Download Chapter Nine

Chapter Nine
9.1
a. The null hypothesis is a claim about a population parameter that is assumed to be true until it is
declared false.
b. An alternative hypothesis is a claim about a population parameter that will be true if the null
hypothesis is false.
c. The critical point(s) divides the whole area under a distribution curve into rejection and nonrejection regions.
d. The significance level, denoted by α, is the probability of making a Type I error, that is, the
probability of rejecting the null hypothesis when it is actually true.
e. The nonrejection region is the area where the null hypothesis is not rejected.
f. The rejection region is the area where the null hypothesis is rejected.
g. A hypothesis test is a two-tailed test if the rejection regions are in both tails of the distribution
curve; it is a left-tailed test if the rejection region is in the left tail; and it is a right-tailed test if the
rejection region is in the right tail.
h. Type I error: A type I error occurs when a true null hypothesis is rejected. The probability of
committing a Type I error, denoted by α is: α = P(H0 is rejected | H0 is true)
Type II error: A Type II error occurs when a false null hypothesis is not rejected. The probability of
committing a Type II error, denoted by β, is: β = P(H0 is not rejected |H0 is false)
9.3
A hypothesis test is a two-tailed test if the sign in the alternative hypothesis is "≠ "; it is a left-tailed
test if the sign in the alternative hypothesis is " < " (less than); and it is a right-tailed test if the sign in
the alternative hypothesis is " > " (greater than). Table 9.3 on page 405 of the text describes these
relationships.
9.5
a. Left-tailed test
b. Right-tailed test
9.7
a. Type II error
b. Type I error
9.9
a. H0: μ = 20 hours;
H1: μ ≠ 20 hours;
a two-tailed test
b. H0: μ = 10 hours;
H1: μ > 10 hours;
a right-tailed test
c. H0: μ = 3 years;
H1: μ ≠ 3 years;
a two-tailed test
159
c. Two-tailed test
160
9.11
Chapter Nine
d. H0: μ = $1000;
H1: μ < $1000;
a left-tailed test
e. H0: μ = 12 minutes;
H1: μ > 12 minutes;
a right-tailed test
For a two-tailed test, the p–value is twice the area in the tail of the sampling distribution curve beyond
the observed value of the sample test statistic.
For a one-tailed test, the p–value is the area in the tail of the sampling distribution curve beyond the
observed value of the sample test statistic.
9.13
a. Step 1:
H0: µ = 46;
H1: µ≠ 46;
A two-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
s x = s / n = 9.7 /
40 = 1.53370467
z = ( x – μ) / s x = (49.60–46) / 1.53370467 = 2.35
From the normal distribution table, area to the right of z = 2.35 is .5 – .4906 = .0094 approximately.
Hence, p–value = 2(.0094) = .0188
b. Step 1:
H0: µ = 26;
H1: µ < 26;
A left-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
s x = s / n = 4.3 /
33 = .74853392
z = ( x – μ) / s x = (24.3–26) / .74853392 = –2.27
From the normal distribution table, area to the left of z = –2.27 is .5 –.4884 = .0116 approximately.
Hence, p–value = .0116
c. Step 1:
H0: µ = 18;
H1: µ > 18;
A right-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
s x = s / n = 7.8 /
55 = 1.05175179
z = ( x – μ) / s x = (20.50 – 18) / 1.05175179 = 2.38
From the normal distribution table, area to the right of z = 2.38 is .5 – .4913 = .0087 approximately.
Hence, p –value = .0087
TI-83: If you have raw data, first enter your data in a list. Otherwise just select Stat, highlight TESTS,
highlight 1: Z Test, and press the ENTER key. If you have raw data choose Data and press the ENTER
key, and then using the down arrow to move to enter the requested values and the list name, but be sure
to keep Freq set at 1. On the other hand if you have summary statistics, instead of Data highlight Stats
and press the ENTER key. Using the down arrow, scroll down to start entering the requested values. In
this example we do not have raw data, so will follow the latter method. We select Stat, scroll down
enter for part a, µ0 = 46, σ = s = 9.7, x  49.60, n = 40, for µA highlighting  µ0, highlighting
Calculate, and pressing the ENTER key. The results are shown below where p is the p value we seek.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
161
Z − Test
µ  46
z = 2.347257645
p = .0189121347
x  29.6
n = 40
MINITAB: Select the following Stat, Basic Statistics, and 1-Sample Z to generate a pop-up menu. In
the pop-up menu, click beside the words Summarized data as long as you do indeed have summary
data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word
standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on
Options to generate another pop-up menu where you make sure it says the correct Alternative
hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than
or less than, then use the down arrow to highlight the left tailed or right tailed µ1. Once the correct µ1 is
in the box, then click on OK. This returns you to the original pop-up menu where you also click on OK
and the resulting confidence interval appears in the Session section. The pop-up menus as well as the
resulting confidence interval are shown below. The results are shown below where p is the p value we
seek.
Excel: Open up KADDSTAT and select KADD, Hypothesis Testing, One Sample, Population mean
using Z. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if
you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this
question, then click beside user input other wise click beside the other choice and enter your range of
cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is
unknown. In the box beside cell: type in the address on your worksheet where you want the answer to
appear. Finally click on OK. For part a, the results and the pop-up menu are shown below.
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Chapter Nine
Caution: For parts b and c, when using technology be sure to select the correct tail for your one tailed
test instead the “NOT EQUAL”to used with the two tail test in these instructions which is only
appropriate for part a.
9.15
a. Step 1:
H0: µ = 72;
Hl: µ > 72;
A right-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
sx = s / n = 6 /
36 = 1.0
z = ( x – μ) / s x = (74.07 – 72) / 1.0 = 2.07
The area under the standard normal curve to the right of z = 2.07 is .5 – .4808 = .0192.
Hence p–value = .0192
b. For α = .01, do not reject H0, since p–value > .01.
c. For α = .025, reject H0, since p–value < .025.
9.17
H0: µ = 16.3 weeks;
s x = s / n = 4.2 /
H1: µ >16.3 weeks;
A right-tailed test.
400 = .21
Test statistic: z = ( x – μ) / s x = (16.9 – 16.3) / .21 = 2.86
The area under the standard normal curve to the right of z = 2.86 is .5 – .4979 = .0021 approximately.
Hence, p–value = .0021
For α = .02, reject H0, since p–value < .02.
