Upload
rushabh-vora
View
254
Download
3
Embed Size (px)
Citation preview
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 1/46
Chapter 10: Statistical Inferences About Two Populations 1
Chapter 10Statistical Inferences about Two Populations
LEARNING OBJECTIVES
The general focus of Chapter 10 is on testing hypotheses and constructing confidenceintervals about parameters from two populations, thereby enabling you to
1. Test hypotheses and construct confidence intervals about the difference in two population means using the z statistic.2. Test hypotheses and establish confidence intervals about the difference in two
population means using the t statistic.3. Test hypotheses and construct confidence intervals about the difference in two
related populations.4. Test hypotheses and construct confidence intervals about the difference in two
population proportions.5. Test hypotheses and construct confidence intervals about two population
variances.
CHAPTER TEACHING STRATEGY
The major emphasis of chapter 10 is on analyzing data from two samples. Thestudent should be ready to deal with this topic given that he/she has tested hypotheses andcomputed confidence intervals in previous chapters on single sample data.
In this chapter, the approach as to whether to use a z statistic or a t statistic for analyzing the differences in two sample means is the same as that used in chapters 8and 9. When the population variances are known, the z statistic can be used. However, if the population variances are unknown and sample variances are being used, then thet test is the appropriate statistic for the analysis. It is always an assumption underlyingthe use of the t statistic that the populations are normally distributed. If sample sizes aresmall and the population variances are known, the z statistic can be used if the
populations are normally distributed.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 2/46
Chapter 10: Statistical Inferences About Two Populations 2
In conducting a t test for the difference of two means from independent populations, there are two different formulas given in the chapter. One version of thistest uses a "pooled" estimate of the population variance and assumes that the populationvariances are equal. The other version does not assume equal population variances and is
simpler to compute. In doing hand calculations, it is generally easier to use the “pooled”variance formula because the degrees of freedom formula for the unequal varianceformula is quite complex. However, it is good to expose students to both formulas sincecomputer software packages often give you the option of using the “pooled” that assumesequal population variances or the formula for unequal variances.
A t test is also included for related (non independent) samples. It is important thatthe student be able to recognize when two samples are related and when they areindependent. The first portion of section 10.3 addresses this issue. To underscore the
potential difference in the outcome of the two techniques, it is sometimes valuable toanalyze some related measures data with both techniques and demonstrate that the results
and conclusions are usually quite different. You can have your students work problemslike this using both techniques to help them understand the differences between the twotests (independent and dependent t tests) and the different outcomes they will obtain.
A z test of proportions for two samples is presented here along with an F test for two population variances. This is a good place to introduce the student to the F distribution in preparation for analysis of variance in Chapter 11. The student will beginto understand that the F values have two different degrees of freedom. The F distributiontables are upper tailed only. For this reason, formula 10.14 is given in the chapter to beused to compute lower tailed F values for two-tailed tests.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 3/46
Chapter 10: Statistical Inferences About Two Populations 3
CHAPTER OUTLINE
10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Means
using the z Statistic (Population Variances Known)Hypothesis TestingConfidence IntervalsUsing the Computer to Test Hypotheses about the Difference in Two
Population Means Using the z Test
10.2 Hypothesis Testing and Confidence Intervals about the Difference in Two Means:Independent Samples and Population Variances Unknown
Hypothesis TestingUsing the Computer to Test Hypotheses and Construct Confidence
Intervals about the Difference in Two Population Means Using the t
TestConfidence Intervals
10.3 Statistical Inferences For Two Related PopulationsHypothesis TestingUsing the Computer to Make Statistical Inferences about Two Related
PopulationsConfidence Intervals
10.4 Statistical Inferences About Two Population Proportions, p1- p2
Hypothesis TestingConfidence IntervalsUsing the Computer to Analyze the Difference in Two Proportions
10.5 Testing Hypotheses About Two Population VariancesUsing the Computer to Test Hypotheses about Two Population Variances
KEY TERMS
Dependent Samples Independent Samples F Distribution Matched-Pairs Test F Value Related Measures
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 4/46
Chapter 10: Statistical Inferences About Two Populations 4
SOLUTIONS TO PROBLEMS IN CHAPTER 10
10.1 Sample 1 Sample 2 x
1 = 51.3 x
2 = 53.2 s12 = 52 s2
2 = 60 n1 = 31 n2 = 32
a) H o: µ1 - µ2 = 0Ha: µ1 - µ2 < 0
For one-tail test, α = .10 z .10 = -1.28
z =3260
3152
)0()2.533.51()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= -1.01
Since the observed z = -1.01 > z c = -1.28, the decision is to fail to rejectthe null hypothesis .
b) Critical value method:
z c =
2
2
2
1
2
1
2121 )()(
nn
x xc
σ σ
µ µ
+
−−−
-1.28 =3260
3152
)0()( 21
+
−− c x x
( x 1 - x 2)c = -2.41
c) The area for z = -1.01 using Table A.5 is .3438.The p-value is .5000 - .3438 = .1562
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 5/46
Chapter 10: Statistical Inferences About Two Populations 5
10.2 Sample 1 Sample 2 n1 = 32 n2 = 31 x 1 = 70.4 x 2 = 68.7 σ 1 = 5.76 σ 2 = 6.1
For a 90% C.I., z .05 = 1.645
2
22
1
21
21 )(nn
z x xσ σ
+±−
(70.4) – 68.7) + 1.64531
1.63276.5 22
+
1.7 ± 2.46
-.76 < µ1 - µ2 < 4.16
10.3 a) Sample 1 Sample 2
x 1 = 88.23 x 2 = 81.2 σ 1
2 = 22.74 σ 22 = 26.65
n1 = 30 n2 = 30
Ho: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
For two-tail test, use α /2 = .01 z .01 = + 2.33
z =30
65.2630
74.22
)0()2.8123.88()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= 5.48
Since the observed z = 5.48 > z .01 = 2.33, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 6/46
Chapter 10: Statistical Inferences About Two Populations 6
b)2
22
1
21
21 )(nn
z x xσ σ
+±−
(88.23 – 81.2) + 2.3330
65.2630
74.22 +
7.03 + 2.99
4.04 < µ < 10.02
This supports the decision made in a) to reject the null hypothesis becausezero is not in the interval.
