You say you did paired t-tests on the original data, before dichotomizing it, and that males increased significantly from the old form to the new but the female change was not significant. Unfortunately, that can not be taken as showing that the male change was bigger than the female change. You need to do an independent-groups t-test on the two sets of change scores. (Better yet, you could replace all the t-tests by confidence intervals for the corresponding means and mean differences, which would give you more information.)
For the dichotomized data, the situation is similar.
You have two contingency table, one for males and one for females.
Males
Yes No Total
Yes Myy Myn My.
No Mny Mnn Mn.
Total M.y M.n M.. = M = total number of Males
Females
Yes No Total
Yes Fyy Fyn Fy.
No Fny Fnn Fn.
Total F.y F.n F.. = F = total number of Females
For each table, the analog of the paired t-test is the McNemar test,
http://en.wikipedia.org/wiki/McNemar%27s_test
I know of no simple standard test of the difference between the changes in endorsement rates, but if all of Myn, Mny, Myy+Mnn, Fyn, Fny, Fyy+Fnn are "large" then an asymptotic test might be justified.
You can use a chi-squared test in your example with different sample sizes. Your "another verb type" would be verbs that are not oral verbs, i.e. all the other verbs
Suppose in your example, $10$ of the $82$ verbs in sample one were oral verbs and $72$ were not, while $20$ of the $89$ verbs in sample two were oral verbs and $69$ were not. Then the table for your four cell chi-squared test could look like
10 72 | 82
20 69 | 89
__ ___ ___
|
30 141 | 171
and in R you might get
chisq.test(rbind(c(10, 72), c(20, 69)))
# Pearson's Chi-squared test with Yates' continuity correction
#
# data: rbind(c(10, 72), c(20, 69))
# X-squared = 2.4459, df = 1, p-value = 0.1178
so this example would not be statistically significant
Best Answer
After the significant chi-squared test on the four groups, you may want to do ad hoc tests comparing Group A with other groups. Because you did not show your actual data, I will show how to do this for fictitious data, which may be somewhat similar to yours.
Suppose you have a table of numbers of sales as follows:
In R:
In R, a chi-squared test of homogeneity rejects the null hypothesis that sales of gender are homogeneous across groups at significance level $0.1%$ with P-value $0.0003 < 0.001 = 0.1\%.$ The Yates continuity correction is declined (parameter 'cor=F') on account of reasonably large counts.
The chi-squared test compares the observed counts in
TBLwith counts (based on marginal totals) that would be expected under the null hypothesis of homogeneity. The Pearson residuals can show where disagreement of observed and expected counts is greatest; look especially for residuals with largest absolute values. Here it seems that the greatest contribution to the relatively large chi-squared statistic comes from groups A and D.You can do an ad hoc test to compare these two groups by selecting only columns 1 and 4 of
TBL. In order to avoid 'false discovery' from repeated analyses on the same data, ad hoc tests should be conducted at a smaller significance level than the main chi-squared test. Here it is clear that Groups A and D differ. Specifically, numbers of sales by males are larger in A and sales by females are larger in D.+Note:_ Another, essentially equivalent, version of the chi-squared test in R is 'prop.test' as follows: It shows proportions of sales by makes in each group. The proportions $0.542$ and $0.381$ were shown to be significantly different by the ad hoc chi-squared test above.