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For 2x2 contingency table, we can test independence of row and column variables by using Fisher's exact test where we assume all marginals are fixed.

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| n11 | n12 |
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| n21 | n22 |
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Under the null of independence, I've read (so many times) the distribution of $n_{11}$(See the table above) is the hypergeometric distribution.

My question is where I used the information of nullity.

Any comments and discussion are welcome and appreciated.

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The hypergeometric results when randomly drawing without replacement from a collections of two kinds of objects.

The classic example is drawing $n$ balls from a (well-mixed) urn containing $w$ white balls and $b$ black balls (where $w+b=N$), and counting the number of white balls ($K$) among the $n$ that were drawn, which will then follow the hypergeometric pmf:

$$P(K=k) = {\frac {{\binom {n}{k}}{\binom {N-n}{w-k}}}{\binom {N}{w}}}$$

How does this relate to the 2x2 table? Write the information like so:

              White    Black    
  Drawn         K       n-K        n
 Not Drawn     w-K     b+K-n      N-n

                w        b         N

and we see the 2x2 table of Fisher's exact test.

So where did we use independence? It's in the very definition of the hypergeometric ("randomly drawing"/"well-mixed" in the above is used to imply the independence required). Without that the probability calculation to yield the hypergeometric pmf would not hold.

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  • $\begingroup$ Thanks, Glen_b. Thus, you mean that some of N balls could be randomly selected UNDER independence. I got it. $\endgroup$
    – inmybrain
    Jun 12 '16 at 12:14
  • $\begingroup$ @inmybrain The "randomly selected" in this context literally means "at each step of ball selection, every ball remaining in the urn has an equal chance of being selected next". The fact that (as a result) the probabilities don't relate to the order in which earlier balls were selected is where the independence comes from. $\endgroup$
    – Glen_b
    Jun 12 '16 at 12:46

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