**P.Mean: Interpretations for a two by two table (created 2008-10-18)**.

I have a question. I have 2X2 table (below) and it shows a significant chi-square value for the overall test. I would now like to see where those differences occur using a method similar to a post-hoc analysis in ANOVA? Can this be done in SAS or other statistical software and, if so, how do I do it?There are four cells in your table, but they are not independent of one another. So any conclusion you base on a single row of the table would be identical to the conclusion that you would base on a single column. So, for example, you might note that among female receivers, the probability of a female giver is 39%. That is significantly higher than the comparable probability among male receivers (19%). You could change this to probabilities for male givers, and the result would be 61% and 81%, respectively. Since the probabilities of a male giver among either female or male receivers and the probabilities of a female giver among either female or male receivers is complementary (they add up to 1), there is no point to discussing both probabilities. Just pick one, and the reverse result for the complementary probability is implied.

You could do this instead by focusing on the column percents, and again you would see that a reverse relationship occurs among the complementary probabilities.

There is no need, then, for a post hoc test. Just pick a single pair of row or column percents and describe it carefully.

If you are aware of the degrees of freedom associated with the Chi-square test, you will note that there is only one degree of freedom. This is a pretty big hint that no further post hoc analysis is needed.

This work is licensed under a Creative Commons Attribution 3.0 United States License. This page was written by Steve Simon and was last modified on 2010-04-01. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Probability concepts.