两独立样本率优效性检验¶
两样本率分别用 \(\hat{p}_1\) 和 \(\hat{p}_2\) 表示，优效界值用 \(\delta\) 表示。
对于高优指标（\(\delta > 0\)），统计学假设如下：
\[
\begin{align}
H_0 &: p_1 - p_2 \leqslant \delta \\
H_1 &: p_1 - p_2 \gt \delta
\end{align}
\]
对于低优指标（\(\delta < 0\)），统计学假设如下：
\[
\begin{align}
H_0 &: p_1 - p_2 \geqslant \delta \\
H_1 &: p_1 - p_2 \lt \delta
\end{align}
\]
以下推导过程在边界条件 \(p_1 - p_2 = \delta\) 下进行。
Z-Test Pooled¶
假设 \(\bar{p}\) 表示合并总体率，则：
\[
\bar{p} = \frac{n_1 \hat{p}_1 + n_2 \hat{p}_2}{n_1 + n_2}
\]
在 \(H_0\) 成立时，两样本的方差可以用 \(\bar{p}\) 来表示：
\[
\operatorname{Var}(\hat{p}_1 - \hat{p}_2) = \frac{\bar{p}(1-\bar{p})}{n_1} + \frac{\bar{p}(1-\bar{p})}{n_2} = \bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\]
Warning
事实上，\(H_0\) 成立时，其边界条件 \(p_1 = p_2 + \delta\)，两组总体率不相等，强行将两组率进行合并是不合适的，实际应用中建议使用 Z-test Unpooled。
可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \sim N(0,1)
\]
在 \(H_1\) 成立时，两样本率差的方差如下：
\[
\operatorname{Var}(\hat{p}_1 - \hat{p}_2) = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}
\]
可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\sim N\left(\frac{p_1 - p_2 - \delta}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}, \
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\]
\[
\begin{align}
\text{Power} & = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p_1 - p_2 - \delta}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2-\delta)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right) \\
\end{align}
\]
\[
\begin{align}
\text{Power} & = P\left(z' < z_{\alpha}\right) \\
& = \Phi\left(\frac{z_{\alpha} - \frac{p_1 - p_2 - \delta}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = \Phi\left(\frac{z_{\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2-\delta)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right)
\end{align}
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \pm (p_1-p_2-\delta)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出
\[
n_2 = \frac{\left(z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p})(1/k+1)} + z_{1-\beta} \sqrt{p_1(1-p_1)/k + p_2(1-p_2)}\right)^2}{(p_1-p_2-\delta)^2}
\]
\[
n_1 = k n_2
\]
Z-Test Pooled with Continuity Correction¶
在 Z-Test Pooled 的基础上，添加校正项。
定义：
\[
c = \frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)
\]
校正项的符号由检验方向决定，左侧检验时，校正项为 \(+c\)，右侧检验时，校正项为 \(-c\)。
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta \pm c}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta \pm c}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\sim N\left(\frac{p_1 - p_2 - \delta \pm c}{\sqrt{\bar{p}(1-\bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}},
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\]
\[
\begin{align}
\text{Power} = P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2-\delta-c)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right)
\end{align}
\]
\[
\begin{align}
\text{Power} = P\left(z' < z_{\alpha}\right)
= \Phi\left(\frac{z_{\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - (p_1-p_2-\delta+c)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\right)
\end{align}
\]
Z-Test Unpooled¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N\left(\frac{p_1 - p_2 - \delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{p_1-p_2-\delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha}\right)
= \Phi\left(z_{\alpha} - \frac{p_1-p_2-\delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \right)
\]
样本量公式推导
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} \pm \frac{(p_1 - p_2 - \delta)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \left[ p_1(1-p_1)/k + p_2(1-p_2) \right]}{(p_1-p_2-\delta)^2}
\]
\[
n_1 = k n_2
\]
Z-Test Unpooled with Continuity Correction¶
在 Z-Test Unpooled 的基础上，添加校正项。
定义：
\[
c = \frac{1}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)
\]
校正项的符号由检验方向决定，左侧检验时，校正项为 \(+c\)，右侧检验时，校正项为 \(-c\)。
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta \pm c}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta \pm c}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N\left(\frac{p_1 - p_2 - \delta + c}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{p_1-p_2-\delta-c}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha}\right)
= \Phi\left(z_{\alpha} - \frac{p_1-p_2-\delta+c}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \right)
\]