Download 4th Ed Chap 08 PG 239-266

Chapter 8
Hypothesis Testing for the Mean
and Variance of a Population
CHAPTER OVERVIEW AND OBJECTIVES
Anytime a sample is taken to check the value of a population
parameter, sampling error will be present. In other words, it is not
reasonable to expect X to exactly equal the true mean, although it
should be close. But how close is "close enough"? This chapter presents statistical methods for determining how close is close enough,
along with the consequences of that determination. By the end of the
chapter, the student should be able to:
1. Discuss what is meant by the terms "statistically significant
difference" and "hypothesis test".
2. Testing on the mean and variance of a population.
3. Discuss the implication of a given decision resulting from a
hypothesis test.
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Instructor's Manual
Chapter 8 Glossary
alternative hypothesis.
A statement in contradiction to the null
hypothesis; the researcher is attempting to determine whether this
statement can be supported.
critical value.
A value selected from an appropriate table in order to
define the rejection region for a statistical test of hypothesis.
This value depends on the significance level of the test ().
null hypothesis.
A statement (equality or inequality) concerning a
population parameter; the researcher wishes to discredit this
statement.
one-tailed test.
A test of hypothesis in which the null hypothesis is
rejected if the value of the test statistic lies in a particular
tail of the corresponding distribution.
p-value.
The value of  at which the hypothesis test procedure changes
conclusions for a given set of data.
power.
For a specified value of the population parameter, the
probability of rejecting the null hypothesis.
rejection region.
Values of the test statistic for which the null
hypothesis is rejected.
significance level ().
The probability of making a Type I error.
This
value is selected prior to obtaining the sample.
test statistic.
A function of the sample observations that provides the
value that is used in determining whether to reject or fail to
reject the null hypothesis.
236
Chapter 8
two-tailed test.
237
A test of hypothesis in which the null hypothesis is
rejected if the value of the test statistic lies in either tail of
the corresponding distribution.
Type I error.
The error that you make by rejecting the null hypothesis
(H0) when, in fact, it is true.
The probability of this error
occurring is the predetermined significance level, .
Type II error.
The error that you make by failing to reject the null
hypothesis (H0) when, in fact, it is false.
The probability of
this error occurring for a specified value of the population
parameter is ß.
237
238
8.1
Instructor's Manual
a)
H0: µ = 100
b)
Ha: µ  100
c)
(Fail to reject H0, H0 is true), (Fail to reject H0, H0 is
false), (Reject H0, H0 is true), (Reject H0, H0 is false).
d)
No, the hypothesis test procedure does not prove that the
claim is right or wrong.
8.2
8.3
a)
Type I error
b)
Type II error
c)
Type II error
a)
False.
These probabilities are conditional probabilities and
are conditioned on different events.
8.4
Increasing  will decrease ß.
b)
False.
c)
True.
d)
False.
a)
H0: Average quality measurement is greater than or equal to
Power is equal to 1-ß.
The critical region is smaller if  is made smaller.
specification.
Ha: Average quality measurement is less than specification.
b)
H0: Average income is less than or equal to $60,000.
Ha: Average income is greater than $60,000.
c)
H0: Median number of children is equal to 2.
Ha: Median number of children is not equal to 2.
d)
H0: Average age of a CEO is not greater than 55.
Ha: Average age of a CEO is greater than 55.
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Chapter 8
8.5
H0: µ = 50.1
Ha: µ =
/ 50.1
Z *  ( x  50.1) /( / n )  (53.2  50.1) /(4 / 36)
Since Z *  4.65  Z.025
8.6
239
 (3.1)(4 / 36 )=4.65
 1.96, reject H0 .
Ha: µ  7500
H0: µ = 7500
- = 7712
x
Z *  ( x  0 ) /( s / n )
s = 1096.31
 (7712  7500) /(1096.31/ 50)
 1.37
Since Z* = 1.37 < Z.025 = 1.96, fail to reject H0.
There is insufficient evidence that the average cost differs from
$7,500.
8.7
H0: µ = 3.1
Ha: µ =
/ 3.1
Z *  ( x   ) /( / n )  (2.4  3.1) /(1.1/ 30)  ( 0.7 /(1.1/ 30))
 3.486
Since Z* = -3.486 < -Z.05 = -1.645, reject H0.
