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Chapter 10: Statistical Inferences About Two Populations 1
Chapter 10 Statistical Inferences about Two Populations
LEARNING OBJECTIVES
The general focus of Chapter 10 is on testing hypotheses and constructing confidence intervals 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 when the differences are normally distributed. 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 when the two populations are normally distributed.
CHAPTER TEACHING STRATEGY
The major emphasis of chapter 10 is on analyzing data from two samples. The student should be ready to deal with this topic given that he/she has tested hypotheses and computed confidence intervals in previous chapters on single sample data.
The z test for analyzing the differences in two sample means is presented here.
Conceptually, this is not radically different than the z test for a single sample mean shown initially in Chapter 7. In analyzing the differences in two sample means where the population variances are unknown, if it can be assumed that the populations are normally distributed, a t test for independent samples can be used. There are two different
Chapter 10: Statistical Inferences About Two Populations 2
formulas given in the chapter to conduct this t test. One version uses a "pooled" estimate of the population variance and assumes that the population variances are equal. The other version does not assume equal population variances and is simpler to compute. However, the degrees of freedom formula for this version is quite complex.
A t test is also included for related (non independent) samples. It is important that
the student be able to recognize when two samples are related and when they are independent. 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 to analyze some related measures data with both techniques and demonstrate that the results and conclusions are usually quite different. You can have your students work problems like this using both techniques to help them understand the differences between the two tests (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 begin to understand that the F values have two different degrees of freedom. The F distribution tables are upper tailed only. For this reason, formula 10.14 is given in the chapter to be used to compute lower tailed F values for two-tailed tests.
CHAPTER OUTLINE
10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Means
using the z Statistic (population variances known) Hypothesis Testing Confidence Intervals
Using 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 Testing Using the Computer to Test Hypotheses and Construct Confidence Intervals about the Difference in Two Population Means Using the t Test Confidence Intervals 10.3 Statistical Inferences For Two Related Populations Hypothesis Testing Using the Computer to Make Statistical Inferences about Two Related Populations Confidence Intervals
Chapter 10: Statistical Inferences About Two Populations 3
10.4 Statistical Inferences About Two Population Proportions, p1 – p2 Hypothesis Testing Confidence Intervals Using the Computer to Analyze the Difference in Two Proportions 10.5 Testing Hypotheses About Two Population Variances Using the Computer to Test Hypotheses about Two Population Variances
KEY TERMS Dependent Samples Independent Samples F Distribution Matched-Pairs Test F Value Related Measures
SOLUTIONS TO PROBLEMS IN CHAPTER 10 10.1 Sample 1 Sample 2
x 1 = 51.3 x 2 = 53.2 σ1
2 = 52 σ22 = 60
n1 = 32 n2 = 32
a) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 < 0 For one-tail test, α = .10 z.10 = -1.28
z =
32
60
32
52
)0()2.533.51()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = -1.02
Since the observed z = -1.02 > zc = -1.645, the decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 4
b) Critical value method:
zc =
2
22
1
21
2121 )()(
nn
xx c
σσ
µµ
+
−−−
-1.645 =
32
60
32
52
)0()( 21
+
−− cxx
(x 1 - x 2)c = -3.08 c) The area for z = -1.02 using Table A.5 is .3461.
The p-value is .5000 - .3461 = .1539
10.2 Sample 1 Sample 2
n1 = 32 n2 = 31
x 1 = 70.4 x 2 = 68.7 σ1 = 5.76 σ2 = 6.1
For a 90% C.I., z.05 = 1.645
2
22
1
21
21 )(nn
zxxσσ +±−
(70.4) – 68.7) + 1.64531
1.6
32
76.5 22
+
1.7 ± 2.465 -.76 < µ1 - µ2 < 4.16
Chapter 10: Statistical Inferences About Two Populations 5
10.3 a) Sample 1 Sample 2
x 1 = 88.23 x 2 = 81.2 σ1
2 = 22.74 σ22 = 26.65
n1 = 30 n2 = 30 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 For two-tail test, use α/2 = .01 z.01 = + 2.33
z =
30
65.26
30
74.22
)0()2.8123.88()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = 5.48
Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the null hypothesis.
b) 2
22
1
21
21 )(nn
zxxσσ
+±−
(88.23 – 81.2) + 2.3330
65.26
30
74.22 +
7.03 + 2.99 4.04 < µµµµ < 10.02
This supports the decision made in a) to reject the null hypothesis because zero is not in the interval.
10.4 Computers/electronics Food/Beverage
x 1 = 1.96 x 2 = 3.02 σ1
2 = 1.0188 σ22 = 0.9180
n1 = 50 n2 = 50 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 For two-tail test, α/2 = .005 z.005 = ±2.575
Chapter 10: Statistical Inferences About Two Populations 6
z =
50
9180.0
50
0188.1
)0()02.396.1()()(
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = -5.39
Since the observed z = -5.39 < zc = -2.575, the decision is to reject the null hypothesis.