10.4 Computers/electronics Food/Beverage
x 1 = 1.96 x 2 = 3.02 σ 1
2 = 1.0188 σ 22 = .9180
n1 = 50 n2 = 50
Ho: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
For two-tail test, α /2 = .005 z .005 = ±2.575
z =509180.0
500188.1
)0()02.396.1()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= -5.39
Since the observed z = -5.39 < z c = -2.575, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 7/46
Chapter 10: Statistical Inferences About Two Populations 7
10.5 A Bn1 = 40 n2 = 37 x 1 = 5.3 x 2 = 6.5σ 1
2 = 1.99 σ 22 = 2.36
For a 95% C.I., z .025 = 1.96
2
22
1
21
21 )(nn
z x xσ σ
+±−
(5.3 – 6.5) + 1.9637
36.24099.1
+
-1.2 ± .66 -1.86 < µ < -.54
The results indicate that we are 95% confident that, on average, Plumber B does between 0.54 and 1.86 more jobs per day than Plumber A. Since zero does not liein this interval, we are confident that there is a difference between Plumber A andPlumber B.
10.6 Managers Specialty n1 = 35 n2 = 41 x 1 = 1.84 x 2 = 1.99σ 1 = .38 σ 2 = .51
for a 98% C.I., z .01 = 2.33
2
22
1
21
21 )(nn
z x xσ σ
+±−
(1.84 - 1.99) ± 2.334151.
3538. 22
+
-.15 ± .2384
-.3884 < µ1 - µ2 < .0884
Point Estimate = -.15
Hypothesis Test:
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 8/46
Chapter 10: Statistical Inferences About Two Populations 8
1) H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
2) z =
2
22
1
21
2121 )()(
nn
x x
σ σ
µ µ
+
−−−
3) α = .02
4) For a two-tailed test, z .01 = + 2.33. If the observed z value is greater than 2.33or less than -2.33, then the decision will be to reject the null hypothesis.
5) Data given above
6) z =41
)51(.35
)38(.
)0()99.184.1(22
+
−−
= -1.47
7) Since z = -1.47 > z .01 = -2.33, the decision is to fail to reject the nullhypothesis .
8) There is no significant difference in the hourly rates of the two groups.
10.7 1996 2006 x 1 = 190 x 2 = 198 σ 1 = 18.50 σ 2 = 15.60 n1 = 51 n2 = 47 α = .01
H0: µ 1 - µ 2 = 0Ha: µ 1 - µ 2 < 0For a one-tailed test, z .01 = -2.33
z =47
)60.15(51
)50.18(
)0()198190()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= -2.32
Since the observed z = -2.32 > z .01 = -2.33, the decision is to fail to reject the nullhypothesis .
10.8 Seattle Atlanta
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 9/46
Chapter 10: Statistical Inferences About Two Populations 9
n1 = 31 n2 = 31 x 1 = 2.64 x 2 = 2.36σ 1
2 = .03 σ 22 = .015
For a 99% C.I., z .005 = 2.575
2
22
1
21
21 )(nn
z x xσ σ
+±−
(2.64-2.36) ± 2.57531015.
3103.
+
.28 ± .10 .18 < µ < .38
Between $ .18 and $ .38 difference with Seattle being more expensive.
10.9 Canon Pioneer x 1 = 5.8 x 2 = 5.0 σ 1 = 1.7 σ 2 = 1.4 n1 = 36 n2 = 45
Ho: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
For two-tail test, α /2 = .025 z .025 = ±1.96
z =45
)4.1(36
)7.1(
)0()0.58.5()()(2
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= 2.27
Since the observed z = 2.27 > z c = 1.96, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 10/46
Chapter 10: Statistical Inferences About Two Populations 10
10.10 A B x 1 = 8.05 x 2 = 7.26 σ 1 = 1.36 σ 2 = 1.06 n1 = 50 n2 = 38
Ho: µ1 - µ2 = 0Ha: µ1 - µ2 > 0
For one-tail test, α = .10 z .10 = 1.28
z =38
)06.1(50
)36.1(
)0()26.705.8()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= 3.06
Since the observed z = 3.06 > z c = 1.28, the decision is to reject the nullhypothesis .
10.11 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17
Sample 1 Sample 2 n1 = 8 n2 = 11
x 1 = 24.56 x 2 = 26.42
s12 = 12.4 s22 = 15.8
For one-tail test, α = .01 Critical t .01,17 = -2.567
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
=111
81
2118)10(8.15)7(4.12
)0()42.2656.24(
+−+
+
−−
=
-1.05
Since the observed t = -1.05 > t .01,19 = -2.567, the decision is to fail to reject thenull hypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 11/46
Chapter 10: Statistical Inferences About Two Populations 11
10.12 a) H o: µ1 - µ2 = 0 α =.10Ha: µ1 - µ2 ≠ 0 df = 20 + 20 - 2 = 38
Sample 1 Sample 2
n1 = 20 n2 = 20 x 1 = 118 x 2 = 113 s1 = 23.9 s2 = 21.6
For two-tail test, α /2 = .05 Critical t .05,38 = 1.697 (used df=30)
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
=
t =201
201
22020)19()6.21()19()9.23()0()113118(
22
+−+
+
−−
= 0.69
Since the observed t = 0.69 < t .05,38 = 1.697, the decision is to fail to rejectthe null hypothesis .
b)2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±− =
(118 – 113) + 1.697201
201
22020)19()6.21()19()9.23( 22
+−+
+
5 + 12.224
-7.224 < µ 1 - µ 2 < 17.224
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 12/46
Chapter 10: Statistical Inferences About Two Populations 12
10.13 H o: µ1 - µ2 = 0 α = .05Ha: µ1 - µ2 > 0 df = n1 + n2 - 2 = 10 + 10 - 2 = 18
Sample 1 Sample 2 n1 = 10 n2 = 10 x 1 = 45.38 x 2 = 40.49 s1 = 2.357 s2 = 2.355
For one-tail test, α = .05 Critical t .05,18 = 1.734
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
=
t =101
101
21010)9()355.2()9()357.2(
)0()49.4038.45(22
+−+
+
−−
= 4.64
Since the observed t = 4.64 > t .05,18 = 1.734, the decision is to reject thenull hypothesis .
