I have a 2x2 contingency table and when I performed a chi-square test for association, a p-value of around 0.1 was obtained. One of the cell has an expected count of 4.7 which is less than 5 and so the Fisher's exact test is probably more appropriate. However, would it not also be okay to test for the null hypothesis that the odds ratio = 1? Is the log odds ratio test also subject to the rule of thumb where the expected cell count should be greater than 5? Would I expect to find a smaller p-value (p < 0.1) from a test for odds ratio = 1?
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[This](https://stats.stackexchange.com/questions/14226/given-the-power-of-computers-these-days-is-there-ever-a-reason-to-do-a-chi-squa/14230#14230) question and answer address the "expected cell count less than 5" myth in the analysis of contingency tables. It seems like you can simply perform a chi-square test and obtain accurate p values by applying a correction described in one of the links. I can access the paper if you are not able to and walk you through the solution. – Demetri Pananos Jun 09 '20 at 12:58
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@DemetriPananos Thanks for the comment and the link. I guess I'm more concerned with the second part of the question that is, why isn't the log odds ratio test used more since it is basically testing the same thing? It is much easier to compute the SE of log(OR) and would it not produce the same p-value for most scenarios? – David Young Jun 09 '20 at 13:06
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I can't say exactly why one test is used more than other, but I highly suspect it is either to do with interpretability and/or familiarity. It can be difficult to describe to people what an odds ratio is, and the chi square test is just so easy. All these tests should be asymptotically equivalent, so as you get more data it won't matter much which test you use. However, I can't say in the finite sample case. – Demetri Pananos Jun 09 '20 at 13:12