单样本率差异性检验¶
Z-test using S(P0)¶
\[
\begin{align}
H_0 & : p = p_0 \\
H_1 & : p \neq p_0
\end{align}
\]
在 \(H_0\) 成立时，样本比例 \(\hat{p}\) 服从近似正态分布：
\[
\hat{p} \sim N\left(p_0,\ \frac{p_0(1-p0)}{n}\right)
\]
那么：
\[
E(\hat{p} - p_0) = E(\hat{p}) - p_0 = p_0 - p_0 = 0
\]
\[
Var(\hat{p} - p_0) = Var(\hat{p}) = \frac{p_0(1-p_0)}{n}
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} \sim N(0, 1)
\]
拒绝 \(H_0\) 的条件：
\[
z > z_{1-\alpha/2}
\]
在 \(H_1\) 成立时，可构建统计量：
\[
z' = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{\sqrt{n}(\hat{p} - p_0)}{\sqrt{p_0(1-p_0)}}
\]
根据中心极限定理，当 \(n\) 较大时，样本比例 \(\hat{p}\) 满足：
\[
\sqrt{n}(\hat{p} - p_1) \xrightarrow{d} N(0, p_1(1-p_1))
\]
即 \(\sqrt{n}\left(\hat{p} - p_1\right)\) 依分布收敛到均值为 \(0\)，方差为 \(p_1(1-p_1)\) 的正态分布，进而有：
\[
z' = \frac{\sqrt{n}(\hat{p} - p_0)}{\sqrt{p_0(1-p_0)}} = \frac{\sqrt{n}(\hat{p} - p_1) + \sqrt{n}(p_1 - p_0)}{\sqrt{p_0(1-p_0)}} \xrightarrow{d} N\left( \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_0(1-p_0)}}, \frac{p_1(1-p_1)}{p_0(1-p_0)} \right)
\]
根据检验类型分类讨论：
单侧检验，\(p_1 > p_0\)
\[
\begin{align}
P\left(z' > z_{1-\alpha} \right)
& = 1 - \Phi\left( \frac{z_{1-\alpha} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} - \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \beta
\end{align}
\]
单侧检验，\(p_1 < p_0\)
\[
\begin{align}
P\left(z' < -z_{1-\alpha} \right)
& = \Phi\left( \frac{-z_{1-\alpha} - \frac{\sqrt{n} (p_1-p_0)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} + \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \beta
\end{align}
\]
双侧检验，\(p_1 \neq p_0\)
\[
\begin{align}
P\left( \left| z' \right| > z_{1-\alpha/2} \right)
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < -z_{1-\alpha/2} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right)
+ \Phi\left( \frac{-z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} - \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) +
1 - \Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} + \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 2 - \left[
\Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} - \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) +
\Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} + \sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right)
\right] \\
& = 1 - \beta
\end{align}
\]
以单侧检验为例，推导样本量计算公式
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} - \sqrt{n}|p_1-p_0|}{\sqrt{p_1(1-p_1)}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left( z_{1-\alpha} \sqrt{p_0(1-p_0)} + z_{1-\beta} \sqrt{p_1(1-p_1)} \right)^2}{\left( p_1 - p_0 \right)^2}
\]
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 \(c\)：
\[
z = \frac{\hat{p} - p_0 + c}{\sqrt{p_0(1-p_0)/n}} \sim N(0, 1)
\]
其中：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } \hat{p} \gt p_0 \\
\frac{1}{2n} & , \text{if } \hat{p} \lt p_0 \\
0 & , \text{if } \left| \hat{p} - p_0 \right| \le \frac{1}{2n}
\end{cases}
\]
在 \(H_1\) 成立时，可构建统计量：
\[
z' = \frac{\hat{p} - p_0 + c}{\sqrt{p_0(1-p_0)/n}} = \frac{\sqrt{n}(\hat{p} - p_0 + c)}{\sqrt{p_0(1-p_0)}}
\]
根据中心极限定理，当 \(n\) 较大时，样本比例 \(\hat{p}\) 满足：
\[
\sqrt{n}(\hat{p} - p_1) \xrightarrow{d} N(0, p_1(1-p_1))
\]
即 \(\sqrt{n}\left(\hat{p} - p_1\right)\) 依分布收敛到均值为 \(0\)，方差为 \(p_1(1-p_1)\) 的正态分布，进而有：