10.14 H o: µ1 - µ2 = 0 α =.01Ha: µ1 - µ2 ≠ 0 df = 18 + 18 - 2 = 34
Sample 1 Sample 2 n1 = 18 n2 = 18 x 1 = 5.333 x 2 = 9.444 s1
2 = 12 s22 = 2.026
For two-tail test, α /2 = .005 Critical t .005,34 = ±2.75 (used df=30)
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
n sn s
x x
+−+
−+−
−−− µ µ
=
t =181
181
2181817)026.2()17(12
)0()444.9333.5(
+−+
+
−−
= -4.66
Since the observed t = -4.66 < t .005,34 = -2.75, reject the null hypothesis.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 13/46
Chapter 10: Statistical Inferences About Two Populations 13
b) For 98% confidence, t .01, 30 = 2.457
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±− =
(5.333 – 9.444) + 2.457181
181
21818)17)(026.2()17)(12(
+−+
+
-4.111 + 2.1689
-6.2799 < µ 1 - µ 2 < -1.9421
10.15 Peoria Evansville n1 = 21 n2 = 26
1 x = 116,900 2 x = 114,000 s1 = 2,300 s2 = 1,750 df = 21 + 26 – 2
90% level of confidence, α /2 = .05 t .05,45 = 1.684 (used df = 40)
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±− =
(116,900 – 114,000) + 1.684261
211
22621)25()1750()20()2300( 22
+−+
+=
2,900 + 994.62
1905.38 < µ 1 - µ 2 < 3894.62
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 14/46
Chapter 10: Statistical Inferences About Two Populations 14
10.16 H o: µ1 - µ2 = 0 α = .10Ha: µ1 - µ2 ≠ 0 df = 12 + 12 - 2 = 22
Co-op Internsn1 = 12 n2 = 12 x 1 = $15.645 x 2 = $15.439
s1 = $1.093 s2 = $0.958
For two-tail test, α /2 = .05Critical t .05,22 = ± 1.717
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
n sn s
x x
+
−+
−+−
−−− µ µ
=
t =121
121
21212)11()958.0()11()093.1(
)0()439.15645.15(22
+−+
+
−−
= 0.49
Since the observed t = 0.49 < t .05,22 = 1.717, the decision is to fail reject the nullhypothesis .
90% Confidence Interval: t .05,22 = ± 1.717
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±− =
(15.645 – 15.439) + 1.717121
121
21212)11()958.0()11()093.1( 22
+−+
+=
0.206 + 0.7204
-0.5144 < µ 1 - µ 2 < 0.9264
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 15/46
Chapter 10: Statistical Inferences About Two Populations 15
10.17 Let Boston be group 1
1) H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0
2) t =
2121
22
212
1
2121
112
)1()1()()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
3) α = .01
4) For a one-tailed test and df = 8 + 9 - 2 = 15, t .01,15 = 2.602. If the observedvalue of t is greater than 2.602, the decision is to reject the null hypothesis.
5) Boston Dallas n1 = 8 n2 = 9
x 1 = 47 x 2 = 44 s1 = 3 s2 = 3
6) t =91
81
15)3(8)3(7
)0()4447(22
++−−
= 2.06
7) Since t = 2.06 < t .01,15 = 2.602, the decision is to fail to reject the nullhypothesis.
8) There is no significant difference in rental rates between Boston and Dallas.
10.18 nm = 22 nno = 20 x m = 112 x no = 122
sm = 11 sno = 12
df = nm + nno - 2 = 22 + 20 - 2 = 40
For a 98% Confidence Interval, α /2 = .01 and t .01,40 = 2.423
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±− =
(112 – 122) + 2.423201
221
22022)19()12()21()11( 22
+−+
+
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 16/46
Chapter 10: Statistical Inferences About Two Populations 16
-10 ± 8.60
-$18.60 < µ1 - µ2 < -$1.40
Point Estimate = -$10
10.19 H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
df = n1 + n2 - 2 = 11 + 11 - 2 = 20
Toronto Mexico Cityn1 = 11 n2 = 11 x 1 = $67,381.82 x 2 = $63,481.82
s1 = $2,067.28 s2 = $1,594.25
For a two-tail test, α /2 = .005 Critical t .005,20 = ±2.845
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
=
t =111
111
21111)10()25.594,1()10()28.067,2(
)0()82.481,6382.381,67(22
+−+
+
−−
= 4.95
Since the observed t = 4.95 > t .005,20 = 2.845, the decision is to Reject the nullhypothesis .
10.20 H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0
df = n1 + n2 - 2 = 9 + 10 - 2 = 17
Men Womenn1 = 9 n2 = 10 x 1 = $110.92 x 2 = $75.48 s1 = $28.79 s2 = $30.51
This is a one-tail test, α = .01 Critical t .01,17 = 2.567
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 17/46
Chapter 10: Statistical Inferences About Two Populations 17
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
=
t =
( . . ) ( )
( . ) ( ) ( . ) ( )
1 1 09 2 7 54 8 0
2 87 9 8 3 05 1 99 1 0 2
19
11 0
2 2
− −
+
+ −+
= 2.60
Since the observed t = 2.60 > t .01,17 = 2.567, the decision is to Reject the nullhypothesis .
