Confidence Intervals for Two Proportions

Suppose we have two independent samples and we want to estimate the difference $p_1 - p_2$ between two population proportions. Let $\hat{p}_1 = x_1/n_1$ and $\hat{p}_2 = x_2/n_2$ be the sample proportions, where $x_i$ is the number of successes in sample $i$.

A confidence interval for the difference of two proportions $p_1 - p_2$ is given by

$$(\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}},$$

where $z^*$ is the standard normal critical value for the chosen confidence level. The three values you will use most often are

$$z^*_{90\%} = 1.645, \qquad z^*_{95\%} = 1.96, \qquad z^*_{99\%} = 2.576.$$

The quantity under the square root is the standard error of $\hat{p}_1 - \hat{p}_2,$ and the product $z^* \cdot SE$ is the margin of error.

For example, with $\hat{p}_1 = 0.4,$ $n_1 = 100,$ $\hat{p}_2 = 0.3,$ and $n_2 = 150,$

$$SE = \sqrt{\dfrac{0.4(0.6)}{100} + \dfrac{0.3(0.7)}{150}} = \sqrt{0.0024 + 0.0014} \approx 0.0616.$$

A 95% CI is then $(0.4 - 0.3) \pm 1.96(0.0616) \approx 0.10 \pm 0.121 = (-0.021,\ 0.221).$

This formula assumes the two samples are independent and each sample contains at least 10 successes and 10 failures.