Difference of Two Proportions
Introduced| English |
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| difference of two sample proportions/ˈdɪfrəns ɒv tuː ˈsæmpl prəˈpɔːʃnz/ |
Comparing two groups
- Often we compare two proportions — say, support in group $1$ vs group $2$.
- The statistic is the difference of two sample proportions 两个样本比例之差 $\hat{p}_1 - \hat{p}_2$.
- Its sampling distribution is centered at the true difference: $\mu = p_1 - p_2$.
- It lets us judge whether two groups genuinely differ.
Combining the spreads
- The standard deviation combines both samples' variability:
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$$\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$$
- Note the $+$: variances add (echoing the "combining variables" rule) even for a difference.
- Each group contributes its own $\frac{p(1-p)}{n}$ term.
Check both samples
- Apply the large counts condition to each group: $n_1p_1, n_1(1-p_1), n_2p_2, n_2(1-p_2)$ all $\ge 10$.
- Require independence: the two samples are independent of each other.
- Also the 10% condition for each if sampling without replacement.
- Only then is the normal model for $\hat{p}_1 - \hat{p}_2$ valid.
Using the normal model
- With conditions met, $\hat{p}_1 - \hat{p}_2$ is approximately normal.
- Standardize with a $z$-score: (observed difference $-$ true difference) $\div$ the combined SD.
- Then read probabilities from the standard normal, as before.
- This is the foundation for two-proportion tests and intervals (Unit 6).
Variances add — even for a difference. The SD of $\hat{p}_1-\hat{p}_2$ combines the two $\frac{p(1-p)}{n}$ terms with a $+$, then a square root; you never subtract them. And check the large counts condition in both groups separately (four counts, all $\ge 10$) — one healthy group doesn't excuse a thin one.
$p_1 = 0.5$ ($n_1 = 100$), $p_2 = 0.4$ ($n_2 = 100$).
- Center: $p_1 - p_2 = 0.5 - 0.4 = 0.1$.
- SD: $\sqrt{\dfrac{0.5(0.5)}{100} + \dfrac{0.4(0.6)}{100}} = \sqrt{0.0025 + 0.0024} = \sqrt{0.0049} = 0.07$.
- Counts: all four ($50, 50, 40, 60$) exceed $10$ → normal is OK.
The difference of two sample proportions $\hat{p}_1 - \hat{p}_2$ has mean $p_1 - p_2$ and SD $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$ (variances add). Model it as normal when the large counts and independence conditions hold for both samples.
Distribution of p-hat-1 − p-hat-2
Centered at p1 − p2; the two variances add under the square root.
p1=0.5, p2=0.4. Find the center (mean) of p-hat-1 − p-hat-2.
Mean = p1 − p2 = 0.5 − 0.4 = 0.1.
With p1=0.5,n1=100 and p2=0.4,n2=100: SD = √(0.0025+0.0024). Round to two decimals.
√0.0049 = 0.07.
To get the SD of a difference of two proportions, you subtract the two variances.
Variances ADD, even for a difference, then square-root.
For a normal model of p-hat-1 − p-hat-2, the large counts condition must hold...
All four counts (both groups) must be ≥ 10.
The two samples must be independent of each other for this model.
Independence between samples is required.