  2(2.665)  4.05  5.33  (1.28, 9.38) 19 of the 20 values fall in this interval.
The proportion is .95. This is the same as the .95 we would expect if the data were normal.
x  3s  4.05  3(2.665)  4.05  7.995  (3.945, 12.045) 20 of the 20 values fall in this
interval. The proportion is 1.000. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the data might be normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 5.00 – 2.25 = 2.75.
IQR
2.75

 1.03 This is not close to the 1.3 we would expect if the data were normal. This
2.665
s
method indicates the data may not be normal.
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334
Chapter 6
Finally, using MINITAB, the normal probability plot is:
Probability Plot of PLANTS
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
4.05
2.665
20
0.686
0.062
Percent
80
70
60
50
40
30
20
10
5
1
-5
0
5
PLANTS
10
15
Since the data do not form a straight line, the data may not be normal.
From 3 of the 4 different methods, the indications are that the number of power plants data are not
normal.
b.
The two largest values are 9 and 13. The two lowest values are 1 and 1. Using MINITAB with the
data deleted yields the descriptive statistics:
Descriptive Statistics: Plants2
Variable
Plants2
N
16
Mean
3.688
Median
3.500
TrMean
3.643
Variable
Plants2
Minimum
1.000
Maximum
7.000
Q1
2.000
Q3
5.000
StDev
1.662
SE Mean
0.416
To determine if the mean number of active nuclear power plants operating in all states exceeds 3
(using the reduced data set), we test:
H0: μ = 3
H a: μ > 3
The test statistic is t 
x  μo
s
n

3.68  3
1.662
16
 1.656
The rejection region requires α = .10 in the upper tail of the t-distribution with df = n – 1 = 16 – 1 =
15. From Table V, Appendix B, t.10 = 1.341. The rejection region is t > 1.341.
Since the observed value of the test statistic falls in the rejection region (t = 1.656 > 1.341), H0 is
rejected. There is sufficient evidence to indicate the mean number of active nuclear power plants
operating in all states exceeds 3 at α = .10.
By eliminating the top two and bottom two observations, we have not changed the decision about H0.
c.
It is very dangerous to eliminate data points to satisfy assumptions. The data may, in fact, not be
normal. By eliminating data points, one has changed the kind of data that come from the parent
population. Thus, incorrect decisions could be made.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.63
335
Using MINITAB, the descriptive statistics for the 2 plants are:
Descriptive Statistics: AL1, AL2
Variable
AL1
AL2
N
2
2
N*
0
0
Mean
0.00750
0.0700
SE Mean
0.00250
0.0200
StDev
0.00354
0.0283
Minimum
0.00500
0.0500
Q1
*
*
Median
0.00750
0.0700
Q3
*
*
Maximum
0.01000
0.0900
To determine if plant 1 is violating the OSHA standard, we test:
H0: μ = .004
Ha: μ > .004
The test statistic is t 
x  μo
s
n

.0075  .004
.00354
2
 1.40
Since no α level was given, we will use α = .10. The rejection region requires α = .05 in the upper tail
of the t-distribution with df = n – 1 = 2 – 1 = 1. From Table V, Appendix B, t.10 = 3.078. The rejection
region is t > 3.078.
Since the observed value of the test statistic does not fall in the rejection region
(t = 1.40  3.078), H0 is not rejected. There is insufficient evidence to indicate the OSHA standard
is violated by plant 1 at α = .10.
To determine if plant 2 is violating the OSHA standard, we test:
H0: μ = .004
Ha: μ > .004
The test statistic is t 
x  μo
s
n

.07  .004
.0283
2
 3.30
Since no α level was given, we will use α = .10. The rejection region requires α = .10 in the upper tail of
the t-distribution with df = n – 1 = 2 – 1 = 1. From Table V, Appendix B, t.10 = 3.078. The rejection region
is t > 3.078.
Since the observed value of the test statistic falls in the rejection region (t = 3.30 > 3.078), H0 is rejected.
There is sufficient evidence to indicate the OSHA standard is violated by plant 2 at α = .10.
6.64
b.
First, check to see if n is large enough.
np0  100(.7)  70 and nq0  100(.3)  30
Since both np0  15 and nq0  15 , the normal approximation will be adequate.
H0: p = .70
Ha: p < .70
The test statistic is z =
pˆ  p0
σ pˆ

pˆ  p0
p0 q0
n

.63  .70
.70(.30)
100
= −1.53
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336
Chapter 6
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < −1.645.
Since the observed value of the test statistic does not fall in the rejection region (−1.53  −1.645),
H0 is not rejected. There is insufficient evidence to indicate that the proportion is less than .70 at
α = .05.
c.
p-value = P(z ≤ −1.53) = .5 − .4370 = .0630
Since p is not less than α = .05, H0 is not rejected.
6.65
a.
z=
pˆ  p 0
p 0q 0
n

.83  .9
.9(.1)
100
= −2.33
.7(.3)
.9(.1)
= .0458 as compared to
= .03 in part a. Since
100
100
the denominator in this problem is smaller, the absolute value of z is larger.
b.
The denominator in Exercise 6.64 is
c.
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < −1.645.
Since the observed value of the test statistic falls in the rejection region (z = −2.33 < −1.645), H0 is
rejected. There is sufficient evidence to indicate the population proportion is less than .9 at α = .05.
6.66
d.
The p-value = P(z ≤ −2.33) = .5 − .4901 = .0099 (from Table IV, Appendix B). Since the p-value is
less than α = .05, H0 is rejected.
a.
No. The p-value is the probability of observing your test statistic or anything more unusual if H0 is
true. For this problem, the p-value = .3300/2 = .1650.
Given the true value of the population proportion, p, is .5, the probability of observing a test statistic
of z = .44 or larger is .1650. Since the p-value is not small (p = .1650), there is no evidence to reject
H0. There is no evidence to indicate the population proportion is greater than .5.
b.
If the alternative hypothesis were two-tailed, the p-value would be 2 times the p-value for a onetailed test. For this problem, the p-value = .3300. The probability of observing your test statistic or
anything more unusual if H0 is true is .3300.
There is no evidence to reject H0 for α ≤ .10. There is no evidence to indicate that p ≠ .5 for α ≤ .10.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.67
337
From Exercise 5.40, n = 50 and since p is the proportion of consumers who do not like the snack food, p̂
will be:
p̂ =
Number of 0 's in sample
29
=
= .58
n
50
First, check to see if the normal approximation will be adequate:
np0  50(.5)  25
nq0  50(.5)  25
Since both np0  15 and nq0  15, the normal distribution will be adequate.
a.
H0: p = .5
Ha: p > .5
The test statistic is z =
pˆ  p 0
σ pˆ

pˆ  p 0
p 0q 0
=
n
.58  .5
.5(1  .5)
50
= 1.13
The rejection region requires α = .10 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.10 = 1.28. The rejection region is z > 1.28.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.13  1.28), H0
is not rejected. There is insufficient evidence to indicate the proportion of customers who do not like
the snack food is greater than .5 at α = .10.
b.
6.68
6.69
p-value = P(z ≥ 1.13) = .5 − .3708 = .1292
The sample size is large enough to use the normal approximation if npo  15 and nqo  15 .
a.
npo  900(.975)  877.5  15 and nqo  900(.025)  22.5  15 . Thus, the sample size is large
enough.
b.
npo  125(.01)  1.25  15 and nqo  125(.99)  123.75  15 . Thus, the sample size is not large
enough.
c.
npo  40(.75)  30  15 and nqo  40(.25)  10  15 . Thus, the sample size is not large enough.
d.
npo  15(.75)  11.25  15 and nqo  15(.25)  3.75  15 . Thus, the sample size is not large enough.
e.
npo  12(.62)  7.44  15 and nqo  12(.38)  4.56  15 . Thus, the sample size is not large enough.
a.
The point estimate is pˆ  .42 .
b.
To determine if whether the p has increased since 2005, we test:
H0: p = .30
Ha: p > .30
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338
6.70
6.71
Chapter 6
pˆ  po
The test statistics is z 
d.
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.01 = 2.33. The rejection region is z > 2.33.
e.
Since the observed value of the test statistic falls in the rejection region (z = 16.56 >.2.33), H0 is
rejected. There is sufficient evidence to conclude that p has increased since 2005 at α = .01.
f.
The p-value = P(z ≥ 16.56) = .5 − .5 ≈ 0 is less than .01, H0 is rejected. The conclusion is the
same as that in part e.
a.
pˆ =
po qo
n

.42  .30
c.
.30(.70)
4, 000
 16.56
x 64
=
= .604
n 106
b.
H0: p = .70
Ha: p ≠ .70
c.
The test statistic is z 
d.
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.58. The rejection region is z > 2.58 or z < −2.58.
e.
Since the observed value of the test statistic does not fall in the rejection region (z = −2.16 </ −2.58),
H0 is not rejected. There is insufficient evidence to indicate the true proportion of consumers who
believe “Made in the USA” means 100% of labor and materials are from the United States is
different from .70 at α = .01.
pˆ  p0
p0 q0
n

.604  .70
.70(.30)
106
 2.16
To determine if the sample provides sufficient evidence to indicate that the true percentage of all
firms that announced one or more acquisitions during the year 2000 is less than 30%, we test:
H0: p = .30
Ha: p < .30
x
748

