Solutions to Exercises for Chapter 15
15.1.
H 0 : s 12 = s 22
H1 : s 12 ¹ s 22
Given : n1 = 18,
Test statistic :
F=
s12
s2
2
=
.0196
.0484
s1 = .14,
n2 = 13,
s2 = .22
F, a = .10, F.05,17 ,12 = 0.4201,
= 0.4050
Since 0.4050 < .4201, reject.
With R:
s1 <- .14
n1 <- 18
s2 <- .22
n2 <- 13
f.obs <- s1^2/s2^2
f.obs
#Two-tailed critical values
qf(c(.05,.95),n1-1,n2-1)
> f.obs
[1] 0.4049587
> #Two-tailed critical values
> qf(c(.05,.95),n1-1,n2-1)
[1] 0.4200526 2.5828389
F.95,17 ,12 = 2.583
15.2.
H 0 : r1 = r 2
H1 : r1 ¹ r 2
Given : r1 = .93,
n1 = 80,
z1 = 1.658
r2 = .83,
n2 = 30,
z 2 = 1.188
Test statistic : z, a = .05, critical value = ±1.96
z=
1.658 -1.188
1
=
1
+
80 - 3 30 - 3
.47
.0129 + .0370
=
.47
.2236
Since 2.10 > 1.96, reject.
With R:
r1 <- .93
n1 <- 80
r2 <- .83
n2 <- 30
r1.z <- .5*log(1+r1) - .5*log(1-r1)
r2.z <- .5*log(1+r2) - .5*log(1-r2)
z.obs <- (r1.z - r2.z)/sqrt((1/(n1-3))+(1/(n2-3)))
z.obs
> z.obs
[1] 2.102533
> #Two-Tailed Critical values
> round(qnorm(c(.025,.975)),3)
[1] -1.96
1.96
= 2.10
15.3. This exercise calls for the comparison of two group having different sample sizes. We enter the
data into R and create a data frame. All of our commands appear in Script 15.11. Next, we generate some
descriptive statistics:
> table(f.group)
f.group
white asian
12
25
> tapply(read,f.group,mean)
white asian
24.00 30.32
> tapply(read,f.group,var)
white
asian
34.36364 249.31000
> tapply(read,f.group,sd)
white
asian
5.86205 15.78955
As you can see, there are more than twice as many Asian children as there are White children. The means
appear to be different, but we also note that the variance of the reading scores of Asian children is
approximately seven times that for the White children. Next, we will examine the distributions of scores
for the two separate groups:
> tapply(read,f.group,skewness)
white
asian
0.00974793 0.95786018
> tapply(read,f.group,SEsk)
white
asian
0.6373020 0.4636835
> tapply(read,f.group,kurtosis)
white
asian
-0.72104644
0.01233870
> tapply(read,f.group,SEku)
white
asian
1.2322465 0.9017205
> boxplot(read ~ f.group)
60
50
40
30
20
10
white
asian
The indices for skewness and kurtosis suggest that the Asian scores are positively skewed. This is
confirmed by the boxplot. At this point, we test the assumptions of normality and homogeneity of
variance:
>#Testing assumptions
> tapply(read,f.group,shapiro.test)
$white
Shapiro-Wilk normality test
W = 0.9604, p-value = 0.7894
$asian
Shapiro-Wilk normality test
W = 0.8958, p-value = 0.01492
> #Levene's test
> leveneTest(read ~ f.group, data = data.var)
Levene's Test for Homogeneity of Variance (center = median)
Df F value
group
1
Pr(>F)
5.1466 0.02957
35
> #Fligner-Killeen test
> fligner.test(read ~ f.group)
Fligner-Killeen test of homogeneity of variances
Fligner-Killeen:med chi-squared = 4.9729, df = 1, p-value = 0.02575
The 12 reading scores for the White children are reasonably normally distributed; the 25 scores from the
Asian children are not (p = .01492). Furthermore, both the Levene test and the Fligner–Killeen test
indicate that the two sample variances suggest nonequivalent population variances with p-values of
.02957 and .02575, respectively. Clearly, these data are problematic. As per the instructions, we now look
at the pooled-sample t-test, the separate-sample t-test with the Welch correction, and the Mann–Whitney
U/Wilcoxon test:
> t.test(read ~ f.group,var.equal=T)
Two Sample t-test
data:
read by f.group
t = -1.3349, df = 35, p-value = 0.1905
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-15.931763
3.291763
sample estimates:
mean in group white mean in group asian
24.00
30.32
> t.test(read ~ f.group,var.equal=F)
Welch Two Sample t-test
data:
read by f.group
t = -1.764, df = 33.7, p-value = 0.0868
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-13.603398
0.963398
sample estimates:
mean in group white mean in group asian
24.00
30.32
> wilcox.test(read ~ f.group)
Wilcoxon rank sum test with continuity correction
data:
read by f.group
W = 124.5, p-value = 0.4165
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(x = c(29, 32, 23, 23, 20, 24, 17, 20, 33,
:
cannot compute exact p-value with ties
As you can see from the output, the traditional pooled-sample t-test yields a p-value of .1905, indicating
nonsignificance. The separate-sample t-test with the Welch correction to adjust for nonequivalent
variances yields a p-value of .0868, also not significant, but “closer.” In looking back at the descriptive
statistics, the Asian sample is more than twice the size of the White sample; it also hai the larger variance.
Thus, when using the pooled-sample estimate of variance, the larger variance is weighted more heavily
than the smaller variance, resulting in an overestimate of the population variance, which in turn results in
an underestimate of the t-statistic.
One might be tempted to turn to the nonparametric alternative, the Mann–Whitney U/Wilcoxon
test in this context. It, too, indicates a nonsignificant difference between the two groups. However, recall
that this procedure can only be interpreted in terms of location if the two group distributions are similar in
shape and variability. Given that the charge was to compare the two groups with regard to central
tendency, we would recommend reporting the results from the separate-sample estimate of the t-test.