AP Statistics · Topic 3.12 · Unit 3

AP Stats 3.12: Setting Up a Two-Prop Test

By Jude Wallis · Published

A two-sample z-test for a difference in proportions has null hypothesis p1 equals p2. Verify randomization, the 10% condition, and normality using the combined (pooled) proportion p-hat-c, checking that its expected success and failure counts in both groups are at least 10.

AP Statistics: Unit 3 (topics 3.12). CED topic 3.12 (Setting Up a Test for the Difference Between Two Population Proportions), skills 2.C, 2.E, 4.E.

What topic 3.12 covers

Topic 3.12 sets up a significance test comparing two population proportions. The correct method is the two-sample z-test for the difference between two population proportions. You identify the parameters, state the hypotheses, and verify conditions, saving the test statistic and p-value for topic 3.13. This is the formal-test partner to the confidence interval in topic 3.10, and it uses the same data structure of two independent groups.

Null and alternative hypotheses

The null hypothesis says the two proportions are equal, written H0 ⁣:p1=p2H_0\colon p_1 = p_2 or equivalently H0 ⁣:p1p2=0H_0\colon p_1 - p_2 = 0. The alternative is one-sided, Ha ⁣:p1>p2H_a\colon p_1 > p_2 or Ha ⁣:p1<p2H_a\colon p_1 < p_2, or two-sided, Ha ⁣:p1p2H_a\colon p_1 \ne p_2. Each can be written as a statement about the difference p1p2p_1 - p_2 instead. State the parameters, response variable, and populations in context, and let the research question decide the direction of the alternative.

The pooled proportion and conditions

Because the null assumes the proportions are equal, you estimate that common value by pooling both samples:

p^c=x1+x2n1+n2,\hat{p}_c = \dfrac{x_1 + x_2}{n_1 + n_2},

the combined proportion of successes. The three conditions are randomization (two independent random samples or a randomized experiment), the 10% condition when sampling without replacement, and normality. For normality, check that n1p^cn_1\hat{p}_c, n1(1p^c)n_1(1-\hat{p}_c), n2p^cn_2\hat{p}_c, and n2(1p^c)n_2(1-\hat{p}_c) are all at least 10.

Why pooling belongs to the test

Pooling is the feature that separates the two-proportion test from the two-proportion interval. Under the null, the two groups share one true proportion, so combining their successes gives the best single estimate of it and the most accurate standard error for the test. The interval in topic 3.10 makes no equality assumption, so it keeps each group's own p^\hat{p} instead.

Deciding up front whether you are building an interval or running a test tells you immediately whether to pool, which prevents one of the most common two-proportion errors. It also helps to write the two counts and two sample sizes in a small table before computing, so the successes and totals do not get swapped. A swapped count changes the pooled proportion and every expected count that follows, which can quietly break an otherwise correct setup. Naming the test, stating the parameters, and checking the pooled conditions in that order gives you a reliable routine to repeat on every two-proportion problem.

Pooled proportion and hypotheses

Group 1 has 48 successes in 200 trials; group 2 has 30 successes in 150 trials. Set up a two-sided test and find the pooled proportion for the conditions.

  1. Parameters: p1p_1 and p2p_2 are the true success proportions for the two populations.

  2. Hypotheses: H0 ⁣:p1=p2H_0\colon p_1 = p_2 and Ha ⁣:p1p2H_a\colon p_1 \ne p_2.

  3. Pooled proportion: p^c=48+30200+150=783500.2229\hat{p}_c = \dfrac{48 + 30}{200 + 150} = \dfrac{78}{350} \approx 0.2229.

  4. Normality check: 200×0.222944.6200 \times 0.2229 \approx 44.6, 200×0.7771155.4200 \times 0.7771 \approx 155.4, 150×0.222933.4150 \times 0.2229 \approx 33.4, 150×0.7771116.6150 \times 0.7771 \approx 116.6.

The pooled proportion is about 0.2229, and all four expected counts exceed 10, so the normality condition is met and a two-sample z-test is appropriate.

Frequently asked questions

Why pool the two samples for a test but not for a confidence interval?

The test's null hypothesis assumes p1=p2p_1 = p_2, so the best estimate of that shared proportion combines both samples. A confidence interval assumes no equality, so it keeps each group's own p^\hat{p}.