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314
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
CHAPTER 9
9.1
Decision rule: Reject H0 if Z STAT < – 1.96 or Z STAT > + 1.96.
Decision: Since ZSTAT = − 0.76 is in between the two critical values, do not reject H0.
9.2
To make a decision concerning the null hypothesis, you first determine the critical value of the
test statistic. The critical value divides the non-rejection region from the rejection region. The
critical value approach compares the value of the computed ZSTAT test statistic from Equation (9.1)
to critical values that divide the normal distribution into regions of rejection and non-rejection.
Decision rule is the rule that accepts or rejects the hypothesis, given the critical value. For
instance, at 95% confidence level, 0.025% at each side of the two-tail test is the rejection area.
The Z value for this rejection area is 1.96. Thus, 1.96 is the critical value and the decision rule is
Reject H0 if ZSTAT > +1.96
or if ZSTAT < −1.96;
otherwise, do not reject H0.
9.3
Decision rule: Reject H0 if Z STAT < – 1.645 or Z STAT > + 1.645.
9.4
Decision rule: Reject H0 if Z STAT < – 2.58 or Z STAT > + 2.58.
9.5
Decision: Since Z STAT = – 2.61 is less than the lower critical value of – 2.58, reject H0.
9.6
p-value = 2(1 − .9772) = 0.0456
9.7
Since the p-value of 0.0456 is less than the 0.10 level of significance, the statistical decision is to
reject the null hypothesis.
9.8
p-value = 0.1676
9.9
α is the probability of incorrectly convicting the defendant when he is innocent. β
is the
probability of incorrectly failing to convict the defendant when he is guilty.
9.10
Under the French judicial system, unlike ours in the United States, the null hypothesis assumes
the defendant is guilty, the alternative hypothesis assumes the defendant is innocent. A Type I
error would be not convicting a guilty person and a Type II error would be convicting an innocent
person.
9.11
(a)
(b)
(c)
(d)
9.12
A Type I error is the mistake of approving an unsafe drug. A Type II error is not
approving a safe drug.
The consumer groups are trying to avoid a Type I error.
The industry lobbyists are trying to avoid a Type II error.
To lower both Type I and Type II errors, the FDA can require more information and
evidence in the form of more rigorous testing. This can easily translate into longer time
to approve a new drug.
H0: µ = 32%. The packet contains 32% cashews.
H1: µ ≠ 32%. The packet does not contain 32% cashews.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.13
(a)
(b)
(c)
9.14
(a)
315
H 0:
µ = 12.1 hours
H 1:
µ ≠ 12.1 hours
A Type I error is the mistake of concluding that the mean number of hours studied by
marketing majors at your school is different from the 12.1-hour-per-week benchmark
reported by The Washington Post when in fact it is not any different.
A Type II error is the mistake of not concluding the mean number of hours studied by
marketing majors at your school is different from the 12.1-hour-per-week benchmark
reported by The Washington Post when it is in fact different.
PHStat output:
ZTestofHypothesisfortheMean
Data
NullHypothesisµ=
LevelofSignificance
PopulationStandardDeviation
SampleSize
SampleMean
7500
0.05
1000
64
7250
IntermediateCalculations
StandardErroroftheMean
125.0000
Z TestStatistic
-2.0000
Two-TailTest
LowerCriticalValue
UpperCriticalValue
p -Value
Rejectthenullhypothesis
H 0:
H 1:
µ = 7500.
µ ≠7500.
-1.9600
1.9600
0.0455
The mean life of a large shipment of light bulbs is equal to 7500 hours.
The mean life of a large shipment of light bulbs differs from 7500 hours.
Decision rule: Reject H 0 if |ZSTAT| > 1.96
X −µ
= -2.0000
σ/ n
Decision: Since |ZSTAT| > 1.96, reject H 0 . There is enough evidence to conclude that the
Test statistic: Z STAT =
(b)
mean life of a large shipment of light bulbs differs from 7500 hours.
p-value = 0.0455. If the population mean life of a large shipment of light bulbs is indeed
equal to 7500 hours, the probability of obtaining a test statistic that is more than 2.0
standard error units away from 0 is 0.0455.
© 2015 Pearson Education Ltd.
316
9.14
cont.
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(c)
PHStat output:
ConfidenceIntervalEstimatefortheMean
Data
PopulationStandardDeviation
SampleMean
SampleSize
ConfidenceLevel
1000
7250
64
95%
IntermediateCalculations
StandardErroroftheMean
125.0000
ZValue
-1.9600
IntervalHalfWidth
244.9955
ConfidenceInterval
IntervalLowerLimit
IntervalUpperLimit
9.15
σ
1000
64
(c)
X ± Z a /2
(d)
You are 95% confident that the population mean life of a large shipment of light bulbs is
somewhere between 7005.00 and 7495.00 hours.
Since the 95% confidence interval does not contain the hypothesized value of 7500, you will
reject H 0 . The conclusions are the same.
n
= 7250 ± 1.96
7005.00
7495.00
7005.00 ≤ µ ≤ 7495.00
(a)
X ± Zα / 2
(b)
σ
n
100 ± 1.96 (25/√100) = 104.9, 95.1
Since, hypothesized π = 95 lies within the confidence intervals, H0 is accepted.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.16
(a)
317
PHStat output:
Data
µ=
Null Hypothesis
Level of Significance
Population Standard Deviation
Sample Size
Sample Mean
Intermediate Calculations
Standard Error of the Mean
Z Test Statistic
1
0.01
0.02
50
0.995
0.002828427
-1.767766953
Two-Tail Test
Lower Critical Value
-2.575829304
Upper Critical Value
2.575829304
p-Value
0.077099872
Do not reject the null hypothesis
H 0:
H 1:
µ = 1.
µ ≠1.
The mean amount of water is 1 gallon.
The mean amount of water differs from 1 gallon.
Decision rule: Reject H 0 if |ZSTAT| > 2.5758
(a)
(b)
(c)
X − µ .995 − 1
=
= −1.7678
σ / n .02/ 50
Decision: Since |ZSTAT| < 2.5758, do not reject H 0 . There is not enough evidence to
Test statistic: Z STAT =
conclude that the mean amount of water contained in 1-gallon bottles purchased from a
nationally known water bottling company is different from 1 gallon.
p-value = 0.0771. If the population mean amount of water contained in 1-gallon bottles
purchased from a nationally known water bottling company is actually 1 gallon, the
probability of obtaining a test statistic that is more than 1.7678 standard error units away
from 0 is 0.0771.
PHStat output:
Data
Population Standard Deviation
Sample Mean
Sample Size
Confidence Level
0.02
0.995
50
99%
Intermediate Calculations
Standard Error of the Mean
Z Value
Interval Half Width
0.002828427
-2.5758293
0.007285545
Confidence Interval
Interval Lower Limit
Interval Upper Limit
0.987714455
1.002285545
© 2015 Pearson Education Ltd.
318
9.16
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(c)
cont.
