Interval for a Difference of Proportions
| English | Chinese | Pinyin |
|---|---|---|
| two-sample z-interval | 双样本z区间 | shuāng yàng běn z qū jiān |
Conditions for two samples
- To estimate $p_1 - p_2$, check the conditions for both samples:
- Random: both come from random samples or a randomized experiment.
- 10%: each $n \le 0.10N$ if sampling without replacement.
- Large counts: all four of $n_1\hat{p}_1,\ n_1(1-\hat{p}_1),\ n_2\hat{p}_2,\ n_2(1-\hat{p}_2) \ge 10$.
Estimate and standard error
- The point estimate is the observed difference $\hat{p}_1 - \hat{p}_2$.
- The standard error combines both samples' variability:
-
$$SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$$
- Each sample uses its own $\hat{p}$ here (this is an interval, not a test).
Build the interval
- The two-sample z-interval 双样本z区间 is estimate $\pm$ margin of error:
-
$$(\hat{p}_1 - \hat{p}_2) \pm z^{*}\,SE$$
- $z^{*}$ is the critical value for your confidence level ($1.96$ for $95\%$).
- The result is a range of plausible values for the difference $p_1 - p_2$.
Interpret in context
- "We're $95\%$ confident the true difference $p_1 - p_2$ is between the endpoints."
- Name both groups and what the proportion measures.
- The margin of error is $z^{*}\,SE$, as before.
- The sign of the difference tells you which group is estimated higher.
For a two-proportion interval, use each sample's own $\hat{p}$ in the SE (there's no pooling — that's only for the test in 6.10). And check the large counts condition with four counts, all $\ge 10$; one large sample doesn't cover for a small one. Interpret the interval as being about the difference $p_1 - p_2$, not either proportion alone.
Group $1$: $\hat{p}_1 = 0.60$, $n_1 = 100$. Group $2$: $\hat{p}_2 = 0.50$, $n_2 = 100$. Build a $95\%$ interval.
- Estimate: $0.60 - 0.50 = 0.10$.
- SE: $\sqrt{\dfrac{0.6(0.4)}{100} + \dfrac{0.5(0.5)}{100}} = \sqrt{0.0024 + 0.0025} = \sqrt{0.0049} = 0.07$.
- Interval: $0.10 \pm 1.96(0.07) = 0.10 \pm 0.137 = (-0.037,\ 0.237)$.
After checking random, 10%, and large counts (four counts) for both samples, a two-sample $z$-interval for $p_1 - p_2$ is $(\hat{p}_1 - \hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$. Interpret it as a range of plausible values for the difference.
Interval for p1 − p2
Estimate ± z*·SE, centered at the observed difference.
p̂1 = 0.60, p̂2 = 0.50. Find the point estimate p̂1 − p̂2.
0.60 − 0.50 = 0.10.
With both n = 100: SE = √(0.0024 + 0.0025). Round to two decimals.
√0.0049 = 0.07.
For a two-proportion confidence interval, each sample uses its own p̂ in the standard error.
No pooling for the interval — that's only for the test.
The large counts condition for a difference of proportions requires...
Both groups' successes and failures must each be ≥ 10.
The two-proportion interval estimates...
It's a range of plausible values for the difference.