8.8
Yes, the data support the alternative hypothesis since 3000 falls
outside the confidence interval.
8.9
H0: µ = 2.0
Ha: µ =
/ 2.0
Since the 95% confidence interval does not contain 2.0, reject H0.
8.10
a)
90% confidence interval for µ is
x  1.645( / n ) to x  1.645( / n )
625.35  1.645(20 / 40) to 625.35+1.645 (20/ 40)
625.35 - 5.202 to 625.35 + 5.202
620.148 to 630.552
b)
Since 633 does not lie in the 90% confidence interval, the
decision is to reject H0.
239
240
8.11
Instructor's Manual
a)
n = 100
 = 430
x  Z / 2 s / n to x  Z / 2 s / n
s = 130
430  1.645(130 / 100) to 430  1.645(130 / 100)
408.615 to 451.385
is a 90% confidence interval for the mean.
b)
H0: µ = 500
Ha: µ  500
Z *  ( x  500) /( / n )
 = .10
 (430  500) /(130 / 100)
 5.38
Since Z= -5.38 < -Z.05 = -1.645, reject Ho.
Yes, there is
sufficient evidence to indicate that the true mean selling price
for the early matches differs from $500.
8.12
20  1.96(4.2 / 49)  18.824
20  1.96(4.2 49)  21.176
1    P( Z  (21.176  22) /(4.2 / 49))  P( Z  (18.824  22) /(4.2 49 )) 
P( Z  1.37)  P( Z  5.29)
 .9147  0  .9147
8.13
n = 25
a)
 = 40
µ = 215
µ0 = 200
Z1  Z / 2  (   0 ) /( / n )
 1.645  (215  200)(40 / 25)
 .23
Z 2  Z / 2  (   0 ) /( / n )
 1.645  (215  200)(40 / 25)
 3.52
240
 = .10
Chapter 8
241
Power = P(Z > -.23) + P(Z < - 3.52)
= .5910 + (.5 - .4998)
= .5912
b)
µ0 = 160
Z1  Z / 2  (   0 )
n /
 1.645  (215  160) 25 / 40
 5.23
Z 2  Z / 2  (   0 )
n / )
 1.645  (215  160) 25 / 40
 8.52
Power = P(Z > - 5.23) + P(Z < - 8.52)
 1 + 0
= 1
8.14
H0: µ = 70
Ha: µ =
/ 70
s = 8
n = 25
70  1.96(8 / 25)  66.864
µ
 = .05
70  1.96(8 / 25)  73.136
Z = (73.136-µ)/(8/
25
)
Z = (66.864-µ/(8/
a)
62
6.96
3.04
b)
64
5.71
1.79
c)
68
3.21
- .71
d)
70
1.96
-1.96
e)
72
.71
-3.21
f)
74
- .54
-4.46
g)
76
-1.79
-5.71
Power of the test:
a)
P(Z > 6.96)
+ P(Z < 3.04)
= .5 + .4988 = .9988
b)
P(Z > 5.71)
+ P(Z < 1.79)
= .5 + .4633 = .9633
241
25
)
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Instructor's Manual
c)
P(Z > 3.21)
+ P(Z < -.71)
= 1 - (.4993 + .2611) = .2396
d)
P(Z > 1.96)
+ P(Z < -1.96) = 1 - (.4750 + .4750) = .0500
e)
P(Z > .71)
+ P(Z < -3.21) = 1 - (.2611 + .4993) = .2396
f)
P(Z > -.54)
+ P(Z < -4.46) = .2054 + .5 = .7054
g)
P(Z > -1.79) + P(Z < -5.71) = .4633 + .5 = .9633
The power curve:
µ
Power
62
0.9988
64
0.9633
68
0.2396
70
0.0500
72
0.2396
74
0.7054
76
0.9633
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Chapter 8
POWER
1
0.8
0.6
0.4
0.2
0
62
64
66
68
70
72
74
76
MEAN
8.15
The number of values of the test statistic that result in
rejecting the null hypothesis should be on average equal to
(.05)(100) = 5.
Because of the random rating of the generated
date, the number of rejections will vary.
8.16
a)
The shape of the histogram is approximately normal.