10.5 A B n1 = 40 n2 = 37 x 1 = 5.3 x 2 = 6.5
σ12 = 1.99 σ2
2 = 2.36 For a 95% C.I., z.025 = 1.96
2
22
1
21
21 )(nn
zxxσσ
+±−
(5.3 – 6.5) + 1.9637
36.2
40
99.1 +
-1.2 ± .66 -1.86 < µµµµ < -.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 lie in this interval, we are confident that there is a difference between Plumber A and Plumber B.
10.6 Managers Specialty
n1 = 35 n2 = 41
x 1 = 1.84 x 2 = 1.99 σ1 = .38 σ2 = .51
for a 98% C.I., z.01 = 2.33
2
22
1
21
21 )(nn
zxxσσ
+±−
Chapter 10: Statistical Inferences About Two Populations 7
(1.84 - 1.99) ± 2.3341
51.
35
38. 22
+
-.15 ± .2384 -.3884 < µ1 - µ2 < .0884 Point Estimate = -.15
Hypothesis Test:
1) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0
2) z =
2
22
1
21
2121 )()(
nn
xx
σσ
µµ
+
−−−
3) α = .02
4) For a two-tailed test, z.01 = + 2.33. If the observed z value is greater than 2.33 or 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 null hypothesis.
8) There is no significant difference in the hourly rates of the two groups.
Chapter 10: Statistical Inferences About Two Populations 8
10.7 1994 2001 x 1 = 190 x 2 = 198 σ1 = 18.50 σ2 = 15.60 n1 = 51 n2 = 47 α = .01 H0: µ1 - µ2 = 0 Ha: µ1 - µ2 < 0 For a one-tailed test, z.01 = -2.33
z =
47
)60.15(
51
)50.18(
)0()198190()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = -2.32
Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the null hypothesis.
10.8 Seattle Atlanta
n1 = 31 n2 = 31
x 1 = 2.64 x 2 = 2.36 σ1
2 = .03 σ22 = .015
For a 99% C.I., z.005 = 2.575
2
22
1
21
21 )(nn
zxxσσ
+±−
(2.64-2.36) ± 2.57531
015.
31
03. +
.28 ± .10 .18 < µµµµ < .38
Between $ .18 and $ .38 difference with Seattle being more expensive.
Chapter 10: Statistical Inferences About Two Populations 9
10.9 Canon Pioneer
x 1 = 5.8 x 2 = 5.0 σ1 = 1.7 σ2 = 1.4 n1 = 36 n2 = 45 Ho: µ1 - µ2 = 0 Ha: 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
xx
σσ
µµ = 2.27
Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis. 10.10 A B x 1 = 8.05 x 2 = 7.26 σ1 = 1.36 σ2 = 1.06 n1 = 50 n2 = 38 Ho: µ1 - µ2 = 0 Ha: 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
xx
σσ
µµ = 3.06
Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 10
10.11 Ho: µ1 - µ2 = 0 α = .01 Ha: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17 Sample 1 Sample 2
n1 = 8 n2 = 11
x 1 = 24.56 x 2 = 26.42 s1
2 = 12.4 s22 = 15.8
For one-tail test, α = .01 Critical t.01,19 = -2.567
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
11
1
8
1
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 the null hypothesis.
10.12 a) Ho: µ1 - µ2 = 0 α =.10 Ha: 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
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
t =
20
1
20
1
22020
)19()6.21()19()9.23(
)0()42.2656.24(22
+−+
+−−
= 0.69
Since the observed t = 0.69 < t.05,38 = 1.697, the decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 11
b) 2121
22
212
121
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±− =
(118 – 113) + 1.69720
1
20
1
22020
)19()6.21()19()9.23( 22
+−+
+
5 + 12.224
-7.224 < µµµµ1 - µµµµ2 < 17.224
10.13 Ho: µ1 - µ2 = 0 α = .05 Ha: 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
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
t =
10
1
10
1
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 the null hypothesis.
