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I am trying to understand the meaning of the null distribution in Fisher Exact Test.

Suppose I have a contigency table:

        Stat1   Stat2
Group1     18       2
Group2      3      40

If I want to know if Stat1 is enriched significantly in Group1, have I to use two-sided test? Is it possible that two-sided test gives me also the probability that Stat1 is enriched in Group2 (another alternative to the null hypothesis)?

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You want a one-sided test from the phrasing of your question. Your “null hypothesis”: Stat1's proportion of Group1 is greater than Stat2's proportion of Group1, is directional. You point out the reason why you might not want the two-sided test.

On the other hand, if you do this, you are essentially saying that you know for a certainty that if there is a difference, it will be in favor of Stat1. If that's not something you really know, then you are being a bit disingenuous, even arguably dishonest. Some people (including me) would look at this as an admission of statistical weakness.

The other question might be whether this would be a typical construction of a Fisher's Exact Test. "Two-sided" is the default for R's fisher.test function, but you can specify other alternatives.

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From what I understood, the null hypothesis is that the probability of belonging to group1 or group2 is not correlated with the value of stat1 or stat2, and the one-sided distribution considers the more extreme arrangements of table (i.e. with lower probability of happening by chance) than the one considered.

If that is so, what is the two-sided equivalent? In my understanding all extreme cases were covered for the first calculation, of one-sided probability, including the one you propose. But correct me if wrong, I am not very well acquainted with Fisher exact test.

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