Chapter 07 - Hypothesis Testing
STUDENT SOLUTION MANUAL
Business Statistics in Practice, Third Canadian Edition
by
Bowerman, Aitken Schermer, Johnson, & O’Connell
CHAPTER 7
Hypothesis Testing
7.3
[LO1, LO2]
a.
H 0 : µ ≤ 42 versus H a : µ > 42.
b.
Type I: decide the mean is > 42 (customers satisfied) when it is really ≤ 42
Type II: decide the mean is ≤ 42 (customers not satisfied) when it is really > 42
7.5
7.7
7.11
[LO1, LO2]
a.
H 0 : µ = 3 versus H a : µ ≠ 3, where µ = mean diameter.
b.
Type I: decide µ ≠ 3 (assign team) when µ = 3 (team is not needed)
Type II: decide µ = 3 (do not assign team) when µ ≠ 3 (team may be needed)
[LO1, LO2]
a.
H 0 : µ ≤ 15 versus H a : µ > 15 where µ = mean temperature of waste water.
b.
Type I: decide µ > 15 (shut down) when µ ≤ 15 (water is cool enough, no shutdown
needed)
Type II: decide µ ≤ 15 (do not shut down) when µ > 15 (water is too warm, shutdown
needed)
c.
Set α = .05 to make the probability of a Type II error smaller.
[LO3]
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-1
Chapter 07 - Hypothesis Testing
a.
z=
18 − 20
= −2
7
49
b.
Since –2 > –2.33, Do not reject H 0 with α =.01.
7.13
c.
z = –2. p-value = P(z < -2) = .0228
d.
Since .0228 is less than .10 and .05, but not less than .01 and 001, reject H 0 at α = .10
and .05, but not at α = .01 or .001.
e.
Since p-value = .0228 is less than .05, there is strong evidence against H 0 .
[LO1, LO3]
a.
H 0 : µ ≤ 42 versus H a : µ > 42
b.
Reject Points
z.10 = 1.28
z.05 = 1.645
z.01 = 2.33
z.001 = 3.09
Since 1.28<1.645<2.33<2.91<3.09, reject H 0 with α =.10, .05, .01, but not with .001.
c.
p-value = 1 − .9982 = .0018
Since p-value=.0018 is less than .10, .05, and .01, reject H 0 at those levels of α, but not
with α =.001.
d.
7.15
Since p-value = .0018 is less than .01, there is very strong evidence against H 0 .
[LO1, LO3]
a.
H 0 : µ ≤ 15 versus H a : µ > 15.
b.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-2
Chapter 07 - Hypothesis Testing
Since 2.41 > 1.645 we reject H 0 for α = 0.05. p-value = 1 – 0.9920 = 0.0080 and since
0.0080 < 0.05 we draw the same conclusion so the plant should be shut down and the
cooling system repaired.
7.17
[LO3]
reject point = 1.645, p-value = 1 – 0.9990 = 0.001 so we reject H 0 and we conclude the plant
should be shut down and the cooling system repaired.
7.19
[LO1, LO3]
a.
H 0 : µ = 500 versus H a : µ ≠ 500.
b.
For α = .01, zα / 2 =z .005 = 2.576
When x = 501.56, z = 3.00, reject H 0 ; readjust.
p-value = 2(1 – .9987) = .0026<.01, reject H 0 , readjust.
When x = 498.75, z = –2.40, do not reject H 0 : do not readjust.
p-value = 2(1 – .9918) = .0164, not less than .01, do not reject H 0 , do not readjust.
When x = 500.63, z = 1.21, do not reject H 0 ; do not readjust.
p-value=2(1 – .8849) = .2302, not less than .01, do not reject H 0 , do not readjust.
When x = 498.125, z = -3.61, reject H 0 ; readjust.
p-value is approximately 0.0003 < .001 reject H 0 , readjust.
7.23
[LO1, LO4]
H 0 : p = .3 versus H a : p ≠ .3.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-3
Chapter 07 - Hypothesis Testing
a.
z=
.2 −.3
= −2.18
.3(.7)
100
− z .01 / 2 = − z .005 = −2.575
Since –2.575 < –2.18, do not reject H 0 .
7.25
b.
p-value = 2P(z > 2.18) = .0292
c.
Reject H 0 at α = .10 and .05, but not at α = .01 or .001.
[LO1, LO4]
a.
H 0 : p ≤ .25 versus H a : p > .25.
b.
pˆ =
146
= .365
400
z.001 = 3.09
Since 5.31 > 3.09, reject H 0 at α = .10, .05, .01, .001; extremely strong evidence.
7.27
c.
p-value = P(z > 5.31) = 0.0000 is less than .001; reject H 0 at all given values of α.
d.
Probably, .365 is significantly higher than .25 in terms of shares.
[LO1, LO4]
H0: p = 0.5 Ha: p < 0.5
p-value = P(z < -3.81) = 0.00007
Since -3.81 < -3.090 or since 0.00007 < 0.001, there is extremely strong evidence that p < 0.5.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-4
Chapter 07 - Hypothesis Testing
7.29
[LO1, LO4]
a.
b.
c.
7.33
b.
Probably, pˆ = .79 is far below the claimed .95.
5−0
= 10 and z.05 = 1.645
.5
Reject H 0 at α = .05 . Conclude µ1 > µ 2 .
z=
5−4
= 2 and z.05 = 1.645
.5
p-value = P(z > 2) = .0228
Reject H 0 at α =.10, .05 , but not at α = .01 or .001.
z=
[LO1, LO5]
a.
b.
c.