10.14 Ho: µ1 - µ2 = 0 α =.01 Ha: 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
Chapter 10: Statistical Inferences About Two Populations 12
For two-tail test, α/2 = .005 Critical t.005,34 = ±2.457 (used df=30)
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
t =
18
1
18
1
21818
17)026.2()17(12
)0()444.9333.5(
+−+
+−−
= -4.66
Since the observed t = -4.66 < t.005,34 = -2.457
Reject the null hypothesis
b) 2121
22
212
121
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±− =
(5.333 – 9.444) + 2.45718
1
18
1
21818
)17)(026.2()17)(12( +−+
+
-4.111 + 2.1689 -6.2799 < µµµµ1 - µµµµ2 < -1.9421 10.15 Peoria Evansville n1 = 21 n2 = 26 1x = 86,900 2x = 84,000 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
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±− =
(86,900 – 84,000) + 1.68426
1
21
1
22621
)25()1750()20()2300( 22
+−+
+ =
2,900 + 994.62 1905.38 < µµµµ1 - µµµµ2 < 3894.62
Chapter 10: Statistical Inferences About Two Populations 13
10.16 Ho: µ1 - µ2 = 0 α = .05 Ha: µ1 - µ2 ≠ 0!= 0 df = 21 + 26 - 2 = 45
Peoria Evansville n1 = 21 n2 = 26 x 1 = $86,900 x 2 = $84,000 s1 = $2,300 s2 = $1,750 For two-tail test, α/2 = .025 Critical t.025,45 = ± 2.021 (used df=40)
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
t =
26
1
21
1
22621
)25()750,1()20()300,2(
)0()000,84900,86(22
+−+
+−−
= 4.91
Since the observed t = 4.91 > t.025,45 = 2.021, the decision is to reject the null hypothesis.
10.17 Let Boston be group 1 1) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0
2) t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ
3) α = .01
4) For a one-tailed test and df = 8 + 9 - 2 = 15, t.01,15 = 2.602. If the observed value 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
Chapter 10: Statistical Inferences About Two Populations 14
6) t =
9
1
8
1
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 null hypothesis. 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
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±− =
(112 – 122) + 2.42320
1
22
1
22022
)19()12()21()11( 22
+−+
+
-10 ± 8.63 -$18.63 < µ1 - µ2 < -$1.37 Point Estimate = -$10 10.19 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 df = n1 + n2 - 2 = 11 + 11 - 2 = 20 Toronto Mexico City n1 = 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
Chapter 10: Statistical Inferences About Two Populations 15
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ =
t =
11
1
11
1
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 null hypothesis.
10.20 Toronto Mexico City n1 = 11 n2 = 11
x 1 = $67,381.82 x 2 = $63,481.82 s1 = $2,067.28 s2 = $1,594.25 df = n1 + n2 - 2 = 11 + 11 - 2 = 20 For a 95% Level of Confidence, α/2 = .025 and t.025,20 = 2.086
2121
22
212
121
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±− =
($67,381.82 - $63,481.82) ± (2.086) 11
1
11
1
21111
)10()25.594,1()10()28.067,2( 22
+−+
+
3,900 ± 1,641.9 2,258.1 < µ1 - µ2 < 5,541.9
Chapter 10: Statistical Inferences About Two Populations 16
10.21 Ho: D = 0 Ha: D > 0 Sample 1 Sample 2 d 38 22 16 27 28 -1 30 21 9 41 38 3 36 38 -2 38 26 12 33 19 14 35 31 4 44 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
45.6011.7 −=−
n
sDd
d
= 3.31
Since the observed t = 3.31 > t.01,8 = 2.896, the decision is to reject the null hypothesis. 10.22 Ho: D = 0 Ha: D ≠ 0 Before After d 107 102 5 99 98 1 110 100 10 113 108 5 96 89 7 98 101 -3 100 99 1 102 102 0 107 105 2 109 110 -1 104 102 2 99 96 3 101 100 1
Chapter 10: Statistical Inferences About Two Populations 17
n = 13 d = 2.5385 sd=3.4789 α = .05 df = n - 1 = 13 - 1 = 12 For a two-tail test and α/2 = .025 Critical t.025,12 = ±2.179
t =
13
4789.305385.2 −=−
n
sDd
d
= 2.63
Since the observed t = 2.63 > t.025,12 = 2.179, the decision is to reject the null hypothesis.
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
std d±
40.56 ± (2.518)22
58.26
40.56 ± 14.27 26.29 < D < 54.83 10.24 Before After d 32 40 -8 28 25 3 35 36 -1 32 32 0 26 29 -3 25 31 -6 37 39 -2 16 30 -14 35 31 4
Chapter 10: Statistical Inferences About Two Populations 18
n = 9 d = -3 sd = 5.6347 α = .025 df = n - 1 = 9 - 1 = 8 For 90% level of confidence and α/2 = .025, t.05,8 = 1.86
t = n
std d±
t = -3 + (1.86) 9
6347.5 = -3 ± 3.49
-0.49 < D < 6.49 10.25 City Cost Resale d Atlanta 20427 25163 -4736 Boston 27255 24625 2630 Des Moines 22115 12600 9515 Kansas City 23256 24588 -1332 Louisville 21887 19267 2620 Portland 24255 20150 4105 Raleigh-Durham 19852 22500 -2648 Reno 23624 16667 6957 Ridgewood 25885 26875 - 990 San Francisco 28999 35333 -6334 Tulsa 20836 16292 4544
d = 1302.82 sd = 4938.22 n = 11, df = 10 α = .01 α/2 = .005 t.005,10= 3.169
n
std d± = 1302.82 + 3.169
11
22.4938 = 1302.82 + 4718.42
-3415.6 < D < 6021.2
Chapter 10: Statistical Inferences About Two Populations 19
10.26 Ho: D = 0 Ha: D < 0 Before After d 2 4 -2 4 5 -1 1 3 -2 3 3 0 4 3 1 2 5 -3 2 6 -4 3 4 -1 1 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
t =
9
716.10778.1 −−=−
n
sDd
d
= -3.11
Since the observed t = -3.11 < t.05,8 = -1.86, the decision is to reject the null hypothesis.
