单样本率差异性检验¶
样本率用 \(\hat{p}\) 表示，总体率用 \(p\) 表示。
对于双侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : p = p_0 \\
H_1 & : p \neq p_0
\end{align}
\]
对于左单侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : p \geqslant p_0 \\
H_1 & : p \lt p_0
\end{align}
\]
对于右单侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : p \leqslant p_0 \\
H_1 & : p \gt p_0
\end{align}
\]
以下推导过程在边界条件 \(p = p_0\) 下进行。
z-test using s(p0)¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}} \sim N\left(\frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}, \frac{p(1-p)}{p_0(1-p_0)}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha/2} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
+ \Phi\left(\frac{z_{\alpha/2} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(\frac{z_{\alpha} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
单侧检验样本量公式推导
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \pm \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}} = \frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} \pm \sqrt{n}(p-p_0)}{\sqrt{p(1-p)}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left(z_{1-\alpha}\sqrt{p_0(1-p_0)} + z_{1-\beta}\sqrt{p(1-p)}\right)^2}{\left(p-p_0\right)^2}
\]
z-test using s(p0) with continuity correction¶
在 z-test using s(p0) 的基础上加入校正项 \(c\)：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } \hat{p} \gt p_0 \\
\frac{1}{2n} & , \text{if } \hat{p} \lt p_0 \\
0 & , \text{if } \left| \hat{p} - p_0 \right| \leqslant \frac{1}{2n}
\end{cases}
\]
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0+c}{\sqrt{p_0(1-p_0)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0+c}{\sqrt{p_0(1-p_0)/n}} \sim N\left(\frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}, \frac{p(1-p)}{p_0(1-p_0)}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha/2} - \frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
+ \Phi\left(\frac{z_{\alpha/2} - \frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(\frac{z_{\alpha} - \frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha} - \frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
\]
z-test using s(phat)¶
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量:
\[
z' = \frac{\hat{p}-p_0}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N\left(\frac{p-p_0}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right)
= 1 - \Phi\left(z_{1-\alpha/2} - \frac{p-p_0}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(z_{\alpha/2} - \frac{p-p_0}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(z_{\alpha} - \frac{p-p_0}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0}{\sqrt{p(1-p)/n}}\right)
\]
单侧检验样本量公式推导
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} \pm \frac{p-p_0}{\sqrt{p(1-p)/n}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left(z_{1-\alpha}+z_{1-\beta}\right)^2 p(1-p)}{\left(p-p_0\right)^2}
\]
z-test using s(phat) with continuity correction¶
在 z-test using s(phat) 的基础上加入校正项 \(c\)：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } \hat{p} \gt p_0 \\
\frac{1}{2n} & , \text{if } \hat{p} \lt p_0 \\
0 & , \text{if } \left| \hat{p} - p_0 \right| \leqslant \frac{1}{2n}
\end{cases}
\]
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p}-p_0+c}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p}-p_0+c}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N\left(\frac{p-p_0+c}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right)
= 1 - \Phi\left(z_{1-\alpha/2} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(z_{\alpha/2} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(z_{\alpha} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
\]