(d)
9.17
(a)
X ± Z a /2
σ
n
= .995 ± 2.5758
.02
50
0.9877 ≤ µ ≤ 1.0023
You are 99% confident that population mean amount of water contained in 1-gallon bottles
purchased from a nationally known water bottling company is somewhere between 0.9877
and 1.0023 gallons.
Since the 99% confidence interval does contain the hypothesized value of 1, you will not
reject H 0 . The conclusions are the same.
(a)
PHStat output:
Z Test of Hypothesis for the Mean
Data
Null Hypothesis
m=
1
Level of Significance
0.01
Population Standard Deviation
0.012
Sample Size
50
Sample Mean
0.995
Intermediate Calculations
Standard Error of the Mean
0.001697056
Z Test Statistic
-2.946278255
Two-Tail Test
Lower Critical Value
-2.575829304
Upper Critical Value
2.575829304
p-Value
0.003216229
Reject the null hypothesis
H 0:
H 1:
µ = 1.
µ ≠1.
The mean amount of water is 1 gallon.
The mean amount of water differs from 1 gallon.
Decision rule: Reject H 0 if |ZSTAT| > 2.5758
X −µ
.995 − 1
=
= −2.9463
σ / n .012/ 50
Decision: Since |ZSTAT| > 2.5758, reject H 0 . There is enough evidence to
Test statistic: Z STAT =
(b)
conclude that the mean amount of water contained in 1-gallon bottles purchased
from a nationally known water bottling company is different from 1 gallon.
p-value = 0.0032. If the population mean amount of water contained in 1-gallon
bottles purchased from a nationally known water bottling company is actually 1
gallon, the probability of obtaining a test statistic that is more than 2.9463
standard error units away from 0 is 0.0032.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.17
cont.
(a)
(c)
319
PHStat output:
Confidence Interval Estimate for the Mean
Data
Population Standard Deviation
Sample Mean
Sample Size
Confidence Level
Intermediate Calculations
Standard Error of the Mean
Z Value
Interval Half Width
0.001697056
-2.5758293
0.004371327
Confidence Interval
Interval Lower Limit
Interval Upper Limit
0.990628673
0.999371327
X ± Z a /2
(d)
(b)
0.012
0.995
50
99%
σ
n
= .995 ± 2.5758
.012
50
0.9906 ≤ µ ≤ 0.9994
You are 99% confident that population mean amount of water contained in 1-gallon
bottles purchased from a nationally known water bottling company is somewhere
between 0.9906 and 0.9994 gallon.
Since the 99% confidence interval does not contain the hypothesized value of 1
gallon, you will reject H 0 . The conclusions are the same.
The smaller population standard deviation results in a smaller standard error of the Z test
and, hence, smaller p-value. Instead of not rejecting H 0 as in Problem 9.16, you end up
rejecting H 0 in this problem.
X – µ 56 – 50
= 2.00
=
S
12
16
n
9.18
t STAT =
9.19
d.f. = n – 1 = 16 – 1 = 15
9.20
For a two-tailed test with a 0.05 level of confidence, the critical values are
9.21
Since t STAT = 2.00 is between the critical bounds of
9.22
No, you should not use the t test to test the null hypothesis that µ = 60 on a population that is leftskewed because the sample size (n = 16) is less than 30. The t test assumes that, if the underlying
population is not normally distributed, the sample size is sufficiently large to enable the test to be
valid. If sample sizes are small (n < 30), the t test should not be used because the sampling
distribution does not meet the requirements of the Central Limit Theorem.
± 2.1315.
± 2.1315, do not reject H0.
© 2015 Pearson Education Ltd.
320
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
9.23
Yes, you may use the t test to test the null hypothesis that µ = 60 even though the population is
left-skewed because the sample size is sufficiently large (n = 160). The t test assumes that, if the
underlying population is not normally distributed, the sample size is sufficiently large to enable
the test statistic t to be influenced by the Central Limit Theorem.
9.24
PHStat output:
t Test for Hypothesis of the Mean
Data
3.7
µ=
Null Hypothesis
Level of Significance
0.05
Sample Size
64
Sample Mean
3.57
Sample Standard Deviation
0.8
Intermediate Calculations
Standard Error of the Mean
0.1
Degrees of Freedom
63
t Test Statistic
-1.3
Two-Tail Test
Lower Critical Value
-1.9983405
Upper Critical Value
1.9983405
p-Value
0.1983372
Do not reject the null hypothesis
H 0 : µ = 3.7
(a)
H1 : µ ≠ 3.7
Decision rule: Reject H 0 if |tSTAT| > 1.9983 d.f. = 63
X − µ 3.57 - 3.7
Test statistic: t STAT =
=
= -1.3
S / n 0.8/ 64
Decision: Since |tSTAT| < 1.9983, do not reject H 0 . There is not enough evidence to
conclude that the population mean waiting time is different from 3.7 minutes at the
0.05 level of significance.
(b)
The sample size of 64 is large enough to apply the Central Limit Theorem and, hence,
you do not need to be concerned about the shape of the population distribution
when conducting the t-test in (a). In general, the t test is appropriate for this
sample size except for the case where the population is extremely skewed or
bimodal.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.25
(a)
H0 : π = 80,000 H1 : π ≠ 80,000
Decision rule: Reject H0 if |tSTAT| > 2.0096 or p-value < 0.05 d.f. = 49
𝑡"#$# =
𝑋−𝜇
70,000 − 80,000
=
= 7.07
𝑆
10,000
𝑛
50
p-value = 0.000
(b)
Decision: Since |tSTAT| > 2.0096 and the p-value of 0.000 < 0.05, reject H0.
There is sufficient evidence to conclude that average life is different.
© 2015 Pearson Education Ltd.
321
322
9.26
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
PHStat output:
(a)
H0 : π = 420.99 H1 : π ≠ 420.99
Decision rule: Reject H0 if |tSTAT| > 2.0096 or p-value < 0.05 d.f. = 49
𝑡"#$# =
(b)
𝑋−𝜇
450 − 420.99
=
= 2.05
𝑆
100
𝑛
50
p-value = 0.0456
Decision: Since |tSTAT| > 2.0096 and the p-value of 0.0456 < 0.05, reject H0.
There is sufficient evidence to conclude that average property prices are different.
Since the results do not provide sufficient evidence to prove that the property prices are
not different, the manager can conclude that the published prices do not prevail in the
current market scenario, and he can look for options to buy at a different price.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.27
323
PHStat output:
t Test for Hypothesis of the Mean
Data
µ=
Null Hypothesis
Level of Significance
Sample Size
Sample Mean
Sample Standard Deviation
Intermediate Calculations
Standard Error of the Mean
Degrees of Freedom
t Test Statistic
200
0.05
18
195.3
21.4
5.044028372
17
-0.931794917
Two-Tail Test
Lower Critical Value
-2.109815559
Upper Critical Value
2.109815559
p-Value
0.364487846
Do not reject the null hypothesis
(a)
H 0 : µ = 200
H1 : µ ≠ 200
Decision rule: Reject H 0 if |tSTAT| > 2.1098 or p-value < 0.05
X − µ 195.3 − 200
Test statistic: tSTAT =
=
= −0.9318
S / n 21.4 / 18
p-value = 0.3645
Decision: Since |tSTAT| < 2.1098 and the p-value of 0.3645 > 0.05, do not reject H 0 . There
(b)
is not enough evidence to conclude that the population mean tread wear index is different
from 200.