9.19
H0: µ ≥ 14 hours;
s x = s / n = 3.0 /
H1 : µ < 14 hours;
A left-tailed test.
200 = .21213203
Test statistic: z = ( x – μ) / s x = (13.75 – 14) / .21213203 = –1.18
The area under the standard normal curve to the left of z = –1.18 is .5 – .3810 = .1190
Hence, p–value = .1190.
For α = .05, do not reject H0, since p–value > .05.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
9.21
a.
H0: µ = 10 minutes;
H1: µ < 10 minutes;
s x = s / n = 3.75 /
100 = .375
163
A left-tailed test.
z = ( x – μ) / s x = (9.25 – 10) / .375 = –2.00
The area under the standard normal curve to the left of z = –2.00 is .5 – .4772 = .0228.
Hence, p–value = .0228
b. Do not reject H0 for α = .02, since p–value > .02.
Reject H0 for α = .05, since p–value < .05.
9.23
a. H0: µ = 32 ounces;
 x   / n = .15 /
H1: µ ≠ 32 ounces;
A two-tailed test.
35 = .02535463
z = ( x – μ) /  x  (31.90 – 32) / .02535463 = –3.94
The area under the standard normal curve to the left of z = –3.94 is approximately .5 –.5 = 0.
Hence, p–value = 2(.00) = .00 approximately
b. For α = .01, reject H0, since p–value < .01.
For α = .05, reject H0, since p–value < .05.
Thus in either case, the inspector will stop the machine.
9.25
The level of significance in a test of hypothesis is the probability of making a Type I error. It is the
area under the probability distribution curve where we reject H0.
9.27
The critical value of z separates the rejection region from the nonrejection region and is found from a
table such as the standard normal distribution table. The observed value of z is the value calculated for
a sample statistic such as x .
9.29
a. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58.
The nonrejection region lies between z = –2.58 and z = 2.58.
b. The rejection region lies to the left of z = –2.58.
The nonrejection region lies to the right of z = –2.58.
c. The rejection region lies to the right of z = 1.96.
The nonrejection region lies to the left of z = 1.96.
Note: To find the critical value or values of z to create the rejection and nonrejection regions one
method is to follow the procedures outlined in Chapter 6 of the text or in problems like 6.53 in Chapter
6 of the solutions manual.
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Chapter Nine
9.31
If H0 is not rejected, the difference between the hypothesized value of μ and the observed value of x is
"statistically not significant".
9.33
a. .10
b. .02
c. .005
9.35
n = 90, x = 15, s = 4, and s x = s / n = 4 / 90 = .42163702
a. Critical value: z = –2.33
Observed value: z = ( x – μ) / s x = (15 – 20) / .42163702 = –11.86
b. Critical value: z = –2.58 and 2.58.
Observed value: z = ( x – μ) / s x = (15 – 20) / .42163702 = –11.86
Note: To find the observed value of z and the critical value of z follow the procedures outlined in
Chapter 6 of the text or in problems 6.17 and 6.53 of the solutions manual.
9.37
a. The rejection region lies to the left of z = –2.33.
The nonrejection region lies to the right of z = –2.33.
b. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58.
The nonrejection region lies between z = –2.58 and z = 2.58.
c. The rejection region lies to the right of z = 2.33.
The nonrejection region lies to the left of z = 2.33.
9.39
a. n = 100, x = 43, s = 5, and s x = s / n = 5 / 100 = .50
Critical value: z = –1.96
Test statistic: z = ( x – μ) / s x =(43 – 45) / .50 = –4.00; Reject H0.
b. n = 100, x = 43.8, s = 7, and s x = s / n = 7 / 100 = .70
Critical value: z = –1.96
Test statistic: z = ( x – μ) / s x = (43.8 – 45) / .70 = –1.71; Do not reject H0.
Comparing parts a and b shows that two samples selected from the same population can yield opposite
conclusions on the same test of hypothesis.
TI-83: If you have raw data, first enter your data in a list. Otherwise just select Stat, then TESTS, then
1: Z Test, and finally pressing the ENTER key. If you have raw data choose Data and press the
ENTER key, and then using the down arrow to move to enter the requested values and the list name,
but be sure to keep Freq set at 1. On the other hand if you have summary statistics, instead of Data
highlight Stats and press the ENTER key. Using the down arrow, scroll down to start entering the
requested values. In this example we do not have raw data, so will follow the latter method. We select
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
165
Stat, scroll down, and enter for part a, µ0 = 45, σ = s = 5, x  43, n = 100. Then we highlight < 45 for
µA, highlight Calculate, and press the ENTER key. The results are shown below where p is the p value
we seek to compare with α. If we make out comparison in terms of z, we calculate the critical value or
values of z first using the methods describe previously. Next we compare the critical value of z with
the observed value of z reported below. Since the observed z is less than the critical value of z we do
not reject the null hypothesis for part b, but the observed z is more than the critical value of z so we can
reject the null hypothesis in part a.
Z − Test
µ< 45
z= −4
p = 3.1686035E −5
x  43
n = 100
Z − Test
µ< 45
z = − 1.714285714
p = .043238098
x  43.8
n = 100
MINITAB: Select the following Stat, Basic Statistics, and 1-Sample Z to generate a pop-up menu. In
the pop-up menu, click beside the words Summarized data as long as you do indeed have summary
data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word
standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on
Options to generate another pop-up menu where you enter in the confidence level or 1− α and then
make sure it says the correct Alternative hypothesis. By default µ1 is set to “not equal to”, if your
alternative is something else, like greater than or less than, then use the down arrow to highlight a left
tailed or right tailed µ1. Once the correct µ1 is in the box, then click on OK. This returns you to the
original pop-up menu where you also click on OK and the resulting confidence interval appears in the
Session section. The pop-up menus for part a as well as the results of the hypothesis tests for both parts
are shown below. Since the observed z is less than the critical value of z we do not reject the null
hypothesis for part b, but the observed z is more than the critical value of z so we can reject the null
hypothesis for part a.
166
Chapter Nine
Excel: Open up KADDSTAT and select KADD, Hypothesis Testing,, One Sample, Population mean
using Z. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if
you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this
question, then click beside user input other wise click beside the other choice and enter your range of
cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is
unknown. In the box beside cell: type in the address on your worksheet where you want the answer to
appear. For part a, the results and the pop-up menu are shown below, while for part b only the results
are displayed. Since the observed z is less than the critical value of z we do not reject the null
hypothesis for part b, but for part a the observed z is more than the critical value of z so we can reject
the null hypothesis.
