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)?


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.