# Qualitative analysis

Which test can I use for smaller data sets (<5 in a single cell, <20 in group, although N>500) but for larger grids (3x4, 4x5 etc.)?

Chi-squared test can be used for larger grids but not for smaller data sets as you have to use the Fischer or Yates correction. Fischer or Yates correction however seem to be reserved for 2x2 grids.

I was searching answer to this problem for ages and cannot find an answer.

• Details matter. First, to assess the applicability of the chi-squared distribution to testing independence in tables, the actual cell counts don't matter at all: what matters are the expected counts. Second, even when many of the expected counts are less than $5$ (a standard rule-of-thumb threshold), when there are many cells that doesn't matter much, either. Thus, the answer in your case ought to depend on the dimensions of your table, the number of cells, and their expected values. Could you provide that information? – whuber Mar 3 '20 at 19:38
• I do have plenty of tables to do, most of them however are at most 5x6. As far as expected counts go, do you mean expected counts in the case when there would be no correlation? So for example in a case of a 600 N and 5x6 table it would be 600 / 30 = 20 for each cell? – John Teres Mar 3 '20 at 20:43
• Usually the expected counts are obtained by multiplying the estimated marginal probabilities; they are rarely the same in every cell. – whuber Mar 3 '20 at 21:01
• Oh, I get you now.The lowest expected count is 0.0217 in this particular table (5x4). Then they go up with the highest one being 177.77. 12 out of 20 have value below 5. – John Teres Mar 3 '20 at 22:50
• That is a concern, then. You can't use the chi-squared distribution to compute a p-value. Depending on your assumptions you can use Fisher's Exact Test or a permutation test. – whuber Mar 3 '20 at 23:15