10.27 Before After d 255 197 58 230 225 5 290 215 75 242 215 27 300 240 60 250 235 15 215 190 25 230 240 -10 225 200 25 219 203 16 236 223 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
Chapter 10: Statistical Inferences About Two Populations 20
n
std d±
28.09 ± (2.764) 11
813.25 = 28.09 ± 21.51
6.58 < D < 49.60 10.28 H0: D = 0
Ha: 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
sDd
d
= 3.86
Since the observed t = 3.86 > t.01,26 = 2.479, the decision is to reject the null hypothesis.
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
std d±
75 + 1.72521
30 = 75 ± 11.29
63.71 < D < 86.29 10.30 Ho: D = 0 Ha: 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
Chapter 10: Statistical Inferences About Two Populations 21
t =
15
9.1085.2 −−=−
n
sDd
d
= -5.81
Since the observed t = -5.81 < t.005,14 = -2.977, the decision is to reject the null hypothesis.
10.31 a) Sample 1 Sample 2 n1 = 368 n2 = 405 x1 = 175 x2 = 182
368
175ˆ
1
11 ==
n
xp = .476
405
182ˆ
2
22 ==
n
xp = .449
773
357
405368
182175
21
21 =++=
++
=nn
xxp = .462
Ho: p1 - p2 = 0 Ha: p1 - p2 ≠ 0 For two-tail, α/2 = .025 and z.025 = ±1.96
+
−−=
+⋅
−−−=
405
1
368
1)538)(.462(.
)0()449.476(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = 0.75
Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the null hypothesis. b) Sample 1 Sample 2 p̂ 1 = .38 p̂ 2 = .25 n1 = 649 n2 = 558
558649
)25(.558)38(.649ˆˆ
21
2211
++=
++
=nn
pnpnp = .32
Chapter 10: Statistical Inferences About Two Populations 22
Ho: p1 - p2 = 0 Ha: p1 - p2 > 0 For a one-tail test and α = .10, z.10 = 1.28
+
−−=
+⋅
−−−=
558
1
649
1)68)(.32(.
)0()25.38(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = 4.83
Since the observed z = 4.83 > zc = 1.28, the decision is to reject the null hypothesis. 10.32 a) n1 = 85 n2 = 90 p̂ 1 = .75 p̂ 2 = .67 For a 90% Confidence Level, z.05 = 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.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(
n
qp
n
qpzpp +±−
(.19 - .17) + 1.961300
)83)(.17(.
1100
)81)(.19(. + = .02 ± .03
-.01 < p1 - p2 < .05
Chapter 10: Statistical Inferences About Two Populations 23
c) n1 = 430 n2 = 399 x1 = 275 x2 = 275
430
275ˆ
1
11 ==
n
xp = .64
399
275ˆ
2
22 ==
n
xp = .69
For an 85% Confidence Level, α/2 = .075 and z.075 = 1.44
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.64 - .69) + 1.44399
)31)(.69(.
430
)36)(.64(. + = -.05 ± .047
-.097 < p1 - p2 < -.003
d) n1 = 1500 n2 = 1500 x1 = 1050 x2 = 1100
1500
1050ˆ
1
11 ==
n
xp = .70
1500
1100ˆ
2
22 ==
n
xp = .733
For an 80% Confidence Level, α/2 = .10 and z.10 = 1.28
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.70 - .733) ± 1.281500
)267)(.733(.
1500
)30)(.70(. + = -.033 ± .02
-.053 < p1 - p2 < -.013 10.33 H0: pm - pw = 0 Ha: 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
pnpnp = .652
Chapter 10: Statistical Inferences About Two Populations 24
+
−−=
+⋅
−−−=
481
1
374
1)348)(.652(.
