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
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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
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
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
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
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
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 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.
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
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
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)
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
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
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.