 .269
n 2, 778
pˆ  po
.269  .30
The test statistic is z 

 3.57
.30(.70)
po qo
2, 778
n
The point estimate is pˆ 
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645.
Since the observed value of the test statistic falls in the rejection region (z = −3.57 < −1.645), H0 is
rejected. There is sufficient evidence to indicate that the true percentage of all firms that
announced one or more acquisitions during the year 2000 is less than 30% at α = .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.72
a.
The population parameter of interest is p = proportion of items that had the wrong price scanned at
California Wal-Mart stores.
b.
To determine if the true proportion of items scanned at California Wal-Mart stores with the wrong
price exceeds the 2% NIST standard, we test:
339
H0: p = .02
Ha: p > .02
c.
The test statistic is z 
pˆ  po
po qo
n

.083  .02
.02(.98)
1000
 14.23
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
d.
Since the observed value of the test statistic falls in the rejection region (z = 14.23 > 1.645), H0 is
rejected. There is sufficient evidence to indicate that the true proportion of items scanned at
California Wal-Mart stores with the wrong price exceeds the 2% NIST standard at α = .05. This
means that the proportion of items with wrong prices at California Wal-Mart stores is much higher
than what is allowed.
e.
In order for the inference to be valid, the sampling distribution of p̂ must be approximately normal.
For this assumption to be valid, both np0  15 and nq0  15 .
np0  1000(.02)  20
nq0  1000(.98)  980
Since np0  15 and nq0  15 , we can assume the distribution of p̂ is approximately normal.
6.73
a.
To determine whether the true proportion of toothpaste brands with the ADA seal of verifying
effective decay prevention is less than .5, we test:
H0: p = .5
Ha: p < .5
6.74
b.
From the printout, the p-value is p = .231.
c.
Since the observed value of p is greater than α = .10, H0 is not rejected. There is insufficient
evidence to indicate the true proportion of toothpaste brands with the ADA seal of verifying effective
decay prevention is less than .5 at α = .10.
a.
Let p = proportion of vacation-home owners who are minorities in 2003.
pˆ 
x
46

 .111
n 416
To determine if the percentage of vacation-home owners in 2006 who are minorities is larger than
6%, we test:
H0: p = .06
Ha: p > .06
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Chapter 6
The test statistic is z 
pˆ  po
po qo
n

.111  .06
.06(.94)
416
 4.38
The rejection region requires α = .01 in the upper tail of the z-distribution. From
Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic falls in the rejection region (z = 4.38 > 2.33), H0 is
rejected. There is sufficient evidence to indicate that the true percentage of vacation-home owners in
2006 who are minorities is larger than 6% at α = .01.
6.75
b.
Since the return rate of the questionnaire was so small compared to the number sent out, one should
be very skeptical of the results. It would be fairly unusual that the sample of returned questionnaires
would be representative of the entire population.
a.
Let p = proportion of middle-aged women who exhibit skin improvement after using the cream. For
x 24
this problem, pˆ  
 .727 .
n 33
First we check to see if the normal approximation is adequate:
np0  33(.6)  19.8 nq0  33(.4)  13.2
Since nq0 = 13.2 is less than 15, the assumption of normality may not be valid. We will will go ahead
and perform the test.
To determine if the cream will improve the skin of more than 60% of middle-aged women, we test:
H0: p = .60
Ha: p > .60
The test statistic is z 
pˆ  p0
p0 q0
n

.727  .60
.60(.40)
33
 1.49
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.49  −1.645),
H0 is not rejected. There is insufficient evidence to indicate the cream will improve the skin of more
than 60% of middle-aged women at α = .05.
b.
The p-value is p  P ( z  1.49)  (.5  .4319)  .0681 . (Using Table IV, Appendix B.) Since the pvalue is greater than α = .05, H0 is not rejected. There is insufficient evidence to indicate the cream
will improve the skin of more than 60% of middle-aged women at α = .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.76
341
Let p = proportion of firms in violation of the new 4-day rule for reporting material changes.
pˆ 
x
23

 .050
n 462
To determine if the percentage of firms in violation of the new 4-day rule for reporting material changes is
less than 10%, we test:
H0: p = .10
Ha: p < .10
The test statistic is z 
pˆ  po

po qo
n
.050  .10
.10(.90)
462
 3.58
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix B,
z.01 = 2.33. The rejection region is z < −2.33.
Since the observed value of the test statistic falls in the rejection region (z = −3.58 < −2.33), Ho is rejected.
There is sufficient evidence to indicate that the true percentage of firms in violation of the new 4-day rule
for reporting material changes is less than 10% at α = .01.
6.77
a.
To determine if the GSR for all scholarship athletes at Division I institutions differs from 60%, we
test:
H0: p = .60
Ha: p ≠ .60
The point estimate is pˆ 
The test statistic is z 
x 315

 .63
n 500
pˆ  po
po qo
n

.63  .60
.60(.40)
500
 1.37
The rejection region requires α /2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.575. The rejection region is z < −2.575 or z > 2.575.
Since the observed value does not fall in the rejection region (z = 1.37  2.575), H0 is not rejected.
There is insufficient evidence to conclude that the GSR for all scholarship athletes at Division I
institution differs from 60% at α = .01.
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342
Chapter 6
b.
To determine if the GSR for all male basketball players at Division I institutions differs from 58%,
we test:
H0: p = .58
Ha: p ≠ .58
The point estimate is pˆ 
The test statistic is z 
x
84

 .42
n 200
pˆ  po
po qo
n

.42  .58
.58(.42)
200
 4.58
The rejection region requires α /2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.575. The rejection region is z < −2.575 or z > 2.575.
Since the observed value falls in the rejection region (z = −4.58 < −2.575), H0 is rejected. There is
sufficient evidence to conclude that the GSR for all male basketball players at Division I institutions
differs from 58% α = .01.
6.78
Let p = proportion of students choosing the three-grill display so that Grill #2 is a compromise between a
more desirable and a less desirable grill.
pˆ 
x 85

 .685
n 124
To determine if the proportion of students choosing the three-grill display so that Grill #2 is a compromise
between a more desirable and a less desirable grill is greater than .167, we test:
H0: p = .167
Ha: p > .167
The test statistic is z 
pˆ  po
po qo
n

.685  .167
.167(.833)
124
 15.47
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B,
z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 15.47 > 1.645), H0 is rejected.
There is sufficient evidence to indicate that the true proportion of students choosing the three-grill display
so that Grill #2 is a compromise between a more desirable and a less desirable grill is greater than .167 at
α = .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.79
343
To minimize the probability of a Type I error, we will select α = .01.
First, check to see if the normal approximation is adequate:
np0  100(.5)  50
nq0  100(.5)  50
Since both np0  .15 and nq0  .15 , the normal distribution will be adequate
p̂ =
x
56
=
= .56
n
100
To determine if more than half of all Diet Coke drinkers prefer Diet Pepsi, we test:
H0: p = .5
Ha: p > .5
The test statistic is z =
p̂  p 0
=
p 0q 0
n
.56  .5
.5(.5)
100
= 1.20
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix B,
z.01 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.20 >/ 2.33), H0 is not
rejected. There is insufficient evidence to indicate that more than half of all Diet Coke drinkers prefer Diet
Pepsi at α = .01.
Since H0 was not rejected, there is no evidence that Diet Coke drinkers prefer Diet Pepsi.
6.80
a.
The power of a test increases when:
1.
2.
3.
6.81
The distance between the null and alternative values of μ increases.
The value of α increases.
The sample size increases.
b.
The power of a test is equal to 1 − β. As β increases, the power decreases.
a.
By the Central Limit Theorem, the sampling distribution of x is
approximately normal with
μ x = μ = 500 and
σx 
b.
σ
n

100
25
= 20.
x0 = μ0 + zα σ x = μ0 + zα
Thus, x0 = 500 + 1.645
σ where z = z = 1.645 from Table IV, Appendix B.
α
.05
n
100
25
= 532.9
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344
Chapter 6
c.
The sampling distribution of x is approximately normal by the
Central Limit Theorem with μ x = μ = 550 and
σx =
d.
σ
n

100
25
= 20.

532.9  550 
β = P( x0 < 532.9 when μ = 550) = P  z 
 = P(z < −.86)

100 / 25 
= .5 − .3051 = .1949
e.
6.82
Power = 1 − β = 1 − .1949 = .8051
From Exercise 6.81 we want to test H0: μ = 500 against Ha: μ > 500 using α = .05, σ = 100, n = 25, and
x = 532.9.
a.

