Proportion Test

Used to test whether the sample proportion $p̂$ from a population differs significantly from a claimed or hypothesized population proportion $p_0$.

Hypotheses Setup

Let $p$ the true population proportion, $\hat{p} = \frac{x}{n}$. Here $x$ = number of successes. $n$ = sample size.

Then:

$$H_0 : p = p_0$$

Alternative hypothesis depends on the problem:

$$H_1 : p < p_0 \quad \text{(left-tailed)}$$ $$H_1 : p > p_0 \quad \text{(right-tailed)}$$ $$H_1 : p \neq p_0 \quad \text{(two-tailed)}$$

Test Statistic

Normal Approximation

Used when $n$ is large and both $np_0 \ge 5$ and $n(1 − p_0) \ge 5$, the sampling distribution of p̂ is approximately normal.

So, under H₀:

$$Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \quad \sim N(0, 1)$$

• For a two-tailed test, reject $H_0$ if $|Z| ≥ Z_{1-\alpha/2}$
• For a right-tailed test, reject $H_0$ if $Z ≥ Z_{1-\alpha}$
• For a left-tailed test, reject $H_0$ if $Z ≤ −Z_{1-\alpha}$

The confidence interval for $p$ is:

$$\hat{p} \pm Z_{1-\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$$

If $p_0$ lies outside the confidence interval, reject $H_0$ at significance level $\alpha$.

Binomial Test

$$X \sim \text{Binomial}(n, p_0)$$

For tailed tests, calculate p-value directly by summing over binomial probabilities. $H_0$ is rejected if $p\text{-value} \le \alpha$.