Where:

• $\hat{p}$ = sample proportion
• $p_0$ = hypothesized population proportion
• $n$ = sample size

Conditions (must check all):

1. Random sample: Data collected randomly
2. Independence: Observations are independent (or $n \leq 0.1N$)
3. Large counts: Both $np_0 \geq 10$ and $n(1-p_0) \geq 10$

Worked Example: A company claims 80% of customers are satisfied. In a random sample of 150 customers, 115 were satisfied. Test at $\alpha = 0.05$.

• $\hat{p} = 115/150 = 0.767$
• $z = \frac{0.767 - 0.80}{\sqrt{\frac{0.80(0.20)}{150}}} = \frac{-0.033}{0.0327} = -1.01$
• From z-table: p-value = 0.312 (two-tailed)
• Conclusion: Fail to reject $H_0$ (insufficient evidence)

Two-Sample z-Test for Difference of Proportions

When to use: Comparing two populations; testing $p_1 - p_2 = 0$

Test statistic: $z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

Where:

• $\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2}$ (pooled proportion)

Conditions:

1. Random samples from both populations
2. Independence: Both samples independent; each $\leq 10\%$ of population
3. Large counts: $n_1\hat{p}_c \geq 10$, $n_1(1-\hat{p}_c) \geq 10$, etc.

Common Mistakes

❌ Using $\hat{p}$ instead of $p_0$ in SE for one-sample test ❌ Forgetting to pool proportions in two-sample test ❌ Using t-distribution for proportions (always use z) ❌ Not checking conditions before testing

Decision Rule

• If $|z| > z_{\alpha/2}$, reject $H_0$
• If p-value $< \alpha$, reject $H_0$

AP Exam Tip

State conditions first. Graders award partial credit for checking them. Name the test: "z-test for a proportion" or "two-sample z-test for difference of proportions."