Introduction to Categorical Data Analysis

2.4 CHI-SQUARED TESTS OF INDEPENDENCE 37
The marginal probabilities then determine the joint probabilities. To test H0, we
identify μij = nπij = nπi+π+j as the expected frequency. Here, μij is the expected
value of nij assuming independence. Usually, {πi+} and {π+j } are unknown, as is
this expected value.
To estimate the expected frequencies, substitute sample proportions for the
unknown marginal probabilities, giving
ˆμij = npi+p+j = n
� �� �
ni+ n+j
=
n n
ni+n+j
n
This is the row total for the cell multiplied by the column total for the cell, divided
by the overall sample size. The {ˆμij } are called estimated expected frequencies. They
have the same row and column totals as the observed counts, but they display the
pattern of independence.
For testing independence in I × J contingency tables, the Pearson and likelihoodratio
statistics equal
X 2 = � (nij −ˆμij ) 2
ˆμij
, G 2 = 2 � � �
nij
nij log
ˆμij
(2.8)
Their large-sample chi-squared distributions have df = (I − 1)(J − 1).
The df value means the following: under H0, {πi+} and {π+j } determine the
cell probabilities. There are I − 1 nonredundant row probabilities. Because they sum
to 1, the first I − 1 determine the last one through πI+ = 1 − (π1+ +···+πI−1,+).
Similarly, there are J − 1 nonredundant column probabilities. So, under H0, there are
(I − 1) + (J − 1) parameters. The alternative hypothesis Ha merely states that there
is not independence. It does not specify a pattern for the IJ cell probabilities. The
probabilities are then solely constrained to sum to 1, so there are IJ − 1 nonredundant
parameters. The value for df is the difference between the number of parameters
under Ha and H0, or
df = (IJ − 1) −[(I − 1) + (J − 1)] =IJ − I − J + 1 = (I − 1)(J − 1)
2.4.4 Example: Gender Gap in Political Affiliation
Table 2.5, from the 2000 General Social Survey, cross classifies gender and political
party identification. Subjects indicated whether they identified more strongly with the
Democratic or Republican party or as Independents. Table 2.5 also contains estimated
expected frequencies for H0: independence. For instance, the first cell has ˆμ11 =
n1+n+1/n = (1557 × 1246)/2757 = 703.7.
The chi-squared test statistics are X 2 = 30.1 and G 2 = 30.0, with df = (I − 1)
(J − 1) = (2 − 1)(3 − 1) = 2. This chi-squared distribution has a mean of df = 2
and a standard deviation of √ (2df) = √ 4 = 2. So, a value of 30 is far out in the
right-hand tail. Each statistic has a P -value < 0.0001. This evidence of association