Comparing Two Proportions

The interval

For two independent random samples, the confidence interval for $p_1-p_2$ has the usual shape, estimate $\pm$ (critical value)(standard error), with the standard error built from the two sample proportions:

Conditions: both samples random and independent of each other; each sample at most $10\%$ of its population; and at least $10$ successes and $10$ failures in each sample. Interpret as "we are $95\%$ confident that the difference in the true proportions ($p_1-p_2$) lies in the interval" — and say which group is subtracted from which.

The test and why it pools

The null hypothesis is $H_0:p_1=p_2$ (equivalently $p_1-p_2=0$). If the two proportions really are equal, the best estimate of that common value uses all the successes:

The large-counts check for the test uses $\hat p_c$: $n_1\hat p_c$, $n_1(1-\hat p_c)$, $n_2\hat p_c$, $n_2(1-\hat p_c)$ all at least $10$. Example: $45$ of $150$ patients improve on a drug and $30$ of $150$ on a placebo. Then $\hat p_c=75/300=0{,}25$, the standard error is $\sqrt{0{,}25\cdot0{,}75\cdot(2/150)}=0{,}05$ and $z=(0{,}30-0{,}20)/0{,}05=2{,}0$.

Interval and test disagree slightly

Because the interval uses separate $\hat p$'s while the test pools, a $95\%$ interval that just barely excludes $0$ and a two-sided test at $\alpha=0{,}05$ can, rarely, give opposite verdicts. When asked, report the test for a decision and the interval for the size of the difference.