532.9  575 
β = P( x0 < 532.9 when μ = 575) = P  z 
 = P(z < −2.11)

100 / 25 
= .5 − .4826 = .0174
(Using Table IV, Appendix B)
6.83
b.
Power = 1 − β = 1 − .0174 = .9826
c.
In Exercise 6.81, β = .1949 and the power is .8051. The value of β has decreased in this exercise
since μ = 575 is further from the hypothesized value than μ = 550. As a result, the power of the test
in this exercise has increased (when β decreases, the power of the test increases).
a.
The sampling distribution of x will be approximately normal (by the Central Limit Theorem) with
σ
15
μ x = μ = 75 and σ x 

= 2.143.
49
n
b.
The sampling distribution of x will be approximately normal (by the Central Limit Theorem) with
σ
15
μ x = μ = 70 and σ x 

= 2.143.
49
n
c.
First, find x0 = μ0 − zα σ x = μ0 − zα
Thus, x0 = 75 − 1.28
15
49
σ where z = 1.28 from Table IV, Appendix B.
.10
n
= 72.257

72.257  70 
Now, find β = P( x0 > 72.257 when μ = 70) = P  z 


15 / 49 
= P(z > 1.05) = .5 − .3531 = .1469
d.
Power = 1 − β = 1 − .1469 = .8531
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Inferences Based on a Single Sample: Tests of Hypothesis
6.84
a.
345
From Exercise 6.83, we want to test H0: μ = 75 against Ha: μ < 75 using α = .10, σ = 15, n = 49, and
x = 72.257. Using Table IV, Appendix B:
If μ = 74,

72.257  74 
β = P( x0 > 72.257 when μ = 74) = P  z 
 = P(z > −.81)

15 / 49 
= .5 + .2910 = .7910
If μ = 72,

72.257  72 
μ = P( x0 > 72.257 when μ = 72) = P  z 
 = P(z > .12)

15 / 49 
= .5 − .0478 = .4522
If μ = 70,
β = P( x0 > 72.257 when μ = 70) = .1469
(Refer to Exercise 6.83, part c.)
If μ = 68,

72.257  68 
β = P( x0 > 72.257 when μ = 68) = P  z 
 = P(z > 1.99)

15 / 49 
= .5 − .4767 = .0233
If μ = 66,

72.257  66 
β = P( x0 > 72.257 when μ = 66) = P  z 
 = P(z > 2.92)

15 / 49 
= .5 − .4982 = .0018
In summary,
μ
β
74
.7910
72
.4522
70
.1469
68
.0233
66
.0018
b.
c.
Looking at the graph, β is approximately .62 when μ = .73.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
346
Chapter 6
d.
Power = 1 − β
Therefore,
μ
β
Power
74
.7910
.2090
72
.4522
.5478
70
.1469
.8531
68
.0233
.9767
66
.0018
.9982
The power curve starts out close to 1 when μ = 66 and decreases as μ increases, while the β curve is
close to 0 when μ = 66 and increases as μ increases.
6.85
e.
As the distance between the true mean μ and the null hypothesized mean μ0 increases, β decreases
and the power increases. We can also see that as β increases, the power decreases.
a.
The sampling distribution of x will be approximately normal (by the Central Limit Theorem) with
σ
1.2
μ x = μ = 30 and σ x 

= .109.
121
n
b.
The sampling distribution of will be approximately normal (CLT) with μ x = μ = 29.8
and σ x 
c.
σ
n

1.2
121
= .109.
First, find xO,L = μ0 − zα/2 σ x = μ0 − zα/2
σ
n
where z.05/2 = z.025 = 1.96 from Table IV, Appendix B.
Thus, xO,L = 30 − 1.96
1.2
121
= 29.79
xO,U = μ0 + zα/2 σ x = μ0 + zα/2
σ = 30 + 1.96 1.2 = 30.21
n
121
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Inferences Based on a Single Sample: Tests of Hypothesis
347
Now, find β = P(29.79 < x < 30.21 when μ = 29.8)
 29.79  29.8
30.21  29.8 
= P
z

 1.2 / 121
1.2 / 121 
= P(−.09 < z < 3.76)
= .0359 + .5 = .5359
d.
 29.79  30.4
30.21  30.4 
z
β = P(29.79 < x < 30.21 when μ = 30.4) = P 

 1.2 / 121
1.2 / 121 
= P(−5.59 < z < −1.74)
= .5 − .4591 = .0409
6.86
a.
To determine if the mean size of California homes exceeds the national average, we test:
H0: μ = 2230
Ha: μ > 2230
The test statistic is z =
x  μ0
σx

2347  2230
257 / 100
= 4.55
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic falls in the rejection region (z = 4.55 > 2.33), H0 is
rejected. There is sufficient evidence to indicate the mean size of California homes exceeds the
national average at α = .01.
b.
To compute the power, we must first set up the rejection regions in terms of .
 s 
 257 
 2, 230  2.33 
= 2,289.88
x0  μ0  zα σ x  μ0  2.33 
 n 
 100 
We would reject H0 if x > 2,289.88
The power of the test when μ = 2,330 would be:


x  μa 
2, 289.88  2, 330 
Power = P( x > 2289.88⏐μ = 2,330) = P  z  0
 P z 


σx 


257 / 100
= P(z > −1.56) = .5 + .4406 = .9406
c.
The power of the test when μ = 2,280 would be:


x  μa 
2, 289.88  2, 280 
Power = P( > 2289.88⏐μ = 2,280) = P  z  0
 P z 


σx 


257 / 100
= P(z > 0.38) = .5 − .1480 = .3520
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348
6.87
Chapter 6
a.
We have failed to reject H0 when it is not true. This is a Type II error.
To compute β, first find:
x0 = μ0 − zα σ x = μ0 − zα
σ where z = 1.645 from Table IV, Appendix B.
.05
n
.01
Thus, x0 = 5.0 − 1.645
100
= 4.998355
Then find:

4.998355  4.9975 
β = P( x0 > 4.998355 when μ = 4.9975) = P  z 


.01/ 100
= P(z > .86) = .5 − .3051 = .1949
6.88
b.
We have rejected H0 when it is true. This is a Type I error. The probability of a Type I error is
α = .05.
c.
A departure of .0025 below 5.0 is μ = 4.9975. Using a, β when μ = 4.9975 is .1949. The power of
the test is 1 − β = 1 − .1949 = .8051
a.
To determine if the mean mpg for 2009 Honda Civic autos is greater than 36 mpg, we test:
H0: μ = 36
Ha: μ > 36
b.
The test statistic is z =
x  μ0
σx

38.3  36
= 2.54
6.4 / 50
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 2.54 > 1.645), H0 is
rejected. There is sufficient evidence to indicate that the mean mpg for 2009 Honda Civic autos is
greater than 36 mpg at α = .05.
We must assume that the sample was a random sample.
c.
First find:
x0 = μ0 + zα σ x = μ0 + zα
Thus, x0 = 36 + 1.645
σ where z = 1.645 from Table IV, Appendix B.
α
n
6.4
50
= 37.49
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Inferences Based on a Single Sample: Tests of Hypothesis
349
For μ = 36.5:

37.49  36.5 
Power = P( x > 37.49│μ = 36.5) = P  z 
 = P(z > 1.09)

6.4 / 50 
= .5 − .3621 = .1379
For μ = 37:

37.49  37 
Power = P( x > 37.49│μ = 37) = P  z 
 = P(z > .54)

6.4 / 50 
= .5 − .2054 = .2946
For μ = 37.5:

37.49  37.5 
Power = P( x > 37.49│μ = 37.5) = P  z 
 = P(z > −.01 )

6.4 / 50 
= .5 + .0041 = .5040
For μ = 38:

37.49  38 
Power = P( x > 37.49│μ = 38) = P  z 
 = P(z > −.56)

6.4 / 50 
= .5 + .2123 = .7123
For μ = 38.5:

37.49  38.5 
Power = P( x > 37.49│μ = 38.5) = P  z 
 = P(z > −.1.12)

6.4 / 50 
= .5 + .3686 = .8686
The plot is:
Scatterplot of Power vs Mu
0.9
0.8
0.7
0.6
Power
d.
0.5
0.4
0.3
0.2
0.1
0.0
0
36.5
37.0
37.5
µ
38.0
38.5
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350
Chapter 6
e.
From the plot, the power is approximately .60.
For μ  37.75 :

37.49  37.75 
Power  P ( x  37.49 | μ  37.75)  P  z 
  P ( z  .29)  .5  .1141  .6141
6.4 50 

f.
From the plot, the power is approximately 1.
For μ  41:

37.49  41 
Power  P ( x  37.49 | μ  41)  P  z 
  P( z  3.88)
6.4 50 

 .5  .5  1.0
If the true value of μ is 41, the approximate probability that the test will fail to reject H0 is
1 − 1 = 0.
6.89
First, find x0 such that P( x < x0 ) = .05.

x  10 
P( x < x0 ) = P  z  0
 = P(z < z0) = .05.

1.2 / 48 
From Table IV, Appendix B, z0 = −1.645.
Thus, z0 =
x0  10
1.2 / 48
 x0 = −1.645(.173) + 10 = 9.715
The probability of a Type II error is:

9.715  9.5 
β = P( x ≥ 9.715│μ = 9.5) = P  z 
 = P(z ≥ 1.24) = .5 − .3925 = .1075

1.2 / 48 
6.90
Using Table VI, Appendix B:
a.
For n = 12, df = n − 1 = 12 − 1 = 11
P(χ2 > χ 02 ) = .10  χ 02 = 17.2750
b.
For n = 9, df = n − 1 = 9 − 1 = 8
P(χ2 > χ 02 ) = .05  χ 02 = 15.5073
c.
For n = 5, df = n − 1 = 5 − 1 = 4
P(χ2 > χ 02 ) = .025  χ 02 = 11.1433
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.91
6.92
a.
df = n − 1 = 16 − 1 = 15; reject H0 if χ2 < 6.26214 or χ2 > 27.4884
b.
df = n − 1 = 23 − 1 = 22; reject H0 if χ2 > 40.2894
c.
df = n − 1 = 15 − 1 = 14; reject H0 if χ2 > 21.0642
d.
df = n − 1 = 13 − 1 = 12; reject H0 if χ2 < 3.57056
e.
df = n − 1 = 7 − 1 = 6; reject H0 if χ2 < 1.63539 or χ2 > 12.5916
f.
df = n − 1 = 25 − 1 = 24; reject H0 if χ2 < 13.8484
a.
It would be necessary to assume that the population has a normal distribution.
b.
H0: σ2 = 1
Ha: σ2 > 1
The test statistic is χ2 =
(n  1) s 2
σ 02

