Testing a proportion, normal approximation or exact?

Question

You observe k successes in n independent trials (say a strategy that was directionally right k out of n days) and want to test whether the true success rate equals some p0. When is the normal-approximation z-test for a proportion valid, and when should you use the exact binomial test instead?

Accepted Answer

The z-test for a proportion. Under H0: p = p0, the sample proportion \hat{p} = k/n is approximately normal, giving z = \frac{\hat{p} - p0}{\sqrt{p0(1 - p0)/n}}. Compare z to normal critical values. This leans on the Central Limit Theorem approximating a Binomial(n, p0) count by a normal. When the approximation is valid. The binomial is well approximated by a normal when both expected counts are reasonably large, a common rule of thumb is n p0 \ge 10 \quad 	ext{and} \quad n(1 - p0) \ge 10. Near p0 = 0.5 this kicks in quickly; near 0 or 1 the binomial is skewed and you need a larger n. A conti…

Testing a proportion, normal approximation or exact?

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