10.21 H o: D = 0Ha: D > 0
Sample 1 Sample 2 d 38 22 1627 28 -130 21 941 38 336 38 -238 26 1233 19 1435 31 444 35 9
n = 9 d =7.11 sd=6.45 α = .01
df = n - 1 = 9 - 1 = 8
For one-tail test and α = .01, the critical t .01,8 = ±2.896
t =9
45.6 011.7−
=−
n
s
Dd
d = 3.31
Since the observed t = 3.31 > t .01,8 = 2.896, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 18/46
Chapter 10: Statistical Inferences About Two Populations 18
10.22 H o: D = 0Ha: D ≠ 0
Before After d 107 102 5
99 98 1110 100 10113 108 5
96 89 798 101 -3
100 99 1102 102 0107 105 2
109 110 -1104 102 299 96 3
101 100 1
n = 13 d = 2.5385 sd=3.4789 α = .05df = n - 1 = 13 - 1 = 12
For a two-tail test and α /2 = .025 Critical t .025,12 = ±2.179
t =13
4789.3
05385.2 −=−
n s
Dd d = 2.63
Since the observed t = 2.63 > t .025,12 = 2.179, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 19/46
Chapter 10: Statistical Inferences About Two Populations 19
10.23 n = 22 d = 40.56 sd = 26.58
For a 98% Level of Confidence, α /2 = .01, and df = n - 1 = 22 - 1 = 21
t .01,21 = 2.518
n
st d d ±
40.56 ± (2.518)22
58.26
40.56 ± 14.27
26.29 < D < 54.83
10.24 Before After d 32 40 -828 25 335 36 -132 32 026 29 -325 31 -637 39 -216 30 -1435 31 4
n = 9 d = -3 sd = 5.6347 α = .025df = n - 1 = 9 - 1 = 8
For 90% level of confidence and α /2 = .05, t .05,8 = 1.86
t =n
st d d ±
t = -3 + (1.86) 9
6347.5 = -3 ± 3.49
-6.49 < D < 0.49
10.25 City Cost Resale d
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 20/46
Chapter 10: Statistical Inferences About Two Populations 20
Atlanta 20427 25163 -4736Boston 27255 24625 2630Des Moines 22115 12600 9515Kansas City 23256 24588 -1332
Louisville 21887 19267 2620Portland 24255 20150 4105Raleigh-Durham 19852 22500 -2648Reno 23624 16667 6957Ridgewood 25885 26875 - 990San Francisco 28999 35333 -6334Tulsa 20836 16292 4544
d = 1302.82 sd = 4938.22 n = 11, df = 10
α = .01 α /2 = .005 t .005,10 = 3.169
n
st d d ± = 1302.82 + 3.169
11
22.4938= 1302.82 + 4718.42
-3415.6 < D < 6021.2
10.26 H o: D = 0Ha: D < 0
Before After d 2 4 -24 5 -11 3 -23 3 04 3 12 5 -32 6 -43 4 -11 5 -4
n = 9 d =-1.778 sd=1.716 α = .05 df = n - 1 = 9 - 1 = 8
For a one-tail test and α = .05, the critical t .05,8 = -1.86
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 21/46
Chapter 10: Statistical Inferences About Two Populations 21
t =9
716.10778.1 −−=−
n s
Dd d = -3.11
Since the observed t = -3.11 < t .05,8 = -1.86, the decision is to reject the nullhypothesis .
10.27 Before After d 255 197 58230 225 5290 215 75242 215 27300 240 60
250 235 15215 190 25230 240 -10225 200 25219 203 16236 223 13
n = 11 d = 28.09 sd=25.813 df = n - 1 = 11 - 1 = 10
For a 98% level of confidence and α /2=.01, t .01,10 = 2.764
n
st d d ±
28.09 ± (2.764) 11
813.25 = 28.09 ± 21.51
6.58 < D < 49.60
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 22/46
Chapter 10: Statistical Inferences About Two Populations 22
10.28 H 0: D = 0Ha: D > 0 n = 27 df = 27 – 1 = 26 d = 3.17 sd = 5
Since α = .01, the critical t .01,26 = 2.479
t =27
5071.3 −
=−
n
s
Dd
d = 3.86
Since the observed t = 3.86 > t .01,26 = 2.479, the decision is to reject the nullhypothesis .
10.29 n = 21 d = 75 sd = 30 df = 21 - 1 = 20
For a 90% confidence level, α /2=.05 and t .05,20 = 1.725
n
st d d ±
75 + 1.72521
30= 75 ± 11.29
63.71 < D < 86.29
10.30 H o: D = 0Ha: D ≠ 0
n = 15 d = -2.85 sd = 1.9 α = .01 df = 15 - 1 = 14
For a two-tail test, α /2 = .005 and the critical t .005,14 = + 2.977
t =159.1
085.2 −−=−
n
s Dd
d = -5.81
Since the observed t = -5.81 < t .005,14 = -2.977, the decision is to reject the nullhypothesis .
10.31 a) Sample 1 Sample 2
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 23/46
Chapter 10: Statistical Inferences About Two Populations 23
n1 = 368 n2 = 405 x1 = 175 x2 = 182
368175
ˆ1
11
==
n x
p = .476405182
ˆ2
22
==
n x
p = .449
773357
405368182175
21
21 =+
+=
+
+=
nn x x
p = .462
Ho: p1 - p2 = 0Ha: p1 - p2 ≠ 0
For two-tail, α /2 = .025 and z .025 = ±1.96
+
−−=
+⋅
−−−=
4051
3681)538)(.462(.
)0()449.476(.
11
)()ˆˆ(
1
2121
nnq p
p p p p z
= 0.75
Since the observed z = 0.75 < z c = 1.96, the decision is to fail to reject the nullhypothesis .
b) Sample 1 Sample 2 p̂ 1 = .38 p̂ 2 = .25 n1 = 649 n2 = 558
558649 )25(.558)38(.649ˆˆ
21
2211++=++= nn pn pn p = .32
Ho: p1 - p2 = 0Ha: p1 - p2 > 0
For a one-tail test and α = .10, z .10 = 1.28
+
−−=
+⋅
−−−=
5581
6491
)68)(.32(.
)0()25.38(.
11
)()ˆˆ(
1
2121
nnq p
p p p p z
= 4.83
Since the observed z = 4.83 > z c = 1.28, the decision is to reject the nullhypothesis .
10.32 a) n1 = 85 n2 = 90 p̂ 1 = .75 p̂ 2 = .67
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 24/46
Chapter 10: Statistical Inferences About Two Populations 24
For a 90% Confidence Level, z .05 = 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.75 - .67) ± 1.64590
)33)(.67(.85
)25)(.75(.+ = .08 ± .11
-.03 < p1 - p2 < .19
b) n1 = 1100 n2 = 1300 p̂ 1 = .19 p̂ 2 = .17
For a 95% Confidence Level, α /2 = .025 and z .025 = 1.96
2
22
1
1121 ˆˆˆˆ)ˆˆ(
nq p
nq p z p p +±−
(.19 - .17) + 1.961300
)83)(.17(.1100
)81)(.19(.+ = .02 ± .03
-.01 < p1 - p2 < .05
c) n1 = 430 n2 = 399 x1 = 275 x2 = 275
430275ˆ
1
11
==
n x
p = .64399275
ˆ2
22
==
n x
p = .69
For an 85% Confidence Level, α /2 = .075 and z .075 = 1.44
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.64 - .69) + 1.44399
)31)(.69(.