351
6(4.84)
= 29.04
1
The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = n − 1 = 7 − 1 =
2
6. From Table VI, Appendix B, χ.05
= 12.5916. The rejection region is χ2 > 12.5916.
Since the observed value of the test statistic falls in the rejection region (29.04 > 12.5916), H0 is
rejected. There is sufficient evidence to indicate that the variance is greater than 1 at α = .05.
c.
H0: σ2 = 1
Ha: σ2 ≠ 1
The test statistic is χ2 =
(n  1) s 2
σ 02

6(4.84)
= 29.04
1
The rejection region requires α/2 = .025 in the upper tail of the χ2 distribution with df = n − 1 = 7 − 1
2
2
= 6. From Table VI, Appendix B, χ.975
= 1.237347 and χ.025
= 14.4494. The rejection region is χ2
2
< 1.237347 or χ > 14.4494.
Since the observed value of the test statistic falls in the rejection region (29.04 > 14.4494), H0 is
rejected. There is sufficient evidence to indicate that the variance is not equal to 1 at α = .05.
6.93
a.
H0: σ2 = 1
Ha: σ2 > 1
The test statistic is χ2 =
(n  1) s 2
σ 02

(100  1)4.84
= 479.16
1
The rejection region requires α = .05 in the upper tail of the χ2 distribution with df =
2
n − 1 = 100 − 1 = 99. From Table VI, Appendix B, χ.05
≈ 124.324. The rejection region is
2
χ > 124.324.
Since the observed value of the test statistic falls in the rejection region (χ2 = 479.16 > 124.324), H0
is rejected. There is sufficient evidence to indicate the variance is larger than 1 at α = .05.
b.
In part b of Exercise 6.92, the test statistic was χ2 = 29.04. The conclusion was to reject H0 as it was
in this problem.
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352
6.94
Chapter 6
Some preliminary calculations are:
 x 

2
s2 =
x
2
n
n 1

302
7 = 7.9048
7 1
176 
To determine if σ2 < 1, we test:
H0: σ2 = 1
Ha: σ2 < 1
The test statistic is χ2 =
(n  1) s 2
σ 02

(7  1)7.9048
= 47.43
1
The rejection region requires α = .05 in the lower tail of the χ2 distribution with df = n − 1 = 7 − 1 = 6.
2
From Table VI, Appendix B, χ.95
= 1.63539. The rejection region is χ2 < 1.63539.
Since the observed value of the test statistic does not fall in the rejection region (χ2 = 47.43  1.63539),
H0 is not rejected. There is insufficient evidence to indicate the variance is less than 1.
6.95
6.96
a.
The rejection region requires α/2 = .01/2 = .005 in each tail of the χ2 distribution with
2
2
df = n – 1 = 46 – 1 = 45. From Table VI, Appendix B, χ.005
≈ 73.12795 and χ.995
≈ 24.3486. The
rejection region is χ2 < 24.3486 or χ2 > 73.12795.
b.
The test statistic is χ 2 
c.
Since the observed value of the test statistic does not fall in the rejection region
(χ2 = 63.7245  24.3486 and χ2 = 63.7245  73.12795), H0 is not rejected. There is
insufficient evidence to indicate the variance is different from 100 at α = .01.
To determine if the breaking strength variance of the new adhesive is less than the variance of the
standard composite adhesive, σ2 = .25, we test:
a.
(n  1) s 2
σ o2

(46  1)11.92
 63.7245 .
100
H0: σ2 = .25
Ha: σ2 < .25
b.
The rejection region requires α = .01 in the lower tail of the χ2 distribution with df = n – 1 = 10 – 1 =
2
9. From Table VI, Appendix B, χ.99
= 2.087912. The rejection region is χ2 < 2.087912.
c.
The test statistic is χ 2 
d.
Since the observed value of the test statistic does not fall in the rejection region
(χ2 = 7.6176  2.087912), H0 is not rejected. There is insufficient evidence to indicate the breaking
strength variance of the new adhesive is less than the variance of the standard composite adhesive,
σ2 = .25 at α = .01.
e.
We must assume that the distribution of the breaking strengths is approximately normal and that a
random sample was selected from this population.
(n  1) s 2
σ o2

(10  1).462
 7.6176 .
.25
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
a.
Let σ2 = weight variance of tees. To determine if the weight variance differs from .000004 (injection
mold process is out-of-control), we test:
H0: σ2 = .000004
Ha: σ2 ≠ .000004
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tees
Variable
Tees
N
40
Mean
0.25248
Median
0.25300
TrMean
0.25256
Variable
Tees
Minimum
0.24700
Maximum
0.25600
Q1
0.25100
Q3
0.25400
The test statistic is χ 2 
(n  1) s 2
σ 02

StDev
0.00223
SE Mean
0.00035
(40  1)(.00223)2
 48.49
.000004
The rejection region requires α/2 = .01/2 = .005 in each tail of the χ 2 distribution with df = n – 1 =
2
2
 66.7659 and χ.995
 20.7065. The rejection
40 – 1 = 39. From Table VI, Appendix B, χ.005
region is χ 2 > 66.7659 or χ 2 < 20.7065.
Since the observed value of the test statistic does not fall in the rejection region ( χ 2 = 49.49 
66.7659 and χ 2 = 49.49  20.7065), H0 is not rejected. There is insufficient evidence to indicate
the injection mold process is out-of-control at α = .01.
c.
We must assume that the distributions of the weights of tees is approximately normal. Using
MINITAB, a histogram of the data is:
10
Frequency
6.97
353
5
0
0.247 0.248 0.249 0.250 0.251 0.252 0.253 0.254 0.255 0.256
Tees
The data look fairly mound-shaped, so the assumption of normality seems to be reasonably satisfied.
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354
6.98
Chapter 6
a.
To determine if the standard derivation of all grassland pixels is less than 22, we test:
H0: σ 2 = 222
Ha: σ < 22
2
b.
6.99
H0: σ 2 = 484
OR
Ha: σ 2 < 484
2
From the printout, the p-value is .105. Since the p-value is not less than a = .10 (p = .105  .10), H0
is not rejected. There is insufficient evidence to indicate the standard deviation of all grassland pixels
is less than 22 at a = .10.
To determine if the diameters of the ball bearings are more variable when produced by the new process,
test:
H0: σ2 = .00156
Ha: σ2 > .00156
The test statistic is χ2 =
(n  1) s 2
σ 02

99(.00211)
= 133.90
.00156
The rejection region requires use of the upper tail of the χ2 distribution with df = n − 1 = 100 − 1 = 99.
We will use df = 100 ≈ 99 due to the limitations of the table. From Table VI, Appendix B,
2
2
χ.025
= 129.561 < 133.90 < 135.807 = χ.010
. The p-value of the test is between .010 and .025. The
decision made depends on the desired α. For α < .010, there is not enough evidence to show that the
variance in the diameters is greater than .00156; the reverse decision would be made for α ≥ .025.
6.100
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: GASTURBINE
Variable
GASTURBINE
N
67
N*
0
Mean
11066
Variable
GASTURBINE
Maximum
16243
SE Mean
195
StDev
1595
Minimum
8714
Q1
9918
Median
10656
Q3
11842
To determine if the heat rates of the augmented gas turbine engine are more variable than the heat rates of
the standard gas turbine engine, we test:
H0: σ2 = 1,5002
Ha: σ2 > 1,5002
The test statistic is χ 2 
(n  1) s 2
σ o2

(67  1)1, 5952
1, 5002
 74.625 .
The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = n – 1 = 67 – 1 = 66.
2
From Table VI, Appendix B, χ.05
≈ 85.95148. The rejection region is χ2 > 85.95148.
Since the observed value of the test statistic does not fall in the rejection region (χ2 = 74.625  85.95148),
H0 is not rejected. There is insufficient evidence to indicate the heat rates of the augmented gas turbine
engine are more variable than the heat rates of the standard gas turbine engine at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.101
355
a.
Since the sample mean of 3.85 is not that far from the value of 3.5, a large standard deviation would
indicate that the value 3.85 is not very many standard deviations from 3.5.
b.
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z > 2.33.
The test statistic is z 
x  μo
σx

3.85  3.5
σ
137
To reject H0, z > 2.33. Thus, we need to find σ so z > 2.33.
z
3.85  3.5
σ
137
 2.33  3.85  3.5  2.33
σ
137
 .35  .199065σ  1.758  σ
Thus, the largest value of σ for which we will reject H0 is 1.758.
c.
To determine if σ < 1.758, we test:
H0: σ2 = 1.7582
Ha: σ2 < 1.7582
The test statistic is χ 2 
(n  1) s 2
σ o2

