Proportion Test | Sahithyan's Notes

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

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

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)

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 $\alpha$ .

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

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