Variances add, even when the proportions subtract

Take independent random samples from two populations with proportions $p_1$ and $p_2$, of sizes $n_1$ and $n_2$, and compute $\hat{p}_1 - \hat{p}_2$. Across all such pairs of samples,

$$\mu_{\hat{p}_1 - \hat{p}_2} = p_1 - p_2.$$

The difference of two unbiased estimators is unbiased for the difference of the parameters, so the sampling distribution is centered exactly where it should be.

The order of subtraction is a choice, and it has to be stated and then kept. If group 1 is the new checkout system and group 2 the old, then $\hat{p}_1 - \hat{p}_2 = 0.46 - 0.36 = 0.10$ says the new system's rate is 10 percentage points higher. Reversing the order gives $-0.10$, which says exactly the same thing about the world and the opposite thing if the sign is read carelessly. Define the order once, at the top, in words.

Because the samples are independent, the variances add:

$$\sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}.$$

The plus sign is the whole content of the formula, and it survives the minus sign in the difference. Each group contributes its own uncertainty to the comparison, so a difference is more variable than either proportion alone, not less. Topic 2.9 gave the reason: $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$ for independent $X$ and $Y$.

With $\hat{p}_1 = 0.46$ from $n_1 = 300$ and $\hat{p}_2 = 0.36$ from $n_2 = 250$,

$$\sqrt{\frac{0.46(0.54)}{300} + \frac{0.36(0.64)}{250}} = \sqrt{0.000828 + 0.000922} \approx 0.0418.$$

Three wrong routes produce numbers that look plausible. Subtracting the variances gives a negative value under the radical here, $-0.0000936$, which is impossible. Adding the two standard deviations gives $0.0288 + 0.0304 = 0.0591$, too large. And pooling the samples into one group and using a single $\sqrt{\frac{p(1-p)}{n}}$ discards the two-group structure entirely.

Every condition from the one-sample case appears twice, and one new condition appears once:

1. Random: both samples are random samples, or the subjects were randomly assigned to two treatments.
2. 10%: each sample is at most 10% of its own population, checked separately.
3. Large counts: all four counts are at least 10. Here $138$ and $162$ in group 1, $90$ and $160$ in group 2.
4. Independence between the groups: the two samples must not be linked.

That last one is the condition with no one-sample counterpart, and it is the one a study design can violate invisibly. Measuring the same 200 people before and after a change produces two proportions that are paired, not independent, because the same person appears in both. The two-sample machinery does not apply to that design, and the variances-add formula is exactly what fails: paired measurements move together, so their difference is less variable than independence would predict.

When all four counts clear 10, the sampling distribution of $\hat{p}_1 - \hat{p}_2$ is approximately normal, so a probability question about the difference is a z-score calculation:

$$z = \frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}.$$

The value $p_1 - p_2 = 0$ has special standing in the topics ahead: it is the claim that the two populations have the same proportion. When the difference's distribution is centered at 0 and a sample difference lands several standard deviations away, that is evidence the two proportions differ, which is the entire content of the two-proportion test.

One interpretation habit carries through: name the object. "In repeated pairs of independent random samples of 300 and 250, the difference in sample proportions typically falls about 0.042 from the true difference." The phrase difference in sample proportions is what keeps the sentence attached to the statistic rather than to either group.