)0()70.59(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = -3.35
Since the observed z = -3.35 < z.05 = -1.645, the decision is to reject the null hypothesis. 10.34 n1 = 210 n2 = 176 1p̂ = .24 2p̂ = .35 For a 90% Confidence Level, α/2 = .05 and z.05 = + 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.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
pnpnp = .529
Ho: p1 - p2 = 0 Ha: p1 - p2 ≠ 0 For two-tail test, α/2 = .10 and zc = ±1.28
+
−−=
+⋅
−−−=
89
1
56
1)471)(.529(.
)0()56.48(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = -0.94
Since the observed z = -0.94 > zc = -1.28, the decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 25
10.36 A B
n1 = 35 n2 = 35 x1 = 5 x2 = 7
35
5ˆ
1
11 ==
n
xp = .14
35
7ˆ
2
22 ==
n
xp = .20
For a 98% Confidence Level, α/2 = .01 and z.01 = 2.33
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.14 - .20) ± 2.3335
)80)(.20(.
35
)86)(.14(. + = -.06 ± .21
-.27 < p1 - p2 < .15 10.37 H0: p1 – p2 = 0 Ha: 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
pnpnp = .0738
+
−−=
+⋅
−−−=
915
1
780
1)9262)(.0738(.
)0()06.09(.
11
)()ˆˆ(
1
2121
nnqp
ppppZ = 2.35
Since the observed z = 2.35 > z.05 = 1.645, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 26
10.38 n1 = 850 n2 = 910 p̂1 = .60 p̂ 2 = .52
For a 95% Confidence Level, α/2 = .025 and z.025 = + 1.96
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.60 - .52) + 1.96910
)48)(.52(.
850
)40)(.60(. + = .08 + .046
.034 < p1 – p2 < .126 10.39 H0: σ1
2 = σ22 α = .01 n1 = 10 s1
2 = 562 Ha: σ1
2 < σ22 n2 = 12 s2
2 = 1013 dfnum = 12 - 1 = 11 dfdenom = 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 the null hypothesis.
10.40 H0: σ12 = σ2
2 α = .05 n1 = 5 S1 = 4.68 Ha: σ1
2 ≠ σ22 n2 = 19 S2 = 2.78
dfnum = 5 - 1 = 4 dfdenom = 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 the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 27
10.41 City 1 City 2
1.18 1.08 1.15 1.17 1.14 1.14 1.07 1.05 1.14 1.21 1.13 1.14 1.09 1.11 1.13 1.19 1.13 1.12 1.03 1.13 n1 = 10 df1 = 9 n2 = 10 df2 = 9 s1
2 = .0018989 s22 = .0023378
H0: σ1
2 = σ22 α = .10 α/2 = .05
Ha: σ12 ≠ σ2
2 Upper tail critical F value = F.05,9,9 = 3.18 Lower tail critical F value = F.95,9,9 = 0.314
F = 0023378.
0018989.2
2
21 =
s
s = 0.81
Since the observed F = 0.81 is greater than the lower tail critical value of 0.314 and less than the upper tail critical value of 3.18, the decision is to fail
to reject the null hypothesis.
10.42 Let Houston = group 1 and Chicago = group 2 1) H0: σ1
2 = σ22
Ha: σ12 ≠ σ2
2
2) F = 2
2
21
s
s
3) α = .01 4) df1 = 12 df2 = 10 This is a two-tailed test The critical table F values are: F.005,12,10 = 5.66 F.995,10,12 = .177
Chapter 10: Statistical Inferences About Two Populations 28
If the observed value is greater than 5.66 or less than .177, the decision will be to reject the null hypothesis.
5) s1
2 = 393.4 s22 = 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 H0: σ12 = σ2
2 α = .05 n1 = 12 s1 = 7.52 Ha: σ1
2 > σ22 n2 = 15 s2 = 6.08
dfnum = 12 - 1 = 11 dfdenom = 15 - 1 = 14
The critical table F value is F.05,10,14 = 5.26
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 the null hypothesis.
10.44 H0: σ12 = σ2
2 α = .01 n1 = 15 s12 = 91.5
Ha: σ12 ≠ σ2
2 n2 = 15 s22 = 67.3
dfnum = 15 - 1 = 14 dfdenom = 15 - 1 = 14
The critical table F values are: F.005,12,14 = 4.43 F.995,14,12 = .226
F = 3.67
5.912
2
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 is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 29
10.45 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 For α = .10 and a two-tailed test, α/2 = .05 and z.05 = + 1.645 Sample 1 Sample 2
1x = 138.4 2x = 142.5
σ1 = 6.71 σ2 = 8.92 n1 = 48 n2 = 39
z =
39
)92.8(
48
)71.6(
)0()5.1424.138()()(2
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = -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 in the means of the two populations.