Results for part a:
9.41
a. Step 1:
H0: μ = 80;
H1: μ ≠ 80;
A two-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
For α = .10, the critical values of z are –1.65 and 1.65.
Step 4:
 x =  / n = 15 / 33 = 2.61116484
z = ( x – μ) /  x = (76.5 – 80) / 2.61116484 = –1.34
Step 5:
b. Step 1:
Do not reject H0.
H0: μ = 32;
H1: μ < 32;
A left-tailed test.
Results for part b:
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
Step 2:
Since n > 30, use the normal distribution.
Step 3:
For α = .01, the critical value of z is –2.33.
Step 4:
s x = s / n = 7.4 / 75 = .85447840
167
z = ( x – μ) / s x = (26.5 – 32) / .85447840 = –6.44
Step 5:
c. Step 1:
Reject H0.
H0: μ = 55;
H1: μ > 55;
A right-tailed test.
Step 2:
Since n > 30, use the normal distribution.
Step 3:
For α = .05, the critical value of z is 1.65.
Step 4:
s x = s / n = 4 / 40 = .63245553
z = ( x – μ) / s x =(60.5 – 55) / .63245553 = 8.70
Step 5:
9.43
Reject H0.
Step 1:
H0: μ = 16.3 weeks;
H1: μ > 16.3 weeks;
Step 2:
Since n > 30, use the normal distribution.
Step 3:
For α = .02, the critical value of z is 2.05.
Step 4:
s x = s / n = 4.2 / 400 = .21
A right-tailed test.
z = ( x – μ) / s x =(16.9 – 16.3) / .21 = 2.86
Step 5:
9.45
Reject H0. Conclude the mean duration of unemployment exceeds 16.3 weeks.
H0: μ = $2.4 million;
H1: μ > $2.4 million;
A right-tailed test.
For α = .025, the critical value of z is 1.96.
s x = s / n = .5 / 32 = ..08838835
Test statistic: z = ( x – μ) / s x = (2.6 – 2.4) / .08838835 = 2.26
Reject H0. Conclude the mean annual cash compensation of CEOs exceeds $2.4 million.
9.47
a. H0: µ = 38.1 years;
H1: µ < 38.1 years;
A left-tailed test.
For α = .01, the critical value of z is –2.33.
sx = s / n = 8 /
700 = .30237158
Test statistic: z = ( x – μ) / s x = (37 – 38.1) / .30237158 = –3.64
Reject H0. Conclude that the mean age of motor cycle owners is less than 38.1 years.
b. The Type I error in this case would be to conclude that the mean age of motor cycle owners is less
than 38.1 years when it is actually equal to 38.1 years . P(Type I error) = α = .01
168
9.49
Chapter Nine
a. H0: µ ≥ $35,000;
H1: µ < $35,000;
A left-tailed test.
For α = .01, the critical value of z is –2.33.
s x = s / n = 5400 / 150 = $440.90815370
Test statistic: z = ( x – μ) / s x = (33,400 – 35,000) / 440.90815370 = –3.63
Reject H0 and conclude that the company should not open a restaurant in this area.
b. If α = 0, there can be no rejection region. Thus, we cannot reject H0.
9.51
a. H0: μ ≥ 8 hours;
H1: μ < 8 hours;
A left-tailed test
For α = .01, the critical value of z is –2.33.
s x = s / n = 2.1 /
200 = .14849242
Test statistic: z = ( x – μ) / s x = (7.68 – 8) / .14849242 = –2.15
Do not reject H0. Conclude that the claim is true.
b. For α = .025, the critical value of z is –1.96. From part a, the value of the test statistic is –2.15.
Hence, we reject H0 and conclude that the claim is false.
The decisions in parts a and b are different. The results of this sample are not very conclusive, since
raising the significance level from .01 to .025 reverses the decision.
9.53
H1: µ ≠ 32 ounces;
H0: µ = 32 ounces;
A two-tailed test.
For α = .02, the critical values of z are –2.33 and 2.33.
 x =  / n = .15 /
35 = .02535463
Test statistic: z = ( x – μ) /  x = (31.90 – 32) / .02535463 = –3.94
Reject H0; the machine needs to be adjusted.
9.55
To make a hypothesis test about the mean textbook costs of college freshmen, first we will take a
sample of 30 or more freshmen and collect the information on how much each of them spends on
textbooks. Then, using the formulas learned in Chapter 3 for the mean and standard deviation for
sample data, we will calculate the sample mean and sample standard deviation. Then we will
determine the significance level and make the test of hypothesis.
9.57
a. Area in each tail = α / 2 = .02 / 2 = .01 and df = n – 1 = 20 – 1 = 19
Rejection region: t < –2.539 and t > 2.539
Nonrejection region: –2.539 < t < 2.539
b. Area in the left tail = α = .01 and df = n –1 = 16 – 1 = 15
Rejection region: t < –2.602
Nonrejection region: t > –2.602
c. Area in the right tail = α = .05 and df = n – 1 = 18 – 1 = 17
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
169
Rejection region: t > 1.740
Nonrejection region: t < 1.740
Note: you can use technology to solve this problem using the approach similar to the ones describes in
problems for the z distribution, only this time selecting the t distribution instead. However, the simplest
way to solve this problem is with the t distribution table provided in the textbook inside the back cover.
9.59
a. Area in the right tail = α = .01 and df = n – 1 = 25 – 1 = 24
Critical value: t = 2.492
s x = s / n = 7.5 /
25 = 1.50
Observed value: t = ( x – μ) / s x = (58.5 – 55) / 1.50 = 2.333
b. Area in each tail = α / 2 = .01 / 2 = .005 and df = n–1 = 25–1 = 24
Critical values: t = –2.797 and 2.797
Observed value: t = ( x – μ) / s x = (58.5 – 55) / 1.50 = 2.333
TI-83: If you have raw data, first enter your data in a list. Otherwise just select the following Stat,
TESTS, and 2: T Test, followed by pressing the ENTER key. If you have raw data choose Data and
press the ENTER key, and then using the down arrow to move to enter the requested values and the list
name, but be sure to keep Freq set at 1. On the other hand if you have summary statistics, instead of
Data highlight Stats and press the ENTER key. Using the down arrow, scroll down to start entering the
requested values. In this example we do not have raw data, so will follow the latter method. We select
Stat, scroll down, and enter µ0 = 55, σ = s =7.5, x  58.5, n = 25. Next we highlight > 55 for part a and
≠ 55 for part b, highlight Calculate, and press the ENTER key. The results are shown below where p is
the p value we seek to compare with α or we can compare the critical value of t with the observed
value of t reported below.
