单样本率等效性检验¶
样本率用 \(\hat{p}\) 表示，总体率用 \(p\) 表示，下等效界值用 \(\delta_1\) 表示，上等效界值用 \(\delta_2\) 表示，\(\delta_1 < 0\)，\(\delta_2 > 0\)。
\[
\begin{align}
H_{01} &: p - p_0 \leqslant \delta_1 \;\text{或}\; H_{02}: p - p_0 \geqslant \delta_2 \\
H_1 \ &: \delta_1 < p - p_0 < \delta_2
\end{align}
\]
以下推导过程在边界条件 \(p - p_0 = \delta_1\) 和 \(p - p_0 = \delta_2\) 下进行。
Z-Test Using S(P0)¶
在 \(H_{01}\) 成立时，可构建 \(z_1\) 统计量：
\[
z_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N(0, 1)
\]
在 \(H_{02}\) 成立时，可构建 \(z_2\) 统计量：
\[
z_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'_1\) 和 \(z'_2\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N\left(\frac{p-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}, \frac{p(1-p)}{(p_0+\delta_1)(1-p_0-\delta_1)}\right)
\]
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N\left(\frac{p-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}, \frac{p(1-p)}{(p_0+\delta_2)(1-p_0-\delta_2)}\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = P(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < z_{\alpha}) \\
& = P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - P(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < z_{\alpha}) \\
& \approx P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - 1 \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}{\sqrt{\frac{p(1-p)}{(p_0+\delta_1)(1-p_0-\delta_1)}}}\right)
+ \Phi\left(\frac{z_{\alpha} - \frac{p-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}{\sqrt{\frac{p(1-p)}{(p_0+\delta_2)(1-p_0-\delta_2)}}}\right) - 1 \\
& = \begin{aligned}[t]
& - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n} - (p-p_0-\delta_1)}{\sqrt{p(1-p)/n}}\right) \\
& + \Phi\left(\frac{z_{\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n} - (p-p_0-\delta_2)}{\sqrt{p(1-p)/n}}\right)
\end{aligned}
\end{align}
\]
Z-Test Using S(P0) with Continuity Correction¶
在 Z-Test Using S(P0) 的基础上加入校正项 \(c_1\) 和 \(c_2\)：
\[
c_1 =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta_1 \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta_1 \\
0 & , \text{if } \left| p - p_0 - \delta_1 \right| \lt \frac{1}{2n}
\end{cases}
\]
\[
c_2 =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta_2 \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta_2 \\
0 & , \text{if } \left| p - p_0 - \delta_2 \right| \lt \frac{1}{2n}
\end{cases}
\]
在 \(H_{01}\) 成立时，可构建 \(z_1\) 统计量：
\[
z_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N(0, 1)
\]
在 \(H_{02}\) 成立时，可构建 \(z_2\) 统计量：
\[
z_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'_1\) 和 \(z'_2\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N\left(\frac{p-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}, \frac{p(1-p)}{(p_0+\delta_1)(1-p_0-\delta_1)}\right)
\]
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N\left(\frac{p-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}, \frac{p(1-p)}{(p_0+\delta_2)(1-p_0-\delta_2)}\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = P(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < z_{\alpha}) \\
& = P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - P(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < z_{\alpha}) \\
& \approx P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - 1 \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}{\sqrt{\frac{p(1-p)}{(p_0+\delta_1)(1-p_0-\delta_1)}}}\right)
+ \Phi\left(\frac{z_{\alpha} - \frac{p-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}{\sqrt{\frac{p(1-p)}{(p_0+\delta_2)(1-p_0-\delta_2)}}}\right) - 1 \\
& = \begin{aligned}[t]
& - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n} - (p-p_0-\delta_1+c_1)}{\sqrt{p(1-p)/n}}\right) \\
& + \Phi\left(\frac{z_{\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n} - (p-p_0-\delta_2+c_2)}{\sqrt{p(1-p)/n}}\right)
\end{aligned}
\end{align}
\]
Z-Test Using S(Phat)¶
在 \(H_{01}\) 成立时，可构建 \(z_1\) 统计量：
\[
z_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_{02}\) 成立时，可构建 \(z_2\) 统计量：
\[
z_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'_1\) 和 \(z'_2\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}, 1\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = P(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < z_{\alpha}) \\
& = P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - P(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < z_{\alpha}) \\
& \approx P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(z_{\alpha} - \frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}\right) - 1 \\
& = - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}\right) + \Phi\left(z_{\alpha} - \frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}\right)
\end{align}
\]
Z-Test Using S(Phat) with Continuity Correction¶
在 Z-Test Using S(Phat) 的基础上加入校正项 \(c\)：
\[
c_1 =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta_1 \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta_1 \\
0 & , \text{if } \left| p - p_0 - \delta_1 \right| \lt \frac{1}{2n}
\end{cases}
\]
\[
c_2 =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta_2 \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta_2 \\
0 & , \text{if } \left| p - p_0 - \delta_2 \right| \lt \frac{1}{2n}
\end{cases}
\]
在 \(H_{01}\) 成立时，可构建 \(z_1\) 统计量：
\[
z_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_{02}\) 成立时，可构建 \(z_2\) 统计量：
\[
z_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'_1\) 和 \(z'_2\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}, 1\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = P(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < z_{\alpha}) \\
& = P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - P(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < z_{\alpha}) \\
& \approx P(z'_1 > z_{1-\alpha}) + P(z'_2 < z_{\alpha}) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(z_{\alpha} - \frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}\right) - 1 \\
& = - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}\right) + \Phi\left(z_{\alpha} - \frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}\right)
\end{align}
\]
参考文献
- Chow S C, Shao J, Wang H, et al. Sample size calculations in clinical research[M]. chapman and hall/CRC, 2017.