Constructing a Confidence Interval for the Difference Between Two Population Proportions

The two-proportion z-interval is $(\hat{p}_1 - \hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$, using each group's own estimate because an interval assumes nothing about either proportion. For 138 of 300 against 90 of 250, the difference is 0.10, the standard error is about 0.0418, and the 95% interval is about $(0.018, 0.182)$. The conditions are random selection or assignment, independence between the groups, the 10% condition for each sample, and all four counts at least 10. The interpretation names the level, states that the interval captures a difference between two population proportions, and gives the order of subtraction.

The signature error is pooling. A pooled proportion is justified only under the assumption that the two populations share one value, which is what a test's null says and what an interval refuses to assume, so a standard error containing $\frac{1}{n_1} + \frac{1}{n_2}$ has imported the very claim being estimated. Alongside it: conditions recited without the four counts, paired designs run through two-sample machinery, and an interpretation that drops the word difference or the order of subtraction, leaving the reader unable to tell which group the interval favors.