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Pooled Proportion — Definition, Formula & Examples

Pooled proportion is the overall success rate you get when you combine two samples into one big group. You use it in a two-proportion z-test, where the null hypothesis assumes both populations share the same true proportion.

Given independent samples with x1x_1 successes out of n1n_1 and x2x_2 successes out of n2n_2, the pooled proportion is p^c=x1+x2n1+n2\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}. It equals the weighted average n1p^1+n2p^2n1+n2\frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2} of the two sample proportions. It estimates the common population proportion pp under H0:p1=p2H_0: p_1 = p_2. It appears in the standard error p^c(1−p^c)(1n1+1n2)\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}.

Key Formula

p^c=x1+x2n1+n2\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}
Where:
  • p^c\hat{p}_c = pooled (combined) sample proportion
  • x1,x2x_1, x_2 = number of successes in sample 1 and sample 2
  • n1,n2n_1, n_2 = size of sample 1 and sample 2

Worked Example

Problem: In a survey, 48 of 200 students at School A and 72 of 200 students at School B ride the bus. Test H0:pA=pBH_0: p_A = p_B against Ha:pA≠pBH_a: p_A \neq p_B. Find the pooled proportion, the standard error, and the z-statistic.
Step 1: Combine the successes and the sample sizes to get the pooled proportion.
p^c=48+72200+200=120400=0.30\hat{p}_c = \frac{48 + 72}{200 + 200} = \frac{120}{400} = 0.30
Step 2: Use the pooled proportion in the standard error, because H0H_0 assumes one shared proportion.
SE=0.30(0.70)(1200+1200)=0.0021≈0.0458SE = \sqrt{0.30(0.70)\left(\frac{1}{200} + \frac{1}{200}\right)} = \sqrt{0.0021} \approx 0.0458
Step 3: The sample proportions are p^A=48/200=0.24\hat{p}_A = 48/200 = 0.24 and p^B=72/200=0.36\hat{p}_B = 72/200 = 0.36. Divide their difference by the standard error.
z=0.24−0.360.0021≈−2.62z = \frac{0.24 - 0.36}{\sqrt{0.0021}} \approx -2.62
Answer: The pooled proportion is 0.30, the standard error is about 0.0458, and the test statistic is z≈−2.62z \approx -2.62.

Why It Matters

On the AP Statistics exam, two-proportion z-tests on free-response questions require the pooled proportion in the standard error. You also use it to check the Large Counts condition. Each of n1p^cn_1\hat{p}_c, n1(1−p^c)n_1(1-\hat{p}_c), n2p^cn_2\hat{p}_c, and n2(1−p^c)n_2(1-\hat{p}_c) must be at least 10.

Common Mistakes

Mistake: Using the pooled proportion to build a confidence interval for p1−p2p_1 - p_2.
Correction: Pooling is justified only when you assume p1=p2p_1 = p_2, which is what a significance test's null hypothesis does. A confidence interval makes no such assumption. It uses the unpooled standard error p^1(1−p^1)n1+p^2(1−p^2)n2\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}.

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