The p-value is 0.3645. If the population mean is indeed 200, the probability of obtaining
a test statistic that is more than 0.9318 standard error units away from 0 in either
direction is 0.3645.
© 2015 Pearson Education Ltd.
324
9.28
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
6.5
0.05
15
7.09
1.406031226
IntermediateCalculations
StandardErroroftheMean
0.3630
DegreesofFreedom
14
t TestStatistic
1.6344
Two-TailTest
LowerCriticalValue
-2.1448
UpperCriticalValue
2.1448
p -Value
0.1245
Donotrejectthenullhypothesis
(a)
H 0 : µ = $6.50 H1 : µ ≠ $6.50
Decision rule: Reject H 0 if |tSTAT| > 2.1448 or p-value < 0.05
Test statistic: tSTAT =
(b)
(c)
(d)
X −µ
= 1.6344
S/ n
Decision: Since |tSTAT| < 2.1448, do not reject H 0 . There is not enough evidence to
conclude that the mean amount spent for lunch is different from $6.50.
The p-value is 0.1245. If the population mean is indeed $6.50, the probability of
obtaining a test statistic that is more than 1.6344 standard error units away from 0 in
either direction is 0.4069.
That the distribution of the amount spent on lunch is normally distributed.
With a sample size of 15, it is difficult to evaluate the assumption of normality. However,
the distribution may be fairly symmetric because the mean and the median are close in
value. Also, the boxplot appears only slightly skewed so the normality assumption does
not appear to be seriously violated.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.29
(a)
(b)
(c)
325
H 0 : µ = 45
H1 : µ ≠ 45
Decision rule: Reject H 0 if |tSTAT| > 2.0555 d.f. = 26
X − µ 43.8889 - 45
Test statistic: t STAT =
=
= −0.2284
S / n 25.2835/ 27
Decision: Since |tSTAT| < 2.0555, do not reject H 0 . There is not enough evidence to
conclude that the mean processing time has changed from 45 days.
The population distribution needs to be normal.
Normal Probability Plot
100
90
80
70
Time
60
50
40
30
20
10
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Z Value
9.30
(d)
.
The normal probability plot indicates that the distribution is not normal and is skewed to
the right. The assumption in (b) has been violated.
(a)
H0 : µ = 2
(b)
d.f. = 49
H1 : µ ≠ 2
Decision rule: Reject H 0 if |tSTAT| > 2.0096
X −µ
2.0007 - 2
Test statistic: t STAT =
=
= 0.1143
S / n 0.0446/ 50
Decision: Since |tSTAT| < 2.0096, do not reject H 0 . There is not enough evidence to
conclude that the mean amount of soft drink filled is different from 2.0 liters.
p-value = 0.9095. If the population mean amount of soft drink filled is indeed 2.0 liters,
the probability of observing a sample of 50 soft drinks that will result in a sample mean
amount of fill more different from 2.0 liters is 0.9095.
© 2015 Pearson Education Ltd.
(c)
Normal Probability Plot
2.15
2.1
2.05
Amount
9.30
cont.
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
2
1.95
1.9
1.85
-3
-2
-1
0
1
2
3
Z Value
(d)
The normal probability plot suggests that the data are rather normally distributed. Hence,
the results in (a) are valid in terms of the normality assumption.
(e)
Time Series Plot
2.15
2.1
2.05
Amount
326
2
1.95
1.9
1.85
1.8
1.75
1
4
7
10 13 16 19 22 25 28 31 34 37 40 43 46 49
The time series plot of the data reveals that there is a downward trend in the amount of
soft drink filled. This violates the assumption that data are drawn independently from a
normal population distribution because the amount of fill in consecutive bottles appears
to be closely related. As a result, the t test in (a) becomes invalid.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
(a)
PHStat output:
Data
µ=
Null Hypothesis
Level of Significance
Sample Size
Sample Mean
Sample Standard Deviation
20
0.05
50
43.04
41.92605736
Intermediate Calculations
Standard Error of the Mean
Degrees of Freedom
t Test Statistic
5.929239893
49
3.885826921
Two-Tail Test
Lower Critical Value
-2.009575199
Upper Critical Value
2.009575199
p-Value
0.000306263
Reject the null hypothesis
H 0 : µ = 20
H1 : µ ≠ 20
Decision rule: Reject H 0 if tSTAT > 2.0096 d.f. = 49
X −µ
43.04 - 20
Test statistic: t STAT =
=
= 3.8858
S / n 41.9261/ 50
Decision: Since tSTAT > 2.0096, reject H 0 . There is enough evidence to conclude that the
(b)
(c)
mean number of days is different from 20.
The population distribution needs to be normal.
Normal Probability Plot
180
160
140
120
Days
9.31
327
100
80
60
40
20
0
-3
-2
-1
0
1
2
3
Z Value
(d)
The normal probability plot indicates that the distribution is skewed to the right.
Even though the population distribution is probably not normally distributed, the result
obtained in (a) should still be valid due to the Central Limit Theorem as a result of the
relatively large sample size of 50.
© 2015 Pearson Education Ltd.
9.32
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
(b)
(c)
H 0 : µ = 8.46
H1 : µ ≠ 8.46
Decision rule: Reject H 0 if |tSTAT| > 2.0106 d.f. = 48
X − µ 8.4209 - 8.46
Test statistic: t STAT =
= -5.9355
=
S / n 0.0461/ 49
Decision: Since |tSTAT| > 2.0106, reject H 0 . There is enough evidence to conclude that
mean widths of the troughs is different from 8.46 inches.
The population distribution needs to be normal.
Normal Probability Plot
8.55
8.5
Width
328
8.45
8.4
8.35
8.3
-3
-2
-1
0
1
2
3
Z Value
(c)
Box-and-whisker Plot
Width
8.3
(d)
8.35
8.4
8.45
8.5
The normal probability plot and the boxplot indicate that the distribution is skewed to the
left. Even though the population distribution is not normally distributed, the result
obtained in (a) should still be valid due to the Central Limit Theorem as a result of the
relatively large sample size of 49.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.33
(a)
329
H0 : µ = 0
H1 : µ ≠ 0
Decision rule: Reject H 0 if |tSTAT| > 1.9842 d.f. = 99
X −µ
- 0.00023
Test statistic: t STAT =
=
= −1.3563
S / n 0.00170/ 100
Decision: Since |tSTAT| < 1.9842, do not reject H 0 . There is not enough evidence to
conclude that the mean difference is different from 0.0 inches.