CLASS
1
2
3
4
5
6
7
8
Frequency Distribution Table
CLASS LIMITS
FREQUENCY
2300 and under 2600
2
2600 and under 2900
5
2900 and under 3200
5
3200 and under 3500
10
3500 and under 3800
11
3800 and under 4100
5
4100 and under 4400
1
4400 and under 4700
1
TOTAL
40
243
243
Frequency Histogram
244
Instructor's Manual
12
10
8
6
4
2
0
2300 and
under 2600
2600 and
under 2900
2900 and
under 3200
3200 and
under 3500
3500 and
under 3800
3800 and
under 4100
4100 and
under 4400
4400 and
under 4700
Class Limits
b)
Z Test for Population Mean
Number of Observations
Population Std. Deviation
Sample Mean
Ho: = 3200
Z*
40
500.000000
3376.914680
Ha:3200
2.24
Reject the null hypothesis since Z* = 2.24 > Z.025 = 1.96.
8.17
a)
µ =
Miles per gallon on a new model
µ0 = Miles per gallon on a new model that the company is
advertising
H0: µ  µ0
Ha: µ < µ0
b)
H0: µ  30
Ha: µ < 30
c)
H0: µ = 15 hours a week
Ha: µ =
/ 15 hours a week
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Chapter 8
8.18
n = 64
H0: µ  106
 = .05
Ha: µ < 106
s
(xi  x)2 / (n  1) 
2016 / 63 
Z = -1.645
32  566
.
- = x/n = 6592/64 = 103
x
Z*=(103-106)/(5.66/ 64)  4.24
Since Z* = -4.24 < -Z.05 = -1.645, reject H0.
8.19
a)
H0: µ = 81
 = .05
- = 85
x
Ha: µ =
/ 81
n = 49
s = 14
Z*=(85-81)/(14/ 49)  2
Since Z* = 2 > Z.025 = 1.96, reject H0.
b)
H0: µ  85
Ha: µ > 85
Z*=(85-85)/(14/ 49)  0
Since Z* = 0 < Z.05 = 1.645, fail to reject H0.
8.20
H0: µ  5
Ha: µ < 5
Z*=(x  5) /( s / n )  (4.6  5) /(1.2 / 46)
 2.26
Since Z* = -2.26 < -Z.05 = -1.645, reject H0.
8.21
n = 100
H0: µ  350
 = 375
s = 150
Ha: µ > 350
Z=(x   ) /( s / n )  (375  350) /(150 / 100)  1.67
Since Z* = 1.67 > Z.05 = 1.645, reject H0.
245
 = .05
245
246
8.22
Instructor's Manual
- = 283
x
n = 60
H0: µ  300
s = 58
 = .05
Ha: µ < 300
Z *  (283  300) /(58 / 60)
 2.27
Since Z* = -2.27 < -1.645, reject H0.
8.23
 = 100
n = 35
H0: µ  280
 = .05
True µ = 240
Ha: µ < 280
z2 = -Z - (µ - µ0)/ ( / n )
= -1.645 - (240 - 280)/ (100 / 35)
= -1.645 - (-40/16.90309)
= .72
P(Z < z2) = P(Z < .72) = .5 + .2642
= .7642 (power of test)
8.24
 = 2.5
n = 35
 = .05
Z1 = Z.05 - (µ - µ0)/ ( / n )
= 1.645 - (7 - 6)/ (2.5 / 35)
= -.72
P(Z > -.72) = .5 + .2642 = .7642
8.25
n = 100
a)
µ0 = 2375
H0: µ  2375
- - µ0)/ ( / n )
Z = (x
 = 250
 = .10
Ha: µ > 2375
= (2434 - 2375)/ (250 / 100)
= 2.34
Since Z* = 2.34 > Z.10 = 1.282, reject H0.
b)
The amount 2434 is significantly larger than 2375.
On a
percentage basis, this amount is 2.5 percent longer lasting.
246
Chapter 8
247
One should ask if consumers would notice the difference in the
life of the cartridges, especially considering that the
standard deviation is assumed to be 250.
c)
The standard deviation of the sample is 243.755, only slightly
less than the assumed 250 pages.
For a truly improved
cartridge, one would expect a decrease in the standard
deviation.
