单样本率等效性检验¶
\[
\begin{align}
H_{01} &: p - p_0 \leqslant \delta_1 \text{ 或 } H_{02}: p - p_0 \geqslant \delta_2 \\
H_1 \ &: \delta_1 < p - p_0 < \delta_2
\end{align}
\]
\(\delta_1\) 和 \(\delta_2\) 为等效性界值，\(\delta_1 < 0\)，\(\delta_2 > 0\)，样本比例用 \(\hat{p}\) 表示。
\[
\operatorname{E}(\hat{p}) = p
\]
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，使用 \(p_0\) 计算样本比例 \(\hat{p}\) 的方差：
\[
\operatorname{Var}(\hat{p}) = \frac{(p_0+\delta_1)(1-p_0-\delta_1)}{n}
\]
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 \(H_1\) 成立时，构建 \(z'_1\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N\left(\frac{p-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}\right)
\]
在 \(H_{02}\) 成立时，使用 \(p_0\) 计算样本比例 \(\hat{p}\) 的方差：
\[
\operatorname{Var}(\hat{p}) = \frac{(p_0+\delta_2)(1-p_0-\delta_2)}{n}
\]
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，构建 \(z'_2\) 统计量：
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N\left(\frac{p-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - 1 \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}\right)
+ \Phi\left(\frac{-z_{1-\alpha} - \frac{p-p_0-\delta_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}\right)
- 1 \\
& = \begin{aligned}[t]
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n} - (p-p_0-\delta_1)}{\sqrt{p(1-p)/n}}\right) + \\
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n} + (p-p_0-\delta_2)}{\sqrt{p(1-p)/n}}\right) - 1
\end{aligned} \\
& = \begin{aligned}[t]
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)} - (p-p_0-\delta_1)\sqrt{n}}{\sqrt{p(1-p)}}\right) \\
& - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)} + (p-p_0-\delta_2)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\end{align}
\]
\(\operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < -z_{1-\alpha}) = 1\) 的证明
略。
Z-test using S(P0) 连续性校正¶
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，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 \(H_1\) 成立时，构建 \(z'_1\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}} \sim N\left(\frac{p-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}\right)
\]
在 \(H_{02}\) 成立时，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，构建 \(z'_2\) 统计量：
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}} \sim N\left(\frac{p-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - 1 \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n}}}\right)
+ \Phi\left(\frac{-z_{1-\alpha} - \frac{p-p_0-\delta_2+c_2}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n}}}\right)
- 1 \\
& = \begin{aligned}[t]
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)/n} - (p-p_0-\delta_1+c_1)}{\sqrt{p(1-p)/n}}\right) + \\
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)/n} + (p-p_0-\delta_2+c_2)}{\sqrt{p(1-p)/n}}\right) - 1
\end{aligned} \\
& = \begin{aligned}[t]
1 & - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_1)(1-p_0-\delta_1)} - (p-p_0-\delta_1+c_1)\sqrt{n}}{\sqrt{p(1-p)}}\right) \\
& - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta_2)(1-p_0-\delta_2)} + (p-p_0-\delta_2+c_2)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\end{align}
\]
Z-test using S(Phat)¶
在 \(H_{01}\) 成立时，使用 \(p\) 计算样本比例 \(\hat{p}\) 的方差：
\[
\operatorname{Var}(\hat{p}) = \frac{p(1-p)}{n}
\]
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 \(H_1\) 成立时，构建 \(z'_1\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}, 1\right)
\]
在 \(H_{01}\) 成立时，使用 \(p\) 计算样本比例 \(\hat{p}\) 的方差：
\[
\operatorname{Var}(\hat{p}) = \frac{p(1-p)}{n}
\]
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，构建 \(z'_2\) 统计量：
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}, 1\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(-z_{1-\alpha} - \frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}\right)
- 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1}{\sqrt{p(1-p)/n}}\right) +
1 - \Phi\left(z_{1-\alpha} + \frac{p-p_0-\delta_2}{\sqrt{p(1-p)/n}}\right) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{(p-p_0-\delta_1)\sqrt{n}}{\sqrt{p(1-p)}}\right) - \Phi\left(z_{1-\alpha} + \frac{(p-p_0-\delta_2)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{align}
\]
Z-test using S(Phat) 连续性校正¶
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，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 \(H_1\) 成立时，构建 \(z'_1\) 统计量：
\[
z'_1 = \frac{\hat{p}-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}, 1\right)
\]
在 \(H_{01}\) 成立时，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，构建 \(z'_2\) 统计量：
\[
z'_2 = \frac{\hat{p}-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}, 1\right)
\]
计算检验效能：
\[
\begin{align}
\text{Power}
& = \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cap \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - \operatorname{Pr}(z'_1 > z_{1-\alpha} \ \cup \ z'_2 < -z_{1-\alpha}) \\
& = \operatorname{Pr}(z'_1 > z_{1-\alpha}) + \operatorname{Pr}(z'_2 < -z_{1-\alpha}) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(-z_{1-\alpha} - \frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}\right)
- 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta_1+c_1}{\sqrt{p(1-p)/n}}\right) +
1 - \Phi\left(z_{1-\alpha} + \frac{p-p_0-\delta_2+c_2}{\sqrt{p(1-p)/n}}\right) - 1 \\
& = 1 - \Phi\left(z_{1-\alpha} - \frac{(p-p_0-\delta_1+c_1)\sqrt{n}}{\sqrt{p(1-p)}}\right) - \Phi\left(z_{1-\alpha} + \frac{(p-p_0-\delta_2+c_2)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{align}
\]