T − Test
µ> 55
t = 2.333333333
p = .0141686084
x  58.5
Sx = 7.5
n = 25
T − Test
µ ≠ 55
t = − 1.714285714
p = .0283372167
x  58.5
Sx = 7.5
n = 25
MINITAB: Select the following Stat, Basic Statistics, and 1-Sample t to generate a pop-up menu. In
the pop-up menu, click beside the words Summarized data as long as you do indeed have summary
data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word
standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on
Options to generate another pop-up menu where you make sure it says the correct Alternative
170
Chapter Nine
hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than
or less than, then use the down arrow to highlight left a tailed or right tailed µ1. Once the correct µ1 is
in the box, then click on OK. This returns you to the original pop-up menu where you also click on OK
and the resulting confidence interval appears in the Session section. The pop-up menus for part a as
well as the results for both parts of this problem are shown below.
Excel: Open up KADDSTAT and select KADD, Hypothesis Testing,, One Sample, Population mean
using t. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if
you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this
question, then click beside user input other wise click beside the other choice and enter your range of
cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is
unknown. In the box beside cell: type in the address on your worksheet where you want the answer to
appear. For part a, the results and the pop-up menu are shown below, while for part b only the results
are displayed.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
Results for part a:
9.61
171
Results for part b:
a. The rejection region lies to the left of t = –2.539.
The nonrejection region lies to the right of t = –2.539.
b. The rejection region lies to the left of t = –2.861 and to the right of t = 2.861.
The nonrejection region lies between t = –2.861 and t = 2.861
c. The rejection region lies to the right of t = 2.539.
The nonrejection region lies to the left of t = 2.539.
9.63
a. Step 1:
Step 2:
H0: µ = 80;
H1: µ ≠ 80;
A two-tailed test.
The sample size is small (n < 30), the population is normally distributed and the
population standard deviation σ, is unknown. Hence, we use the t distribution to make
the test.
Step 3:
df = n – 1 = 25 – 1 = 24 and α / 2 = .01 / 2 = .005
Critical values: t = –2.797 and 2.797
Step 4:
sx = s / n = 8 /
25 = 1.60
t = ( x – μ) / s x = (77 – 80) / 1.60 = –1.875
Step 5:
Do not reject H0.
b. Steps 1, 2, and 3 are the same as for part a.
Step 4:
sx = s / n = 6 /
25 = 1.20
t = ( x – μ) / s x = (86 – 80) / 1.20 = 5.000
Step 5: Reject H0.
Comparing parts a and b shows that two samples selected from the same population can yield opposite
conclusions on the same test of hypothesis.
Note: Using technology follow the instructions outline in 9.59, and then compare your α with the
172
Chapter Nine
p value or the critical value of t with its observed value. This will allow you to determine whether to
reject the null hypothesis or not.
9.65
Hl: µ ≠ 24;
a. H0: µ = 24;
A two-tailed test.
df = n– 1 = 25 – 1 = 24 and α / 2 = .01 / 2 = .005
The critical values of t are –2.797 and 2.797.
s x = s / n = 4.9 /
25 = .98
t = ( x – μ) / s x = (28.5–24) / .98 = 4.592
b. H0 :µ = 30;
Reject H0.
H1 : µ < 30:
A left-tailed test.
df = n–1 = 16–1 = 15 and α = .025
The critical value of t is –2.131.
s x = s / n = 6.6 / 16 = 1.65
c.
t = ( x – μ) / s x = (27.5 – 30) / 1.65 = –1.515
Do not reject H0.
H0: µ = 18,
A right tailed test.
H1: µ > 18;
df = n – 1 = 20 – 1 = 19 and α = .10
The critical value of t is 1.328.
sx = s / n = 8 /
20 = 1.78885438
t = ( x – μ) / s x = (22.5 – 18) / 1.78885438 = 2.516
9.67
H0: µ = 69.5 inches;
H1: µ ≠ 69.5 inches;
Reject H0.
A two-tailed test.
df = n –1 = 25 –1 = 24 and α / 2 = .01 / 2 = .005
The critical values of t are –2.797 and 2.797.
s x = s / n = 2.1 /
25 = .420
t = ( x – μ) / s x =(70.25–69.5) / .420 = 1.786
9.69
H0: µ ≤ 7 hours;
H1: µ > 7 hours;
Do not reject H0.
A right-tailed test.
df = n – 1 = 20 – 1 = 19 and α = .025
The critical value of t is 2.093.
s x = s / n = 2.3 /
9.71
20 = .51429563
t = ( x – μ) / s x = (10.5 – 7) / .51429563 = 6.805
Reject H0. The president's claim is not true.
H0: µ ≤ 30 calories;
A right-tailed test.
H1: µ > 30 calories;
df = n – 1 = 16 – 1 = 15 and α = .05
The critical value of t is 1.753.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
173
s x = s / n = 3 / 16 = .750
9.73
t = ( x – μ) / s x = (32–30) / .750 = 2.667
Reject H0. The manufacturer's claim is false.
a. H0: µ ≤ 45 minutes;
A right-tailed test.
H1: µ > 45 minutes;
df = n –1 = 20 –1 = 19 and α = .01
The critical value of t is 2.539.
sx = s / n = 3 /
20 = .67082039
t = ( x – μ) / s x = (49.50 – 45) / .67082039 = 6.708
Reject H0. The mean drying time for these paints is more than 45 minutes.
b. The Type I error would occur if the mean drying time for these paints is 45 minutes or less, but we
conclude otherwise. The probability of such an error is .01 here.
9.75
a. If α = 0, there is no rejection region. Hence, the decision must be: “Do not reject H0”.
b. H0: µ ≥ 1200 words;
H1: µ < 1200 words;
A left-tailed test.
df = n – 1 = 25 – 1 = 24 and α = .05
Critical value: t = –1.711
s x = s / n = 85 /
25 = 17.00
t = ( x – μ) / s x = (1125 – 1200) / 17.00 = –4.412
Reject H0. Conclude that the claim of the business school is false.