\[
z' = \frac{\sqrt{n}(\hat{p} - p_1) + \sqrt{n}(p_1 - p_0 + c)}{\sqrt{p_0(1-p_0)}} \xrightarrow{d} N\left( \frac{\sqrt{n}(p_1-p_0+c)}{\sqrt{p_0(1-p_0)}}, \frac{p_1(1-p_1)}{p_0(1-p_0)} \right)
\]
根据检验类型分类讨论：
单侧检验，\(p_1 > p_0\)
\[
\begin{align}
P\left(z' > z_{1-\alpha} \right)
& = 1 - \Phi\left( \frac{z_{1-\alpha} - \frac{\sqrt{n} \left(p_1-p_0-\frac{1}{2n} \right)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} - \sqrt{n} \left(p_1-p_0-\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \beta
\end{align}
\]
单侧检验，\(p_1 < p_0\)
\[
\begin{align}
P\left(z' < -z_{1-\alpha} \right)
& = \Phi\left( \frac{-z_{1-\alpha} - \frac{\sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha} + \frac{\sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} + \sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \beta
\end{align}
\]
双侧检验，\(p_1 \neq p_0\)
\[
\begin{align}
P\left( \left| z' \right| > z_{1-\alpha/2} \right)
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < -z_{1-\alpha/2} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha/2} - \frac{\sqrt{n} \left(p_1-p_0-\frac{1}{2n} \right)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right)
+ \Phi\left( \frac{-z_{1-\alpha/2} - \frac{\sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_0(1-p_0)}}}{\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}} \right) \\
& = 1 - \Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} - \sqrt{n} \left(p_1-p_0-\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right) +
1 - \Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} + \sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 2 - \left[
\Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} - \sqrt{n} \left(p_1-p_0-\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right) +
\Phi\left( \frac{z_{1-\alpha/2} \sqrt{p_0(1-p_0)} + \sqrt{n} \left(p_1-p_0+\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} \right)
\right] \\
& = 1 - \beta
\end{align}
\]
以单侧检验为例，推导样本量计算公式
根据标准正态分布分位数的定义：
\[
\frac{z_{1-\alpha} \sqrt{p_0(1-p_0)} - \sqrt{n} \left( \left| p_1-p_0 \right| - \frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} = z_{\beta} \Rightarrow z_{1-\alpha} \sqrt{p_0(1-p_0)} + z_{1-\beta} \sqrt{p_1(1-p_1)} = \sqrt{n} \left(| p_1 - p_0| - \frac{1}{2n}\right)
\]
令 \(A = z_{1-\alpha} \sqrt{p_0(1-p_0)} + z_{1-\beta} \sqrt{p_1(1-p_1)}\)，\(\delta = | p_1 - p_0|\)，则：
\[
A = \sqrt{n} \left(\delta - \frac{1}{2n}\right) \Rightarrow A\sqrt{n} = \delta n - \frac{1}{2} \Rightarrow \delta n - A\sqrt{n} - \frac{1}{2} = 0
\]
令 \(x = \sqrt{n}\)，则：
\[
\delta x^2 - Ax - \frac{1}{2} = 0 \Rightarrow x = \frac{A \pm \sqrt{A^2 + 2\delta}}{2\delta}
\]
由于 \(A \gt 0\)，\(\delta \gt 0\)，故取正根：
\[
x = \frac{A + \sqrt{A^2 + 2\delta}}{2\delta}
\]
\[
n = x^2 = \left( \frac{A + \sqrt{A^2 + 2\delta}}{2\delta} \right)^2 \tag{2.1}
\]
引入未经校正的样本量 \(n'\)：
\[
n' = \frac{A^2}{\delta^2}
\]
即 \(A = \delta\sqrt{n'}\)，代入式 \((2.1)\)：
\[
\begin{align}
n & = \left( \frac{\delta\sqrt{n'} + \sqrt{\delta^2 n' + 2\delta}}{2\delta} \right)^2 \\
& = \frac{\delta^2 \left(\sqrt{n'} + \sqrt{n' + \frac{2}{\delta}}\right)^2}{4\delta^2} \\
& = \frac{n'}{4} \left( 1 + \sqrt{1 + \frac{2}{\delta n'}}\right)^2