8.26
a)
n = 200
- = .001506
x
H0: µ  .0016
s = .000587
 = .01
Ha = µ < .0016
b)
Type I error is worse.
The type I error would be concluding
c)
that the plant is meeting OSHA standards when it is not.
- - µ0)/ ( s / n )
Z* = (x
= (.001506 - .0016)/ (.000587 / 200)
= -2.26
Since Z* = -2.26 > -2.33, fail to reject H0.
There is insufficient evidence
to indicate that the plant meets OSHA standards.
8.27
8.28
8.29
H0: µ = 100
Ha: µ =
/ 100
p-value = .03
a)
 = .1 > p-value = .03.
b)
 = .05 > p-value = .03.
Therefore, reject H0.
c)
 = .01 < p-value = .03.
Therefore, fail to reject H0.
a)
Not significant.
b)
Significant.
c)
Not significant.
d)
Inconclusive.
a)
p-value = 2(.5 - .4943) = .0114
b)
p-value = .5 - .4943 = .0057
Therefore, reject H0.
247
248
8.30
Instructor's Manual
c)
p-value = 2(.5 - .4693) = .0614
d)
p-value = .5 - .4693 = .0307
In a statistical sense, there is sufficient evidence to say that
the new drug takes effect in less time than the old drug.
In a
practical sense, the time of 3.5 minutes is almost the same as the
time of 3.7 minutes.
The .2 minute (12 second) difference may not
be enough of a difference to make a doctor decide to use the new
drug just because of the time difference.
8.31
n = 60
- = 925
x
s = 200
H0: µ = 975
Ha: µ  975
- - µ0)/ ( / n )
Z = (x
= (925 - 975)/ (200 / 60)
= -1.94
p-value = 2(.5 - .4738) = .0524
The p-value is the largest value of  for which you would fail to
reject the null hypothesis.
b)
Since the p-value is larger than .05, the conclusion is to
fail to reject the null hypothesis.
8.32
a)
H0: µ  175
Ha: µ < 175
- - µ)/ ( / n ) = (172 - 175)/ (8 / 70) = -3.14
Z* = (x
p-value = .5 - .4992 = .0008
b)
The p-value is the smallest significance level that could be
assigned and have the null hypothesis rejected.
248
Chapter 8
8.33
249
H0: µ  3
Ha: µ < 3
- - 3)/ ( s / n ) = (2.75 - 3)/ (1.5 / 60)
Z* = (x
= (-.25/ (1.5 / 60) ) = -1.29
p-value = .5 - .4015 = .0985
If  = .05, the decision is to fail to reject H0.
8.34
a)
- = 62
x
n = 50
s = 20
H0: µ = 70
Ha: µ =
/ 70
- - µ0)/ ( s / n )
Z* = (x
= (62 - 70)/ (20 / 50)
= -8/2.8284
= -2.83
p-value is 2(.5 - .4977) = .0046
b)
No, since H0 is rejected at a significance level of .01.
99% confidence interval for the mean is
62 ± 2.58 (20 / 50)
54.7 to 69.3
8.35
a)
Using the general rule with regard to p-values, the conclusion
would be to fail to reject the null hypothesis. The p-value is
.205.
Z Test for Population Mean
Number of Observations
Population Std. Deviation
Sample Mean
Ho :   75
Z*
60
23.754108
77.528183
Ha :   75
0.824414
P  Z  Z *
0.204852
Z Critical,  =0.05
1.644853
249
250
Instructor's Manual
b)
There is insufficient evidence to conclude that the average
cost of good shoplifted by those customers that have
shoplifted is greater than $75.
8.36
a)
Z Test for Population Mean
Number of Observations
Population Std. Deviation
Sample Mean
Ho:  80
Z*
P[Z  Z*]
50
10.152207
77.087320
Ha: < 80
Z Critical,  =0.05
-1.644853
-2.028697
0.021245
Using the p-value rule-of-thumb, the results are inconclusive.
b)
At a significance level of .10, .05, and .01, we would reject
the null hypothesis, reject the null hypothesis, and fail to
8.37
a)
reject the null hypothesis, respectively.