(b)
(c)
(d)
X ±t
S
⎛ 0.001696 ⎞
= -0.00023 ± 1.9842⎜⎜
⎟⎟
n
100 ⎠
⎝
-0.0005665 ≤
µ ≤ 0.0001065
You are 95% confident that the mean difference is somewhere between -0.0005665 and
0.0001065 inches.
Since the 95% confidence interval does not contain 0, you do not reject the null
hypothesis in part (a). Hence, you will make the same decision and arrive at the same
conclusion as in (a).
In order for the t test to be valid, the data are assumed to be independently drawn from a
population that is normally distributed. Since the sample size is 100, which is considered
quite large, the t distribution will provide a good approximation to the sampling
distribution of the mean as long as the population distribution is not very skewed.
Box-and-whisker Plot
Error
-0.006
-0.004
-0.002
0
0.002
0.004
0.006
The boxplot suggests that the data has a distribution that is skewed slightly to the right.
Given the relatively large sample size of 100 observations, the t distribution should still
provide a good approximation to the sampling distribution of the mean.
9.34
(a)
H 0 : µ = 5.5
H1 : µ ≠ 5.5
Decision rule: Reject H 0 if |tSTAT| > 2.680 d.f. = 49
X − µ 5.5014 - 5.5
Test statistic: t STAT =
=
= 0.0935
S / n 0.1058/ 50
Decision: Since |tSTAT| < 2.680, do not reject H 0 . There is not enough evidence to
conclude that the mean amount of tea per bag is different from 5.5 grams.
s
0.1058
= 5.5014 ± 2.6800 ⋅
n
50
5.46<µ < 5.54
(b)
X ±t⋅
(c)
With 99% confidence, you can conclude that the population mean amount of tea per bag
is somewhere between 5.46 and 5.54 grams.
The conclusions are the same.
© 2015 Pearson Education Ltd.
330
9.35
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
60
0.05
30
79
57.78079862
IntermediateCalculations
StandardErroroftheMean
10.5493
DegreesofFreedom
29
t TestStatistic
1.8011
Two-TailTest
LowerCriticalValue
-2.0452
UpperCriticalValue
2.0452
p -Value
0.0821
Donotrejectthenullhypothesis
H 0 : µ = 60
H1 : µ ≠ 60
Decision rule: Reject H 0 if p-value < 0.05.
Test statistic: tSTAT =
(b)
(c)
X −µ
= 1.8011
S
n
Decision: Since p-value = 0.0821 > 0.05, do not reject H 0 . There is not enough evidence
to conclude that population mean time Facebook time is different from 60 minutes.
The population distribution needs to be normal.
You could construct charts and observe their appearance. For small- or moderate-sized
data sets, construct a stem-and-leaf display or a boxplot. For large data sets, plot a
histogram or polygon. You could also compute descriptive numerical measures and
compare the characteristics of the data with the theoretical properties of the normal
distribution. Compare the mean and median and see if the interquartile range is
approximately 1.33 times the standard deviation and if the range is approximately 6 times
the standard deviation. Evaluate how the values in the data are distributed. Determine
whether approximately two-thirds of the values lie between the mean and ±1 standard
deviation. Determine whether approximately four-fifths of the values lie between the
mean and ±1.28 standard deviations. Determine whether approximately 19 out of every
20 values lie between the mean ±2 standard deviations.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.35
cont.
331
(d)
Boxplot
Minutes
0
50
100
150
200
250
NormalProbabilityPlot
300
250
Minutes
200
150
100
50
0
-3
-2
-1
0
ZValue
1
2
3
Both the boxplot and the normal probability plot indicate that the distribution is rightskewed. Hence, the t-test in (a) might not be valid.
9.36
p-value = 1 − 0.9772 = 0.0228
9.37
Since the p-value = 0.0228 is less than
9.38
One-tail hypothesis test would test whether the population mean is less than or more than a
specified value. When the entire rejection region is contained in one tail of the sampling
distribution of the test statistic, the test is called a one-tail test, or directional test. See illustration
9.5 on page 358.
9.39
Since the p-value = 0.0838 is greater than
9.40
p-value = P(Z < 1.38) = 0.9162
9.41
Since the p-value = 0.9162 > 0.01, do not reject the null hypothesis.
9.42
t = 2.7638
α = 0.05, reject H0.
α = 0.01, do not reject H0.
© 2015 Pearson Education Ltd.
332
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
9.43
Since tSTAT = 2.39 < 2.7638, do not reject H0.
9.44
t = -2.5280
9.45
Since tSTAT = -1.15 > -2.5280, do not reject H0.
9.46
(a)
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
3900
0.05
100
3975
275
IntermediateCalculations
StandardErroroftheMean
27.5000
DegreesofFreedom
99
t TestStatistic
2.7273
Upper-TailTest
UpperCriticalValue
1.6604
p -Value
0.0038
Rejectthenullhypothesis
CalculationsArea
H 0 : µ ≤ 36.5
H1 : µ > 36.5
Decision rule: Reject H 0 if p-value < 0.05 d.f. = 99
X −µ
Test statistic: tSTAT =
= 2.7273
S/ n
p-value = 0.0038
Decision: Since the p-value of 0.0038 < 0.05, reject H 0 . There is enough evidence to
(b)
conclude that the population mean bus miles is more than 3,900 bus miles.
The p-value is 0.0038. If the population mean is indeed no more than 3900 hours, the
probability of obtaining a test statistic that is more than 2.7273 is 0.0038.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.47
(a)
H0 : π = 32 H1 : π ≠ 32
Decision rule: Reject H0 if p-value < 0.05 d.f. = 49
𝑡"#$# =
(b)
𝑋−𝜇
30 − 32
=
= −1.42
𝑆
10
𝑛
50
p-value = 0.0818
Decision: Since |tSTAT| > 2.0096 and the p-value of 0.0818 < 0.05, reject H0.
There is no enough evidence that the population average is more than 32 minutes.
Since the value of t-stat is negative, the rejection area will lie in the lower tail.
© 2015 Pearson Education Ltd.
333
334
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
9.48
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
25
0.01
355
23.05
16.83
IntermediateCalculations
StandardErroroftheMean
0.8932
DegreesofFreedom
354
t TestStatistic
-2.1831
(a)
Lower-TailTest
LowerCriticalValue
-2.3369
p -Value
0.0148
Donotrejectthenullhypothesis
H0: µ ≥ 25
H1: µ < 25
Decision rule: If p-value < 0.01, reject H0.
Test statistic: tSTAT =
(b)
9.49
H 0:
H 1:
(a)
(c)
(d)
X –µ
= = -2.1831
S
n
p-value = 0.0148
Decision: Since p-value = 0.0148 > 0.01, do not reject H0. There is not enough evidence
to conclude the population mean wait time is less than 25 minutes.
The probability of obtaining a test statistic that is -2.1831 or lower when the null
hypothesis is true is essentially 0.0148.
µ ≥ 25 min.
µ < 25 min.
The mean delivery time is not less than 25 minutes.