10.46 Sample 1 Sample 2 1x = 34.9 2x = 27.6
σ12 = 2.97 σ2
2 = 3.50 n1 = 34 n2 = 31 For 98% Confidence Level, z.01 = 2.33
2
22
1
21
21 )(nn
zxxσσ
+±−
(34.9 – 27.6) + 2.3331
50.3
34
97.2 + = 7.3 + 1.04
6.26 < µµµµ1 - µµµµ2 < 8.34
Chapter 10: Statistical Inferences About Two Populations 30
10.47 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 Sample 1 Sample 2
1x = 2.06 2x = 1.93 s1
2 = .176 s22 = .143
n1 = 12 n2 = 15 This is a one-tailed test with df = 12 + 15 - 2 = 25. The critical value is
t.05,25 = 1.708. If the observed value is greater than 1.708, the decision will be to reject the null hypothesis.
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ
t =
15
1
12
1
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, the decision is to fail to reject the null hypothesis. The mean for population one is not significantly greater than the mean for population two.
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, t35,.025 = 2.042
2121
22
212
121
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±−
(74.6 – 70.9) + 2.04220
1
20
1
22020
)19()6.21()19()9.23( 22
+−+
+
3.7 + 2.22 1.48 < µµµµ1 - µµµµ2 < 5.92
Chapter 10: Statistical Inferences About Two Populations 31
10.49 Ho: D = 0 α = .01 Ha: 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 decision will be to reject the null hypothesis.
t =
21
01.1016.1 −−=−
n
sDd
d
= -5.26
Since the observed value of t = -5.26 is less than the critical t value of -2.528, the decision is to reject the null hypothesis. The population difference is less
than zero. 10.50 Respondent Before After d
1 47 63 -16 2 33 35 - 2 3 38 36 2 4 50 56 - 6 5 39 44 - 5 6 27 29 - 2 7 35 32 3 8 46 54 - 8 9 41 47 - 6
d = -4.44 sd = 5.703 df = 8 For a 99% Confidence Level, α/2 = .005 and t8,.005 = 3.355
n
std d± = -4.44 + 3.355
9
703.5 = -4.44 + 6.38
-10.82 < D < 1.94 10.51 Ho: p1 - p2 = 0 α = .05 α/2 = .025 Ha: 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 decision will be to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 32
Sample 1 Sample 2 x1 = 345 x2 = 421 n1 = 783 n2 = 896
896783
421345
21
21
++=
++
=nn
xxp = .4562
783
345ˆ
1
11 ==
n
xp = .4406
896
421ˆ
2
22 ==
n
xp = .4699
+
−−=
+⋅
−−−=
896
1
783
1)5438)(.4562(.
)0()4699.4406(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = -1.20
Since the observed value of z = -1.20 is greater than -1.96, the decision is to fail to reject the null hypothesis. There is no significant difference in the population
proportions.
10.52 Sample 1 Sample 2 n1 = 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(
n
qp
n
qpzpp +±−
(.71 - .67) + 2.575378
)33)(.67(.
409
)29)(.71(. + = .04 ± .085
-.045 < p1 - p2 < .125
10.53 H0: σ12 = σ2
2 α = .05 n1 = 8 s12 = 46
Ha: σ12 ≠ σ2
2 n2 = 10 S22 = 37
dfnum = 8 - 1 = 7 dfdenom = 10 - 1 = 9 The critical F values are: F.025,7,9 = 4.20 F.975,9,7 = .238
Chapter 10: Statistical Inferences About Two Populations 33
If the observed value of F is greater than 4.20 or less than .238, then the decision will be to reject the null hypothesis.
F = 37
462
2
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 no significant difference in the variances of the two populations.
10.54 Term Whole Life
x t = $75,000 x w = $45,000 st = $22,000 sw = $15,500 nt = 27 nw = 29 df = 27 + 29 - 2 = 54 For a 95% Confidence Level, α/2 = .025 and t.025,40 = 2.021 (used df=40)
2121
22
212
121
11
2
)1()1()(
nnnn
nsnstxx +
−+−+−
±−
(75,000 – 45,000) + 2.02129
1
27
1
22927
)28()500,15()26()000,22( 22
+−+
+
30,000 ± 10,220.73 19,779.27 < µ1 - µ2 < 40,220.73 10.55 Morning Afternoon d 43 41 2 51 49 2 37 44 -7 24 32 -8 47 46 1 44 42 2 50 47 3 55 51 4 46 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
Chapter 10: Statistical Inferences About Two Populations 34
n
std d±
-0.444 + (1.86) 9
447.4 = -0.444 ± 2.757
-3.201 < D < 2.313
10.56 Let group 1 be 1990 Ho: p1 - p2 = 0 Ha: p1 - p2 < 0 α = .05 The critical table z value is: z.05 = -1.645 n1 = 1300 n2 = 1450 1p̂ = .447 2p̂ = .487
14501300
)1450)(487(.)1300)(447(.ˆˆ
21
2211
++=
++
=nn
pnpnp = .468
+
−−=
+⋅
−−−=
1450
1
1300
1)532)(.468(.