- - µ0)/ ( s / n ) = (56  50) /(24 15)  .968
t* = (x
Since t* = .968 < t.025,14 = 2.145, fail to reject H0.
b)
t* = (56 - 50)/ (24 / 15) = .968
Since t* = .968 < t.05,14 = 1.761, fail to reject H0.
c)
t* = (111.6 - 113.7)/ (2.5 / 30) = -4.6
Since -4.6 < -t.05,29 = -1.699, reject H0.
d)
t* = (78 - 85)/ (25 / 25) = -1.4
Since
8.38
t* = -1.4 < t.025,24 = 2.064, fail to reject H0.
a)
.01 = 2(.005) < p-value < 2(.01) = .02
b)
.005 < p-value < .01
c)
 2(.10) = .2
p-value =
d)
 .10
p-value =
250
Chapter 8
8.39
H0: µ = 20
 = .05
Ha: µ =
/ 20
n = 10
251
s = 3.66
- = 16.4
x
Note s2 = (3506 - (1842/10))/9 = 13.378
t* = (18.4 - 20)/ (3.66 / 10) = -1.38
Since t* = -1.38 < t.025,9 = 2.262, fail to reject H0.
8.40
a)
H0: µ0  15
- = 13.8
x
Ha: µ < 15
n = 25
s = 4
t* = (13.8 - 15)/ (4 / 25) = -1.5
Since t* = -1.5 > -t.05,24 = -1.711, fail to reject H0.
b)
8.41
The parent population is approximately normally distributed.
n = 20
- = 6.36
x
H0: µ = 6
Ha: µ  6
s = .6021
 = .05
t* = (6.36 - 6)/ (.6021/ 20)
= 2.67
Since t* = 2.67 > t.025,19 = 2.093, reject the null hypothesis.
8.42
H0: µ  1000
Ha: µ > 1000
- - 100)/ ( s / n ) = (1018 - 1000)/ (30 / 21) = 2.75
t* = (x
Since t* = 2.75 > t.05,20 = 1.725, reject H0.
8.43
H0: µ  180
Ha: µ > 180
- - 180)/(s/ n ) = (186 - 180)/(10.2/ 26 ) = 3.0
t* = (x
p < .005
Since the p-value is small, reject H0.
8.44
H0: µ = 16
Ha: µ =
/ 16
n = 10
- = 15.890
x
s = .307
- - 16)/(s/ n ) = (15.890 - 16)/(.307/ 10 ) = -1.13
t* = (x
Since t* = -1.13 > -t.025,9 = -2.262, fail to reject H0.
251
252
8.45
Instructor's Manual
a)
The null hypothesis could be rejected at a significance level
of .094271.
t Test for Population Mean
Number of Observations
25
Sample Standard Deviation
36.916889
Sample Mean
209.992600
b)
Ho :   200
Ha :   200
T*
P[T  T*]
1.353391
0.094271
T Critical, =0.05
1.710882
The standard assumption is that the data come from a normally
distributed disbution.
8.46 a)
By the general rule for p-values, the conclusion would be to
reject the null hypothesis.
The p-value is .002813.
t Test for Population Mean
Number of Observations
25
Sample Standard Deviation
7.540222
Sample Mean
25.018716
Ho: = 20
Ha:  20
b)
T*
2 * P[T  |T*|] two tail
3.327963
0.002813
| T Critical |,  =0.05
2.063898
The histogram indicates that the data are
approximately normally distributed.
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Chapter 8
253
253
254
8.47
8.48
Instructor's Manual
a)
2
2* < .95,10
= 3.94030
or
2
2* > .05,10
= 18.3070
b)
2
2* < .975,30
= 16.7908
or
2
2* > .025,30
= 46.98
c)
2
2* < .995,7
= .989265
or
2
2* > .005,7
= 20.2777
d)
2
2* < .975,80
= 57.1532
or
2
2* > .025,80
= 106.629
H0:   11.3
Ha:  > 11.3
2* = (n - 1)s2)/(20) = (25 - 1)(12.8)2/(11.3)2 = 30.795
2
Since 2* = 30.795 < .10,24
= 33.196, fail to reject H0.
