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：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha/2}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{\left[d+(\hat{p}_1-\hat{p}_2)-1\right]^2} \\
n_1 & = k n_2
\end{align}
\]
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\
\text{U} & = 1
\end{align}
\]
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：
\[
\begin{align}
n_2 & = \frac{z_{1-\alpha}^2 \left[\hat{p}_1(1-\hat{p}_1)/k + \hat{p}_2(1-\hat{p}_2)\right]}{d^2} \\
n_1 & = k n_2
\end{align}
\]
\[
\begin{align}
\text{L} & = -1 \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}
\end{align}
\]
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。
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：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - \operatorname{max}\left(\text{L}, -1\right)
\]
\[
\begin{align}
\text{L} & = \hat{p}_1 - \hat{p}_2 - z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} - \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right) \\
\text{U} & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - \operatorname{max}\left(\text{L}, -1\right)
\]
\[
\begin{align}
\text{L} & = -1 \\
\text{U} & = \hat{p}_1 - \hat{p}_2 + z_{1 - \alpha} \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} + \frac{1}{2} \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = \operatorname{min}\left(\text{U}, 1\right) - (\hat{p}_1-\hat{p}_2)
\]
Newcombe-Wilson¶
先使用 Wilson Score 方法计算两组各自的 Wilson 区间，再代入 Newcombe 混合误差框架构建率差的置信区间。
设 \(\hat{p}_1\) 的 Wilson 区间为 \((L_1, U_1)\)，\(\hat{p}_2\) 的 Wilson 区间为 \((L_2, U_2)\)。
\[
\begin{align}
L & = \hat{p}_1 - \hat{p}_2 - \sqrt{\left(\hat{p}_1 - L_1\right)^2 + \left(U_2 - \hat{p}_2\right)^2} \\
U & = \hat{p}_1 - \hat{p}_2 + \sqrt{\left(U_1 - \hat{p}_1\right)^2 + \left(\hat{p}_2 - L_2\right)^2}
\end{align}
\]
置信区间宽度：
\[
d = U - L
\]
\[
\begin{align}
L & = \hat{p}_1 - \hat{p}_2 - \sqrt{\left(\hat{p}_1 - L_1\right)^2 + \left(U_2 - \hat{p}_2\right)^2} \\
U & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - L
\]
\[
\begin{align}
L & = -1 \\
U & = \hat{p}_1 - \hat{p}_2 + \sqrt{\left(U_1 - \hat{p}_1\right)^2 + \left(\hat{p}_2 - L_2\right)^2}
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = U - (\hat{p}_1-\hat{p}_2)
\]
Newcombe-Wilson with Continuity Correction¶
先使用 Wilson Score 连续性校正 方法计算两组各自的 Wilson 区间，再代入 Newcombe 混合误差框架构建率差的置信区间。
设 \(\hat{p}_1\) 的 Wilson 区间为 \((L_1, U_1)\)，\(\hat{p}_2\) 的 Wilson 区间为 \((L_2, U_2)\)。
Wilson Score 连续性校正置信区间溢出问题
使用 Wilson Score 连续性校正方法计算单组率置信区间时，若 \(L_i < 0\) 则截断为 \(0\)，若 \(U_i > 1\) 则截断为 \(1\)。
\[
\begin{align}
L & = \hat{p}_1 - \hat{p}_2 - \sqrt{\left(\hat{p}_1 - L_1\right)^2 + \left(U_2 - \hat{p}_2\right)^2} \\
U & = \hat{p}_1 - \hat{p}_2 + \sqrt{\left(U_1 - \hat{p}_1\right)^2 + \left(\hat{p}_2 - L_2\right)^2}
\end{align}
\]
置信区间宽度：
\[
d = U - L
\]
\[
\begin{align}
L & = \hat{p}_1 - \hat{p}_2 - \sqrt{\left(\hat{p}_1 - L_1\right)^2 + \left(U_2 - \hat{p}_2\right)^2} \\