430
)36)(.64(.+ = -.05 ± .047
-.097 < p1 - p2 < -.003
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 25/46
Chapter 10: Statistical Inferences About Two Populations 25
d) n1 = 1500 n2 = 1500 x1 = 1050 x2 = 1100
15001050
ˆ1
11
==
n x
p = .7015001100ˆ
2
22
==
n x
p = .733
For an 80% Confidence Level, α /2 = .10 and z .10 = 1.28
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.70 - .733) ± 1.281500
)267)(.733(.1500
)30)(.70(.+ = -.033 ± .02
-.053 < p1 - p2 < -.013
10.33 H 0: pm - pw = 0Ha: pm - pw < 0 nm = 374 nw = 481 p̂ m = .59 p̂ w = .70
For a one-tailed test and α = .05, z .05 = -1.645
481374)70(.481)59(.374ˆˆ
++=
++=
wm
wwmm
nn pn pn
p = .652
+
−−=
+⋅
−−−=
4811
3741
)348)(.652(.
)0()70.59(.
11
)()ˆˆ(
1
2121
nnq p
p p p p z
= -3.35
Since the observed z = -3.35 < z .05 = -1.645, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 26/46
Chapter 10: Statistical Inferences About Two Populations 26
10.34 n1 = 210 n2 = 176 1ˆ p = .24 2
ˆ p = .35
For a 90% Confidence Level, α /2 = .05 and z .05 = + 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(n
q pn
q p z p p +±−
(.24 - .35) + 1.645176
)65)(.35(.210
)76)(.24(.+ = -.11 + .0765
-.1865 < p1 – p2 < -.0335
10.35 Computer Firms Banks p̂ 1 = .48 p̂ 2 = .56 n1 = 56 n2 = 89
8956)56(.89)48(.56ˆˆ
21
2211
++=
++=
nn pn pn
p = .529
Ho: p1 - p2 = 0Ha: p1 - p2 ≠ 0
For two-tail test, α /2 = .10 and z c = ±1.28
+
−−=
+⋅
−−−=
891
561
)471)(.529(.
)0()56.48(.
11
)()ˆˆ(
1
2121
nnq p
p p p p z
= -0.94
Since the observed z = -0.94 > z c = -1.28, the decision is to fail to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 27/46
Chapter 10: Statistical Inferences About Two Populations 27
10.36 A Bn1 = 35 n2 = 35
x1 = 5 x2 = 7
355ˆ
1
11
==
n x
p = .14357
ˆ2
22
==
n x
p = .20
For a 98% Confidence Level, α /2 = .01 and z .01 = 2.33
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.14 - .20) ± 2.3335
)80)(.20(.
35
)86)(.14(.+ = -.06 ± .21
-.27 < p1 - p2 < .15
10.37 H 0: p1 – p2 = 0Ha: p1 – p2 ≠ 0
α = .10 p̂ 1 = .09 p̂ 2 = .06 n1 = 780 n2 = 915
For a two-tailed test, α /2 = .05 and z .05 = + 1.645
915780)06(.915)09(.780ˆˆ
21
2211
++=
++=
nn pn pn
p = .0738
+
−−=
+⋅
−−−=
9151
7801
)9262)(.0738(.
)0()06.09(.
11
)()ˆˆ(
1
2121
nnq p
p p p p Z
= 2.35
Since the observed z = 2.35 > z .05 = 1.645, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 28/46
Chapter 10: Statistical Inferences About Two Populations 28
10.38 n1 = 850 n2 = 910 p̂ 1 = .60 p̂ 2 = .52
For a 95% Confidence Level, α /2 = .025 and z .025 = + 1.96
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.60 - .52) + 1.96910
)48)(.52(.850
)40)(.60(.+ = .08 + .046
.034 < p1 – p2 < .126
10.39 H 0: σ 12 = σ 2
2 α = .01 n1 = 10 s12 = 562
Ha: σ 12 < σ 2
2 n2 = 12 s22 = 1013
df num = 12 - 1 = 11 df denom = 10 - 1 = 9
Table F .01,10,9 = 5.26
F =562
10132
1
22 =
s
s= 1.80
Since the observed F = 1.80 < F .01,10,9 = 5.26, the decision is to fail to reject thenull hypothesis .
10.40 H 0: σ 12 = σ 2
2 α = .05 n1 = 5 s1 = 4.68Ha: σ 1
2 ≠ σ 22 n2 = 19 s2 = 2.78
df num = 5 - 1 = 4 df denom = 19 - 1 = 18
The critical table F values are: F .025,4,18 = 3.61 F .95,18,4 = .277
F = 2
2
22
21
)78.2(
)68.4(=
s
s= 2.83
Since the observed F = 2.83 < F .025,4,18 = 3.61, the decision is to fail to reject thenull hypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 29/46
Chapter 10: Statistical Inferences About Two Populations 29
10.41 City 1 City 2
3.43 3.333.40 3.423.39 3.393.32 3.303.39 3.463.38 3.393.34 3.363.38 3.443.38 3.373.28 3.38
n1 = 10 df 1 = 9 n2 = 10 df 2 = 9 s1
2 = .0018989 s22 = .0023378
H0: σ 12 = σ 2
2 α = .10 α /2 = .05Ha: σ 1
2 ≠ σ 22
Upper tail critical F value = F .05,9,9 = 3.18
Lower tail critical F value = F .95,9,9 = 0.314
F =0023378.0018989.
22
21 =
s
s= 0.81
Since the observed F = 0.81 is greater than the lower tail critical value of 0.314and less than the upper tail critical value of 3.18, the decision is to failto reject the null hypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 30/46
Chapter 10: Statistical Inferences About Two Populations 30
10.42 Let Houston = group 1 and Chicago = group 2
1) H 0: σ 12 = σ 2
2 Ha: σ 1
2 ≠ σ 22
2) F = 22
21
s s
3) α = .01
4) df 1 = 12 df 2 = 10 This is a two-tailed test
The critical table F values are: F .005,12,10 = 5.66 F .995,10,12 = .177
If the observed value is greater than 5.66 or less than .177, the decision will beto reject the null hypothesis.
5) s12 = 393.4 s2
2 = 702.7
6) F =7.702
4.393= 0.56
7) Since F = 0.56 is greater than .177 and less than 5.66,the decision is to fail to reject the null hypothesis .