(137  1)1.52
1.7582
 99.011 .
The rejection region requires α = .01 in the lower tail of the χ2 distribution with
df = n – 1 = 137 – 1 = 136. Since there are no values in the table with df > 100, we will use
MINITAB to compute the p-value of the test statistic.
Cumulative Distribution Function
Chi-Square with 136 DF
x
99.011
P( X <= x )
0.0072496
Since the p-value is less than α (p-value = 0.0072496 < α = .01), H0 is rejected. There is sufficient
evidence to indicate the standard deviation of the scores is less than 1.758 at α = .01.
6.102
For a large sample test of hypothesis about a population mean, no assumptions are necessary because the
Central Limit Theorem assures that the test statistic will be approximately normally distributed. For a
small sample test of hypothesis about a population mean, we must assume that the population being
sampled from is normal. The test statistic for the large sample test is the z statistic, and the test statistic for
the small sample test is the t statistic.
6.103
The smaller the p-value associated with a test of hypothesis, the stronger the support for the alternative
hypothesis. The p-value is the probability of observing your test statistic or anything more unusual, given
the null hypothesis is true. If this value is small, it would be very unusual to observe this test statistic if the
null hypothesis were true. Thus, it would indicate the alternative hypothesis is true.
6.104
The elements of the test of hypothesis that should be specified prior to analyzing the data are: null
hypothesis, alternative hypothesis, and rejection region based on α.
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356
Chapter 6
6.105
There is not a direct relationship between α and β. That is, if α is known, it does not mean β is known
because β depends on the value of the parameter in the alternative hypothesis and the sample size.
However, as α decreases, β increases for a fixed value of the parameter and a fixed sample size. Thus if α
is very small, β will tend to be large.
6.106
α = P(Type I error) = P(rejecting H0 when it is true). Thus, if rejection of H0 would cause your
firm to go out of business, you would want this probability or α to be small.
6.107
a.
H0: μ = 80
Ha: μ < 80
The test statistic is t 
x  μ0
s/ n

72.6  80
19.4 / 20
= -7.51
The rejection region requires α = .05 in the lower tail of the t-distribution with df = n − 1 = 20 − 1 =
19. From Table V, Appendix B, t.05 = 1.729. The rejection region is t < −1.729.
Since the observed value of the test statistic falls in the rejection region (−7.51 < −1.729), H0 is
rejected. There is sufficient evidence to indicate that the mean is less than 80 at α = .05.
b.
H0: μ = 80
Ha: μ ≠ 80
The test statistic is t 
x  μ0
s/ n

72.6  80
19.4 / 20
= −7.51
The rejection region requires α/2 = .01/2 = .005 in each tail of the t-distribution with df = n − 1 =
20 − 1 = 19. From Table V, Appendix B, t.005 = 2.861. The rejection region is t < −2.861 or
t > 2.861.
Since the observed value of the test statistic falls in the rejection region (−7.51 < −2.861), H0 is
rejected. There is sufficient evidence to indicate that the mean is different from 80 at α = .01.
6.108
a.
H0: μ = 8.3
Ha: μ ≠ 8.3
The test statistic is z 
x  μ0
σx

8.2  8.3
.79 / 175
= −1.67
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic does not fall in the rejection region (−1.67  −1.96), H0
is not rejected. There is insufficient evidence to indicate that the mean is different from 8.3 at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
b.
357
H0: μ = 8.4
Ha: μ ≠ 8.4
The test statistic is z 
x  μ0

σx
8.2  8.4
.79 / 175
= −3.35
The rejection region is the same as part b, z < −1.96 or z > 1.96.
Since the observed value of the test statistic falls in the rejection region (−3.35 < −1.96), H0 is
rejected. There is sufficient evidence to indicate that the mean is different from 8.4 at α = .05.
c.
H0: σ = 1
Ha: σ ≠ 1
H0: σ2 = 1
or
Ha: σ2 ≠ 1
The test statistic is χ 2 
(n  1) s 2
σ 02

(175  1)(.79) 2
 108.59
1
The rejection region requires α/2 = .05/2 = .025 in each tail of the χ 2 distribution with df = n – 1 =
2
2
 129.561 and χ.975
 74.2219. The rejection
175 – 1 = 174. From Table VI, Appendix B, χ.025
region is χ 2 > 129.561 or χ 2 < 74.2219.
Since the observed value of the test statistic does not fall in the rejection region ( χ 2 = 108.59 
129.561 and χ 2 = 108.59  74.2219), H0 is not rejected. There is insufficient evidence to indicate
the variance differs from 1 at α = .05.
d.
In part a, the rejection region is z < −1.96 or z > 1.96. In terms of x , the rejection region would be:
z
z
x  μ0
σx
x  μ0
σx
 1.96 
xU  8.3
.79
 1.96 
175
 .117  xU  8.3  xU  8.417
xL  8.3
.79
175
 .117  xL  8.3  xL  8.183
Based on x , the rejection region would be: Reject H0 if x < 8.183 or x > 8.417
The power of the test is the probability the test statistic falls in the rejection region, given the
alternative hypothesis is true. In this case, we will let μa = 8.5.
Power  P ( x  8.183 | μa  8.5)  P ( x  8.417 | μa  8.5)


8.183  8.5 
8.417  8.5 
 P z 
  P z 

.79 175 
.79 175 


 P( z  5.31)  P( z  1.39)  (.5  .5)  (.5  .4177)  .9177
(Using Table IV, Appendix B)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
358
6.109
Chapter 6
a.
H0: p = .35
Ha: p < .35
The test statistic is z =
pˆ  p 0
p 0q 0

.29  .35
.35(.65)
200
n
= −1.78
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < −1.645.
Once the observed value of the test statistic falls in the rejection region (z = −1.78 < −1.645), H0 is
rejected. There is sufficient evidence to indicate p < .35 at α = .05.
b.
H0: p = .35
Ha: p ≠ .35
The test statistic is z = −1.78 (from a).
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, t.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic does not fall in the rejection region (z = −1.78  −1.96),
H0 is not rejected. There is insufficient evidence to indicate p is different from .35 at α = .05.
6.110
6.111
a.
The p-value = .1288 = P(t ≥ 1.174). Since the p-value is not very small, there is no evidence to reject
H0 for α ≤ .10. There is no evidence to indicate the mean is greater than 10.
b.
We must assume that a random sample was selected from a population that is normally distributed.
c.
For the alternative hypothesis Ha: μ ≠ 10, the p-value is 2 times the p-value for the one-tailed test.
The p-value = 2(.1288) = .2576. There is no evidence to reject H0 for α ≤ .10. There is no evidence
to indicate the mean is different from 10.
a.
H0: σ2 = 30
Ha: σ2 > 30
The test statistic is χ2 =
(n  1) s 2
σ 02

(41  1)(6.9) 2
= 63.48
30
The rejection region requires α = .05 in the upper tail of the χ2 distribution with
2
= 55.7585. The rejection region is χ2 > 55.7585.
df = n − 1 = 40. From Table VI, Appendix B, χ .05
Since the observed value of the test statistic falls in the rejection region (χ2 = 63.48 > 55.7585), H0 is
rejected. There is sufficient evidence to indicate the variance is larger than 30 at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
b.
359
H0: σ2 = 30
Ha: σ2 ≠ 30
The test statistic is χ2 = 63.48 (from part a).
The rejection region requires α/2 = .05/2 = .025 in each tail of the χ2 distribution with df = n − 1 =
2
2
= 59.3417 and χ .975
= 24.4331. The rejection region is χ2 <
40. From Table VI, Appendix B, χ .025
24.4331 or χ2 > 59.3417.
Since the observed value of the test statistic falls in the rejection region (χ2 = 63.48 > 59.3417), H0 is
rejected. There is sufficient evidence to indicate the variance is not 30 at α = .05.
6.112
6.113
a.
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table
IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33.
b.
The test statistic is z 
c.
Since the observed value of the test statistic falls in the rejection region (z = 23.31 > 2.33),
H0 rejected. There is sufficient evidence to indicate the mean number of semester hours of
college credit taken by all first-time candidates for the CPA exam is greater than 140 hours
at α = .01.
a.
b.
pˆ 
x  μ0
σx

141.31  140
17.77
100, 000
 23.31
x 320

 .519
n 616
To determine whether the value of p has changed since 1999, we test:
H0: p = .62
Ha: p ≠ .62
c.
The test statistic is z 
pˆ  po
po qo
n

.519  .62
.62(.38)
616
 5.16
d.
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
e.
Since the observed value of the test statistic falls in the rejection region
(z = −5.16 < −1.96), H0 is rejected. There is sufficient evidence to indicate the true proportion of
businesses that suffered unauthorized use of computer systems in 2006 is different than .62 at α =
.05.
f.
The p-value is
p = P(z ≤ -5.16) + P(z ≥ 5.16) = (.5 - .5) + (.5 - .5) = 0 + 0 = 0
(Using Table IV, Appendix B)
Since the observed value of p is less than α = .05, H0 is rejected. This is the same conclusion as in
part e.
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360
6.114
Chapter 6
a.
b.
c.
The population parameter of interest is p = proportion of all television viewers with access to cableTV who agree with the statement “Overall, I find the quality of news on cable networks to be better
than news on the ABC, CBS, and NBC networks.
pˆ 
x 248

 .496
n 500
To determine if the true proportion of TV-viewers who find cable news to be better quality than
network news differs from .50, we test:
H0: p = .50
Ha: p  .50
d.
The test statistic is z 
pˆ  p0
p0 q0
n

.496  .50
.50(.50)
500
 0.18
The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645 or z < −1.645.
Since the observed value of the test statistic does not fall in the rejection region
(z = −0.18  −1.645), H0 is not rejected. There is insufficient evidence to indicate the true
proportion of TV-viewers who find cable news to be better quality than network news differs from
.50 at α = .10.
e.
The sample size is large enough to use the normal approximation if npo  15 and nqo  15 .
npo  500(.5)  250  15 and nqo  500(.5)  250  15 . Thus, the normal distribution will be
adequate.
6.115
a.
To determine if the average high technology stock is riskier than the market as a whole, we test:
H0: μ = 1
Ha: μ > 1
b.
The test statistic is t =
x  μ0
s/ n
The rejection region requires α = .10 in the upper tail of the t-distribution with df = n − 1 = 15 − 1 =
14. From Table V, Appendix B, t.10 = 1.345. The rejection region is t > 1.345.
c.
We must assume the population of beta coefficients of technology stocks is normally distributed.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
d.
The test statistic is t =
x  μ0
s/ n

