Hi, I am looking for an advice from someone experienced in analysis of discrete data with Poisson distribution (e.g. counts) in R.
I have 3 variables, which could be coded either to create contingency table (2x2x2), or in 0/1 (no/yes) and I would like to know, if there is a significant dependence of one of them on the other two (or their interaction).
If I use solution recommended for contingency tables, I have only information about overall significance and cannot see which of the “independent” variables contributes on the result. Moreover, R warns me the Chi-square test could be problematic for this design. What is even worse, the result is not the same for the two possibilities, i.e. summary(table_x), and chisq.test(table_x). I have also find solution from the MASS library in loglm() function, but the results are total mess for me.
I have tried to solve it with the help of GLM too, but it would not satisfy me either. In the coding for family = poisson the results are compared with a “control group,” but I do not have any meaningful combination to be “control” and I would like to see the differences among all variables. On the other hand, for family = binomial version I am not able to find a significant result even in the most dissimilar setup possible.
Any other ways how to solve this problem in R?
(PS: total N is 24 in my dataset)
Your N is too tiny for a chisq test. you have 8 cells. Chisq should have at least 5 expected observations per cell or it will fail. You have 3. Fisher's exact test is what you want to use here.
https://stat.ethz.ch/R-manual/R-patched/library/stats/html/fisher.test.html
Well, this is a part of the problem, but it does not explain differences in both Chi-square tests suggesting the computing mechanism differs between summary() and chisq.test().
Moreover, fisher.test() works just for 2D table (2x2), not 3D (2x2x2) as mine. (And if I take my testing data as 2D table, the results are the same for both Fisher and Chi-square).
chisq.test applies a continuity correction to the analysis by default. try running chisq.test(table, correct =FALSE) . This should give you the same results.
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