8) There is no significant difference in the variances of number of days between Houston and Chicago.
10.43 H 0: σ 12 = σ 2
2 α = .05 n1 = 12 s1 = 7.52Ha: σ 1
2 > σ 22 n2 = 15 s2 = 6.08
df num = 12 - 1 = 11 df denom = 15 - 1 = 14
The critical table F value is F .05,10,14 = 2.60
F = 2
2
22
21
)08.6()52.7(
= s
s
= 1.53
Since the observed F = 1.53 < F .05,10,14 = 2.60, the decision is to fail to reject thenull hypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 31/46
Chapter 10: Statistical Inferences About Two Populations 31
10.44 H 0: σ 12 = σ 2
2 α = .01 n1 = 15 s12 = 91.5
Ha: σ 12 ≠ σ 2
2 n2 = 15 s22 = 67.3
df num = 15 - 1 = 14 df denom = 15 - 1 = 14 The critical table F values are: F .005,12,14 = 4.43 F .995,14,12 = .226
F =3.675.91
22
21 =
s
s= 1.36
Since the observed F = 1.36 < F .005,12,14 = 4.43 and > F .995,14,12 = .226, the decision isto fail to reject the null hypothesis .
10.45 H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0
For α = .10 and a two-tailed test, α /2 = .05 and z .05 = + 1.645
Sample 1 Sample 2 x 1 = 138.4 x 2 = 142.5σ 1 = 6.71 σ 2 = 8.92
n1 = 48 n2 = 39
z =39
)92.8(48
)71.6(
)0()5.1424.138()()(2
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= -2.38
Since the observed value of z = -2.38 is less than the critical value of z = -1.645,the decision is to reject the null hypothesis . There is a significant difference inthe means of the two populations.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 32/46
Chapter 10: Statistical Inferences About Two Populations 32
10.46 Sample 1 Sample 2 x 1 = 34.9 x 2 = 27.6σ 1
2 = 2.97 σ 22 = 3.50
n1 = 34 n2 = 31
For 98% Confidence Level, z .01 = 2.33
2
22
1
21
21 )(n
s
n
s z x x +±−
(34.9 – 27.6) + 2.333150.3
3497.2
+ = 7.3 + 1.04
6.26 < µ 1 - µ 2 < 8.34
10.47 H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0
Sample 1 Sample 2 x 1= 2.06 x 2 = 1.93
s12 = .176 s2
2 = .143 n1 = 12 n2 = 15 α = .05
This is a one-tailed test with df = 12 + 15 - 2 = 25. The critical value ist .05,25 = 1.708. If the observed value is greater than 1.708, the decision will be toreject the null hypothesis.
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
t =151
121
25)14)(143(.)11)(176(.
)0()93.106.2(
++
−−
= 0.85
Since the observed value of t = 0.85 is less than the critical value of t = 1.708, thedecision is to fail to reject the null hypothesis . The mean for population one isnot significantly greater than the mean for population two.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 33/46
Chapter 10: Statistical Inferences About Two Populations 33
10.48 Sample 1 Sample 2 x 1 = 74.6 x 2 = 70.9 s1
2 = 10.5 s22 = 11.4
n1 = 18 n2 = 19
For 95% confidence, α /2 = .025.Using df = 18 + 19 - 2 = 35, t 30,.025 = 2.042
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±−
(74.6 – 70.9) + 2.042191
181
21918)18)(4.11()17)(5.10(
+−+
+
3.7 + 2.22
1.48 < µ 1 - µ 2 < 5.92
10.49 H o: D = 0 α = .01Ha: D < 0
n = 21 df = 20 d = -1.16 sd = 1.01
The critical t .01,20 = -2.528. If the observed t is less than -2.528, then the decisionwill be to reject the null hypothesis.
t =2101.1
016.1 −−=−
n s
Dd d = -5.26
Since the observed value of t = -5.26 is less than the critical t value of -2.528, thedecision is to reject the null hypothesis . The population difference is lessthan zero.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 34/46
Chapter 10: Statistical Inferences About Two Populations 34
10.50 Respondent Before After d 1 47 63 -162 33 35 - 23 38 36 24 50 56 - 6
5 39 44 - 56 27 29 - 27 35 32 38 46 54 - 89 41 47 - 6
d = -4.44 sd = 5.703 df = 8For a 99% Confidence Level, α /2 = .005 and t 8,.005 = 3.355
n
st d d ± = -4.44 + 3.355
9
703.5= -4.44 + 6.38
-10.82 < D < 1.94
10.51 H o: p1 - p2 = 0 α = .05 α /2 = .025Ha: p1 - p2 ≠ 0 z .025 = + 1.96
If the observed value of z is greater than 1.96 or less than -1.96, then the decisionwill be to reject the null hypothesis.
Sample 1 Sample 2 x1 = 345 x2 = 421n1 = 783 n2 = 896
896783421345
21
21
+
+=
+
+=
nn x x
p = .4562
783345ˆ
1
11
==
n x
p = .4406896421
ˆ2
22
==
n x
p = .4699
+
−−=
+⋅
−−−=
8961
7831
)5438)(.4562(.
)0()4699.4406(.11
)()ˆˆ(
1
2121
nnq p
p p p p z = -1.20
Since the observed value of z = -1.20 is greater than -1.96, the decision is to fail toreject the null hypothesis . There is no significant difference.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 35/46
Chapter 10: Statistical Inferences About Two Populations 35
10.52 Sample 1 Sample 2n1 = 409 n2 = 378 p̂ 1 = .71 p̂ 2 = .67
For a 99% Confidence Level, α /2 = .005 and z .005 = 2.575
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.71 - .67) + 2.575378
)33)(.67(.409
)29)(.71(.+ = .04 ± .085
-.045 < p1 - p2 < .125
10.53 H 0: σ 12 = σ 2
2 α = .05 n1 = 8 s12 = 46
Ha: σ 12 ≠ σ 2
2 n2 = 10 s22 = 37
df num = 8 - 1 = 7 df denom = 10 - 1 = 9The critical F values are: F .025,7,9 = 4.20 F .975,9,7 = .238
If the observed value of F is greater than 4.20 or less than .238, then the decisionwill be to reject the null hypothesis.
