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。
Exact Test¶
在 \(H_0\) 成立时，设 \(X_0 \sim b(n, p_0)\)，寻找满足以下条件的 \(k_1\) 和 \(k_2\)：
\[
k_1 = \sup \left\{ k \;\middle|\; \sum_{i=0}^{k} P(X_0 = i) \leqslant \alpha/2 \right\}
\]
\[
k_2 = \inf \left\{ k \;\middle|\; \sum_{i=k}^{n} P(X_0 = i) \leqslant \alpha/2 \right\}
\]
在 \(H_1\) 成立时，设 \(X_1 \sim b(n, p)\)，检验效能为：
\[
\text{Power} = \sum_{i=0}^{k_1} P(X_1 = i) + \sum_{i=k_2}^{n} P(X_1 = i)
\]
在 \(H_0\) 成立时，设 \(X_0 \sim b(n, p_0)\)，寻找满足以下条件的 \(k\)：
\[
k = \sup \left\{ k \;\middle|\; \sum_{i=0}^{k} P(X_0 = i) \leqslant \alpha \right\}
\]
在 \(H_1\) 成立时，设 \(X_1 \sim b(n, p)\)，检验效能为：
\[
\text{Power} = \sum_{i=0}^{k} P(X_1 = i)
\]
在 \(H_0\) 成立时，设 \(X_0 \sim b(n, p_0)\)，寻找满足以下条件的 \(k\)：
\[
k = \inf \left\{ k \;\middle|\; \sum_{i=k}^{n} P(X_0 = i) \leqslant \alpha \right\}
\]
在 \(H_1\) 成立时，设 \(X_1 \sim b(n, p)\)，检验效能为：
\[
\text{Power} = \sum_{i=k}^{n} P(X_1 = i)
\]
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：
\[
z' = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}} \sim N\left(\frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}, \frac{p(1-p)}{p_0(1-p_0)}\right)
\]
\[
\begin{align}
\text{Power}
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
+ \Phi\left(\frac{z_{\alpha/2} - \frac{p-p_0}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2}\sqrt{p_0(1-p_0)/n} - (p-p_0)}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(\frac{z_{\alpha/2}\sqrt{p_0(1-p_0)/n} - (p-p_0)}{\sqrt{p(1-p)/n}}\right)
\end{align}
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(\frac{z_{\alpha}\sqrt{p_0(1-p_0)/n} - (p-p_0)}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{p_0(1-p_0)/n} - (p-p_0)}{\sqrt{p(1-p)/n}}\right)
\]
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：
\[
\frac{z_{1-\alpha} \sqrt{p_0(1-p_0)/n} \pm (p-p_0)}{\sqrt{p(1-p)/n}} = z_{\beta}
\]
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 \(c\)：
定义：
\[
c =
\begin{cases}
\frac{1}{2n} & , \text{if } \left| \hat{p} - p_0 \right| \gt \frac{1}{2n} \\
0 & , \text{otherwise}
\end{cases}
\]
校正项的符号由检验方向决定，左侧检验时，校正项为 \(+c\)，右侧检验时，校正项为 \(-c\)。
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p} - p_0 \pm c}{\sqrt{p_0(1-p_0)/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p} - p_0 \pm c}{\sqrt{p_0(1-p_0)/n}} \sim N\left(\frac{p - p_0 \pm c}{\sqrt{p_0(1-p_0)/n}}, \frac{p(1-p)}{p_0(1-p_0)}\right)
\]
\[
\begin{align}
\text{Power}
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2} - \frac{p-p_0-c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right)
+ \Phi\left(\frac{z_{\alpha/2} - \frac{p-p_0+c}{\sqrt{p_0(1-p_0)/n}}}{\sqrt{\frac{p(1-p)}{p_0(1-p_0)}}}\right) \\
& = 1 - \Phi\left(\frac{z_{1-\alpha/2}\sqrt{p_0(1-p_0)/n} - (p-p_0-c)}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(\frac{z_{\alpha/2}\sqrt{p_0(1-p_0)/n} - (p-p_0+c)}{\sqrt{p(1-p)/n}}\right)
\end{align}
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(\frac{z_{\alpha}\sqrt{p_0(1-p_0)/n} - (p-p_0+c)}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(\frac{z_{1-\alpha}\sqrt{p_0(1-p_0)/n} - (p-p_0-c)}{\sqrt{p(1-p)/n}}\right)
\]
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 \(c\)：
定义：
\[
c =
\begin{cases}
\frac{1}{2n} & , \text{if } \left| \hat{p} - p_0 \right| \gt \frac{1}{2n} \\
0 & , \text{otherwise}
\end{cases}
\]
校正项的符号由检验方向决定，左侧检验时，校正项为 \(+c\)，右侧检验时，校正项为 \(-c\)。
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{p} - p_0 \pm c}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{p} - p_0 \pm c}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N\left(\frac{p - p_0 \pm c}{\sqrt{p(1-p)/n}}, 1\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < z_{\alpha/2} \right)
= 1 - \Phi\left(z_{1-\alpha/2} - \frac{p-p_0-c}{\sqrt{p(1-p)/n}}\right)
+ \Phi\left(z_{\alpha/2} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' < z_{\alpha} \right)
= \Phi\left(z_{\alpha} - \frac{p-p_0+c}{\sqrt{p(1-p)/n}}\right)
\]
\[
\text{Power} = P\left(z' > z_{1-\alpha} \right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{p-p_0-c}{\sqrt{p(1-p)/n}}\right)
\]
参考文献
- JH Z. Zar JH. Dichotomous variables[J]. Biostatistical Analysis 5th ed. Upper Saddle River, NJ: Prentice-Hall, 2010: 557-558.
- Chow S C, Shao J, Wang H, et al. Sample size calculations in clinical research[M]. chapman and hall/CRC, 2017.
- Ryan, Thomas. (2013). Sample Size Determination and Power. Sample Size Determination and Power. 10.1002/9781118439241.