$Z$-Test for Proportion
Let $X$ be the observed number of individuals possessing certain attributes (say, number of successes) in a random sample of size $n$ from a large population, then $\hat{p}=\frac{X}{n}$ be the observed proportion of successes. Let $p$ be the population proportion of successes and $q = 1- p$ be the population proportion of failures.
The hypothesis testing problem can be setup as:
| Situation | Hypothesis Testing Problem |
|---|---|
| Situation A : | $H_0: p=p_0$ against $H_a : p < p_0$ (Left-tailed) |
| Situation B : | $H_0: p=p_0$ against $H_a : p > p_0$ (Right-tailed) |
| Situation C : | $H_0: p=p_0$ against $H_a : p \neq p_0$ (Two-tailed) |
Formula
The test statistic for $H_0:p=p_0$ is
$Z=\frac{\hat{p}-p}{SE(\hat{p})}$
where
$SE(\hat{p})= \sqrt{\frac{p(1-p)}{n}}$is standard error of $p$,
The test statistic $Z$ follows standard normal distribution $N(0,1)$