7.37
316
= .79
400
.79 −.95
z=
= −14.68
.95(.05)
400
Reject H 0 at each value of α; extremely strong evidence.
pˆ =
[LO5]
a.
7.35
H 0 : p = .95 versus H a : p < .95.
H 0 : there is no difference in the mean audit delay for the two types of companies
H a : the mean audit delay for public owner-controlled companies is less than that for
manager-controlled companies.
− 10.4 − 0
= −2.0967 and − z.05 = −1.645
4.96
Reject H 0 ; conclude µ 1 < µ 2 .
z=
p-value = P(z < –2.1) = .0179
Reject H 0 at α = .10 and .05, but not at α = .01 or .001. There is strong evidence that
µ1 < µ 2 .
[LO1, LO5]
a.
H 0 : µ1 − µ 2 ≤ 0 versus H a : µ1 − µ 2 > 0 .
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-5
Chapter 07 - Hypothesis Testing
b.
7.39
.13 − 0
= 1.41 and z.05 = 1.645
.092
Do not reject H 0 at α = .05. Cannot conclude µ1 > µ 2 .
z=
[LO1, LO5]
a.
H 0 : µ1 − µ 2 = 0 versus H a : µ1 − µ 2 ≠ 0 .
b.
z=
z .025
8.2 − 7.3
(1.6) 2 (1.4) 2
+
125
175
= 1.96
= 5.06
Since 5.06 > 1.96, reject H 0 at α = .05. Conclude µ1 and µ 2 differ.
c.
7.43
p-value = 2P(z > 5.06) is less than .001. Reject H 0 at each of the given values of α ;
extremely strong evidence.
[LO1, LO6]
H 0 : p1 − p2 ≥ −.12 versus H a : p1 − p2 < −.12 .
p-value = P(z < –2.08) = .5 – .4812 = .0188 < α = .05
Reject H 0 at α = .10, .05, but not at α = .01 or .001; strong evidence.
7.45
[LO1, LO6]
a.
H 0 : p1 − p2 = 0 versus H a : p1 − p2 ≠ 0
b.
pˆ1 =
25
9
25 + 9
= .179, pˆ 2 =
= .15, pˆ =
= .17
140
60
140 + 60
1 
 1
s pˆ 1 − pˆ 2 = (.17)(.83)
+  = .058
 140 60 
.179 − .15
z=
= .50
.058
p-value = 2P(z > .5) = 2(.5 – .1915) = .617 > α = .10
Do not reject H 0 at any of the α-values; little or no evidence that p1 and p2 differ.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-6
Chapter 07 - Hypothesis Testing
7.47
[LO6]
(a)
Canada is more positive about the economy than either the US or Britain. Canada and
Australia are equally positive about their respective economies.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-7
Chapter 07 - Hypothesis Testing
(b)
All three countries are significantly more optimistic about the future than would seem to be
warranted by their enthusiasm for their economy…
7.53
[LO1, LO7]
a.
H 0 : µ ≤ 15 versus Ha : µ > 15.
n = 100, σ = 2, α = .025
µ = 15.1
μ
β
15.2
.8315
15.3
.6772
15.4
.4840
15.5
.2946
15.6
.1492
15.7
.0618
15.8
.0207
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-8
Chapter 07 - Hypothesis Testing
7.55
15.9
.0055
16.0
.0012
b.
No, β = .2946 when µ = 15.5. Increase the sample size.
c.
Plot is not included in this manual; power increase.
[LO1, LO7]
H 0 : µ ≤ 15 versus H a : µ ≥ 15, µ 0 = 15
α = .025 and β = .025 for µ a = 15.5.
z* = zα = z.025 = 1.96
z β = z.025 = 1.96
7.61
[LO8]
The effect size is actually a medium-sized effect, suggesting that, all things being equal, the
effect would have been demonstrated to be statistically significant, had you had enough
individuals within your division.
7.63
[LO8]
; small to medium effect (possibly important)
7.65
[LO1, LO4]
a.
H 0 : p ≥ .05 versus H a : p < .05.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-9
Chapter 07 - Hypothesis Testing
b.
18
= .0288
625
.0288 − .05
= −2.43
z=
(.05)(.95)
625
− z.01 = −2.33,− z.001 = −3.09
pˆ =
Since –3.09 < –2.43 ≤ –2.33, reject H 0 at α = .10, .05, .01, but not at α = .001;
very strong evidence.
7.67
7.69
c.
p-value = P(z < –2.43) = .0075
Reject H 0 at α = .10, .05, .01, but not at α = .001.
d.
Probably
[LO3]
a.
Since p-value = .0139, we reject H 0 for α = .1 and .05, but not at α = .01 and .001.
b.
Strong evidence.
[LO1, LO4]
a.
H 0 : p ≤ .60, H a : p > .60
b.
z=
.64 − .60
= 2.58
(.6)(.4)
1000
z.01 = 2.33, z.001 = 3.09
Since 2.33 < 2.58 < 3.09, Reject H 0 at α = .10, .05, .01, but not at .001, very strong evidence
7.71
[LO6]
a.
Wife entitled to half the assets.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-10
Chapter 07 - Hypothesis Testing
Reject at .01, but not at .001. Strong evidence exists.
Pension split evenly.
Reject at all α.
Stock options split evenly.
Reject at all α.
Managing the household.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-11
Chapter 07 - Hypothesis Testing
Reject at all α.
Corporate wife that travels.
Reject at all α.
Lifestyle of corporate wife.
Reject at all α.
b.
Answers will vary.
Student Solution Manual
Business Statistics in Practice, Third Canadian Edition
© 2014 McGraw-Hill Ryerson Limited. All rights reserved.
7-12