4.2.2 Confidence Interval for Difference in Proportion

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Proof: Confidence interval for difference in proportions

Let

• $X\sim \text{Binomial}(n_1,p_1)$ : Binomial distribution with parameters $n_1$ and $p_1$
• $Y\sim \text{Binomial}(n_2,p_2)$ : Binomial distribution with parameters $n_2$ and $p_2$
• $\Delta =p_1-p_2$ : Difference in proportions

Assume $X$ and $Y$ are independent.

Let

• ${\hat{p}}_1=\frac{X}{n_1}$ : Sample proportion for $X$
• ${\hat{p}}_2=\frac{Y}{n_2}$ : Sample proportion for $Y$
• $\hat{\Delta }={\hat{p}}_1-{\hat{p}}_2$ : Sample difference in proportions

Then $E({\hat{p}}_1)=p_1$ and $E({\hat{p}}_2)=p_2$, so $E(\hat{\Delta })=E({\hat{p}}_1-{\hat{p}}_2)=E({\hat{p}}_1)-E({\hat{p}}_2)=p_1-p_2=\Delta$

Thus $\hat{\Delta }$ is an unbiased estimator of $\Delta$.

Additionally, $\mathrm{Var}({\hat{p}}_1)=\frac{p_1(1-p_1)}{n_1}$ and $\mathrm{Var}({\hat{p}}_2)=\frac{p_2(1-p_2)}{n_2}$

By independence, $\mathrm{Var}(\hat{\Delta })=\mathrm{Var}({\hat{p}}_1-{\hat{p}}_2)=\mathrm{Var}({\hat{p}}_1)+\mathrm{Var}({\hat{p}}_2)=\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}$

For sufficiently large $n_1$ and $n_2$, by the Central Limit Theorem, ${\hat{p}}_1$ and ${\hat{p}}_2$ are approximately normally distributed: ${\hat{p}}_1\sim N\left(p_1,\frac{p_1(1-p_1)}{n_1}\right)\text{and}{\hat{p}}_2\sim N\left(p_2,\frac{p_2(1-p_2)}{n_2}\right)$

Therefore, $\hat{\Delta }={\hat{p}}_1-{\hat{p}}_2\sim N\left(p_1-p_2,\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}\right)$

By standardization:

$$\begin{array}{rl}\frac{\hat{\Delta }-\Delta }{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}& \sim N(0,1)\\ \frac{({\hat{p}}_1-{\hat{p}}_2)-(p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}& \sim N(0,1)\end{array}$$

Since $p_1$ and $p_2$ are unknown, we estimate them using ${\hat{p}}_1$ and ${\hat{p}}_2$: $\frac{({\hat{p}}_1-{\hat{p}}_2)-(p_1-p_2)}{\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}}\approx N(0,1)$

Thus the approximate $(1-\alpha )100\%$ confidence interval for $\Delta$ is

$$\begin{array}{rl}1-\alpha & =P\left(-z_{\alpha /2}<\frac{\hat{\Delta }-\Delta }{\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}}<z_{\alpha /2}\right)\\ & =P\left(-z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}<\hat{\Delta }-\Delta <z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\right)\\ & =P\left(\hat{\Delta }-z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}<\Delta <\hat{\Delta }+z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\right)\\ & =P\left(({\hat{p}}_1-{\hat{p}}_2)-z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}<p_1-p_2<({\hat{p}}_1-{\hat{p}}_2)+z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\right)\end{array}$$

From the last result, we can see that the following interval is an approximate $(1-\alpha )100\%$ confidence interval for $\Delta =p_1-p_2$

$$\left(({\hat{p}}_1-{\hat{p}}_2)-z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}},({\hat{p}}_1-{\hat{p}}_2)+z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}\right)$$