8.49
n = 12
a)
s2 = (x2 - (x)2/n)/(n - 1) = (1065 - (107)2/12)/11 = 10.08
90% confidence interval for the population variance is
2
2
(n - 1)s2//2,n-1
to (n - 1)s2/1-/2,n-1
= (12 - 1)(10.08)/19.6751 to (12 - 1)(10.08)/4.5748
= 5.6355 to 24.237
b)
90% confidence interval for the population standard
deviation is 5.6355 to 24.237
c)
H0:   5
2.374 to 4.923
Ha:  < 5
2 = (n - 1)s2/2 = (11)(10.08)/25 = 4.4352
2
Since 2* = 4.4352 < .9,11
= 5.5777, reject H0.
8.50
a)
n = 25
s = 9.8%
s2 = 96.04
95% confidence interval for the population variance
2
2
(n - 1)s2//2,n-1
to (n-1)s2/1-/2,n-1
(24)(96.04)/39.3641 to (24)(96.04)/12.4011
58.554 to 185.867
b)
95% confidence interval for the population standard deviation
is 7.652% to 13.633%
c)
H0:  = 13%
Ha:  =
/ 13%
254
Significance level = .05
Chapter 8
Since 13% lies within the 95% confidence interval for the
standard deviation, fail to reject H0.
255
255
256
8.51
Instructor's Manual
n = 15
a)
 = .10
s = 1.7
(n - 1)s2/2/2,n-1 to (n - 1)s2/21-/2,n-1
14(1.7)2/23.6848 to 14(1.7)2/6.57063
1.7083 to 6.1577
A 90% confidence for the standard deviation is 1.307 to 2.48.
b)
Ho:   3
Ha:  < 3
 = .05
2* = (n - 1)s2/20
= 14(1.7)2/32
= 4.5
Since 2* = 4.5 < 2.95,14 = 6.57063, reject Ho.
c)
The data are assumed to be a random sample from a normally
distributed population.
8.52
2
.05,18
= 28.869 = b
2
.95,18
= 9.39 = a
Therefore, P(9.39  2  28.869) = .90
8.53
H0:   2
Ha:  < 2
2* = ((n - 1)s2/20) = 25(1.85)2/(2)2 = 21.39
p-value > .10.
8.54
H0: 2  3000
Fail to reject H0.
Ha: 2 > 3000
-)2)/(n - 1)]/20 = 45,130/3000 = 15.04
2* = [(n-1) ((xi - x
2
Since 2* = 15.04 < .01,13
= 27.6883, fail to reject H0.
8.55
a)
The first, second, and third histograms are for
ThicknessProcess1 ThicknessProcess2, and ThicknessProcess3.
256
Chapter 8
257
b)ThicknessProcess1 has the most variation, followed by ThicknesProcess2,
and then by ThicknessProcess3.
257
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Instructor's Manual
258
Chapter 8
259
259
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Instructor's Manual
After removing the value of $104.50, the conclusion changes.
Since the p-value = .518806 is greater than the .05 significance
level, fail to reject the null hypothesis.
There is
insufficient evidence to conclude that the standard deviation of
the variable CostTuneUp is greater than $5.
260
Chapter 8
8.57
Only b(Ha: µ  10) and d(Ha: µ > 10) can be acceptable alternative
261
hypotheses since an equal sign cannot appear in the alternative
hypothesis.
8.58
a)
H0: µ  14
b)
Type I error is more serious.
- - 14)/ ( s / n ) = (14.75 - 14/ (1.8 / 50) = 2.95
Z* = (x
c)
Ha: µ > 14
Since Z* = 2.95 > Z.05 = 1.645, reject H0.
8.59
a)
True, since the significance level is defined as the
probability of a Type I error.
b)
True, decreasing the probability of a Type I error increases
the probability of a Type II error.
c)
False, if the p-value is less than the significance level,
then the null hypothesis is rejected.
8.60
d)
True, there are two ways that a wrong conclusion can occur – a
a)
Type I error and a Type II error.
