For a <two-by-two contingency table>, <Fisher's exact test> conditions on both row totals and both column totals. Under the null of no association, the upper-left count $K$ then has the <hypergeometric distribution>
$$
\mathbb P(K=k\mid\text{margins})
=\frac{\binom{c_1}{k}\binom{n-c_1}{r_1-k}}{\binom n{r_1}},
$$
where $r_1$ and $c_1$ are the first row and column totals. A one-sided p-value sums the appropriate hypergeometric tail; a common two-sided p-value sums the probabilities of all feasible tables no more likely under the null than the observed table. Because the conditional distribution is discrete, the p-value is <super-uniform random variable> rather than generally exactly uniform.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 1 / ii / Solution

A random variable $P\in[0,1]$ is super-uniform when $\mathbb P(P\leq\alpha)\leq\alpha$ for every $\alpha\in[0,1]$. A valid p-value is super-uniform under its null hypothesis.


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 Incoming links: Super-uniform random variable