)0()487.447(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = -2.10
Since the observed z = -3.73 is less than z.05 = -1.645, the decision is to reject the
null hypothesis. 1997 has a significantly higher proportion.
10.57 Accounting Data Entry
n1 = 16 n2 = 14
x 1 = 26,400 x 2 = 25,800 s1 = 1,200 s2 = 1,050
H0: σ1
2 = σ22
Ha: σ12 ≠ σ2
2 dfnum = 16 – 1 = 15 dfdenom = 14 – 1 = 13 The critical F values are: F.025,15,13 = 3.05 F.975,15,13 = 0.33
Chapter 10: Statistical Inferences About Two Populations 35
F = 500,102,1
000,440,12
2
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. 10.58 H0: σ1
2 = σ22 α = .01 n1 = 8 n2 = 7
Ha: σ12 ≠ σ2
2 S12 = 72,909 S2
2 = 129,569 dfnum = 6 dfdenom = 7
The critical F values are: F.005,6,7 = 9.16 F.995,7,6 = .11
F = 909,72
569,1292
2
21 =
s
s = 1.78
Since F = 1.95 < F.005,6,7 = 9.16 but also > F.995,7,6 = .11, the decision is to fail to reject the null hypothesis. There is no difference in the variances of the shifts.
10.59 Men Women
n1 = 60 n2 = 41
x 1 = 631 x 2 = 848 σ1 = 100 σ2 = 100 For a 95% Confidence Level, α/2 = .025 and z.025 = 1.96
2
22
1
21
21 )(nn
zxxσσ
+±−
(631 – 848) + 1.9641
100
60
100 22
+ = -217 ± 39.7
-256.7 < µ1 - µ2 < -177.3
Chapter 10: Statistical Inferences About Two Populations 36
10.60 Ho: µ1 - µ2 = 0 α = .01 Ha: 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.005,42 = ±2.704 (used df=40)
t =
2121
22
212
1
2121
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ
t =
24
1
20
1
42
)23()7.2()19()2.3(
)0()89.1453.17(22
++−−
= 2.97
Since the observed t = 2.97 > t.005,42 = 2.704, the decision is to reject the null hypothesis. 10.61 With Fertilizer Without Fertilizer
x 1 = 38.4 x 2 = 23.1 σ1 = 9.8 σ2 = 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()()(2
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = 7.37
Since the observed z = 7.37 > z.01 = 2.33, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 37
10.62 Specialty Discount n1 = 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
qp
n
qpzpp +±−
(.75 - .52) + 1.645500
)48)(.52(.
350
)25)(.75(. + = .23 ± .053
.177 < p1 - p2 < .283
10.63 H0: σ1
2 = σ22 α = .05 n1 = 27 s1 = 22,000
Ha: σ12 ≠ σ2
2 n2 = 29 s2 = 15,500 dfnum = 27 - 1 = 26 dfdenom = 29 - 1 = 28 The critical F values are: F.025,24,28 = 2.17 F.975,28,24 = .46
F = 2
2
22
21
500,15
000,22=s
s = 2.01
Since the observed F = 2.01 < F.025,24,28 = 2.17 and > than F.975,28,24 = .46, the decision is to fail to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 38
10.64 Name Brand Store Brand d 54 49 5 55 50 5 59 52 7 53 51 2 54 50 4 61 56 5 51 47 4 53 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
std d±
4.5 + 1.8958
414.1 = 4.5 ± .947
3.553 < D < 5.447 10.65 Ho: µ1 - µ2 = 0 α = .01 Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40 Wisconsin Tennessee n1 = 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
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ
t =
19
1
23
1
40
)18)(6491.4()22)(9644.9(
)0()7368.71652.69(
++−−
= -2.44
Chapter 10: Statistical Inferences About Two Populations 39
Since the observed t = -2.44 < t.01,40 = -2.423, the decision is to reject the null hypothesis. 10.66 Wednesday Friday d 71 53 18 56 47 9 75 52 23 68 55 13 74 58 16
n = 5 d = 15.8 sd = 5.263 df = 5 - 1 = 4 Ho: D = 0 α = .05 Ha: D > 0 For one-tail test, α = .05 and the critical t.05,4 = 2.132
t =
5
263.508.15 −=−
n
sDd
d
= 6.71
Since the observed t = 6.71 > t.05,4 = 2.132, the decision is to reject the null hypothesis. 10.67 Ho: P1 - P2 = 0 α = .05 Ha: P1 - P2 ≠ 0 Machine 1 Machine 2 x1 = 38 x2 = 21 n1 = 191 n2 = 202
191
38ˆ
1
11 ==
n
xp = .199
202
21ˆ
2
22 ==
n
xp = .104
202191
)202)(104(.)191)(199(.ˆˆ
21
2211
++=
++
=nn
pnpnp = .15
For two-tail, α/2 = .025 and the critical z values are: z.025 = ±1.96
Chapter 10: Statistical Inferences About Two Populations 40
+
−−=
+⋅
−−−=
202
1
191
1)85)(.15(.
