两独立样本率差置信区间¶
两样本率分别用 \(\hat{p}_1\) 和 \(\hat{p}_2\) 表示，两组样本量分别用 \(n_1\) 和 \(n_2\) 表示。
Pearson's Chi-Square¶
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}
\end{align}
\]
置信区间宽度：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - \operatorname{max}\left(\text{L}, -1\right)
\]
样本量的闭式解的分类讨论
设 \(k = n_1 / n_2\)
\(\text{L} < -1, \text{U} \leqslant 1\)
\[
d = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} - (-1)
\]
可解出：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha/2}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{\left[d-(\hat{p}_1-\hat{p}_2)-1\right]^2} \\
n_1 & = k n_2
\end{align}
\]
\(\text{L} < -1, \text{U} > 1\)
\[
d = 1 - (-1) = 2
\]
\(d\) 与样本量无关，无法确定样本量。
\(\text{L} \geqslant -1, \text{U} \leqslant 1\)
\[
d = 2 \cdot z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}
\]
可解出：
\[
\begin{align}
n_2 & = \frac{4 \cdot z_{1-\alpha/2}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{d^2} \\
n_1 & = k n_2
\end{align}
\]
\(\text{L} \geqslant -1, \text{U} > 1\)
\[
d = 1 - \left( \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \right),
\]
可解出：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha/2}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{\left[d+(\hat{p}_1-\hat{p}_2)-1\right]^2} \\
n_1 & = k n_2
\end{align}
\]
\[
\begin{align}
\text{L} & = -1 \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - (\hat{p}_1-\hat{p}_2)
\]
样本量的闭式解的分类讨论
设 \(k = n_1 / n_2\)
\(\text{U} \leqslant 1\)
\[
d = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} - (\hat{p}_1 - \hat{p}_2)
\]
可解出：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{d^2} \\
n_1 & = k n_2
\end{align}
\]
\(\text{U} > 1\)
\[
d = 1 - (\hat{p}_1 - \hat{p}_2)
\]
\(d\) 与样本量无关，无法确定样本量。
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\
\text{U} & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - \operatorname{max}\left(\text{L}, -1\right)
\]
样本量的闭式解的分类讨论
设 \(k = n_1 / n_2\)
\(\text{L} < -1\)
\[
d = \hat{p_1} - \hat{p}_2 - (-1)
\]
\(d\) 与样本量无关，无法确定样本量。
\(\text{L} \geqslant -1\)
\[
d = (\hat{p}_1-\hat{p}_2) - \left( \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \right),
\]
可解出：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{d^2} \\
n_1 & = k n_2
\end{align}
\]
Yate's Chi-Square with Continuity Correction¶
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} - \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right) \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha/2} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} + \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\end{align}
\]
置信区间宽度：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - \operatorname{max}\left(\text{L}, -1\right)
\]
\[
\begin{align}
\text{L} & = -1 \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} + \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - (\hat{p}_1-\hat{p}_2)
\]
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} - \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right) \\
\text{U} & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - \operatorname{max}\left(\text{L}, -1\right)
\]