F =3746
22
21
=
s
s= 1.24
Since the observed F = 1.24 is less than F .025,7,9 =4.20 and greater than F .975,9,7 = .238, the decision is to fail to reject the null hypothesis . There is nosignificant difference in the variances of the two populations.
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 36/46
Chapter 10: Statistical Inferences About Two Populations 36
10.54 Term Whole Life x t = $75,000 x w = $45,000 s t = $22,000 sw = $15,500 nt = 27 nw = 29
df = 27 + 29 - 2 = 54
For a 95% Confidence Level, α /2 = .025 and t .025,50 = 2.009 (used df=50)
2121
22
212
121
112
)1()1()(
nnnn
n sn st x x +
−+
−+−±−
(75,000 – 45,000) + 2.009291
271
22927)28()500,15()26()000,22( 22
+−+
+
30,000 ± 10,160.11
19,839.89 < µ1 - µ2 < 40,160.11
10.55 Morning Afternoon d 43 41 251 49 237 44 -7
24 32 -847 46 144 42 250 47 355 51 446 49 -3
n = 9 d = -0.444 sd =4.447 df = 9 - 1 = 8
For a 90% Confidence Level: α /2 = .05 and t .05,8 = 1.86
n st d d ±
-0.444 + (1.86) 9
447.4 = -0.444 ± 2.757
-3.201 < D < 2.313
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 37/46
Chapter 10: Statistical Inferences About Two Populations 37
10.56 Marketing Accountants n1 = 400 n2 = 450 x1 = 220 x2 = 216
Ho: p1 - p2 = 0
Ha: p1 - p2 > 0 α = .01
The critical table z value is: z .01 = 2.33
1ˆ p =
400220
= .55 2ˆ p =
450216
= .48
450400216220
21
21
+
+=
+
+=
nn x x
p = .513
+
−−=
+⋅
−−−=
4501
4001
)487)(.513(.
)0()48.55(.
11
)()ˆˆ(
1
2121
nnq p
p p p p z = 2.04
Since the observed z = 2.04 is less than z .01 = 2.33, the decision is to fail to rejectthe null hypothesis . There is no significant difference between marketingmanagers and accountants in the proportion who keep track of obligations “intheir head”.
10.57 Accounting Data Entryn1 = 16 n2 = 14
x 1 = 26,400 x 2 = 25,800 s1 = 1,200 s2 = 1,050
H0: σ 12 = σ 2
2
Ha: σ 12 ≠ σ 2
2 α = .05 and α /2 = .025
df num = 16 – 1 = 15 df denom = 14 – 1 = 13
The critical F values are: F .025,15,13 = 3.05 F .975,15,13 = 0.33
F =500,102,1000,440,1
22
21 =
s
s= 1.31
Since the observed F = 1.31 is less than F .025,15,13 = 3.05 and greater than F .975,15,13 = 0.33, the decision is to fail to reject the null hypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 38/46
Chapter 10: Statistical Inferences About Two Populations 38
10.58 Men Women
n1 = 60 n2 = 41 x 1 = 631 x 2 = 848σ 1 = 100 σ 2 = 100
For a 95% Confidence Level, α /2 = .025 and z .025 = 1.96
2
22
1
21
21 )(n
s
n
s z x x +±−
(631 – 848) + 1.9641
100
60
100 22
+ = -217 ± 39.7
-256.7 < µ1 - µ2 < -177.3
10.59 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0 df = 20 + 24 - 2 = 42
Detroit Charlotte n1 = 20 n2 = 24 x 1 = 17.53 x 2 = 14.89 s1 = 3.2 s2 = 2.7
For two-tail test, α /2 = .005 and the critical t .005,40 = ±2.704 (used df=40)
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
t =241
201
42)23()7.2()19()2.3(
)0()89.1453.17(22
++
−−
= 2.97
Since the observed t = 2.97 > t .005,40 = 2.704, the decision is to reject the nullhypothesis .
10.60 With Fertilizer Without Fertilizer
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 39/46
Chapter 10: Statistical Inferences About Two Populations 39
x 1 = 38.4 x 2 = 23.1 σ 1 = 9.8 σ 2 = 7.4 n1 = 35 n2 = 35
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
For one-tail test, α = .01 and z .01 = 2.33
z =35
)4.7(35
)8.9(
)0()1.234.38()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= 7.37
Since the observed z = 7.37 > z .01 = 2.33, the decision is to reject the nullhypothesis .
10.61 Specialty Discountn1 = 350 n2 = 500 p̂ 1 = .75 p̂ 2 = .52
For a 90% Confidence Level, α /2 = .05 and z .05 = 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ( n
q pn
q p z p p +±−
(.75 - .52) + 1.645500
)48)(.52(.350
)25)(.75(.+ = .23 ± .053
.177 < p1 - p2 < .283
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 40/46
Chapter 10: Statistical Inferences About Two Populations 40
10.62 H 0: σ 12 = σ 2
2 α = .01 n1 = 8 n2 = 7Ha: σ 1
2 ≠ σ 22 s1
2 = 72,909 s22 = 129,569
df num = 6 df denom = 7The critical F values are: F .005,6,7 = 9.16 F .995,7,6 = .11
F =909,72569,129
22
21 =
s
s= 1.78
Since F = 1.78 < F .005,6,7 = 9.16 but also > F .995,7,6 = .11, the decision is to fail toreject the null hypothesis . There is no difference in the variances of the shifts.
10.63 Name Brand Store Brand d 54 49 555 50 559 52 753 51 254 50 461 56 551 47 453 49 4
n = 8 d = 4.5 sd=1.414 df = 8 - 1 = 7
For a 90% Confidence Level, α /2 = .05 and t .05,7 = 1.895
n
st d d ±
4.5 + 1.8958
414.1= 4.5 ± .947
3.553 < D < 5.447
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 41/46
Chapter 10: Statistical Inferences About Two Populations 41
10.64 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40
Wisconsin Tennesseen1 = 23 n2 = 19 x 1 = 69.652 x 2 = 71.7368 s1
2 = 9.9644 s22 = 4.6491
For one-tail test, α = .01 and the critical t .01,40 = -2.423
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
t =191
231
40)18)(6491.4()22)(9644.9(
)0()7368.71652.69(
++
−−
= -2.44
Since the observed t = -2.44 < t .01,40 = -2.423, the decision is to reject the nullhypothesis .