The mean delivery time is less than 25 minutes.
Decision rule: If tSTAT < – 1.6896, reject H0.
Test statistic: t STAT =
(b)
CalculationsArea
X – µ 22.4 – 25
= – 2.6
=
S
6
n
36
Decision: Since tSTAT = – 2.6 is less than –1.6896, reject H0. There is enough evidence to
conclude the population mean delivery time has been reduced below the previous value
of 25 minutes.
Decision rule: If p value < 0.05, reject H0.
p value = 0.0068
Decision: Since p value = 0.0068 is less than α = 0.05, reject H0. There is enough evidence
to conclude the populationmean delivery time has been reduced below the previous value of
25 minutes.
The probability of obtaining a sample whose mean is 22.4 minutes or less when the null
hypothesis is true is 0.0068.
The conclusions are the same.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.50
335
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
70
0.01
55
75
9
IntermediateCalculations
StandardErroroftheMean
1.2136
DegreesofFreedom
54
t TestStatistic
4.1201
Upper-TailTest
UpperCriticalValue
2.3974
p -Value
0.0001
Rejectthenullhypothesis
(a)
H0: µ ≤ $70
H1: µ > $70
Decision rule: If the p-value < 0.01, reject H0.
Test statistic: tSTAT =
(b)
CalculationsArea
X –µ
= 4.1201
S
n
p-value = 0.0001
Decision: Since the p-value is less than 0.01, reject H0. There is enough evidence to
conclude that the mean one-time gift donation is greater than $70.
When the null hypothesis is true, the probability of obtaining a sample whose test statistic
is 4.1201 or more is 0.0001.
© 2015 Pearson Education Ltd.
336
9.51
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
4
0.05
100
3.25
2.7
IntermediateCalculations
StandardErroroftheMean
0.2700
DegreesofFreedom
99
t TestStatistic
-2.7778
Lower-TailTest
LowerCriticalValue
-1.6604
p -Value
0.0033
Rejectthenullhypothesis
H0: µ ≥ 4 min.
H1: µ < 4 min.
Decision rule: If tSTAT < – 1.6604, reject H0.
Test statistic: tSTAT =
(b)
(c)
(d)
CalculationsArea
X –µ
= -2.7778
S
n
Decision: Since tSTAT = -2.7778 is less than –1.6604, reject H0. There is enough evidence
to conclude the population mean waiting time is less than 4 minutes.
Decision rule: If p value < 0.05, reject H0.
p value = 0.0033
Decision: Since p value = 0.0033 is less than α = 0.05, reject H0. There is enough evidence
to conclude the population mean waiting time is less than 4 minutes.
The probability of obtaining a sample whose mean is 3.25 minutes or less when the null
hypothesis is true is 0.0033.
The conclusions are the same.
X 88
= 0.22
=
n 400
9.52
p=
9.53
Z STAT =
p −π
π (1 − π )
or Z STAT =
=
0.22 − 0.20
= 1.00
0.20(0.8)
n
400
X − nπ
88 − 400(0.20 )
= 1.00
=
nπ (1 − π )
400(0.20 )(0.80 )
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.54
H0: π = 0.20
H1: π ≠ 0.20
Decision rule: If Z < – 1.96 or Z > 1.96, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
=
0.22 − 0.20
0.20(0.8)
400
n
= 1.00
Decision: Since Z = 1.00 is between the critical bounds of
9.55
337
(a)
± 1.96, do not reject H0.
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.4000
0.05
25
60
IntermediateCalculations
SampleProportion
0.416666667
StandardError
0.0632
ZTestStatistic
0.2635
Two-TailTest
LowerCriticalValue
-1.9600
UpperCriticalValue
1.9600
p -Value
0.7921
Donotrejectthenullhypothesis
H0: π = 0.4
H1: π ≠ 0.4
Decision rule: p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= 0.2635
p-value = 0.7921
n
Decision: Since p-value = 0.7921 > 0.05, do not reject H0. There is not enough evidence that
the proportion of full-time students at the university who are employed is different from the
national norm of 0.40.
© 2015 Pearson Education Ltd.
338
9.55
cont.
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(b)
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.4000
0.05
32
60
IntermediateCalculations
SampleProportion
0.533333333
StandardError
0.0632
ZTestStatistic
2.1082
Two-TailTest
LowerCriticalValue
-1.9600
UpperCriticalValue
1.9600
p -Value
0.0350
Rejectthenullhypothesis
H0: π = 0.4
H1: π ≠ 0.4
Decision rule: p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= 2.1082
p-value = 0.0350
n
Decision: Since p-value = 0.0350 < 0.05, reject H0. There is enough evidence that the
proportion of full-time students at the university who are employed is different from the
national norm of 0.40.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.56
(a)
H0: π ≤ 0.5
H1: π > 0.5
Decision rule: p-value < 0.05, reject H0.
(b)
Z STAT =
p −π
π (1 − π )
n
p = 235/500 = 0.47
𝑍"#$# =
𝑜. 47 − 0.50
0.50 (0.50)/500
= −1.34
p = 0.0899
Since, p > 0.05, do not reject H0. There is not enough evidence that the awareness
regarding the test is greater than 50%.
© 2015 Pearson Education Ltd.
339
340
9.56
cont.
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(c)
𝑍"#$# =
(d)
𝑜. 47 − 0.50
0.50 (0.50)/100
= 0.6
p = 0.2743
Since, p > 0.05, do not reject H0. There is not enough evidence that the awareness
regarding the test is greater than 50%.
The Centre for Disease Control and Prevention, using the results, can presume that the
awareness regarding the test is not greater than 50%. Thus, the centre can take
appropriate measures to spread the awareness to reach the expected awareness level.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.57
341
PHStat output:
Z Test of Hypothesis for the
Proportion
Data
Null Hypothesis
π=
Level of Significance
Number of Successes
Sample Size
0.55
0.05
9
17
Intermediate Calculations
Sample Proportion
0.529411765
Standard Error
0.12065995
Z Test Statistic
-0.170630232
Two-Tail Test
Lower Critical Value
-1.959963985
Upper Critical Value
1.959963985
p-Value
0.864514525
Do not reject the null hypothesis
H0: π = 0.55 H1: π ≠ 0.55
Decision rule: If ZSTAT < – 1.96 or ZSTAT > 1.96, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
n
=
0.5294 − 0.55
0.55 (1 − 0.55 )
17
= −0.1706
Decision: Since ZSTAT = −0.1706 is between the two critical bounds, do not reject H0. There is not
enough evidence that the proportion of females in this position at this medical center is different
from what would be expected in the general workforce.
© 2015 Pearson Education Ltd.
342
9.58
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.3500
0.05
328
801
IntermediateCalculations
SampleProportion
0.40948814
StandardError
0.0169
ZTestStatistic
3.5298
Two-TailTest
LowerCriticalValue
-1.9600
UpperCriticalValue
1.9600
p -Value
0.0004
Rejectthenullhypothesis
H0: π = 0.35
H1: π ≠ 0.35
Decision rule: If ZSTAT < – 1.96 or ZSTAT > 1.96 or if p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= 3.5298
n
Decision: Since ZSTAT = 3.5298 is greater than the upper critical bound of 1.96, reject H0.