- ± 1.96 ( s / n )
x
11.9 ± 1.96 (1.4 / 50)
11.9 ± .388 (i.e., 11.512 to 12.288)
8.61
b)
Since 10 does not lie in this interval, reject H0.
a)
11 - 1.96 (3.1/ 40) = 11 - 1.624 = 9.376
11 + 1.96 (3.1/ 40) = 11 + 1.624 = 12.624
1 - ß = P(Z < (9.376 – 10) /(3.1/ 40) )
+ P(Z > (12.624 - 10)/ (3.1/ 40) )) = P(Z < -.75) + P(Z > 3.17)
= (.5 - .2734) + (.5 - .4992) = .2266 + .0008 = .2274
b)
9.5 - 1.96(3.1/5) = 9.5 - 1.215 = 8.285
9.5 + 1.96(3.1/5) = 9.5 + 1.215 = 10.715
1 - ß = P(Z < (8.285 - 10)/(3.1/5)) + P(Z > (10.715
261
262
Instructor's Manual
- 10)/(3.1/5)) = P(Z < -2.77) + P(Z > 1.15)
= .0028 + .1251 = .1279
c)
8 - 1.96 (3.1/ 40) = 8 - .96 = 7.04
8 + 1.96 (3.1/ 40) = 8 + .96 = 8.96
1 - ß = P(Z < (7.04-10)/ (3.1/ 40) )
+ P(Z > (8.96-10)/ (3.1/ 40) )
= P(Z < -6.04) + P(Z > -2.12) = 0 + .5 + .4830
= 0 + .9830 = .9830
8.62
a)
H0: µ  9
n = 23
Ha: µ < 9
s = 1.2
- = 8.5
x
t* = (8.5 - 9)/ (1.2 / 23) = -1.998
Since t* = -1.998 < -t.1,22 = -1.321, reject H0.
The time required to produce a batch of cylinders has reduced
significantly.
b)
The reduction in the sample mean by 30 seconds needs to be
considered in relation to the cost of the new machine and
other similar factors.
8.63
H0: µ = 4
Ha: µ =
/ 4
Z* = (3.7 - 4)/(.606/ 50 ) = -3.5
The p-value is 2(.5 - .4998) = .0004.
8.64
Reject H0.
a)
p-value = 2(.5 - .4846) = 2(.0154) = .0308
b)
p-value = 2(.5 - .4115) = 2(.0885) = .177
c)
p-value is between 2(.01) and 2(.025).
d)
p-value is between 2(.05) and 2(.1).
262
.02 < p < .05
.1 < p < .2
Chapter 8
8.65
a)
H0: µ = 4
Ha: µ  4
- - µ)/ ( s / n )
Z* = (x
= (4.2 - 4)/ (.8 / 61)
= 1.95
b)
Since Z* = 1.95 < Z.025 = 1.96, fail to reject H0.
- - µ)/ ( s / n )
t* = (x
= (4.2 - 4)/ (.8 / 61)
= 1.95
Since t* = 1.95 < t.025,60 = 2.0, fail to reject H0.
The critical value in part (b) is larger by .04.
8.66
a)
H0: µ  300
Ha: µ > 300
Z* = (305 - 300)/ (70 / 200)
= 1.01
The p-value = .5 - .3438 = .1562
b)
8.67 a)
Fail to reject H0 for  = .01,  = .05, or  = .10.
- ± 1.96 ( s / n )
95% confidence interval for µ is x
14.5 ± 1.96 (1.4 / 80)
b)
14.193 to 14.807
H0: µ  14
Ha: µ > 14
- - µ)/ ( s / n ) = (14.5 - 14)/ (1.4 80) = 3.19
Z* = (x
Since Z* = 3.19 > Z.05 = 1.645, reject H0.
8.68
n = 50
H0: µ = 50,000
- = 46,700
x
s = 14,500
Ha: µ =
/ 50,000
Z* = (46,700 - 50,000)/(14,500/ 50 )
= -1.61
p-value = 2(.0537) = .1074 > .10
263
263
264
Instructor's Manual
By the rule-of-thumb for p-value, fail to reject H0.
There is
insufficient evidence to indicate that the mean loan amount differs
from 50,000.
8.69
- = 14.2
x
n = 20
H0: µ  15
s = 3.3498
 = .10
Ha: µ < 15
t* = ( - µ)/ ( s / n )
= (14.2 - 15)/(3.3498/ 20 )
= -1.068
Since t* = -1.068 > t.10,19 = -2.539, fail to reject H0
8.70
H0: µ  .75
Ha: µ > .75
- - .75)/ ( s / n ) = (.80 - .75)/(.12/ 45 )
Z* = (x
= (.05/(.12/ 45 )) = 2.795
Since Z* = 2.795 > Z.10 = 1.28, reject H0.