\end{align}
\]
故：
\[
n = \frac{n'}{4} \left( 1 + \sqrt{1 + \frac{2}{|p_1 - p_0| n'}}\right)^2
\]
Z-test using S(Phat)¶
\[
\begin{align}
H_0 & : p = p_0 \\
H_1 & : p \neq p_0
\end{align}
\]
在 \(H_0\) 成立时：
\[
E(\hat{p} - p_0) = E(\hat{p}) - p_0 = p_0 - p_0 = 0
\]
\[
Var(\hat{p} - p_0) = Var(\hat{p}) = \frac{\hat{p}(1-\hat{p})}{n}
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{p} - p_0}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N(0, 1)
\]
拒绝 \(H_0\) 的条件：
\[
z > z_{1-\alpha/2}
\]
在 \(H_1\) 成立时，可构建统计量:
\[
z' = \frac{\hat{p} - p_0}{\sqrt{\hat{p}\left(1 - \hat{p}\right) / n}} = \frac{\sqrt{n}\left(\hat{p} - p_0\right)}{\sqrt{\hat{p}\left(1 - \hat{p}\right)}}
\]
根据中心极限定理，当 \(n\) 较大时，样本比例 \(\hat{p}\) 满足：
\[
\sqrt{n}\left(\hat{p} - p_1\right) \xrightarrow{d} N\left(0, p_1(1-p_1)\right)
\]
即 \(\sqrt{n}\left(\hat{p} - p_1\right)\) 依分布收敛到均值为 \(0\)，方差为 \(p_1(1-p_1)\) 的正态分布，进而有：
\[
\sqrt{n}\left(\hat{p} - p_0\right) = \sqrt{n}\left(\hat{p} - p_1\right) + \sqrt{n}\left(p_1 - p_0\right) \xrightarrow{d} N\left(\sqrt{n}(p_1-p_0), p_1(1-p_1)\right)\tag{3.1}
\]
根据大数定律和连续映射定理：
\[
\sqrt{\hat{p}\left(1 - \hat{p}\right)} \xrightarrow{p} \sqrt{p_1\left(1-p_1\right)} \tag{3.2}
\]
由 \((3.1), (3.2)\)，基于 Slutsky 定理：
\[
z' \xrightarrow{d} N\left(\frac{\sqrt{n}(p_1 - p_0)}{\sqrt{p_1(1-p_1)}}, 1\right)
\]
根据检验类型分类讨论：
单侧检验，\(p_1 > p_0\)
\[
P\left(z' > z_{1-\alpha} \right) = 1 - \Phi\left( z_{1-\alpha} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) = 1 - \beta
\]
单侧检验，\(p_1 < p_0\)
\[
P\left(z' < -z_{1-\alpha} \right)
= \Phi\left( -z_{1-\alpha} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right)
= 1 - \Phi\left( z_{1-\alpha} + \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right)
= 1 - \beta
\]
双侧检验，\(p_1 \neq p_0\)
\[
\begin{align}
P\left( \left| z' \right| > z_{1-\alpha/2} \right)
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < -z_{1-\alpha/2} \right) \\
& = 1 - \Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right)
+ \Phi\left( -z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) +
1 - \Phi\left( z_{1-\alpha/2} + \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) \\
& = 2 - \left[
\Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right) +
\Phi\left( z_{1-\alpha/2} + \frac{\sqrt{n}(p_1-p_0)}{\sqrt{p_1(1-p_1)}} \right)
\right] \\
& = 1 - \beta
\end{align}
\]
以单侧检验为例，推导样本量计算公式
根据标准正态分布分位数的定义：
\[
z_{1-\alpha/2} - \frac{\sqrt{n} \left| p_1 - p_0 \right|}{\sqrt{p_1(1-p_1)}} = z_{\beta}
\]
可解出：
\[
n = \frac{\left( z_{1-\alpha/2} + z_{1-\beta} \right)^2 p_1(1-p_1)}{\left(p_1-p_0\right)^2}
\]
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 \(c\)：
\[
z = \frac{\hat{p} - p_0 + c}{\sqrt{\hat{p}(1-\hat{p})/n}} \sim N(0, 1)
\]
其中：
\[
c =
\begin{cases}
- \frac{1}{2n} & , \text{if } \hat{p} \gt p_0 \\
\frac{1}{2n} & , \text{if } \hat{p} \lt p_0 \\
0 & , \text{if } \left| \hat{p} - p_0 \right| \le \frac{1}{2n}
\end{cases}
\]
在 \(H_1\) 成立时，可构建统计量：
\[
z' = \frac{\hat{p} - p_0 + c}{\sqrt{\hat{p}(1-\hat{p})/n}} = \frac{\sqrt{n}(\hat{p} - p_0 + c)}{\sqrt{\hat{p}(1-\hat{p})}}
\]
根据中心极限定理，当 \(n\) 较大时，样本比例 \(\hat{p}\) 满足：
\[
\sqrt{n}(\hat{p} - p_1) \xrightarrow{d} N(0, p_1(1-p_1))
\]
