Editing Binomial test (section)

==Usage==
A binomial test is a [[Statistical hypothesis testing|statistical hypothesis test]] used to determine whether the proportion of successes in a sample differs from an expected proportion in a binomial distribution. It is useful for situations when there are two possible outcomes (e.g., success/failure, yes/no, heads/tails), i.e., where repeated experiments produce [[binary data]]. If one assumes an underlying probability <math>\pi_0</math> between 0 and 1, the null hypothesis is

: <math>H_0\colon\pi=\pi_0</math>

For a sample of size <math>n</math>, we would expect <math>n\pi_0</math> successes. The formula of the [[binomial distribution]] gives the probability of those <math>n</math> samples instead producing <math>k</math> successes:

: <math>\Pr(X=k)=\binom{n}{k}\pi_0^k(1-\pi_0)^{n-k}</math>

Suppose that we want to test the alternative hypothesis

: <math>H_{A}\colon\pi<\pi_0</math>

i.e., we suspect that the actual probability of success is lower than <math>\pi_0</math>. Then our <math>p</math>-value would be computed using a one-tailed test; specifically, we compute the probability of seeing an outcome as extreme as, or more extreme (i.e., less likely), than <math>k</math>:

: <math>p = \sum_{i=0}^k\Pr(X=i)=\sum_{i=0}^k\binom{n}{i}\pi_0^i(1-\pi_0)^{n-i}</math>

An analogous computation can be done if we're testing if <math>\pi>\pi_0</math> using the summation of the range from <math>k</math> to <math>n</math> instead.

Calculating a <math>p</math>-value for a two-tailed test is slightly more complicated, since a binomial distribution isn't symmetric if <math>\pi_0\neq 0.5</math>. This means that we can't just double the <math>p</math>-value from the one-tailed test. Recall that we want to consider events that are as extreme, or more extreme, than the one we've seen, so we should consider the probability that we would see an event that is as, or less, likely than <math>X=k</math>. Let <math>\mathcal{I}=\{i\colon\Pr(X=i)\leq \Pr(X=k)\}</math> denote all such events. Then the two-tailed <math>p</math>-value is calculated as,

: <math>p = \sum_{i\in\mathcal{I}}\Pr(X=i)=\sum_{i\in\mathcal{I}}\binom{n}{i}\pi_0^i(1-\pi_0)^{n-i}</math>