)0()104.199(.
11
)()ˆˆ(
1
2121
nnqp
ppppz = 2.64
Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null hypothesis. 10.68 Construction Telephone Repair n1 = 338 n2 = 281 x1 = 297 x2 = 192
338
297ˆ
1
11 ==
n
xp = .879
281
192ˆ
2
22 ==
n
xp = .683
For a 90% Confidence Level, α/2 = .05 and z.05 = 1.645
2
22
1
1121
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp +±−
(.879 - .683) + 1.645281
)317)(.683(.
338
)121)(.879(. + = .196 ± .054
.142 < p1 - p2 < .250 10.69 Aerospace Automobile n1 = 33 n2 = 35
x 1 = 12.4 x 2 = 4.6 σ1 = 2.9 σ2 = 1.8 For a 99% Confidence Level, α/2 = .005 and z.005 = 2.575
2
22
1
21
21 )(nn
zxxσσ
+±−
(12.4 – 4.6) + 2.57535
)8.1(
33
)9.2( 22
+ = 7.8 ± 1.52
6.28 < µ1 - µ2 < 9.32
Chapter 10: Statistical Inferences About Two Populations 41
10.70 Discount Specialty
x 1 = $47.20 x 2 = $27.40 σ1 = $12.45 σ2 = $9.82 n1 = 60 n2 = 40 Ho: µ1 - µ2 = 0 α = .01 Ha: µ1 - µ2 ≠ 0 For two-tail test, α/2 = .005 and zc = ±2.575
z =
40
)82.9(
60
)45.12(
)0()40.2720.47()()(22
2
22
1
21
2121
+
−−=
+
−−−
nn
xx
σσ
µµ = 8.86
Since the observed z = 8.86 > zc = 2.575, the decision is to reject the null hypothesis. 10.71 Before After d 12 8 4 7 3 4 10 8 2 16 9 7 8 5 3
n = 5 d = 4.0 sd = 1.8708 df = 5 - 1 = 4 Ho: D = 0 α = .01 Ha: D > 0 For one-tail test, α = .01 and the critical t.01,4 = 3.747
t =
5
8708.100.4 −=−
n
sDd
d
= 4.78
Since the observed t = 4.78 > t.01,4 = 3.747, the decision is to reject the null hypothesis.
Chapter 10: Statistical Inferences About Two Populations 42
10.72 Ho: µ1 - µ2 = 0 α = .01 Ha: 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
11
2
)1()1(
)()(
nnnn
nsns
xx
+−+
−+−
−−− µµ
t =
6
1
10
1
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 null hypothesis. 10.73 A t test was used to test to determine if Hong Kong has significantly different
rates than Bombay. Let group 1 be Hong Kong. Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0
n1 = 19 n2 = 23 x1 = 130.4 x 2 = 128.4 S1 = 12.9 S2 = 13.9 α = .01 t = 0.48 with a p-value of .634 which is not significant at of .05. There is not enough evidence in these data to declare that there is a difference in the average rental rates of the two cities. 10.74 H0: D = 0 Ha: D ≠ 0
This is a related measures before and after study. Fourteen people were involved in the study. Before the treatment, the sample mean was 4.357 and after the
Chapter 10: Statistical Inferences About Two Populations 43
treatment, the mean was 5.214. The higher number after the treatment indicates that subjects were more likely to “blow the whistle” after having been through the treatment. The observed t value was –3.12 which was more extreme than two-tailed table t value of + 2.16 causing the researcher to reject the null hypothesis. 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.
10.75 The point estimates from the sample data indicate that in the northern city the market share is .3108 and in the southern city the market share is .2701. The point estimate for the difference in the two proportions of market share are .0407. Since the 99% confidence interval ranges from -.0394 to +.1207 and zero is in the interval, any hypothesis testing decision based on this interval would result in failure to reject the null hypothesis. Alpha is .01 with a two-tailed test. This is underscored by a calculated 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.
10.76 A test of differences of the variances of the populations of the two machines is
being computed. The hypotheses are: H0: σ1
2 = σ22
Ha: σ12 ≠ σ2
2
Twenty-six pipes were measured for sample one and twenty-six pipes were measured for sample two. The observed F = 1.79 is not significant at α = .05 for a two-tailed test since the associated p-value is .0758. There is no significant difference in the variance of pipe lengths for pipes produced by machine A versus machine B.