Introduction to Probability and Statistics for Engineers and Scientists

498 Chapter 11:Goodness of Fit Tests and Categorical Data Analysis

as Republicans, and 32 women who classified themselves as Independents; that is,N 11 =
68,N 12 =56, andN 13 =32. Similarly,N 21 =52,N 22 =72, andN 23 =20.
Use these data to test the hypothesis that a randomly chosen individual’s gender and
political affiliation are independent.

SOLUTION From the above data, we obtain that the six values ofnpˆiqˆj=NiMj/nare as
follows:

N 1 M 1

n

=

156 × 120

300

=62.40

N 1 M 2

n

=

156 × 128

300

=66.56

N 1 M 3

n

=

156 × 52

300

=27.04

N 2 M 1

n

=

144 × 120

300

=57.60

N 2 M 2

n

=

144 × 128

300

=61.44

N 2 M 3

n

=

144 × 52

300

=24.96

The value of the test statistic is thus

TS=

(68−62.40)^2

62.40

+

(56−66.56)^2

66.56

+

(32−27.04)^2

27.04

+

(52−57.60)^2

57.60

+

(72−61.44)^2

61.44

+

(20−24.96)^2

24.96

=6.433

Since (r−1)(s−1)=2, we must compare the value ofTSwith the critical valueχ.05,2^2.
From Table A2

χ.05,2^2 =5.991

SinceTS≥5. 991, the null hypothesis is rejected at the 5 percent level of significance.
That is, the hypothesis that gender and political affiliation of members of the population
are independent is rejected at the 5 percent level of significance. ■

The results of the test of independence of the characteristics of a randomly chosen
member of the population can also be obtained by computing the resultingp-value. If
the observed value of the test statistic isT=t, then the significance levelαtest would