1.23  1
.37 / 15
361
= 2.41
Since the observed value of the test statistic falls in the rejection region (t = 2.41 > 1.345), H0 is
rejected. There is sufficient evidence to indicate the mean high technology stock is riskier than the
market as a whole at α = .10.
e.
From Table V, Appendix B, with df = n − 1 = 15 − 1 = 14, .01 < P(t ≥ 2.41) < .025. Thus, .01 < pvalue < .025. The probability of observing this test statistic, t = 2.41, or anything more unusual is
between .01 and .025. Since this probability is small, there is evidence to indicate the null hypothesis
is false for α = .05.
f.
To determine if the variance of the stock beta values differs from .15, we test:
Ho: σ2 = .15
Ha: σ2 ≠ .15
The test statistic is χ 2 
(n  1) s 2
σ o2

(15  1).372
 12.7773 .
.15
The rejection region requires α/2 = .05/2 = .025 in each tail of the χ2 distribution with
2
2
= 5.62872 and χ.025
= 26.1190.
df = n – 1 = 15 – 1 = 14. From Table VI, Appendix B, χ.975
2
2
The rejection region is χ > 26.1190 or χ < 5.62872.
Since the observed value of the test statistic does not fall in the rejection region
(χ2 = 12.7773  26.1190 and χ2 = 12.7773  5.62875), H0 is not rejected. There is
insufficient evidence to indicate the variance of the stock beta values differs from .15 at α = .05.
6.116
a.
Let p = proportion of time the camera correctly detects liars. The null hypothesis would be:
H0: p = .75
b.
6.117
A Type I error would be to conclude the camera cannot correctly identify liars 75% of the time when,
in fact, it can. A Type II error would be to conclude the camera can correctly identify liars 75% of the
time when, in fact, it cannot.
To determine whether the true mean pouring temperature differs from the target setting, we test:
H0: μ = 2,550
Ha: μ ≠ 2,550
From the printout, the test statistic is t = 1.210 and the p-value is .257.
Since the p-value is greater than α = .01, H0 is not rejected. There is insufficient evidence to indicate the
true mean pouring temperature is different from 2,550 degrees at α = .01.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
362
6.118
Chapter 6
a.
Let p = proportion of shoppers using cents-off coupons. To determine if the proportion of shoppers
using cents-off coupons exceeds .65, we test:
H0: p = .65
Ha: p > .65
The test statistic is z 
pˆ  p0
p0 q0
n

.72  .65
.65(.35)
1, 000
 4.64
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 4.64 > 1.645), H0 is
rejected. There is sufficient evidence to indicate the proportion of shoppers using cents-off coupons
exceeds .65 at α = .05.
b.
The sample size is large enough if the np0  15 and nq0  15 .
np0  1000(.65)  650
nq0  1000(.35)  350
Since both np0  15 and nq0  15, the normal distribution will be adequate.
6.119
c.
The p-value is p  P( z  4.64)  (.5  .5)  .0 . (Using Table IV, Appendix B.) Since the p-value is
smaller than α = .05, H0 is rejected. There is sufficient evidence to indicate the proportion of
shoppers using cents-off coupons exceeds .65 at α = .05.
a.
The hypotheses would be:
H0: Individual does not have the disease
Ha: Individual does have the disease
b.
A Type I error would be: Conclude the individual has the disease when in fact he/she does not. This
would be a false positive test.
A Type II error would be: Conclude the individual does not have the disease when in fact he/she
does. This would be a false negative test.
c.
If the disease is serious, either error would be grave. Arguments could be made for either error being
more grave. However, I believe a Type II error would be more grave: Concluding the individual
does not have the disease when he/she does. This person would not receive critical treatment, and
may suffer very serious consequences. Thus, it is more important to minimize β.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
6.120
363
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Tunnel
Variable
Tunnel
N
10
Mean
989.8
Median
970.5
TrMean
987.9
Variable
Tunnel
Minimum
735.0
Maximum
1260.0
Q1
862.5
Q3
1096.8
StDev
160.7
SE Mean
50.8
To determine whether peak hour pricing succeeded in reducing the average number of vehicles attempting
to use the Lincoln Tunnel during the peak rush hour, we test:
H0: μ = 1,220
Ha: μ < 1,220
The test statistic is t =
x  μ0
s/ n

989.8  1, 220
160.7 / 10
= −4.53
Since no α is given, we will use α = .05. The rejection region requires α = .05 in the lower tail of the
t-distribution with df = n − 1 = 10 − 1 = 9. From Table V, Appendix B, t.05 = 1.833. The rejection region
is t < −1.833.
Since the observed value of the test statistic falls in the rejection region (t = −4.53 < −1.833), H0 is rejected.
There is sufficient evidence to indicate that peak hour pricing succeeded in reducing the average number of
vehicles attempting to use the Lincoln Tunnel during the peak rush hour at α = .05.
6.121
To determine if the true standard deviation of the point-spread errors exceed 15 (variance exceeds 225), we
test:
H0: σ2 = 225
Ha: σ2 > 225
The test statistic is χ2 =
(n  1) s 2
σ 02

(240  1)13.32
= 187.896
225
The rejection region requires α in the upper tail of the χ2 distribution with df = n − 1 = 240 − 1 = 239.
The maximum value of df in Table VI is 100. Thus, we cannot find the rejection region using Table VII.
Using a statistical package, the p-value associated with χ2 = 187.896 is .9938.
Since the p-value is so large, there is no evidence to reject H0. There is insufficient evidence to indicate
that the true standard deviation of the point-spread errors exceeds 15 for any reasonable value of α.
(Since the observed variance (or standard deviation) is less than the hypothesized value of the variance (or
standard deviation) under H0, there is no way H0 will be rejected for any reasonable value of α.)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
364
6.122
Chapter 6
a.
To determine if the true mean number of pecks at the blue string is less than 7.5, we test:
H0: μ = 7.5
Ha: μ < 7.5
The test statistic is z 
x  μ0
σx

1.13  7.5
2.21
72
 24.46
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV,
Appendix B, z.01 = 2.33. The rejection region is z < −2.33.
Since the observed value of the test statistic falls in the rejection region (z = −24.46 < −2.33), H0 is
rejected. There is sufficient evidence to indicate the true mean number of pecks at the blue string is
less than 7.5 at α = .01.
6.123
b.
From Exercise 5.100, the 99% confidence interval is (.46, 1.80). Since the hypothesized value of the
mean (μ = 7.5) does not fall in the confidence interval, it is not a likely candidate for the true value of
the mean. Thus, you would reject it. This agrees with the conclusion in part a.
a.
First, check to see if n is large enough:
np0  132(.5)  66
nq0  132(.5)  66
Since both np0  15 and nq0  15 , the normal distribution will be adequate.
To determine if there is evidence to reject the claim that no more than half of all manufacturers are
dissatisfied with their trade promotion spending, we test:
H0: p = .5
Ha: p > .5
The test statistic is z =
p̂  p 0
p 0q 0
n
=
.36  .5
.5(.5)
132
= 3.22
The rejection region requires α = .02 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.02 = 2.05. The rejection region is z > 2.05.
Since the observed value of the test statistic does not fall in the rejection region (z = 3.22  2.05),
H0 is not rejected. There is insufficient evidence to reject the claim that no more than half of all
manufacturers are dissatisfied with their trade promotion spending at α = .02.
b.
The observed significance level is p-value = P(z ≥ 3.22 ) ≈ .5 + .5 = 1. Since this p-value is so
large, H0 will not be rejected for any reasonable value of α.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
c.
First, we must define the rejection region in terms of p̂ .
p̂ = p0 + zα σ p̂ = .5 + 2.05
6.124
a.
365
.5(.5)
= .589
132



.589  .55 
 = P(z < .90) = .5 + .3159 = .8159
β = P( p̂ < .589│p = .55) = P  z 

.55(.45) 


132 
p̂ = 24/40 = .6
To determine if the proportion of shoplifters turned over to police is greater than .5, we test:
H0: p = .5
Ha: p > .5
The test statistic is z =
pˆ  p0
p0 q0
n

.6  .5
.5(.5)
40
= 1.26
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.26  1.645),
H0 is not rejected. There is insufficient evidence to indicate the proportion of shoplifters turned over
to police is greater than .5 at α = .05.
b.
To determine if the normal approximation is appropriate, we check:
np0  40(.5)  20
nq0  40(.5)  20
Since both np0  15 and nq0  15 , the normal distribution will be adequate.
c.
The observed significance level of the test is p-value = P(z ≥ 1.26) = .5 − .3962 = .1038.
(Using Table IV, Appendix B)
The probability of observing the value of our test statistic or anything more unusual if the true value
of p is .5 is .4038. Since this p-value is so large, there is no evidence to reject H0. There is no
evidence to indicate the true proportion of shoplifters turned over to police is greater than .5.
d.
Any value of α that is greater than the p-value would lead one to reject H0. Thus, for this problem,
we would reject H0 for any value of α > .1038.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
366
6.125
Chapter 6
a.
A Type II error is concluding the percentage of shoplifters turned over to police is 50% when in
fact, the percentage is higher than 50%.
b.
First, calculate the value of p̂ that corresponds to the border between the acceptance region and the
rejection region.
P( pˆ  po )  P ( z  zo )  .05.
From Table IV, Appendix B, zo = 1.645
pˆ  po  1.645σ pˆ  .5  1.645
.5(.5)
 .5  .1300  .6300
40
β  P ( pˆ  .6300 when p  .55)



.6300-.55 
  P( z  1.02)  .5  .3461  .8461
 P z 

.55(.45) 


40 
c.
If n increases, the probability of a Type II error would decrease.
First, calculate the value of p̂ that corresponds to the border between the acceptance region and the
rejection region.
P( pˆ  po )  P ( z  zo )  .05.
From Table IV, Appendix B, zo = 1.645
pˆ  po  1.645σ pˆ  .5  1.645
.5(.5)
 .5  .082  .582
100
β  P ( pˆ  .582 when p  .55)


.582 -.55
 P z 

.55(.45)

100
6.126
a.