10.65 Wednesday Friday d 71 53 1856 47 975 52 2368 55 1374 58 16
n = 5 d = 15.8 sd = 5.263 df = 5 - 1 = 4
Ho: D = 0 α = .05Ha: D > 0
For one-tail test, α = .05 and the critical t .05,4 = 2.132
t =5
263.508.15 −
=−
n
s
Dd
d = 6.71
Since the observed t = 6.71 > t .05,4 = 2.132, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 42/46
Chapter 10: Statistical Inferences About Two Populations 42
10.66 H o: p1 - p2 = 0 α = .05Ha: p1 - p2 ≠ 0
Machine 1 Machine 2 x1 = 38 x2 = 21
n1 = 191 n2 = 202
19138ˆ
1
11 ==
n x
p = .19920221
ˆ2
22
==
n x
p = .104
202191)202)(104(.)191)(199(.ˆˆ
21
2211
++=+
+=nn
pn pn p = .15
For two-tail, α /2 = .025 and the critical z values are: z .025 = ±1.96
+
−−=
+⋅
−−−=
2021
1911
)85)(.15(.)0()104.199(.
11)()ˆˆ(
1
2121
nnq p
p p p p z = 2.64
Since the observed z = 2.64 > z c = 1.96, the decision is to reject the nullhypothesis .
10.67 Construction Telephone Repair
n1 = 338 n2 = 281 x1 = 297 x2 = 192
338297ˆ
1
11
==
n x
p = .879281192ˆ
2
22
==
n x
p = .683
For a 90% Confidence Level, α /2 = .05 and z .05 = 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(
nq p
nq p
z p p +±−
(.879 - .683) + 1.645281
)317)(.683(.338
)121)(.879(.+ = .196 ± .054
.142 < p1 - p2 < .250
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 43/46
Chapter 10: Statistical Inferences About Two Populations 43
10.68 Aerospace Automobile n1 = 33 n2 = 35 x 1 = 12.4 x 2 = 4.6 σ 1 = 2.9 σ 2 = 1.8
For a 99% Confidence Level, α /2 = .005 and z .005 = 2.575
2
22
1
21
21 )(nn
z x xσ σ
+±−
(12.4 – 4.6) + 2.57535
)8.1(33
)9.2( 22
+ = 7.8 ± 1.52
6.28 < µ1 - µ2 < 9.32
10.69 Discount Specialty x 1 = $47.20 x 2 = $27.40
σ 1 = $12.45 σ 2 = $9.82 n1 = 60 n2 = 40
Ho: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0
For two-tail test, α /2 = .005 and z c = ±2.575
z =40
)82.9(60
)45.12(
)0()40.2720.47()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
x x
σ σ
µ µ
= 8.86
Since the observed z = 8.86 > z c = 2.575, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 44/46
Chapter 10: Statistical Inferences About Two Populations 44
10.70 Before After d 12 8 4
7 3 410 8 216 9 7
8 5 3
n = 5 d = 4.0 sd = 1.8708 df = 5 - 1 = 4
Ho: D = 0 α = .01Ha: D > 0
For one-tail test, α = .01 and the critical t .01,4 = 3.747
t =5
8708.100.4 −
=−
n
s
Dd
d = 4.78
Since the observed t = 4.78 > t .01,4 = 3.747, the decision is to reject the nullhypothesis .
10.71 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0 df = 10 + 6 - 2 = 14
A B___ n1 = 10 n2 = 6
x 1 = 18.3 x 2 = 9.667 s1
2 = 17.122 s22 = 7.467
For two-tail test, α /2 = .005 and the critical t .005,14 = ±2.977
t =
2121
22
212
1
2121
112
)1()1(
)()(
nnnnn sn s
x x
+−+
−+−
−−− µ µ
t =61
101
14)5)(467.7()9)(122.17(
)0()667.93.18(
++
−−
= 4.52
Since the observed t = 4.52 > t .005,14 = 2.977, the decision is to reject the nullhypothesis .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 45/46
Chapter 10: Statistical Inferences About Two Populations 45
10.72 A t test was used to test to determine if Hong Kong has significantly differentrates than Mumbai. Let group 1 be Hong Kong.
Ho: µ1 - µ2 = 0
Ha: µ1 - µ2 ≠ 0n1 = 19 n2 = 23 x 1 = 130.4 x 2 = 128.4 s1 = 12.9 s2 = 13.9 98% C.I. and α /2 = .01
t = 0.48 with a p-value of .634 which is not significant at of .05. There is notenough evidence in these data to declare that there is a difference in the averagerental rates of the two cities.
10.73 H 0: D = 0Ha: D ≠ 0
This is a related measures before and after study. Fourteen people were involvedin the study. Before the treatment, the sample mean was 4.357 and after thetreatment, the mean was 5.214. The higher number after the treatment indicatesthat subjects were more likely to “blow the whistle” after having been through thetreatment. The observed t value was –3.12 which was more extreme than two-tailed table t value of + 2.16 and as a result, the researcher rejects the nullhypothesis. This is underscored by a p-value of .0081 which is less than α = .05.The study concludes that there is a significantly higher likelihood of “blowing the
whistle” after the treatment.
10.74 The point estimates from the sample data indicate that in the northern city themarket share is .31078 and in the southern city the market share is .27013. The
point estimate for the difference in the two proportions of market share are .04065. Since the 99% confidence interval ranges from -.03936 to +.12067 andzero is in the interval, any hypothesis testing decision based on this interval wouldresult in failure to reject the null hypothesis. Alpha is .01 with a two-tailed test.
This is underscored by an observed z value of 1.31 which has an associated p-value of .191 which, of course, is not significant for any of the usual values of α .
8/3/2019 Ken Black QA 5th chapter 10 Solution
http://slidepdf.com/reader/full/ken-black-qa-5th-chapter-10-solution 46/46
Chapter 10: Statistical Inferences About Two Populations 46
10.75 A test of differences of the variances of the populations of the two machines is being computed. The hypotheses are:
H0: σ 12 = σ 2
2
Ha: σ 12 > σ 2
2
Twenty-six pipes were measured for sample one and twenty-eight pipes weremeasured for sample two. The observed F = 2.0575 is significant at α = .05 for aone-tailed test since the associated p-value is .034787. The variance of pipelengths for machine 1 is significantly greater than the variance of pipe lengths for machine 2.