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We are working on a bioinformatics problem. We have a collection (say $U$) of kmers or small substrings of length $k$ of some set of sequences. We have two different experiments. Only a subset of kmers say, $A,B \subseteq U$, is enriched in each of the two experiments, respectively. We want to show $A \cap B$ is statistically significant.

We were trying to use Fisher's exact test (FET) on a contingency table with entries $|A \cap B|, |A \setminus B|, |B \setminus A|, |U \setminus (A \cup B)|$. But it seems that this FET has the null hypothesis: “probability of a kmer getting enriched is same in the two experiments”.

However, for our purpose it should be the alternate hypothesis.

Is there a different way to use FET for our purpose?

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  • $\begingroup$ Not a categorical data expert, so this is in a comment. You've hit a core issue in null-hypothesis significance testing. I think you could get at what you want by combining a permutation test with a two one-sided test approach. You have a two-part null: Determine a difference in probabilities that would be substantively meaningful, and the null is that the population difference is greater than that in either direction. Then use a permutation test and use the resulting distribution to establish whether the null(s) can be rejected. $\endgroup$ – Patrick Malone May 29 '18 at 16:45
  • $\begingroup$ Thank you @PatrickMalone. I am quite new in statistical significance testing. Could you please point me to some examples where this approach of combining permutation test and two one-sided test has been applied? $\endgroup$ – Soumitra May 29 '18 at 17:16
  • $\begingroup$ If I get you right, you must have 2x2 frequency table with sides A (1 vs 0) and B (1 vs 0). Cell (1,1) collects k-mers which belong to both A and B. You want to test that frequency f in (1,1) is bigger than expected there under randomness (that would be f_in_row1*f_in_col1 / f_total). So, you test that the residual in (1,1) is significantly positive. But in your situation of 2x2 table an increase in (1,1) automatically means same-size inrease in (0,0) and same decrease in (1,0) and (0,1), because marginal totals are fixed. $\endgroup$ – ttnphns May 29 '18 at 18:56
  • $\begingroup$ (cont.) Therefore sig. testing of cell (1,1) is equivalent to testing any of the 4 cells, i.e. to testing all the table for the sig. of association/independence. It could be chi-square test, but you preferred Fisher exact test. It is all right. I don't see why you are in doubts. $\endgroup$ – ttnphns May 29 '18 at 18:57
  • $\begingroup$ Sorry, I'm just speculating here. $\endgroup$ – Patrick Malone May 29 '18 at 19:09

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