You conclude that there is enough evidence the proportion of all LinkedIn members who
plan to spend at least $1,000 on consumer electronics in the coming year is different from
35%.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.59
343
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.2000
0.05
135
500
IntermediateCalculations
SampleProportion
0.27
StandardError
0.0179
ZTestStatistic
3.9131
(a)
Upper-TailTest
UpperCriticalValue
1.6449
p -Value
0.0000
Rejectthenullhypothesis
H0: π ≤ 0.2 H1: π > 0.2
Decision rule: If ZSTAT > 1.6449 or p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= 3.9131
n
(b)
Decision: Since ZSTAT = 3.9131 is larger than the critical bound of 1.6449, reject H0. There
is enough evidence to conclude that more than 20% of the customers would upgrade to a
new cellphone.
The manager in charge of promotional programs concerning residential customers can
use the results in (a) to try to convince potential customers to upgrade to a new cellphone
since more than 20% of all potential customers will do so based on the conclusion in (a).
© 2015 Pearson Education Ltd.
344
9.60
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.3100
0.05
28
100
IntermediateCalculations
SampleProportion
0.28
StandardError
0.0462
ZTestStatistic
-0.6487
(b)
Lower-TailTest
LowerCriticalValue
-1.6449
p -Value
0.2583
Donotrejectthenullhypothesis
H0: π ≥ 0.31
H1: π < 0.31
Decision rule: If p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= -0.6487,
p-value = 0.2583
n
Decision: Since p-value = 0.2583 > 0.05, do not reject H0. There is not enough evidence
to show that the percentage is less than 31%.
9.61
The null hypothesis represents the status quo or the hypothesis that is to be disproved. The null
hypothesis includes an equal sign in its definition of a parameter of interest. The alternative
hypothesis is the opposite of the null hypothesis and usually represents taking an action. The
alternative hypothesis includes either a less than sign, a not equal to sign, or a greater than sign in
its definition of a parameter of interest.
9.62
A Type I error represents rejecting a true null hypothesis, while a Type II error represents not
rejecting a false null hypothesis.
9.63
When population standard deviation is not known, which is a real business scenario in many
cases.
9.64
When the sample size is less than 30, it cannot be normally assumed that the population is
normally distributed. Thus, for small sample size, nonparametric tests are more appropriate.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
345
9.65
The p-value is the probability of obtaining a test statistic equal to or more extreme than the result
obtained from the sample data, given that the null hypothesis is true.
9.66
Assuming a two-tailed test is used, if the hypothesized value for the parameter does not fall into
the confidence interval, then the null hypothesis can be rejected.
9.67
The following are the 6-step critical value approach to hypothesis testing: (1) State the null
hypothesis H0. State the alternative hypothesis H1. (2) Choose the level of significance α .
Choose the sample size n. (3) Determine the appropriate statistical technique and corresponding
test statistic to use. (4) Set up the critical values that divide the rejection and nonrejection regions.
(5) Collect the data and compute the sample value of the appropriate test statistic. (6) Determine
whether the test statistic has fallen into the rejection or the nonrejection region. The computed
value of the test statistic is compared with the critical values for the appropriate sampling
distribution to determine whether it falls into the rejection or nonrejection region. Make the
statistical decision. If the test statistic falls into the nonrejection region, the null hypothesis H0
cannot be rejected. If the test statistic falls into the rejection region, the null hypothesis is
rejected. Express the statistical decision in terms of a particular situation.
9.68
The following are the 5-step p-value approach to hypothesis testing: (1) State the null hypothesis,
H0, and the alternative hypothesis, H1. (2) Choose the level of significance, α, and the sample
size, n. (3) Determine the appropriate test statistic and the sampling distribution. (4) Collect the
sample data, compute the value of the test statistic, and compute the p-value. (5) Make the
statistical decision and state the managerial conclusion. If the p-value is greater than or equal to α,
you do not reject the null hypothesis, H0. If the p-value is less than α, you reject the null
hypothesis.
9.69
(a)
H 0 : π = 0.5
(b)
The level of significance is the probability of committing a Type I error, which is the
probability of concluding that the population proportion of web page visitors preferring the
new design is not 0.50 when in fact 50% of the population proportion of web page visitors
prefer the new design. The risk associated with Type II error is the probability of not
rejecting the claim that 50% of the population proportion of web page visitors prefer the
new design when it should be rejected.
If you reject the null hypothesis for a p-value of 0.20, there is a 20% probability that you
may have incorrectly concluded that the population proportion of web page visitors
preferring the new design is not 0.50 when in fact 50% of the population proportion of web
page visitors prefer the new design.
The argument for raising the level of significance might be that the consequences of
incorrectly concluding the proportion is not 50% are deemed not very severe.
Before raising the level of significance of a test, you have to genuinely evaluate whether
the cost of committing a Type I error is really not as bad as you have thought.
If the p-value is actually 0.12, you will be more confident about rejecting the null
hypothesis. If the p-value is 0.06, you will be even more confident that a Type I error is
much less likely to occur.
(c)
(d)
(e)
(f)
H1 : π ≠ 0.5
© 2015 Pearson Education Ltd.
346
9.70
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
(b)
(c)
(d)
A Type I error occurs when a firm is predicted to be a bankrupt firm when it will not.
A Type II error occurs when a firm is predicted to be a non-bankrupt firm when it will go
bankrupt.
The executives are trying to avoid a Type I error by adopting a very stringent decision
criterion. Only firms that show significant evidence of being in financial stress will be
predicted to go bankrupt within the next two years at the chosen level of the possibility of
making a Type I error.
If the revised model results in more moderate or large Z scores, the probability of
committing a Type I error will increase. Many more of the firms will be predicted to go
bankrupt than will go bankrupt. On the other hand, the revised model that results in more
moderate or large Z scores will lower the probability of committing a Type II error because
few firms will be predicted to go bankrupt than will actually go bankrupt.
9.71
(a)
.
H0: µ ≤ 1000 H1: µ > 0.5
Decision rule: Reject H0 if p-value < 0.05 d.f. = 49.
tSTAT =
(b)
(c)
(d)
X − µ 1200 − 1000
=
= 2.83
S
500
n
50
Decision: Since |tSTAT| > 2.0096, reject H0.
Thus, there is not enough evidence to believe that the insurance claim is less than $1000.
p-value = 0.0033
Decision: the p-value of 0.0033 < 0.05, reject H0.
Thus, there is not enough evidence to believe that the insurance claim is less than $1000.
The probability of obtaining a sample whose mean is $1200 or less when the null
hypothesis is true is 0.0033.
