Topic 6.10 Setting Up a Test for the Difference of Two Population Proportions: state the hypotheses about the difference of two proportions, identify the significance level, and verify the conditions for a two-sample z-test using the pooled proportion.

What this topic is asking

The College Board (Topic 6.10) wants you to set up a test for the difference of two proportions: state the hypotheses about $p_1$ and $p_2$, identify the significance level, compute the pooled (combined) proportion $\hat{p}_c$, and check the conditions for the two-sample z-test.

Hypotheses about two proportions

The hypotheses concern the two population proportions, and the null almost always asserts they are equal (a difference of $0$). Define each $p_i$ in words and fix the subtraction order, so the alternative's direction matches the wording: "higher in group $1$" gives $p_1 > p_2$, "different" gives $p_1 \ne p_2$. As always, hypotheses are about parameters, not the sample proportions $\hat{p}_1, \hat{p}_2$.

The pooled proportion and why it appears

This is the defining difference from the two-proportion interval. An interval has no assumption of equality, so it keeps the samples separate (unpooled). A test assumes equality under $H_0$, so it pools all successes over all trials to estimate the single proportion both groups would share if $H_0$ were true, and uses $\hat{p}_c$ in both the condition check and the standard error of the test statistic (Topic 6.11). The pooled value is a weighted average (weighted by sample size), not a plain average of $\hat{p}_1$ and $\hat{p}_2$.

Checking the conditions

The large-counts check uses the pooled $\hat{p}_c$, consistent with reasoning under $H_0$ that the proportions are equal. (Contrast the interval, which used each $\hat{p}_i$ separately.) Verifying all four counts and the between-sample independence is what earns the normal model for the difference.

Try this

Q1. Compute the pooled proportion for $x_1 = 40, n_1 = 100, x_2 = 60, n_2 = 200$. [1 point]

• Cue. $\hat{p}_c = \dfrac{40 + 60}{100 + 200} = \dfrac{100}{300} \approx 0.333$.

Q2. Why does a two-proportion test pool the proportions while the interval does not? [1 point]

• Cue. The test assumes $p_1 = p_2$ under $H_0$, so it estimates one common proportion; the interval makes no such assumption and keeps the samples separate.