# Chi square test confusion

I'm getting confused by the chi squared test of homogeneity, When performing the test, do the samples need to be distributed as multinomial?

• Well, they could be some other things -- such as Poisson for example. You need something with a conditional variance-covariance structure of the right form or otherwise the usual denominator won't be right. – Glen_b Oct 17 '16 at 9:13
• @Glen_b In my case I have 3 independent samples, each sample has a proportion, and I want to test proportions for equality. If the sample had a multinomial distribution then the sum of theses proportions would be 1. Not in my case. Is th chi squared test of homogeneity valid in my case? – Toney Shields Oct 17 '16 at 9:28
• Within each sample the success and failure counts are (at least notionally) binomial, right? – Glen_b Oct 17 '16 at 9:45
• @Glen_b yes they are. – Toney Shields Oct 17 '16 at 10:32

The mention of multinomial made in your other question would be to the distribution within samples. The values across samples would be independent.

There are cases where the samples are dependent in both directions because both margins are being treated as fixed (in the 2x2 case this would be hypergeometric).

There are also cases that could be independent in both directions (Poisson counts); if you condition on the totals in one margin these would become multinomial.

For the chi-squared distribution to work you need the asymptotic multivariate normal to have the right form of dependence such that the chi-square statistic is equivalent to a quadratic form of the right kind to correspond to a sum of squares of independent standard normals of dimension equal to the df for the chi-square.

• Within each sample the success and failure counts are binomial but the response categories are not mutually exclusive. – Toney Shields Oct 17 '16 at 16:09