8.71
a)
H0: µ  6
Ha: µ > 6
t* = ( - µ)/ ( s / n )
= (6.5 - 6)/ (1.3 / 20)
=
1.720
Since t* = 1.720 < t.05,19 = 1.729, fail to reject H0
b)
A small significance level may be more appropriate since a large
significance level may result in the company believing that it
will reach its goal of eliminating jobs when, in fact, it can’t.
8.72
a)
95% confidence interval for 2 is
-)2)/(n - 1))/(.025,21
2
[(n - 1) (((xi - x
)] to
-)2)/(n - 1))/(.975,21
2
[(n - 1) (((xi - x
)]
1.67/35.479 to 1.67/10.283
b)
95% confidence interval for  is
264
.047 to .162
Chapter 8
.047 to
c)
217 to .403
.162
H0: 2 = .07
Ha: 2 =
/ .07
Since .07 lies in the confidence interval .047 to .162, fail
to reject the null hypothesis.
8.73
H0:   20.6
Ha:  < 20.6
2* = ((n - 1)s2)/20 = 6100/(20.6)2 = 6100/424.36 = 14.37
2
Since 2* = 14.37 > .95,17
= 8.67176, fail to reject H0.
8.74
a)
H0:   4
Ha:  > 4
2* = ((n-1)s2)/20 = 23(4.7)2/(4)2 = 31.75
2
Since 2* = 31.75 < .10,23
= 32.0069, fail to reject H0.
b)The distribution of the waiting time must be normally
distributed.
8.75
H0:   0.6
Ha:  > 0.6
2* = ((n - 1)s2)/20 = 20(0.7)2/(.6)2 = 27.22
2
Since 2* = 27.22 < .05,20
= 31.4104, fail to reject H0.
8.76
a)
H0:   20
Ha:  > 20
2* = ((n - 1)s2)/20 = 6900/(20)2 = 17.25
2
Since 2* = 17.25 < .05,17
= 27.5871, fail to reject H0.
b)
Install two new terminals.
c)
p-value is greater than .10.
 = 6.42667
8.77
n = 75
s = 2.41153

a) x - t.05,74 ( s / n ) to x + t.05.74 (s / n )
6.427 - (1.645)(2.41153/ 75 ) to
6.427 + (1.645)(2.41153/ 75 )
A 90% confidence interval on the mean is 5.969 to 6.885.
265
265
266
Instructor's Manual
b)
6.427 - (1.96)(2.41153/ 75 ) to
6.427 + (1.96)(2.41153/ 75 )
A 95% confidence interval on the mean is 5.881 to 6.973.
c)
6.427 - (2.58)(2.41153) 75 ) to
6.427 + (2.58)(2.41153/ 75 )
A 99% confidence interval on the mean is 5.708 to 7.145.
8.78
n = 60
a)
-x= 38
H0: µ  40
s=2
Ha: µ < 40
Z* = (38 - 40)/(2/ 60 = -7.75
p-value 0
For a significance level of .05 or .01, reject H0.
b)
2.025,59  2.025,60 = 83.2976
 = .05
2.975,59  2.975,60 = 40.487
((59)(4)/83.2976)1/2 to ((59)(4)/40.4817)1/2
1.683 to 2.414
Since 2.5 does not lie within the 95% confidence interval,
reject H0.
8.79
a)
H0: µ = 2.6
Ha: µ =
/ 2.6
Since the p-value (.023303) is less than .05, reject H0.
Note that the value of 2.6 falls outside of the 95% confidence
interval.
b)
This is consistent with rejecting H0.
Answer for bootstrap confidence intervals will vary due to
different random samples used in the calculations.
8.80
For a one-tail test (Ha: µ > 70), t* is still 1.751.
Since this
value is positive, the p-value is one-half of the p-value for the
266
Chapter 8
two-tail test.
That is, the p-value = (.0831)/2 = .042.
267
Since the
p-value is less than the significance level of .05, reject H0.
There
is sufficient evidence to conclude that the ratings are higher than
70.
8.81
If the mean or the standard deviation of the population were
changed, the distribution of the chi-square test statistic would
still have approximately the same shape.
8.82
The increase in the power begins to level off for µ > 114 and µ <
86.
267