即 \(\sqrt{n}\left(\hat{p} - p_1\right)\) 依分布收敛到均值为 \(0\)，方差为 \(p_1(1-p_1)\) 的正态分布，进而有：
\[
\sqrt{n}\left(\hat{p} - p_0 + c\right) = \sqrt{n}\left(\hat{p} - p_1\right) + \sqrt{n}\left(p_1 - p_0 + c\right) \xrightarrow{d} N\left(\sqrt{n}(p_1 - p_0 + c), p_1(1-p_1)\right) \tag{4.1}
\]
根据大数定律和连续映射定理：
\[
\sqrt{\hat{p}\left(1 - \hat{p}\right)} \xrightarrow{p} \sqrt{p_1\left(1-p_1\right)} \tag{4.2}
\]
由 \((4.1), (4.2)\)，基于 Slutsky 定理：
\[
z' \xrightarrow{d} N\left(\frac{\sqrt{n}(p_1 - p_0 + c)}{\sqrt{p_1(1-p_1)}}, 1\right)
\]
根据检验类型分类讨论：
单侧检验，\(p_1 > p_0\)
\[
P\left(z' > z_{1-\alpha} \right) = 1 - \Phi\left( z_{1-\alpha} - \frac{\sqrt{n}(p_1-p_0-\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right) = 1 - \beta
\]
单侧检验，\(p_1 < p_0\)
\[
P\left(z' < -z_{1-\alpha} \right)
= \Phi\left( -z_{1-\alpha} - \frac{\sqrt{n}(p_1-p_0+\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right)
= 1 - \Phi\left( z_{1-\alpha} + \frac{\sqrt{n}(p_1-p_0+\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right)
= 1 - \beta
\]
双侧检验，\(p_1 \neq p_0\)
\[
\begin{align}
P\left( \left| z' \right| > z_{1-\alpha/2} \right)
& = P\left(z' > z_{1-\alpha/2} \right) + P\left(z' < -z_{1-\alpha/2} \right) \\
& = 1 - \Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0-\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right)
+ \Phi\left( -z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0+\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right) \\
& = 1 - \Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0-\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right) +
1 - \Phi\left( z_{1-\alpha/2} + \frac{\sqrt{n}(p_1-p_0+\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right) \\
& = 2 - \left[
\Phi\left( z_{1-\alpha/2} - \frac{\sqrt{n}(p_1-p_0-\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right) +
\Phi\left( z_{1-\alpha/2} + \frac{\sqrt{n}(p_1-p_0+\frac{1}{2n})}{\sqrt{p_1(1-p_1)}} \right)
\right] \\
& = 1 - \beta
\end{align}
\]
以单侧检验为例，推导样本量计算公式
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} - \frac{\sqrt{n} \left( |p_1-p_0|-\frac{1}{2n} \right)}{\sqrt{p_1(1-p_1)}} = z_{\beta} \Rightarrow \left(z_{1-\alpha} + z_{1-\beta}\right) \sqrt{p_1(1-p_1)} = \sqrt{n} \left(| p_1 - p_0| - \frac{1}{2n}\right)
\]
令 \(A = \left(z_{1-\alpha} + z_{1-\beta}\right) \sqrt{p_1(1-p_1)}\)，\(\delta = | p_1 - p_0|\)，则：
\[
A = \sqrt{n} \left(\delta - \frac{1}{2n}\right) \Rightarrow A\sqrt{n} = \delta n - \frac{1}{2} \Rightarrow \delta n - A\sqrt{n} - \frac{1}{2} = 0
\]
令 \(x = \sqrt{n}\)，则：
\[
\delta x^2 - Ax - \frac{1}{2} = 0 \Rightarrow x = \frac{A \pm \sqrt{A^2 + 2\delta}}{2\delta}
\]
由于 \(A \gt 0\)，\(\delta \gt 0\)，故取正根：
\[
x = \frac{A + \sqrt{A^2 + 2\delta}}{2\delta}
\]
\[
n = x^2 = \left( \frac{A + \sqrt{A^2 + 2\delta}}{2\delta} \right)^2 \tag{2.1}
\]
引入未经校正的样本量 \(n'\)：
\[
n' = \frac{A^2}{\delta^2}
\]
即 \(A = \delta\sqrt{n'}\)，代入式 \((2.1)\)：
\[
\begin{align}
n & = \left( \frac{\delta\sqrt{n'} + \sqrt{\delta^2 n' + 2\delta}}{2\delta} \right)^2 \\
& = \frac{\delta^2 \left(\sqrt{n'} + \sqrt{n' + \frac{2}{\delta}}\right)^2}{4\delta^2} \\
& = \frac{n'}{4} \left( 1 + \sqrt{1 + \frac{2}{\delta n'}}\right)^2
\end{align}
\]
故：
\[
n = \frac{n'}{4} \left( 1 + \sqrt{1 + \frac{2}{|p_1 - p_0| n'}}\right)^2
\]