9.77
From the given data: n = 10,
 x = 257, and  x 2 = 7341
x =  x / n = 257 / 10 = 25.70 hours.
s
2
 x  ( x) 2 / n =
n 1
H0: µ = 18 hours;
7341  (257 ) 2 / 10
= 9.04372097
10  1
H1: µ ≠ 18 hours;
A two-tailed test.
df = n – 1 = 10 – 1 = 9 and α / 2 = .05 / 2 = .025
Critical values: t = –2.262 and t = 2.262
s x = s / n = 9.04372097 / 10 = 2.85987568
t = ( x – μ) / s x = (25.70 – 18) / 2.85987568 = 2.692
Reject H0. Conclude that the claim of the earlier study is false.
9.79
To make a hypothesis test about the mean amount spent on gas by all customers at the given gas
station, first we will take a sample of less than 30 customers and collect the information on how much
they spent on gas at this gas station. Then, using the formulas learned in Chapter 3 for the mean and
174
Chapter Nine
standard deviation for sample data, we will calculate the sample mean and sample standard deviation.
Assuming the amounts spent on gas by this station’s customers are normally distributed, we will
determine the significance level and make the test. Our hypotheses would be:
H0: µ = $10.90 versus H1: µ ≠ $10.90
9.81
In order to use the normal distribution in a test of hypothesis about a population proportion, both np
and nq must be greater than 5, where p is the value of the population proportion in the null hypothesis
and q = 1 – p.
9.83
a. Yes; np = 30(.65) = 19.5 > 5; and nq = 30(.35) = 10.5 > 5
b. No; np = 70(.05) = 3.5 < 5
c. No; np = 60(.06) = 3.6 < 5
d. Yes; np = 900(.17) = 153 > 5, and nq = 900(.83) = 747 > 5
9.85
a. The rejection region lies to the left of z = –1.96 and to the right of z = 1.96.
The nonrejection region lies between z = –1.96 and z = 1.96.
b. The rejection region lies to the left of z = –2.05.
The nonrejection region lies to the right of z = –2.05.
c. The rejection region lies to the right of z = 1.96.
The nonrejection region lies to the left of z = 1.96.
9.87
a. Area in the left tail = α = .01
Critical value: z = –2.33
 pˆ 
pq
=
n
.63(.37 )
= .03413942
200
Observed value of z =
pˆ  p
 pˆ
=
.60  .63
= –.88
.03413942
b. Area in each tail = α / 2 = .01 / 2 = .005
Critical values: z = –2.58 and 2.58
Observed value of z is –.88 as in part a.
9.89
a. The rejection region lies to the left of z = –2.33.
The nonrejection region lies to the right of z = –2.33.
b. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58.
The nonrejection region lies between z = –2.58 and z = 2.58.
c. The rejection region lies to the right of z = 2.33.
The nonrejection region lies to the left of z = 2.33.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
9.91
a. Step 1:
Step 2:
H0: p = .45;
H1: p < .45;
175
A left-tailed test.
np = 400(.45) = 180 and nq = 400(.55) = 220
Since np > 5 and nq > 5, use the normal distribution.
Step 3:
For α = .025, the critical value of z is –1.96.
Step 4:
 pˆ 
z=
Step 5:
pq
=
n
pˆ  p
 pˆ
=
.45 (. 55 )
= .02487469
400
.42  .45
= –1.21
.02487469
Do not reject H0.
b. Steps 1, 2, and 3 are identical to part a.
pˆ  p
Step 4:
z=
Step 5:
Reject H0.
 pˆ
=
.39  .45
= –2.41
.02487469
The results of parts a and b show that two different samples from the same population can yield
opposite decisions on a test of the same hypothesis.
TI-83: If you have raw data, first enter your data in a list. Otherwise just select the following Stat,
TESTS, and 5: 1−PropZTest, followed by pressing the ENTER key. Now enter the requested values. In
this example p = .45. x is the number of successes in the sample or the sample mean of n* p̂ . If it is not
yet calculated you can enter the values in parenthesis and the calculation will be done for you, so in the
first part of this problem we enter (400*.42) followed by pressing the ENTER key and the calculator
returns a value of 168 for x. Next we enter n = 400, highlight the < symbol, highlight Calculate, and
press the ENTER key. The results are shown below where p is the p value we seek to compare with α
or we can compare the critical value of z with its observed value reported below. The results show we
can not reject the null hypothesis for part a, while for part b we can reject the null hypothesis.
1 −PropZTest
prop < .45
z = -1.206045378
p = .1139000591
p̂ = .42
n = 400
1 −PropZTest
prop < .45
z = -2.412090757
p = .007930662
p̂ = .39
n = 400
MINITAB: Select the following Stat, Basic Statistics, and 1 Proportion to generate a pop-up menu. In
the pop-up menu, click beside the words Summarized data and enter n in the first box, and enter µ or
the value of n* p̂ in the 2nd box. Next click on Options to generate another pop-up menu where you
176
Chapter Nine
enter the proportion from the null hypothesis beside the words Test proportion and make sure it says
the correct Alternative hypothesis. By default µ1 is set to “not equal to”, if your alternative is
something else, like greater than or less than, then use the down arrow to highlight left a tailed or right
tailed µ1. Once the correct µ1 is in the box, click on the empty box below the alternative hypothesis,
and then click on OK. This returns you to the original pop-up menu where you also click on OK and
the resulting confidence interval appears in the Session section. The pop-up menus for part a as well as
the results for both parts of this problem are shown below.
Excel: Open up KADDSTAT and select KADD, Hypothesis Testing, One Sample, Population mean
using t. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if
you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this
question, then click beside user input other wise click beside the other choice and enter your range of
cells. After clicking beside user input if you enter the n in the first box and in the next box enter x
which is number of successes in the sample which is n* p̂ . In the box beside cell: type in the address
on your worksheet where you want the answer to appear. For part a, the results and the pop-up menu
are shown below, while for part b only the results are displayed. The results show we can not reject the
null hypothesis for part a, while for part b we can reject the null hypothesis.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
Results for part a:
9.93
a. Step 1:
Step 2:
H1: p ≠ .57;
H0: p = .57;
A two-tailed test.
np = 800(.57) = 456 and nq = 800(l – .57) = 344
Since np > 5 and nq > 5, use the normal distribution.
