Carrying Out a Two-Proportion Test
The test statistic
- With the pooled SE ready, compute the $z$ test statistic:
-
$$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$$
- The "$-0$" is the null value: $H_0$ says $p_1 - p_2 = 0$.
- $z$ measures how many standard errors the observed difference sits from $0$.
Find the p-value
- Convert $z$ to a p-value using the standard normal, as in 6.5.
- One-sided $H_a$: the tail area in that direction.
- Two-sided $H_a$: double the one-tail area.
- The p-value is the chance of a difference this extreme if the proportions were truly equal.
Make a decision
- Compare the p-value to $\alpha$:
- p $\le \alpha$ → reject $H_0$: convincing evidence the proportions differ.
- p $> \alpha$ → fail to reject $H_0$: not convincing evidence.
- Same decision rule as any test.
Conclude in context
- Write the conclusion about the two population proportions, in context.
- "There is convincing evidence that the two groups' proportions differ" (if rejected).
- Or "there is not convincing evidence of a difference" (if not).
- Name the groups and what the proportion measures.
Keep the test's pooled SE — don't switch to the interval's un-pooled SE mid-problem. The $z$ here divides by $\sqrt{\hat{p}_c(1-\hat{p}_c)(\frac1{n_1}+\frac1{n_2})}$. And match the p-value to $H_a$: one tail for a one-sided alternative, doubled for two-sided. A test and its matching interval should agree — mixing their SEs breaks that.
From 6.10: $\hat{p}_1 = 0.30$, $\hat{p}_2 = 0.20$, pooled SE $\approx 0.061$, two-sided.
- $z = \dfrac{0.30 - 0.20}{0.061} \approx 1.64$.
- p-value (two-sided): $2 \times P(Z > 1.64) \approx 2(0.05) = 0.10$.
- Decide at $\alpha = 0.05$: $0.10 > 0.05$ → fail to reject; not convincing evidence of a difference.
Compute $z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1 + 1/n_2)}}$, find its p-value (doubled for two-sided), and compare to $\alpha$: reject if p $\le \alpha$, else fail to reject. State the conclusion about the two population proportions in context.
Two-tailed p-value
For a two-sided test, double the tail area beyond z.
p̂1 = 0.30, p̂2 = 0.20, pooled SE ≈ 0.061. Compute z = (p̂1−p̂2)/SE. Two decimals.
0.10 / 0.061 ≈ 1.64.
With a two-sided p-value of 0.10 and α = 0.05, the decision is...
0.10 > 0.05, so fail to reject.
The test statistic here divides by the pooled standard error, not the un-pooled one.
The two-proportion test uses the pooled SE from H0.
For a two-sided alternative, the p-value is...
Both directions are extreme, so double.
A conclusion should be stated in context about the two population proportions.
Always tie the conclusion back to the real groups.