单样本率优效性检验¶
对于高优指标（\(\delta > 0\)），统计学假设如下：
\[
\begin{align}
H_0 &: p - p_0 \leqslant \delta \\
H_1 &: p - p_0 \gt \delta
\end{align}
\]
对于低优指标（\(\delta < 0\)），统计学假设如下：
\[
\begin{align}
H_0 &: p - p_0 \geqslant \delta \\
H_1 &: p - p_0 \lt \delta
\end{align}
\]
\(\delta\) 为优效界值，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 \(p - p_0 = \delta\) 下进行。
Z-test using S(P0)¶
在 \(H_0\) 成立时，使用 \(p_0\) 计算样本比例 \(\hat{p}\) 的方差：
\[
\operatorname{Var}(\hat{p}) = \frac{(p_0+\delta)(1-p_0-\delta)}{n}
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0-\delta}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0-\delta}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}} \sim N\left(\frac{p-p_0-\delta}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}\right)
\]
\[
\begin{aligned}
\text{Power}
& = \operatorname{Pr}(z' > z_{1-\alpha}) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)/n} - (p-p_0-\delta)}{\sqrt{p(1-p)/n}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} - (p-p_0-\delta)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
& = \operatorname{Pr}(z' < z_{\alpha}) \\
& = \Phi\left(\frac{z_{\alpha} - \frac{p-p_0-\delta}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)/n} + (p-p_0-\delta)}{\sqrt{p(1-p)/n}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} + (p-p_0-\delta)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\]
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} \pm (p-p_0-\delta)\sqrt{n}}{\sqrt{p(1-p)}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left[z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} + z_{1-\beta}\sqrt{p(1-p)}\right]^2}{\left(p-p_0-\delta\right)^2}
\]
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 \(c\)：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta \\
0 & , \text{if } \left| p - p_0 - \delta \right| \lt \frac{1}{2n}
\end{cases}
\]
在 \(H_0\) 成立时，构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0-\delta+c}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0-\delta+c}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}} \sim N\left(\frac{p-p_0-\delta+c}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}, \frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}\right)
\]
\[
\begin{aligned}
\text{Power}
& = \operatorname{Pr}(z' > z_{1-\alpha}) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0-\delta+c}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)/n} - (p-p_0-\delta+c)}{\sqrt{p(1-p)/n}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} - (p-p_0-\delta+c)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
& = \operatorname{Pr}(z' < z_{\alpha}) \\
& = \Phi\left(\frac{z_{\alpha} - \frac{p-p_0-\delta+c}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}{\frac{\sqrt{p(1-p)/n}}{\sqrt{(p_0+\delta)(1-p_0-\delta)/n}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)/n} + (p-p_0-\delta+c)}{\sqrt{p(1-p)/n}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{(p_0+\delta)(1-p_0-\delta)} + (p-p_0-\delta+c)\sqrt{n}}{\sqrt{p(1-p)}}\right)
\end{aligned}
\]
Z-test using S(Phat)¶
在 \(H_0\) 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 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0-\delta}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0-\delta}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
\text{Power}
= \operatorname{Pr}(z' > z_{1-\alpha})
= 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power}
= \operatorname{Pr}(z' < z_{\alpha})
= \Phi\left(z_{\alpha} - \frac{p-p_0-\delta}{\sqrt{p(1-p)/n}}\right)
= 1 - \Phi\left(z_{1-\alpha} + \frac{p-p_0-\delta}{\sqrt{p(1-p)/n}}\right)
\]
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} \pm \frac{p-p_0-\delta}{\sqrt{p(1-p)/n}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left(z_{1-\alpha}+z_{1-\beta}\right)^2 p(1-p)}{\left(p-p_0-\delta\right)^2}
\]
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 \(c\)：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } p \gt p_0 + \delta \\
\frac{1}{2n} & , \text{if } p \lt p_0 + \delta \\
0 & , \text{if } \left| p - p_0 - \delta \right| \lt \frac{1}{2n}
\end{cases}
\]
在 \(H_0\) 成立时，构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0-\delta+c}{\sqrt{p(1-p)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0-\delta+c}{\sqrt{p(1-p)/n}} \sim N\left(\frac{p-p_0-\delta+c}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
\text{Power}
= \operatorname{Pr}(z' > z_{1-\alpha})
= 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-\delta+c}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power}
= \operatorname{Pr}(z' < z_{\alpha})
= \Phi\left(z_{\alpha} - \frac{p-p_0-\delta+c}{\sqrt{p(1-p)/n}}\right)
= 1 - \Phi\left(z_{1-\alpha} + \frac{p-p_0-\delta+c}{\sqrt{p(1-p)/n}}\right)
\]