Step 3:
For α = .05, the critical values of z are –1.96 and 1.96.
Step 4:
p = .57 and q = 1 – p = 1 – .57 = .43
 pˆ 
z=
Step 5:
b. Step 1:
Step 2:
pq
=
n
pˆ  p
 pˆ
=
.57 (.43)
= .01750357
800
.50  .57
= –4.00
.01750357
Reject the null hypothesis.
H0: p = .26;
H1: p < .26;
A left-tailed test.
np = 400(.26) = 104 and nq = 400(l –.26) = 296
Since np > 5 and nq > 5, use the normal distribution.
Step 3:
For α = .01, the critical value of z is –2.33.
Step 4:
p = .26 and q = 1 – p = 1 –.26 = .74
 pˆ 
z=
Step 5:
c. Step l:
Step 2:
pq
=
n
pˆ  p
 pˆ
=
.26 (. 74 )
= .02193171
400
.23  .26
= –1.37
.02193171
Do not reject the null hypothesis.
H0: p = .84;
Hl: p > .84;
A right-tailed test.
np = 250(.84) = 210 and nq = 250(l –.84) = 40
Since np > 5, and nq > 5, use the normal distribution.
Step 3:
For α = .025, the critical value of is 1.96.
Step 4:
p = .84 and q = 1 –p = 1 –.84 = .16
177
Results for part b:
178
Chapter Nine
 pˆ 
z=
Step 5:
9.95
pq
=
n
pˆ  p
 pˆ
=
.84 (. 16 )
= .02318620
250
.85  .84
= .43
.02318620
Do not reject the null hypothesis.
H0: p = .49;
H1: p > .49;
A right-tailed test.
For α = .025, the critical value of z is 1.96.
 pˆ  pq / n =
(. 49 )(. 51) / 200 = .03534827
p̂ = .52
z=
pˆ  p
 pˆ
=
.52  .49
= .85
.03534827
Do not reject H0. Do not conclude that the current percentage of management and professional jobs
held by women exceeds 49%.
9.97
H0: p = .45;
H1: p > .45;
A right-tailed test.
For α = .01, the critical value of z is 2.33.
p̂ = 248 / 500 = .496
 pˆ  pq / n =
z=
pˆ  p
 pˆ
=
(. 45 )(. 55 ) / 500 = .02224860
.496  .45
= 2.07
.02224860
Do not reject H0. Do not conclude that the proportion of employers monitoring their employees use of
company phones exceeds 45%.
9.99
a. H0: p = .32;
H1: p > .32;
A right-tailed test.
For α = .025, the critical value of z is 1.96.
p̂ = 396 / 1100 = .36
 pˆ  pq / n =
z=
pˆ  p
 pˆ
=
(.32 )(. 68 ) / 1100 = .01406479
.36  .32
= 2.84
.01406479
Reject H0. Conclude that more than 32% of households earning $75,000 or more would struggle to
pay an unexpected bill of $5000.
b. The Type I error would occur if we concluded that more than 32% of households earning $75,000
or more would struggle to pay an unexpected bill of $5000, when actually it is not true. The
probability of making this error is α = .025.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
9.101
a. H0: p ≥ .60;
H1: p < .60;
179
A left-tailed test.
For α = .01, the critical value of z is –2.33.
p̂ = 208 / 400 = .52
 pˆ 
z=
pq
=
n
pˆ  p
 pˆ
=
.60 (. 40 )
= .02449490
400
.52  .60
= –3.27
.02449490
Reject H0. Conclude that the company's claim is not true.
b. If α = 0, there is no rejection region, so we could not reject H0. Thus, we would conclude that the
company's claim is true.
9.103
a. H0: p ≤ .07;
H1: p > .07;
A right-tailed test.
For α = .02, the critical value of z = 2.05.
p̂ = 22 / 200 = .11
 pˆ 
z=
pq
=
n
pˆ  p
 pˆ
=
.07 (.93)
= .01804162
200
.11  .07
= 2.22
.01804162
Reject H0. Conclude that the machine should be stopped.
b. If α = .01, the critical value of z is 2.33. Since the value of the test statistic is 2.22, we do not reject
H0. Thus, our decision differs from that of part a; we conclude that the machine should not be
stopped. The decisions in parts a and b are different. The results of this sample are not very
conclusive, since lowering the significance level from 2% to 1% reverses the decision.
9.105
Select a random sample of 40 students from your school and ask them whether or not they hold off–
campus jobs. From this data calculate p̂ .
Using hypotheses H0: p = .65 and H1: p ≠ .65, choose a level of significance and find the critical values
from the table of the normal distribution.
Then compute  pˆ 
9.107
H0: µ = 40;
(.65 )(. 35 )
pˆ  .65
and z =
and make a decision.
40
 pˆ
H1: µ ≠ 40
sx = s / n = 6 /
64 = .75
z = ( x – μ) / s x = (38.4 – 40) / .75 = –2.13
180
Chapter Nine
a. For α = .02, the critical values of z are –2.33 and 2.33. Hence, do not reject H0.
b. P(Type I error) = α = .02
c. Area to the left of z = –2.13 is .5 – .4834 = .0166. Hence, p–value = 2(.0166) = .0332
If α = .01, do not reject H0 since p–value > .01.
If α = .05, reject H0 since p–value < .05.
9.109
H0: p = .44;
 pˆ 
z=
H1: p < .44
pq
=
n
pˆ  p
 pˆ
=
.44 (. 56 )
= .02339991
450
.39  .44
= –2.14
.02339991
a. For α = .02, the critical value of z is –2.05. Reject H0.
b. P(Type I error) = α = .02
c. Area to the left of –2.14 is .5 –.4838 = .0162. Hence, p–value = .0162
If α = .01, do not reject H0, since p–value > .01.
If α = .025, reject H0, since p–value < .025.
9.111
H0: µ = 1245 cubic feet;
H1: µ < 1245 cubic feet
s x = s / n = 250 / 100 = 25.00
z = ( x – μ) / s x = (1175 – 1245) / 25.00 = –2.80
Area to the left of z = –2.80 is .5 – .4974 = .0026. Hence, p–value = .0026
Reject H0 for α = .025, since p–value < .025.
