两独立样本均值差异性检验¶
对于双侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: \mu_1 = \mu_2 \\
H_1 &: \mu_1 \neq \mu_2
\end{align}
\]
对于左单侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: \mu_1 \geqslant \mu_2 \\
H_1 &: \mu_1 \lt \mu_2
\end{align}
\]
对于右单侧检验，统计学假设如下：
\[
\begin{align}
H_0 &: \mu_1 \leqslant \mu_2 \\
H_1 &: \mu_1 \gt \mu_2
\end{align}
\]
两样本均值分别用 \(\hat{\mu}_1\) 和 \(\hat{\mu}_2\) 表示，两样本方差分别用 \(s_1\) 和 \(s_2\) 表示，两总体方差分别用 \(\sigma_1\) 和 \(\sigma_2\) 表示。
以下推导过程在边界条件 \(\mu_1 = \mu_2\) 下进行。
z 检验¶
假设两组方差相等¶
令 \(\sigma\) = \(\sigma_1\) = \(\sigma_2\)，则：
\[
\operatorname{SD}(\hat{\mu}_1 - \hat{\mu}_2) = \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}
\]
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim N\left(\frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}, 1\right)
\]
\[
\text{Power}
= P\left(z' > z_{1-\alpha/2}\right) + P\left(z' < z_{\alpha/2}\right)
= 1 - \Phi\left(z_{1-\alpha/2} - \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
+ \Phi\left(z_{\alpha/2} - \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(z' < z_{\alpha}\right)
= \Phi\left(z_{\alpha} - \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
= 1 - \Phi\left(z_{1-\alpha} + \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
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：
\[
z_{1-\alpha} \pm \frac{\mu_1 - \mu_2}{\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \sigma^2 \left(\frac{1}{k} + 1\right)}{\left(\mu_1 - \mu_2\right)^2}
\]
\[
n_1 = k n_2
\]
假设两组方差不等¶
\[
\operatorname{SD}(\hat{\mu}_1 - \hat{\mu}_2) = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}
\]
在 \(H_0\) 成立时，可构建 \(z\) 统计量：
\[
z = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \sim N(0, 1)
\]
在 \(H_1\) 成立时，可构建 \(z'\) 统计量：
\[
z' = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \sim N\left(\frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}, 1\right)
\]
\[
\text{Power}
= P\left(z' > z_{1-\alpha/2}\right) + P\left(z' < z_{\alpha/2}\right)
= 1 - \Phi\left(z_{1-\alpha/2} - \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)
+ \Phi\left(z_{\alpha/2} - \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(z' < z_{\alpha}\right)
= \Phi\left(z_{\alpha} - \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)
= 1 - \Phi\left(z_{1-\alpha} + \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)
\]
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：
\[
z_{1-\alpha} \pm \frac{\mu_1 - \mu_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = z_\beta
\]
设 \(n_1 = kn_2\)，由上式可解出
\[
n_2 = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2 \left(\frac{\sigma_1^2}{k} + \sigma_2^2\right)}{\left(\mu_1 - \mu_2\right)^2}
\]
\[
n_1 = k n_2
\]
t 检验¶
假设两组方差相等¶
当两组总体方差相等时，即 \(\sigma_1^2 = \sigma_2^2\) 时，可计算合并方差 \(s_c^2\) ：
\[
s_c^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}
\]
\[
\operatorname{SD}(\hat{\mu}_1 - \hat{\mu}_2) = s_c \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}
\]
在 \(H_0\) 成立时，可构建 \(t\) 统计量：
\[
t = \frac{\hat{\mu}_1 - \hat{\mu}_2}{s_c\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim t(n_1 + n_2 - 2)
\]
在 \(H_1\) 成立时，可构建 \(t'\) 统计量：
\[
t' = \frac{\hat{\mu}_1 - \hat{\mu}_2}{s_c\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\sim t\left(n_1 + n_2 - 2, \frac{\mu_1 - \mu_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
令 \(T(x;v,\lambda)\) 为自由度为 \(v\)，非中心参数为 \(\lambda\) 的非中心 \(t\) 分布的累积分布函数。
\[
\begin{align}
\text{Power}
& = P\left(t' > t_{1-\alpha/2}\right) + P\left(t' < t_{\alpha/2}\right) \\
& = 1 - T\left(t_{1-\alpha/2, n_1 + n_2 - 2}; n_1 + n_2 - 2, \frac{\mu_1 - \mu_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
+ T\left(t_{\alpha/2, n_1 + n_2 - 2}; n_1 + n_2 - 2, \frac{\mu_1 - \mu_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\end{align}
\]
\[
\text{Power}
= P\left(t' < t_{\alpha}\right)
= T\left(t_{\alpha, n_1 + n_2 - 2}; n_1 + n_2 - 2, \frac{\mu_1 - \mu_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(t' > t_{1-\alpha}\right)
= 1 - T\left(t_{1-\alpha, n_1 + n_2 - 2}; n_1 + n_2 - 2, \frac{\mu_1 - \mu_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\right)
\]
假设两组方差不等¶
\[
\operatorname{SD}(\hat{\mu}_1 - \hat{\mu}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
\]
当两总体方差不相等时，即 \(\sigma_1^2 \ne \sigma_2^2\) 时，可使用以下近似 \(t\) 检验进行推导。
Welch 近似 t 检验¶
在 \(H_0\) 成立时，可构建 \(t\) 统计量：
\[
t = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \sim t(v')
\]
其中：
\[
v' = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2(n_1 + 1)} + \frac{s_2^4}{n_2^2(n_2 + 1)}} - 2
\]
在 \(H_1\) 成立时，可构建 \(t'\) 统计量：
\[
t' = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\sim t\left(v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]
令 \(T(x;v,\lambda)\) 为自由度为 \(v\)，非中心参数为 \(\lambda\) 的非中心 \(t\) 分布的累积分布函数。
\[
\begin{align}
\text{Power}
& = P\left(t' > t_{1-\alpha/2}\right) + P\left(t' < t_{\alpha/2}\right) \\
& = 1 - T\left(t_{1-\alpha/2, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
+ T\left(t_{\alpha/2, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\end{align}
\]
\[
\text{Power}
= P\left(t' < t_{\alpha}\right)
= T\left(t_{\alpha, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(t' > t_{1-\alpha}\right)
= 1 - T\left(t_{1-\alpha, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]
Satterthwaite 近似 t 检验¶
在 \(H_0\) 成立时，可构建 \(t\) 统计量：
\[
t = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \sim t(v')
\]
其中：
\[
v' = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2(n_1 - 1)} + \frac{s_2^4}{n_2^2(n_2 - 1)}}
\]
在 \(H_1\) 成立时，可构建 \(t'\) 统计量：
\[
t' = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\sim t\left(v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]
令 \(T(x;v,\lambda)\) 为自由度为 \(v\)，非中心参数为 \(\lambda\) 的非中心 \(t\) 分布的累积分布函数。
\[
\begin{align}
\text{Power}
& = P\left(t' > t_{1-\alpha/2}\right) + P\left(t' < t_{\alpha/2}\right) \\
& = 1 - T\left(t_{1-\alpha/2, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
+ T\left(t_{\alpha/2, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\end{align}
\]
\[
\text{Power}
= P\left(t' < t_{\alpha}\right)
= T\left(t_{\alpha, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]
\[
\text{Power}
= P\left(t' > t_{1-\alpha}\right)
= 1 - T\left(t_{1-\alpha, v'}; v', \frac{\mu_1 - \mu_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\right)
\]