单样本率置信区间¶
渐进正态法¶
样本率 \(\hat{p}\) 服从二项分布，\(E(\hat{p}) = p\)，\(Var(\hat{p}) = p(1-p)/n\)，渐进正态法计算的置信限可能超出 \([0, 1]\)。
\[
\begin{align}
\text{Lower Limit} & = \max\left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}, 0\right) \\
\text{Upper Limit} & = \min\left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}, 1\right)
\end{align}
\]
样本量 \(n\) 闭式解的分类讨论
情况 1. \(U \ge 1, L \gt 0\)
\[
d = 1 - \left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}\right) \Rightarrow n = \frac{z_{1-\alpha/2}^2 p(1-p)}{\left(d+p-1\right)^2}
\]
将上式代入条件：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \ge 1 \Rightarrow d \ge 2(1-p)
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \gt 0 \Rightarrow d \lt 1
\]
因此，
\[
n = \frac{z_{1-\alpha/2}^2 p(1-p)}{\left(d+p-1\right)^2} \ ,\ 2(1-p) \le d \lt 1
\]
情况 2. \(U \ge 1, L \le 0\)
\[
d = 1 - 0 = 1
\]
这种情况下置信区间宽度恒等于 1，与样本量 \(n\) 无关，无实际意义。
情况 3. \(U \lt 1, L \gt 0\)
\[
d = \left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}\right) - \left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}\right) \Rightarrow n = \frac{4 z_{1-\alpha/2}^2 p(1-p)}{d^2}
\]
将上式代入条件：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \lt 1 \Rightarrow d \lt 2(1-p)
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \gt 0 \Rightarrow d \lt 2p
\]
因此，
\[
n = \frac{4 z_{1-\alpha/2}^2 p(1-p)}{d^2} \ ,\ d \lt \max\left\lbrace 2p, 2(1-p) \right\rbrace
\]
情况 4. \(U \lt 1, L \le 0\)
\[
d = \left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}\right) - 0 \Rightarrow n = \frac{z_{1-\alpha/2}^2 p(1-p)}{(d-p)^2}\ 且 \ d \gt p
\]
将上式代入条件：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \lt 1 \Rightarrow d \lt 1
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} \le 0 \Rightarrow d \ge 2p
\]
因此，
\[
n = \frac{z_{1-\alpha/2}^2 p(1-p)}{(d-p)^2} \ ,\ 2p \le d \lt 1
\]
综上，
\[
n =
\begin{cases}
\frac{z_{1-\alpha/2}^2 p(1-p)}{\left(d+p-1\right)^2} &, 2(1-p) \le d \lt 1 \\
\frac{4 z_{1-\alpha/2}^2 p(1-p)}{d^2} &, d \lt \max\left\lbrace 2p, 2(1-p) \right\rbrace \\
\frac{z_{1-\alpha/2}^2 p(1-p)}{(d-p)^2} &, 2p \le d \lt 1
\end{cases}
\]
\[
\begin{align}
\text{Lower Limit} & = 0 \\
\text{Upper Limit} & = \min\left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}, 1\right)
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = \max\left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}}, 0\right) \\
\text{Upper Limit} & = 1
\end{align}
\]
渐进正态法（连续性校正）¶
在 渐进正态法 的基础上添加校正项 \(\frac{1}{2n}\)，置信限仍可能超出 \([0, 1]\)：
\[
\begin{align}
\text{Lower Limit} & = \max\left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n}, 0\right) \\
\text{Upper Limit} & = \min\left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n}, 1\right)
\end{align}
\]
样本量 \(n\) 闭式解的分类讨论
情况 1. \(U \ge 1, L \gt 0\)
\[
d = 1 - \left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n}\right) \Rightarrow \frac{1}{2n} + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - (p + d - 1) = 0