9.113
a. H0: µ = 180 months;
H1: µ < 180 months;
A left-tailed test.
For α = .02, the critical value of z is –2.05.
s x = s / n = 27 /
60 = 3.48568501
z = ( x – μ) / s x = (171 – 180) / 3.48568501 = –2.58
Reject H0. The sample supports the alternative hypothesis that the current mean sentence for such
crimes is less than 180 months.
b. The Type I error would be to conclude that the current mean sentence for such crimes is less than
180 months when in fact it is not. P(Type I error) = .02
c. If α = 0, there is no rejection region and, hence, we cannot reject H0. Thus, our conclusion would
change.
9.115
a. H0: µ ≤ 2400 square feet;
H1: µ > 2400 square feet;
For α = .05, the critical value of z is 1.65.
A right-tailed test.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
s x = s / n = 472 /
181
50 = 66.75088014 square feet
z = ( x – μ) / s x = (2540 – 2400) / 66.75088014 = 2.10
Reject H0. The sample supports the alternative hypothesis that the real estate agents’ claim is false.
b. For α = .01, the critical value of z is 2.33.
From part a, the observed value of z is 2.10. Hence, do not reject H0.
The results of parts a and b show that the sample does not support the alternative hypothesis very
strongly, since lowering the significance level from .05 to .01 reverses the conclusion.
9.117
H0: µ = 8 minutes;
H1: µ < 8 minutes;
A left-tailed test.
For α = .025, the critical value of z is –1.96.
s x = s / n = 2.1 /
32 = .37123106 minute
z = ( x – μ) / s x = (7.5 – 8) / .37123106 = –1.35
Do not reject H0. At the .025 level of significance, the difference between the sample mean and the
hypothesized value of the population mean is small enough to attribute to chance. Thus, the manager's
claim is not justified.
9.119
H0: µ = 25 minutes;
H1: µ≠ 25 minutes;
A two-tailed test.
df = n – 1 = 16 – 1 = 15and α / 2 = .01 / 2 = .005
The critical values of t are –2.947 and 2.947.
s x = s / n = 4.8 / 16 = 1.20
t = ( x – μ) / s x = (27.5 – 25) / 1.20 = 2.083
Do not reject H0.
9.121
a.
H0: µ = 114 minutes;
H1: µ < 114 minutes;
A left-tailed test.
df = n – 1 = 25 –1 = 24 and α = .01
The critical value of t is –2.492.
s x = s / n = 11 /
25 = 2.20 minutes
t = ( x – μ) / s x = (109 – 114) / 2.20 = –2.273
Do not reject H0. The mean time currently spent by all adults with their families is not less than 114
minutes a day.
b. If α = 0, there is no rejection region, so there is no need to go through the five steps of hypothesis
testing. We cannot reject H0.
9.123
H0: µ ≤ 150 calories;
H1:µ > 150 calories
df = n – 1 = 10 – 1 = 9 and α = .025
182
Chapter Nine
The critical value of t is 2.262.
 x = 1527, and  x 2 = 233,663
x =
 x / n = 1527 / 10 = 152.7
2
 x  ( x) 2 / n =
s
n 1
233 ,663  (1527 ) 2 / 10
= 7.37940076
10  1
s x = s / n = 7.37940076 / 10 = 2.33357142
t = ( x – μ) / s x = (152.70 – 150) / 2.33357142 = 1.157
9.125
a. H0: p = .32;
Hl: p > .32;
Do not reject H0.
A right-tailed test.
For α = .02, the critical value of z = 2.05.
pq
=
n
 pˆ 
z=
pˆ  p
.35  .32
= 1.88
.016
=
 pˆ
.32 (. 68 )
= .016
850
Do not reject H0. Do not conclude that the current percentage of faculty who hold this opinion
exceeds 32%.
b. The Type I error would occur if we concluded that the current percentage of the faculty who hold
this opinion exceeds 32%, when in fact it does not. P(Type I error) = .02.
9.127
H0: p = .56;
H1: p < .56;
A left-tailed test.
For α = .02, the critical value of z is –2.05.
p̂ = .50
 pˆ 
z=
pq
=
n
pˆ  p
 pˆ
=
.56 (. 44 )
= .04052982
150
.50  .56
= –1.48
.04052982
Do not reject H0. Do not conclude that the current percentage of attorneys who take work on vacation
is less than 56%.
9.129
a. H0: p ≥ .90;
H1: p < .90;
For α = .02, the critical value of z is –2.05.
 pˆ 
pq
=
n
.90 (. 10 )
= .03162278
90
p̂ = 75 / 90 = .833
A left-tailed test.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
z=
pˆ  p
=
 pˆ
183
.833  .90
= –2.12
.03162278
Reject H0. Conclude that the company's policy is not maintained.
b. If α = 0, there is no rejection region; thus, we cannot reject H0 and cannot conclude that the
company's policy is not maintained.
9.131
H0: p ≥ .90;
 pˆ 
H1: p < .90;
pq
=
n
A left-tailed test.
.90 (. 10 )
= .03162278
90
p̂ = 75 / 90 = .833
z=
pˆ  p
=
 pˆ
.833  .90
= –2.12
.03162278
The area to the left of z = –2.12 under the normal curve is .5 – .4830 = .0170.
Hence,
p–value = .0170
If α = .05, reject H0 since p–value < .05.
If α = .01, do not reject H0 since p–value > .01.
9.133
Since np > 5 and nq > 5, the distribution of p̂ is approximately normal.
For α = .05, the critical value of z is 1.65.
 pˆ 
Now z =
pq
=
n
.04 (. 96 )
= .01718676
130
pˆ  p
 pˆ
, so 1.65 =
pˆ  .04
.01718676
Solving for p̂ yields p̂ = 1.65(.01718676) +.04 = .0684
Thus, we would reject H0 if p̂ > .0684.
Hence, .0684 =
c
c

n 130
Then, c = 130(.0684) = 8.89  9
Therefore, reject H0 and shut down the machine if the number of defectives in a sample of 130 parts is
9 or more.
9.135
First, we must be sure that the cure rate of the new therapy is not lower than that of the old therapy. Let
p be the proportion of all patients cured with the new therapy. We must test the hypotheses:
H0: p = .60;
H1: p < .60
For α = .01, the critical value of z is –2.33.
