Hypothesis Tests for Two Proportions

When we want to compare two population proportions $p_1$ and $p_2,$ we draw an independent sample from each population and test whether the proportions are equal.

From the two populations we have:

• Sample 1: size $n_1$ with $x_1$ successes, so $\hat{p}_1 = \dfrac{x_1}{n_1}.$
• Sample 2: size $n_2$ with $x_2$ successes, so $\hat{p}_2 = \dfrac{x_2}{n_2}.$

The null hypothesis is always

$$H_0\!: p_1 = p_2.$$

Under $H_0,$ both populations share a single common proportion $p.$ Combining both samples, our best estimate of $p$ is the pooled sample proportion

$$\hat{p} = \dfrac{x_1 + x_2}{n_1 + n_2}.$$

The test statistic is then

$$z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}}.$$

Provided $n_1$ and $n_2$ are large enough, $z$ is approximately standard normal under $H_0.$

For a quick illustration, if $n_1 = n_2 = 100$ with $x_1 = 50$ and $x_2 = 40,$ then $\hat{p}_1 = 0.50,$ $\hat{p}_2 = 0.40,$ $\hat{p} = 90/200 = 0.45,$ and

$$z = \dfrac{0.50 - 0.40}{\sqrt{0.45 \cdot 0.55 \cdot (0.01 + 0.01)}} = \dfrac{0.10}{\sqrt{0.00495}} \approx 1.42.$$