\]
令 \(x = \frac{1}{\sqrt{n}}\)，\(A = z_{1 - \alpha/2}\sqrt{p(1-p)}\)，则：
\[
\frac{1}{2}x^2 + Ax - (p + d - 1) = 0
\]
解上述一元二次方程，取正根：
\[
x = -A + \sqrt{A^2 + 2(p + d - 1)}
\]
代入 \(x = \frac{1}{\sqrt{n}}\)，得：
\[
n = \frac{1}{x^2} = \frac{1}{\left(-A + \sqrt{A^2 + 2(p + d - 1)}\right)^2}
\]
将上式代入条件，且根据 \(\frac{1}{2}x^2 + Ax = p + d - 1\)：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n} \ge 1 \Rightarrow p + Ax + \frac{1}{2}x^2 \ge 1 \Rightarrow d \ge 2(1-p)
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n} \gt 0 \Rightarrow p - Ax - \frac{1}{2}x^2 \gt 0 \Rightarrow d \lt 1
\]
因此，
\[
n = \frac{1}{\left(-A + \sqrt{A^2 + 2(p + d - 1)}\right)^2} \ , \ 2(1-p) \le d \lt 1
\]
情况 2. \(U \ge 1, L \le 0\)
\[
d = 1 - 0 = 1
\]
这种情况下置信区间宽度恒等于 1，与样本量 \(n\) 无关，无实际意义。
情况 3. \(U \lt 1, L \gt 0\)
\[
d = \left(p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n}\right) - \left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n}\right)
\]
化简:
\[
d = 2z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{n}
\]
令 \(x = \frac{1}{\sqrt{n}}\)，\(A = z_{1 - \alpha/2}\sqrt{p(1-p)}\)，则：
\[
x^2 + 2Ax - d = 0
\]
解上述一元二次方程，取正根：
\[
x = -A + \sqrt{A^2 + d}
\]
代入 \(x = \frac{1}{\sqrt{n}}\)，得：
\[
n = \frac{1}{x^2} = \frac{1}{\left(-A + \sqrt{A^2 + d}\right)^2}
\]
将上式代入条件，且根据 \(x^2 + 2Ax = d\)：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n} \lt 1 \Rightarrow p + Ax + \frac{1}{2}x^2 \lt 1 \Rightarrow d \lt 2(1-p)
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n} \gt 0 \Rightarrow p - Ax - \frac{1}{2}x^2 \gt 0 \Rightarrow d \lt 2p
\]
因此，
\[
n = \frac{1}{\left(-A + \sqrt{A^2 + d}\right)^2} \ , \ d \lt \min\left\lbrace 2p, 2(1-p) \right\rbrace
\]
情况 4. \(U \lt 1, L \le 0\)
\[
d = \left(p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n}\right) - 0
\]
令 \(x = \frac{1}{\sqrt{n}}\)，\(A = z_{1 - \alpha/2}\sqrt{p(1-p)}\)，则：
\[
\frac{1}{2}x^2 + Ax + p - d = 0
\]
解上述一元二次方程，取正根：
\[
x = -A + \sqrt{A^2 - 2(p - d)}
\]
代入 \(x = \frac{1}{\sqrt{n}}\)，得：
\[
n = \frac{1}{x^2} = \frac{1}{\left(-A + \sqrt{A^2 - 2(p - d)}\right)^2}
\]
将上式代入条件，且根据 \(\frac{1}{2}x^2 + Ax = d - p\)：
\[
p + z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n} \lt 1 \Rightarrow p + Ax + \frac{1}{2}x^2 \lt 1 \Rightarrow d \lt 1
\]
\[
p - z_{1-\alpha/2} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n} \le 0 \Rightarrow p - Ax - \frac{1}{2}x^2 \le 0 \Rightarrow d \ge 2p
\]
因此，
\[
n = \frac{1}{\left(-A + \sqrt{A^2 - 2(p - d)}\right)^2} \ , \ 2p \le d \lt 1
\]
综上，
\[
n =
\begin{cases}
\frac{1}{\left(-A + \sqrt{A^2 + 2(p + d - 1)}\right)^2} &, 2(1-p) \le d \lt 1 \\
\frac{1}{\left(-A + \sqrt{A^2 + d}\right)^2} &, d \lt \min\left\lbrace 2p, 2(1-p) \right\rbrace \\
\frac{1}{\left(-A + \sqrt{A^2 - 2(p - d)}\right)^2} &, 2p \le d \lt 1
\end{cases}
\]
其中，\(A = z_{1 - \alpha/2}\sqrt{p(1-p)}\) 。
\[