184
Chapter Nine
 pˆ 
pq
=
n
.60 (. 40 )
= .03464102
200
From the data: p̂ = x / n = 108 / 200 = .54
Test statistic: z =
pˆ  p
 pˆ
=
.54  .60
= –1.73
.03464102
Do not reject H0. Thus, we cannot conclude that the new therapy has a lower cure rate. Next, we must
see if the new therapy is effective in reducing the number of visits. Let µ be the mean number of visits
required for all patients using the new therapy regime. We must test the hypotheses:
H0: µ = 140 visits;
H1: µ < 140 visits
For α = .01, the critical value of z is –2.33.
s x = s / n = 38 /
200 = 2.68700577
Test statistic: z = ( x – μ) / s x = (132 – 140) / 2.68700577 = –2.98
Reject H0. Conclude that the new therapy regime requires fewer visits on average. Based on the results
of these two hypothesis tests, the health care provider should support the new therapy regime.
9.137
a. Let p be the proportion of all people receiving the new vaccine who contract the disease within a
year. Then the appropriate hypotheses are:
H0: p = .30;
H1: p < .30
b. Let x be the number of people in a sample of 100 inoculated with the new vaccine who contract the
disease within a year. Then, under H0, x is a binomial random variable with n = 100 and p = .30.
Hence,
σ=
µ = np = 100(.30) = 30
npq  100 (.30 )(. 70 ) = 4.58257569
If 84 or more of the 100 people in the sample do not contract the disease, then x < 16. Using a
normal approximation, and correcting for continuity, we need P(x < 16.5).
For x = 16.5: z =
16 .5  30
= –2.95
4.58257569
Thus,   P( x  16.5 p  .30)  P( z  2.95)  .5  .4984  .0016
c. Let x be the number of people in a sample of 20 inoculated with the new vaccine who contract the
disease within a year. Then, under H0, x is a binomial random variable with n = 20 and p = .30.
Using Table IV, Appendix C:
  P( x  3 p  .30) = P(0) + P(l) + P(2) = .0008 +.0068 +.0278 = .0354
9.139
The following are two possible experiments we might conduct to investigate the effectiveness of
middle taillights.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual
185
I. Let: p1 = proportion of all collisions involving cars built since 1984 that were rear–end collisions.
Let: p2 = proportion of all collisions involving cars built before 1984 that were rear–end collisions.
(p2 would be known)
We would test H0: p1 = p2 versus H1: p1 < p2.
We would take a random sample of collisions involving cars built since 1984 and determine the
number that were rear–end collisions. We would have to assume the following:
i. The only change in cars built since 1984 that would reduce rear–end collisions is the middle
taillight.
ii. None of the cars built before 1984 had middle taillights.
iii. People's driving habits, traffic volume, and other variables that might affect rear–end collisions
have not changed appreciably since 1984.
II. Let: µ1 = mean number of rear–end collisions per 1000 cars built since 1984
µ2 = mean number of rear–end collisions per 1000 cars built before 1984 (µ2 would be known)
We would test H0: µ1 = µ2 versus H1: µ1 < µ2.
We could take several random samples of 1000 cars built since 1984. We would determine the
number of rear–end collisions in each sample of 1000 cars.
We would find the mean and standard deviation of these numbers and use them to form the test
statistic.
We would require the same assumptions as those listed for the test in part 1. Also, if we took less
than 30 samples, we would have to assume that the number of rear–end collisions per 1000 cars has
a normal distribution.
Self–Review Test for Chapter Nine
1. a
2. b
3. a
4. b
5. a
6. a
7. a
8. b
9. c
10. a
11. c
12. b
13. d
14. c
15. a
16. b
17. a. Step 1:
H0: µ = $85,900;
H1: µ >85,900;
Step 2:
Since n > 30, use the normal distribution.
Step 3:
s x = s / n = 27,000 /
A right-tailed test.
36 = $4500
z = ( x – μ) / s x = (95,000 – 85,900) / 4500 = 2.02
The area to the right of z = 2.02 under the normal curve is .5 – .4783 = .0217. Hence,
p–value = .0217
b. If α = .01, do not reject H0, since p–value > .01.
If α = .05, reject H0, since p–value < .05.
186
Chapter Nine
18. a. Step 1:
H0: µ =185 minutes;
H1: µ < 185 minutes
Step 2:
Since n > 30, use the normal distribution.
Step 3:
For α = .01, the critical value of z is –2.33.
Step 4
s x = s / n = 12 /
36 = 2
z = ( x – μ) / s x = (179 – 185) / 2 = –3.00
Step 5:
Reject H0. Conclude that the mean durations of games have decreased after the meeting.
b. The Type I error would be to conclude that the mean durations of games have decreased after the
meeting when they are actually equal to the duration of games before the meeting.
P(Type I error) = α = .01
c. If α = 0, there is no rejection region, so do not reject H0.
d. From part a, z = –3.00. The area to the left of z = –3.00 under the normal curve is .5 – .4987 = .0013.
19. a.
Hence, p–value =.0013.
For α = .01, reject H0, since p–value < .01.
H0: µ ≥ 31 months;
H1: µ < 31 months
The critical value of t is: –2.131
s x = s / n = 7.2 / 16 = 1.80
t = ( x – μ) / s x = (25–31) / 1.80 = –3.333
Reject H0. Conclude that the editor's claim is false.
b. The Type I error would be to conclude that the editor's claim is false when it is actually true.
P(Type I error) = α = .025
c. For α = .001, the critical value of t is –3.733. Since the observed value of t = –2.778 is greater than
–3.733, we would not reject H0.
20. a. H0: p = .50;
H1: p < .50;
A left-tailed test.
For α = .05, the critical value of z is –1.65.
 pˆ 
p̂ = 450 / 1000 = .45
z=
pˆ  p
 pˆ
=
.45  .50
= –3.16
.015811388
pq
=
n
.50 (. 50 )
= .015811388
1000
Reject H0. Conclude that the less than 50% have a will.
b. The Type I error would be to conclude that the percentage of adults with wills was less than 50 %
when it is actually 50%. P(Type I error) = α = .05
c. If α = 0, there is no rejection region, so do not reject H0.
21. a. Referring to Problem 20.
The area to the left of z = –3.16 under the normal curve is .5 –.5000 = .0000.
Hence, p–value = .0000
b. If α = .05, reject H0, since p–value < .05
If α = .01, reject H0, since p–value <.01