  P( z  0.64)  .5  .2389  .7389


To determine whether the mean profit change for restaurants with frequency programs is greater than
$1047.34, we test:
H0: μ = 1047.34
Ha: μ > 1047.34
b.
Some preliminary calculations are:
x =
 x  30,113.17
n
12
 x 

= 2,509.43
2
s2 =
x
2
n
n 1

30,113.172
12
= 4,619,331.955
12  1
126, 379, 568.8 
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on a Single Sample: Tests of Hypothesis
s=
367
4, 619, 331.955 = 2149.2631
The test statistic is t =
x  μ0
s/ n
2509.43  1047.34

2149.2631/ 12
= 2.36
The rejection region requires α = .05 in the upper tail of the t-distribution with
df = n − 1 = 12 − 1 = 11. From Table V, Appendix B, t.05 = 1.796. The rejection region is
t > 1.796.
Since the observed value of the test statistic falls in the rejection region (t = 2.36 > 1.796), H0 is
rejected. There is sufficient evidence to indicate the mean profit change for restaurants with
frequency programs is greater than $1047.34 for α = .05.
It appears that the frequency program would be profitable for the company if adopted nationwide.
6.127
a.
To determine if the production process should be halted, we test:
H0: μ = 3
Ha: μ > 3
where μ = mean amount of PCB in the effluent.
The test statistic is z 
x  μ0
σx

3.1  3
.5 / 50
= 1.41
The rejection region requires α =.01 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic does not fall in the rejection region, (z = 1.41  2.33),
H0 is not rejected. There is insufficient evidence to indicate the mean amount of PCB in the effluent
is more than 3 parts per million at α = .01. Do not halt the manufacturing process.
6.128
b.
As plant manager, I do not want to shut down the plant unnecessarily. Therefore, I want α = P(shut
down plant when μ = 3) to be small.
c.
The p-value is p = P(z ≥ 1.41) = .5 − .4207 = .0793. Since the p-value is not less than α = .01, H0 is
not rejected.
a.
A Type II error would be concluding the mean amount of PCB in the air is less than or equal to 3
parts per million when, in fact, it is more than 3 parts per million.
b.
From Exercise 6.127, z =
x0  μ
σ/ n
 x0  z
σ
n
+ μ0  x0 = 2.33
.5
50
+ 3  x0 = 3.165



3.165  3.1 
For μ = 3.1, β = P( x ≤ 3.165) = P  z 
 = P(z ≤ .92) = .5 + .3212 = .8212
.5




50
(from Table IV, Appendix B)
c.
Power = 1 − β = 1 − .8212 = .1788
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368
Chapter 6
d.



3.165  3.2 
For μ = 3.2, β = P( x ≤ 3.165) = P  z 
 = P(z ≤ −.49) = .5 − .1879 = .3121
.5




50
Power = 1 − β = 1 − .3121 = .6879
As the plant's mean PCB departs further from 3, the power increases.
6.129
a.
No, it increases the risk of falsely rejecting H0, i.e., closing the plant unnecessarily.
b.
First, find x0 such that P( x > x0 ) = P(z > z0) = .05.
From Table IV, Appendix B, z0 = 1.645
z=
x0  μ
σ/ n
x0  3
 1.645 =
.5 / 50
 x0 = 3.116
Then, compute:

3.116  3.1 
β = P( x0 ≤ 3.116 when μ = 3.1) = P  z 
 = P(z ≤ .23)

.5 / 50 
= .5 + .0910 = .5910
Power = 1 − β = 1 − .5910 = .4090
6.130
c.
The power of the test increases as α increases.
a.
Some preliminary calculations:
x =
s2 =
s=
 x  79.93 = 15.986
n

x2
5
 x 

2
n

n 1
.00043 = .0207
1, 277.7627 
5 1
79.932
5
= .00043
To determine if the mean measurement differs from 16.01, we test:
H0: μ = 16.01
Ha: μ ≠ 16.01
The test statistic is t =
x  μ0
s/ n

15, 986  16.01
.0207 / 5
= −2.59
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 = 5 −
1 = 4. From Table V, Appendix B, t.025 = 2.776. The rejection region is t < −2.776 or t > 2.776.
Since the observed value of the test statistic does not fall in the rejection region (t = −2.59  −2.776),
H0 is not rejected. There is insufficient evidence to indicate the true mean measurement differs from
16.01 at α = .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
369
b.
We must assume that the sample of measurements was randomly selected from a population of
measurements that is normally distributed.
c.
To determine if the standard deviation of the weight measurements is greater than .01, we test:
H0: σ2 = .012
Ha: σ2 > .012
The test statistic is χ 2 
(n  1) s 2
σ o2

(5  1).0207 2
.012
 16.0684 .
The rejection region requires α = .05 in the upper tail of the χ2 distribution with
2
df = n – 1 = 5 – 1 = 4. From Table VI, Appendix B, χ.05
= 9.48773. The rejection region is
χ2 > 9.48773.
Since the observed value of the test statistic falls in the rejection region (χ2 = 16.0684 > 9.48773), H0
is rejected. There is sufficient evidence to indicate the standard deviation of the weight
measurements is greater than .01 at α = .05.
6.131
a.
To determine if the product brings pain relief to headache sufferers in less than 3.5 minutes, on
average, we test:
H0: μ = 3.5
Ha: μ < 3.5
The test statistic is z =
x  μ0
σx

3.3  3.5
1.1 / 50
= −1.29
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645.
Since the observed value of the test statistic does not fall in the rejection region
(z = −1.29  −1.645), H0 is not rejected. There is insufficient evidence to indicate that the product
brings pain relief to headache sufferers in less than 3.5 minutes, on average, at α = .05.
6.132
b.
The p-value is p = P(z ≤ −1.29) = .5 − .4015 = .0985.
c.
If we are conducting a test of hypothesis where the manufacturer's claim is stated in the alternative
hypothesis (Ha), then small p-values would support the manufacturer's claim. On the other hand, if
we are conducting a test of hypothesis where the manufacturer's claim is stated in the null hypothesis
(H0), then large p-values would support the manufacturer's claim.
a.
Let pi = proportion of first round games won by the ith seed. To determine if the higher seed has a
better than 50-50 chance of winning a first-round game, we test:
H0: pi = .5
Ha: pi > .5 for i = 1, 2, 3, …, 8
The test statistic is zi 
pˆ i  p0
po qo
.
n
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370
Chapter 6
No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail
of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
xi
x
x
x
52
49
41
. Thus, pˆ1  1 
 1 , pˆ 2  2 
 .942 , pˆ 3  3 
 .788 ,
n
n 52
n 52
n 52
x
x
x
x
42
37
36
35
 .808 , pˆ 5  5 
 .712 , pˆ 6  6 
 .692 , pˆ 7  7 
 .673 ,
pˆ 4  4 
n 52
n 52
n 52
n 52
x
22
 .423
pˆ 8  8 
n 52
pˆ i 
The corresponding test statistics are:
z1 
pˆ1  p0
po qo

n
z3 
pˆ 3  p0
po qo

n
z5 
pˆ 5  p0
po qo

n
z7 
pˆ 7  p0
po qo
n

1.00  .5
.5(.5)
52
.788  .5
.5(.5)
52
.712  .5
.5(.5)
52
.673  .5
.5(.5)
52
 7.21 , z2 
pˆ 2  p0
po qo

.942  .5
n
 4.15 , z4 
pˆ 4  p0
po qo

n
 3.06 , z6 
pˆ 6  p0
po qo

n
 2.50 , z8 
pˆ 8  p0
po qo
n

.5(.5)
52
.808  .5
.5(.5)
52
.692  .5
.5(.5)
52
.423  .5
.5(.5)
52
 6.37 ,
 4.44 ,
 2.77 ,
 1.11
For games matching 1 and 16, since the observed value of the test statistic falls in the rejection region
(z1 = 7.21 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #1 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 2 and 15, since the observed value of the test statistic falls in the rejection region
(z2 = 6.37 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #2 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 3 and 14, since the observed value of the test statistic falls in the rejection region
(z3 = 4.15 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #3 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 4 and 13, since the observed value of the test statistic falls in the rejection region
(z4 = 4.44 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #4 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 5 and 12, since the observed value of the test statistic falls in the rejection region
(z5 = 3.06 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #5 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
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Inferences Based on a Single Sample: Tests of Hypothesis
371
For games matching 6 and 11, since the observed value of the test statistic falls in the rejection region
(z6 = 2.77 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #6 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 7 and 10, since the observed value of the test statistic falls in the rejection region
(z7 = 2.50 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #7 seed has a better
than 50-50 chance of winning a first-round game at α = .05.
For games matching 8 and 9, since the observed value of the test statistic does not fall in the rejection
region (z8 = −1.11  1.645), H0 is not rejected. There is insufficient evidence to indicate the #8 seed
has a better than 50-50 chance of winning a first-round game at α = .05.
b.
Let μi = mean margin of victory. To determine if the mean margin of victory is greater than 10
points, we test:
H0: μi = 10
Ha: μi > 10 i = 1, 2, 3, and 4
The test statistic is zi 
xi  μ0
σx
No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail
of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
The test statistics are:
z1 
z3 
x1  μ0
σx
x3  μ0
σx