U & = 1
\end{align}
\]
从样本率差到置信区间下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - L
\]
\[
\begin{align}
L & = -1 \\
U & = \hat{p}_1 - \hat{p}_2 + \sqrt{\left(U_1 - \hat{p}_1\right)^2 + \left(\hat{p}_2 - L_2\right)^2}
\end{align}
\]
从样本率差到置信区间上限的距离：
\[
d = U - (\hat{p}_1-\hat{p}_2)
\]
Farrington and Manning's Score¶
基于得分检验，构建 FMD 统计量：
\[
z_{FMD} = \frac{\hat{p}_1 - \hat{p}_2 - \delta_0}{\sqrt{\frac{\tilde{p}_1(1-\tilde{p}_1)}{n_1} + \frac{\tilde{p}_2(1-\tilde{p}_2)}{n_2}}} \sim N(0, 1)
\]
通过反转该检验得到 \(\delta_0\) 的范围，即为 \(\hat{p}_1 - \hat{p}_2\) 的置信区间。
其中：
\[
\begin{align}
& \tilde{p}_1 = \tilde{p}_2 + \delta_0 \\
& \tilde{p}_2 = 2B\cos(A) - \frac{L_2}{3L_3} \\
& A = \frac{1}{3} \left[\pi + \arccos\left(\frac{C}{B^3}\right)\right] \\
& B = \operatorname{sign}(C) \sqrt{\frac{L_2^2}{9L_3^2} - \frac{L_1}{3L_3}} \\
& C = \frac{L_2^3}{27L_3^3} - \frac{L_1L_2}{6L_3^2} + \frac{L_0}{2L_3} \\
& L_0 = x_{21} \delta_0(1-\delta_0) \\
& L_1 = \left[n_2\delta_0 - N - 2 x_{21}\right] \delta_0 + m_1 \\
& L_2 = \left(N + n_2\right)\delta_0 - N - m_1 \\
& L_3 = N \\
& m_1 = x_{11} + x_{21} \\
& N = n_1 + n_2 \\
& x_{11} = n_1\hat{p}_1 \\
& x_{21} = n_2\hat{p}_2
\end{align}
\]
\[
\begin{align}
z_{FMD}(L) & = z_{1 - \alpha/2} \\
z_{FMD}(U) & = z_{\alpha/2}
\end{align}
\]
置信区间宽度：
\[
d = U - L
\]
\[
\begin{align}
z_{FMD}(L) & = z_{1 - \alpha} \\
U & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - L
\]
\[
\begin{align}
L & = -1 \\
z_{FMD}(U) & = z_{\alpha}
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = U - (\hat{p}_1-\hat{p}_2)
\]
Miettinen and Nurminen's Score¶
在 Farrington and Maning's Score 的基础上增加校正因子 \(N/(N-1)\)，降低方差估计的偏倚。
基于得分检验，构建 MND 统计量：
\[
z_{MND} = \frac{\hat{p}_1 - \hat{p}_2 - \delta_0}{\sqrt{\left(\frac{\tilde{p}_1(1-\tilde{p}_1)}{n_1} + \frac{\tilde{p}_2(1-\tilde{p}_2)}{n_2}\right)\left(\frac{N}{N-1}\right)}} \sim N(0, 1)
\]
通过反转该检验得到 \(\delta_0\) 的范围，即为 \(\hat{p}_1 - \hat{p}_2\) 的置信区间。
其中：
\[
\begin{align}
& \tilde{p}_1 = \tilde{p}_2 + \delta_0 \\
& \tilde{p}_2 = 2B\cos(A) - \frac{L_2}{3L_3} \\
& A = \frac{1}{3} \left[\pi + \arccos\left(\frac{C}{B^3}\right)\right] \\
& B = \operatorname{sign}(C) \sqrt{\frac{L_2^2}{9L_3^2} - \frac{L_1}{3L_3}} \\
& C = \frac{L_2^3}{27L_3^3} - \frac{L_1L_2}{6L_3^2} + \frac{L_0}{2L_3} \\
& L_0 = x_{21} \delta_0(1-\delta_0) \\
& L_1 = \left[n_2\delta_0 - N - 2 x_{21}\right] \delta_0 + m_1 \\
& L_2 = \left(N + n_2\right)\delta_0 - N - m_1 \\
& L_3 = N \\
& m_1 = x_{11} + x_{21} \\
& N = n_1 + n_2 \\
& x_{11} = n_1\hat{p}_1 \\
& x_{21} = n_2\hat{p}_2
\end{align}
\]
\[
\begin{align}
z_{MND}(L) & = z_{1 - \alpha/2} \\
z_{MND}(U) & = z_{\alpha/2}
\end{align}
\]
置信区间宽度：
\[
d = U - L
\]
\[
\begin{align}
z_{MND}(L) & = z_{1 - \alpha} \\
U & = 1
\end{align}
\]
从样本率差到置信下限的距离：
\[
d = (\hat{p}_1-\hat{p}_2) - L
\]
\[
\begin{align}
L & = -1 \\
z_{MND}(U) & = z_{\alpha}
\end{align}
\]
从样本率差到置信上限的距离：
\[
d = U - (\hat{p}_1-\hat{p}_2)
\]