The conclusions are the same.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.72
(a)
347
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
6.5
0.05
60
7.25
1.75
IntermediateCalculations
StandardErroroftheMean
0.2259
DegreesofFreedom
59
t TestStatistic
3.3197
Two-TailTest
LowerCriticalValue
-2.0010
UpperCriticalValue
2.0010
p -Value
0.0015
Rejectthenullhypothesis
H0: µ = $6.50
H1: µ ≠ $6.50
Decision rule: d.f. = 59. If p-value < 0.05, reject H0.
Test statistic: tSTAT =
(b)
(c)
X –µ
= = 3.3197
S
n
p-value = 0.0015
Decision: Since p-value < 0.05, reject H0. There is enough evidence to conclude that the
mean amount spent differs from $6.50
p-value = 0.0015.
Note: The-p value was found using Excel.
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.5
0.05
31
60
IntermediateCalculations
SampleProportion
0.516666667
StandardError
0.0645
ZTestStatistic
0.2582
Upper-TailTest
UpperCriticalValue
1.6449
p -Value
0.3981
Donotrejectthenullhypothesis
© 2015 Pearson Education Ltd.
348
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
9.72
cont.
H0: π ≤ 0.50.
H1: π > 0.50.
Decision rule: If p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= = 0.2582
p-value = 0.3981
n
(d)
Decision: Since p-value > 0.05, do not reject H0. There is not sufficient evidence to
conclude that more than 50% of customers say they "definitely will" recommend the
specialty coffee shop to family and friends.
PHStat output:
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
6.5
0.05
60
6.25
1.75
IntermediateCalculations
StandardErroroftheMean
0.2259
DegreesofFreedom
59
t TestStatistic
-1.1066
Two-TailTest
LowerCriticalValue
-2.0010
UpperCriticalValue
2.0010
p -Value
0.2730
Donotrejectthenullhypothesis
H0: µ = $6.50
H1: µ ≠ $6.50
Decision rule: d.f. = 59. If p-value < 0.05, reject H0.
Test statistic: t =
X – µ 6.25 – 6.5
= -1.1066
=
S
1.75
n
60
p-value = 0.2730
Decision: Since the p-value > 0.05, do not reject H0. There is not enough evidence to
conclude that the mean amount spent differs from $6.50.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.72
cont.
(e)
349
PHStat output:
ZTestofHypothesisfortheProportion
Data
NullHypothesisπ =
LevelofSignificance
NumberofItemsofInterest
SampleSize
0.5
0.05
39
60
IntermediateCalculations
SampleProportion
0.65
StandardError
0.0645
ZTestStatistic
2.3238
Upper-TailTest
UpperCriticalValue
1.6449
p -Value
0.0101
Rejectthenullhypothesis
H0: π ≤ 0.50.
H1: π > 0.50.
Decision rule: If p-value < 0.05, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
= = 2.3238
p-value = 0.0101
n
Decision: Since p-value < 0.05, reject H0. There is sufficient evidence to conclude that
more than 50% of customers say they "definitely will" recommend the specialty coffee
shop to family and friends.
9.73
(a)
H0: µ ≥ $100
H1: µ < $100
Decision rule: d.f. = 74. If tSTAT < – 1.6657, reject H0.
Test statistic: t STAT =
X – µ $93.70 – $100
= – 1.5791
=
S
$34.55
n
75
Decision: Since the test statistic of tSTAT = – 1.5791 is greater than the critical bound of
–1.6657, do not reject H0. There is not enough evidence to conclude that the mean
reimbursement for office visits to doctors paid by Medicare is less than $100.
© 2015 Pearson Education Ltd.
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9.73
cont.
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(b)
H0: π ≤ 0.10. At most 10% of all reimbursements for office visits to doctors paid by
Medicare are incorrect.
H1: π > 0.10. More than 10% of all reimbursements for office visits to doctors paid by
Medicare are incorrect.
Decision rule: If ZSTAT > 1.645, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
=
n
(c)
(d)
0.10(1 − 0.10)
75
= 1.73
Decision: Since ZSTAT = 1.73 is greater than the critical bound of 1.645, reject H0. There is
sufficient evidence to conclude that more than 10% of all reimbursements for office visits
to doctors paid by Medicare are incorrect.
To perform the t-test on the population mean, you must assume that the observed
sequence in which the data were collected is random and that the data are approximately
normally distributed.
H0: µ ≥ $100. The mean reimbursement for office visits to doctors paid by Medicare is at
least $100.
H1: µ < $100. The mean reimbursement for office visits to doctors paid by Medicare is less
than $100.
Decision rule: d.f. = 74. If tSTAT < – 1.6657, reject H0.
X – µ $90 – $100
= – 2.5066
=
S
$34.55
n
75
Test statistic: t STAT =
(e)
0.16 - 0.10
Decision: Since tSTAT = – 2.5066 is less than the critical bound of – 1.6657, reject H0.
There is enough evidence to conclude that the mean reimbursement for office visits to
doctors paid by Medicare is less than $100.
H0: π ≤ 0.10. At most 10% of all reimbursements for office visits to
doctors paid by Medicare are incorrect.
H1: π > 0.10. More than 10% of all reimbursements for office visits to doctors paid by
Medicare are incorrect.
Decision rule: If ZSTAT > 1.645, reject H0.
Test statistic: Z STAT =
p −π
π (1 − π )
n
=
0.20 - 0.10
0.10(1 − 0.10)
75
= 2.89
Decision: Since ZSTAT = 2.89 is greater than the critical bound of 1.645, reject H0. There is
sufficient evidence to conclude that more than 10% of all reimbursements for office visits
to doctors paid by Medicare are incorrect.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
(a)
H0: µ ≥ 5 minutes. The mean waiting time at a bank branch in a commercial district of the
city is at least 5 minutes during the 12:00 p.m. to 1 p.m. peak lunch period.
H1: µ < 5 minutes. The mean waiting time at a bank branch in a commercial district of the
city is less than 5 minutes during the 12:00 p.m. to 1 p.m. peak lunch period.
Decision rule: d.f. = 14. If tSTAT < – 1.7613, reject H0.
Test statistic: t STAT =
(b)
(c)
X – µ 4.2866 – 5.0
= – 1.6867
=
S
1.637985
n
15
Decision: Since tSTAT = – 1.6867 is greater than the critical bound of – 1.7613, do not
reject H0. There is not enough evidence to conclude that the mean waiting time at a bank
branch in a commercial district of the city is less than 5 minutes during the 12:00 p.m. to
1 p.m. peak lunch period.
To perform the t-test on the population mean, you must assume that the observed
sequence in which the data were collected is random and that the data are approximately
normally distributed.
Normal probability plot:
Normal Probability Plot
7
6
5
Waiting Time
9.74
4
3
2
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Z Value
(d)
(e)
351
With the exception of one extreme point, the data are approximately normally
distributed.
Based on the results of (a), the manager does not have enough evidence to make that
statement.