\begin{align}
\text{Lower Limit} & = 0 \\
\text{Upper Limit} & = \min\left(p + z_{1-\alpha} \sqrt{\frac{p(1-p)}{n}} + \frac{1}{2n}, 1\right)
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = \max\left(p - z_{1-\alpha} \sqrt{\frac{p(1-p)}{n}} - \frac{1}{2n}, 0\right) \\
\text{Upper Limit} & = 1
\end{align}
\]
Clopper-Pearson¶
\[
\begin{align}
\text{Lower Limit} & = \left[ 1 + \frac{n - np + 1}{np F_{\frac{\alpha}{2};\ 2np,\ 2(n - np + 1)}} \right]^{-1} \\
\text{Upper Limit} & = \left[ 1 + \frac{n - np}{(np + 1) F_{1-\frac{\alpha}{2};\ 2(np + 1), \ 2(n - np)}} \right]^{-1}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = 0 \\
\text{Upper Limit} & = \left[ 1 + \frac{n - np}{(np + 1) F_{1-\alpha;\ 2(np + 1), \ 2(n - np)}} \right]^{-1}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = \left[ 1 + \frac{n - np + 1}{np F_{\alpha;\ 2np,\ 2(n - np + 1)}} \right]^{-1} \\
\text{Upper Limit} & = 1
\end{align}
\]
Wilson Score¶
\[
\begin{align}
\text{Lower Limit} & = \frac{\left(2np + z_{1-\alpha/2}^2\right) - z_{1-\alpha/2} \sqrt{z_{1-\alpha/2}^2 + 4np(1-p)}}{2\left(n + z_{1-\alpha/2}^2\right)} \\
\text{Upper Limit} & = \frac{\left(2np + z_{1-\alpha/2}^2\right) + z_{1-\alpha/2} \sqrt{z_{1-\alpha/2}^2 + 4np(1-p)}}{2\left(n + z_{1-\alpha/2}^2\right)}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = 0 \\
\text{Upper Limit} & = \frac{\left(2np + z_{1-\alpha}^2\right) + z_{1-\alpha} \sqrt{z_{1-\alpha}^2 + 4np(1-p)}}{2\left(n + z_{1-\alpha}^2\right)}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = \frac{\left(2np + z_{1-\alpha}^2\right) - z_{1-\alpha} \sqrt{z_{1-\alpha}^2 + 4np(1-p)}}{2\left(n + z_{1-\alpha}^2\right)} \\
\text{Upper Limit} & = 1
\end{align}
\]
Wilson Score 连续性校正¶
\[
\begin{align}
\text{Lower Limit} & = \frac{\left(2np + z_{1-\alpha/2}^2 - 1\right) - z_{1-\alpha/2} \sqrt{z_{1-\alpha/2}^2 - \frac{1}{n} + 4np(1-p) + 4p - 2}}{2\left(n + z_{1-\alpha/2}^2\right)} \\
\text{Upper Limit} & = \frac{\left(2np + z_{1-\alpha/2}^2 + 1\right) + z_{1-\alpha/2} \sqrt{z_{1-\alpha/2}^2 - \frac{1}{n} + 4np(1-p) - 4p + 2}}{2\left(n + z_{1-\alpha/2}^2\right)}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = 0 \\
\text{Upper Limit} & = \frac{\left(2np + z_{1-\alpha}^2 + 1\right) + z_{1-\alpha} \sqrt{z_{1-\alpha}^2 - \frac{1}{n} + 4np(1-p) - 4p + 2}}{2\left(n + z_{1-\alpha}^2\right)}
\end{align}
\]
\[
\begin{align}
\text{Lower Limit} & = \frac{\left(2np + z_{1-\alpha}^2 - 1\right) - z_{1-\alpha} \sqrt{z_{1-\alpha}^2 - \frac{1}{n} + 4np(1-p) + 4p - 2}}{2\left(n + z_{1-\alpha}^2\right)} \\
\text{Upper Limit} & = 1
\end{align}
\]
Wilson Score 连续性校正置信区间宽度随样本量 \(n\) 的变化
以 \(p = 0.9\) 为例，绘制双侧 95% 置信区间宽度随样本量 \(n\) 变化的图像如下：
如果将 \(n\) 视为连续性变量，则随着 \(n\) 的增大，置信区间宽度先增大后减小，这可能会给数值求解带来一些麻烦。
若设定置信区间宽度为 \(0.8\)，则理论上存在两个数值解，实际应取较大的解作为样本量估算结果。
brentq 要求求根区间左右两端点处的函数值异号，此时可先用 minimize_scalar 求出区间内的极大值，将极大值点作为求根区间下限，再应用 brentq 进行数值求解。