22.9  10
12.4
52
10.6  10
12.0
52
 7.50 , z2 
 0.36 , z4 
x2  μ0
σx
x4  μ0
σx


17.2  10
11.4
52
10.0  10
12.5
52
 4.55 ,
0
For games matching 1 and 16, since the observed value of the test statistic falls in the rejection region
(z1 = 7.50 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #1 seed wins by more
than 10 points in first-round games at α = .05.
For games matching 2 and 15, since the observed value of the test statistic falls in the rejection region
(z2 = 4.55 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #2 seed wins by more
than 10 points in first-round games at α = .05.
For games matching 3 and 14, since the observed value of the test statistic does not fall in the
rejection region (z3 = 0.36  1.645), H0 is not rejected. There is insufficient evidence to indicate the
#3 seed wins by more than 10 points in first-round games at
α = .05.
For games matching 4 and 13, since the observed value of the test statistic does not fall in the
rejection region (z4 = 0  1.645), H0 is not rejected. There is insufficient evidence to indicate the #4
seed wins by more than 10 points in first-round games at α = .05.
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372
Chapter 6
c.
Let μi = mean margin of victory. To determine if the mean margin of victory is less than 5 points, we
test:
H0: μi = 5
Ha: μi < 5 i = 5, 6, 7, and 8
xi  μ0
The test statistic is zi 
σx
No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail
of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645.
The test statistics are:
z5 
z7 
x5  μ0
σx
x7  μ0
σx


5.3  5
10.4
52
3.2  5
10.5
52
 0.21 , z6 
x6  μ0
 1.24 , z8 
σx

x8  μ0
σx
4.3  5
10.7

52
 .47 ,
2.1  5
11.0
52
 4.65
For games matching 5 and 12, since the observed value of the test statistic does not fall in the
rejection region (z5 = 0.21  −1.645), H0 is not rejected. There is insufficient evidence to indicate
the #5 seed wins by less than 5 points in first-round games at α = .05.
For games matching 6 and 11, since the observed value of the test statistic does not fall in the
rejection region (z6 = −0.47  −1.645), H0 is not rejected. There is insufficient evidence to indicate
the #6 seed wins by less than 5 points in first-round games at α = .05.
For games matching 7 and 10, since the observed value of the test statistic does not fall in the
rejection region (z7 = −1.24  −1.645), H0 is not rejected. There is insufficient evidence to indicate
the #7 seed wins by less than 5 points in first-round games at α = .05.
For games matching 8 and 9, since the observed value of the test statistic falls in the rejection region
(z8 = −4.65 < −1.645), H0 is rejected. There is sufficient evidence to indicate the #8 seed wins by less
than 5 points in first-round games at α = .05.
d.
To determine if the standard deviation of victory margin differs from 11, we test:
H0: σ12 = 112 = 121
Ha: σ12 ≠ 112 = 121
The test statistic is χ i2 
(n  1) si2
σ 02
No α level was given, so we will use α = .05. The rejection region requires α/2 = .05/2 = .025 in
each tail of the χ 2 distribution with df = n – 1 = 52 – 1 = 51. From Table VI, Appendix B,
2
2
χ.025
 71.4202 and χ.975
 32.3574. The rejection region is χ 2 < 32.3574 or χ 2 > 71.4202.
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Inferences Based on a Single Sample: Tests of Hypothesis
373
The test statistics are:
χ12 
χ 32 
χ 52 
χ 72 
(n  1) s12
σ 02
(n  1) s32
σ 02
(n  1) s52
σ 02
(n  1) s72
σ 02

(n  1) s22 (52  1)(11.4)2
(52  1)(12.4)2
 64.808 , χ 22 

 54.777 ,
121
121
σ 02

(n  1) s42 (52  1)(12.5)2
(52  1)(12.0)2

 65.857 ,
 60.694 , χ 42 
121
121
σ 02

(n  1) s62 (52  1)(10.7) 2
(52  1)(10.4)2
 45.588 , χ 62 

 48.256 ,
121
121
σ 02

(n  1) s82 (52  1)(11)2
(52  1)(10.5)2
 46.469 , χ82 

 51.000
121
121
σ 02
For games matching 1 and 16, since the observed value of the test statistic does not fall in the
rejection region ( χ12 = 64.808  71.4202 and χ12 = 64.808  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 2 and 15, since the observed value of the test statistic does not fall in the
rejection region ( χ 22 = 54.777  71.4202 and χ 22 = 54.777  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 3 and 14, since the observed value of the test statistic does not fall in the
rejection region ( χ32 = 60.694  71.4202 and χ32 = 60.694  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 4 and 13, since the observed value of the test statistic does not fall in the
rejection region ( χ 42 = 65.857  71.4202 and χ 42 = 65.857  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 5 and 12, since the observed value of the test statistic does not fall in the
rejection region ( χ52 = 45.588  71.4202 and χ52 = 45.588  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 6 and 11, since the observed value of the test statistic does not fall in the
rejection region ( χ 62 = 48.256  71.4202 and χ 62 = 48.256  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 7 and 10, since the observed value of the test statistic does not fall in the
rejection region ( χ 72 = 46.469  71.4202 and χ 72 = 46.469  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
For games matching 8 and 9, since the observed value of the test statistic does not fall in the rejection
region ( χ82 = 51.000  71.4202 and χ82 = 51.000  32.3574), H0 is not rejected. There is
insufficient evidence to indicate standard deviation of victory margin differs from 11 at α = .05.
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374
Chapter 6
e.
Let μ = mean difference in game outcome and point spread. To determine if the point
spread is a good predictor of the victory margin, we test:
H0: μ = 0
Ha: μ ≠ 0
The test statistic is z 
x  μ0
σx

.7  0
11.3
360
 1.18 .
Since no α was given, we will use α = .05. The rejection region requires α/2 = .05/2 = .025 in each
tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z > 1.96 or
z < −1.96.
Since the observed value of the test statistic does not fall in the rejection region
(z = 1.18  1.96), H0 is not rejected. There is insufficient evidence to indicate there is a difference
in the game outcome and point spread at α = .05. There is no evidence to indicate the point spread is
not a good predictor of the victory margin.
6.133
a.
To determine whether the true mean rating for this instructor-related factor exceeds 4, we test:
H0: μ = 4
Ha: μ > 4
The test statistic is z =
x  μ0
σx

4.7  4
1.62 / 40
= 2.73
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 2.73 > 1.645), H0 is
rejected. There is sufficient evidence to indicate that the true mean rating for this instructor-related
factor exceeds 4 at α = .05.
b.
If the sample size is large enough, one could almost always reject H0. Thus, we might be able to
detect very small differences if the sample size is large enough. This would be statistical
significance. However, even though statistical significance is found, it does not necessarily mean
that there is practical significance. A statistical significance can sometimes be found between the
hypothesized value of a mean and the estimated value of the mean, but, in practice, this difference
would mean nothing. This would be practical significance.
c.
Since the sample size is sufficiently large (n = 40), the Central Limit Theorem indicates that the
sampling distribution of is approximately normal. Also, since the sample size is large, s is a good
estimator of σ. Thus, the analysis used is appropriate.
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Inferences Based on a Single Sample: Tests of Hypothesis
6.134
375
Let p = proportion of patients taking the pill who reported an improved condition.
First we check to see if the normal approximation is adequate:
np0  7000(.5)  3500
nq0  7000(.5)  3500
Since both np0  15 and nq0  15 , the normal distribution will be adequate.
To determine if there really is a placebo effect at the clinic, we test:
H0: p = .5
Ha: p > .5
The test statistic is z =
pˆ  p0
p0 q0

n
.7  .5
.5(.5)
7000
= 33.47
The rejection region requires α = .05 in the upper tail of the z distribution. From Table IV, Appendix B,
z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 33.47 > 1.645), H0 is rejected.
There is sufficient evidence to indicate that there really is a placebo effect at the clinic at α = .05.
6.135
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Candy
Variable
Candy
N
5
Mean
22.000
SE
Mean
0.894
StDev
2.000
Minimum
20.000
Q1
20.500
Median
21.000
Q3
24.000
Maximum
25.000
To give the benefit of the doubt to the students we will use a small value of α. (We do not want to reject H0
when it is true to favor the students.) Thus, we will use α = .001.
We must also assume that the sample comes from a normal distribution. To determine if the mean number
of candies exceeds 15, we test:
H0: μ = 15
Ha: μ > 15
The test statistic is z 
x  μo
σ
n

22  15
2
5
 7.83
The rejection region requires α = .001 in the upper tail of the z-distribution. From Table IV, Appendix B,
z.001 = 3.08. The rejection region is z > 3.08.
Since the observed value of the test statistic falls in the rejection region (z = 7.83 > 3.08),
H0 is rejected. There is sufficient evidence to indicate the mean number of candies exceeds 15 at α = .001.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.