两独立样本率非劣效检验¶
统计学假设：
\[
\begin{align}
H_0 &: p_1 - p_2 \le \delta \\
H_1 &: p_1 - p_2 \gt \delta
\end{align}
\]
其中，\(\delta\) 为非劣效界值，两样本率分别用 \(\hat{p}_1\) 和 \(\hat{p}_2\) 表示。
\[
E(\hat{p}_1 - \hat{p}_2) = p_1 - p_2
\]
\[
Var(\hat{p}_1 - \hat{p}_2) = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}
\]
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：
\[
Var(\hat{p}_1 - \hat{p}_2) = \frac{\bar{p}(1-\bar{p})}{n_1} + \frac{\bar{p}(1-\bar{p})}{n_2} = \bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)
\]
注意
这种做法实际上是站不住脚的，因为在 \(H_0\) 的边界条件下，两组总体率不相等，并不是来自同一个总体，强行合并两组率是不合适的，实际应用中建议使用 Unpooled 方法。
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，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
+
\frac{p_1 - p_2 - \delta}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 - \delta}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}, \
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\end{align}
\]
计算检验效能：
\[
\begin{align}
P\left(\left|z'\right| > z_{1-\alpha}\right) & = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta\right|}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - \left|p_1-p_2-\delta\right|}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} - \left|p_1-p_2-\delta\right|}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} \sqrt{\left(\frac{kp_1+p_2}{k+1}\right) \left(1-\frac{kp_1+p_2}{k+1}\right) \left(\frac{1}{k}+1\right)} + z_{1-\beta} \sqrt{\frac{1}{k}p_1(1-p_1) + p_2(1-p_2)}\right)^2}{(p_1-p_2-\delta)^2}
\]
\[
n_1 = k n_2
\]
Z-Test Pooled 连续性校正¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\cdot
\frac{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
+
\frac{p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}, \
\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\right)
\end{align}
\]
计算检验效能：
\[
\begin{align}
P\left(\left|z'\right| > z_{1-\alpha}\right)
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right|}
{\sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
{\sqrt{\frac{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}}
\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} -
\left|p_1-p_2-\delta-\frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right|}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
& = 1 - \beta
\end{align}
\]
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{\bar{p}(1-\bar{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} -
\left|p_1-p_2-\delta-\frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right|}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
可使用数值方法求解上述方程。
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：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
+
\frac{\left(p_1 - p_2 - \delta\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 - \delta}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1\right)
\end{align}
\]
计算检验效能：
\[
P\left(\left|z'\right| > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta\right|}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
= 1 - \beta
\]
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta\right|}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出：
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \left[ \frac{1}{k}p_1(1-p_1) + p_2(1-p_2) \right]}{(p_1-p_2-\delta)^2}
\]
\[
n_1 = k n_2
\]
Z-Test Unpooled 连续性校正¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}_1 - \hat{p}_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}_1 - \hat{p}_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
\]
根据中心极限定理，当 \(n_1\) 和 \(n_2\) 较大时，满足：
\[
\frac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \sim N(0,1)
\]
进而有：
\[
\begin{align}
z' & = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right) + \left(p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& = \frac{\left(\hat{p}_1 - \hat{p}_2\right) - \left(p_1 - p_2\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}
+
\frac{\left(p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} \\
& \xrightarrow{d}
N\left(\frac{p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}, \ 1
\right)
\end{align}
\]
计算检验效能：
\[
P\left(\left|z'\right| > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right|}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}}\right) \\
= 1 - \beta
\]
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} - \frac{\left|p_1 - p_2 - \delta - \frac{1}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right|}
{\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}} = z_\beta
\]
可使用数值方法求解上述方程。