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Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
(b)
H 0 : µ ≤ 1500
H1 : µ > 1500
Decision rule: Reject H 0 if tSTAT > 1.6991 d.f. = 29
X – µ 1723.4 – 1500
Test statistic: t STAT =
=
= 13.664
S
89.55
n
30
Decision: Since tSTAT = 13.664 > 1.6991, reject H 0 . There is enough evidence to
conclude that the mean force is greater than 1500 pounds.
In order for the t test to be valid, the data are assumed to be independently drawn from
a population that is normally distributed.
(c)
Box-and-whisker Plot
Force
1500
(d)
9.76
(a)
(b)
1550
1600
1650
1700
1750
1800
1850
1900
The boxplot suggests that the distribution of the data is skewed only slightly to the left.
With a sample size of 30, the t distribution will provide a good approximation to the
sampling distribution of the sample mean.
Based on (a), you can conclude that the mean force is greater than 1500 pounds,
especially in light of p-value = 0.0.
H 0 : µ ≥ 0.35
H1 : µ < 0.35
Decision rule: Reject H 0 if tSTAT < −1.690 d.f. = 35
X – µ 0.3167 – 0.35
Test statistic: t STAT =
=
= −1.4735
S
0.1357
n
36
Decision: Since tSTAT > −1.690, do not reject H 0 . There is not enough evidence to
conclude that the mean moisture content for Boston shingles is less than 0.35 pounds per
100 square feet.
p-value = 0.0748. If the population mean moisture content is in fact no less than 0.35
pounds per 100 square feet, the probability of observing a sample of 36 shingles that will
result in a sample mean moisture content of 0.3167 pounds per 100 square feet or less is
.0748.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.76
(c)
cont.
(d)
(e)
353
H 0 : µ ≥ 0.35
H1 : µ < 0.35
Decision rule: Reject H 0 if tSTAT < -1.6973 d.f. = 30
X – µ 0.2735 – 0.35
Test statistic: t STAT =
=
= −3.1003
S
0.1373
n
31
Decision: Since tSTAT < -1.6973, reject H 0 . There is enough evidence to conclude that the
mean moisture content for Vermont shingles is less than 0.35 pounds per 100 square feet.
p-value = 0.0021. If the population mean moisture content is in fact no less than 0.35
pounds per 100 square feet, the probability of observing a sample of 31 shingles that will
result in a sample mean moisture content of 0.2735 pounds per 100 square feet or less is
.0021.
In order for the t test to be valid, the data are assumed to be independently drawn
from a population that is normally distributed. Since the sample sizes are 36 and 31,
respectively, which are considered quite large, the t distribution will provide a good
approximation to the sampling distribution of the mean as long as the population
distribution is not very skewed.
(f)
Box-and-whisker Plot (Boston)
0
0.2
0.4
0.6
0.8
1
0.8
1
Box-and-whisker Plot (Vermont)
0
0.2
0.4
0.6
Both boxplots suggest that the data are skewed slightly to the right, more so for the
Boston shingles. However, the very large sample sizes mean that the results of the t test
are relatively insensitive to the departure from normality.
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Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(a)
(b)
(c)
(d)
(e)
H 0 : µ = 3150
H1 : µ ≠ 3150
Decision rule: Reject H 0 if |tSTAT| > 1.9665 d.f. = 367
X – µ 3124.2147 – 3150
Test statistic: t STAT =
=
= −14.2497
S
34.713
n
368
Decision: Since tSTAT < −1.9665, reject H 0 . There is enough evidence to conclude that
the mean weight for Boston shingles is different from 3150 pounds.
p-value is virtually zero. If the population mean weight is in fact 3150 pounds, the
probability of observing a sample of 368 shingles that will yield a test statistic more
extreme than –14.2497 is virtually zero.
H 0 : µ = 3700
H1 : µ ≠ 3700
Decision rule: Reject H 0 if |tSTAT| > 1.967 d.f. = 329
X – µ 3704.0424 – 3700
Test statistic: t STAT =
=
= 1.571
S
46.7443
n
330
Decision: Since tSTAT < 1.967, do not reject H 0 . There is not enough evidence to
conclude that the mean weight for Vermont shingles is different from 3700 pounds.
p-value = .1171. The probability of observing a sample of 330 shingles that will yield a
test statistic more extreme than 1.571 is 0.1171 if the population mean weight is in fact
3700 pounds.
In order for the t test to be valid, the data are assumed to be independently drawn from a
population that is normally distributed. Since the sample sizes are 368 and 330,
respectively, which are considered large enough, the t distribution will provide a good
approximation to the sampling distribution of the mean even if the population is not
normally distributed.
© 2015 Pearson Education Ltd.
Solutions to End-of-Section and Chapter Review Problems
9.78
(a)
H 0 : µ = 0.3
355
H1 : µ ≠ 0.3
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
0.3
0.05
170
0.26
0.142382504
IntermediateCalculations
StandardErroroftheMean
0.0109
DegreesofFreedom
169
t TestStatistic
-3.2912
Two-TailTest
LowerCriticalValue
-1.9741
UpperCriticalValue
1.9741
p -Value
0.0012
Rejectthenullhypothesis
Decision rule: Reject H 0 if |tSTAT| > 1.9741 d.f. = 169
Test statistic: tSTAT =
(b)
X –µ
= -3.2912,
S
n
p-value = 0.0012
Decision: Since tSTAT < −1.9741, reject H 0 . There is enough evidence to conclude that
the mean granule loss for Boston shingles is different from 0.3 grams.
p-value is virtually zero. If the population mean granule loss is in fact 0.3 grams, the
probability of observing a sample of 170 shingles that will yield a test statistic more
extreme than -3.2912 is 0.0012.
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Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
(c)
H 0 : µ = 0.5
H1 : µ ≠ 0.5
cont.
tTestforHypothesisoftheMean
Data
NullHypothesisµ=
LevelofSignificance
SampleSize
SampleMean
SampleStandardDeviation
0.3
0.05
140
0.22
0.122698672
IntermediateCalculations
StandardErroroftheMean
0.0104
DegreesofFreedom
139
t TestStatistic
-7.9075
Two-TailTest
LowerCriticalValue
-1.9772
UpperCriticalValue
1.9772
p -Value
0.0000
Rejectthenullhypothesis
Decision rule: Reject H 0 if |tSTAT| > 1.9772
Test statistic: tSTAT =
(d)
(e)
d.f. = 139
X –µ
= -7.9075
S
n
Decision: Since tSTAT < −1.977, reject H 0 . There is enough evidence to conclude that the
mean granule loss for Vermont shingles is different from 0.3 grams.
p-value is virtually zero. The probability of observing a sample of 140 shingles that
will yield a test statistic more extreme than -1.977 is virtually zero if the population
mean granule loss is in fact 0.3 grams.
In order for the t test to be valid, the data are assumed to be independently drawn from a
population that is normally distributed. Both normal probability plots indicate that the
data are slightly right-skewed. Since the sample sizes are 170 and 140, respectively,
which are considered large enough, the t distribution will provide a good approximation
to the sampling distribution of the mean even if the population is not normally
distributed.
© 2015 Pearson Education Ltd.