平均数差异的显著性检验

• 对样本平均数之间的差异$(\bar{X}_{1} - \bar{X}_{2})$来检验两总体之间的差异$(\mu_{1} - \mu_{2})$
• 分为3类
  1. 两总体正态，两总体方差已知
  2. 两总体正态，两总体方差未知
  3. 两总体都是非正态

计算

两总体均为正态，且两总体方差已知

• 两个样本之间的差异记为$D_{\bar{X}}$，其服从正态分布
• $\mu_{D_{\bar{X}}} = \mu_{1} - \mu_{2}$
• 对两个样本平均数差异$(\bar{X}_{1} - \bar{X}_{2})$的显著性检验实际上就是对$D_{\bar{X}}$与$\mu_{D_{\bar{X}}} = 0$差异的检验。

独立样本的平均数差异检验

• 标准误 $SE_{D_{\bar{X}}} = \sqrt{\frac{\sigma^{2}_{1}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}$

  proof

  $$\begin{array}{ll} \because & D(\bar{X}) = \sqrt{E(\bar{X} - \mu_{\bar{X}})^{2}} \\ \therefore & SE_{D_{\bar{X}}} = D(\bar{X}_{1} - \bar{X}_{2}) = \sqrt{E(\bar{X}_{1} - \bar{X}_{2} - (\mu_{\bar{X}_{1}} - \mu_{\bar{X})2}))^{2}} \\ & = \sqrt{E((\bar{X}_{1} - \mu_{\bar{X}_{1}}) - (\bar{X}_{2} - \mu_{\bar{X}_{2}}))^{2}} \\ & = \sqrt{E(\bar{X}_{1} - \mu_{\bar{X}_{1}})^{2} + E(\bar{X}_{2} - \mu_{\bar{X}_{2}})^{2} - 2 \cdot E((\bar{X}_{1} - \mu_{\bar{X}_{1}})(\bar{X}_{2} - \mu_{\bar{X}_{2}}))}\\ & = \sqrt{Var(\bar{X}_{1})^{2} + Var(\bar{X}_{2})^{2} - 2\cdot Cov(\bar{X}_{1}, \bar{X}_{2})} \\ \because & \text{The two samples } X_{1} \text{ and } X_{2} \text{ are independent.} \\ \therefore & Cov(\bar{X}_{1}, \bar{X}_{2}) = 0 \\ \therefore & SE_{D_{\bar{X}}} = D(\bar{X}_{1} - \bar{X}_{2}) = \sqrt{Var(\bar{X}_{1}) + Var(\bar{X}_{2})} = \sqrt{\frac{\sigma^{2}_{1}}{n_{1}} + \frac{\sigma^{2}_{2}}{n_{2}}} \\ \end{array}$$

• 使用Z检验 $Z = \frac{D_{\bar{X}} - \mu_{D_{\bar{X}}}}{SE_{D_{\bar{X}}}}$

• 一般情况下要判断两个样本的均值差异是否显著，则$\mu_{D_{\bar{X}}} = 0$

相关样本的平均数差异检验

• 标准误 $SE_{D_{\bar{X}}} = \sqrt{\frac{\sigma_{1}^{2}}{n} + \frac{\sigma^{2}_{2}}{n} - 2r \cdot \frac{\sigma_{1}}{\sqrt{n}} \cdot \frac{\sigma_{2}}{\sqrt{n}}}$

  proof

  $$\begin{array}{ll} \because & D(\bar{X}) = \sqrt{E(\bar{X} - \mu_{\bar{X}})^{2}} \\ \therefore & SE_{D_{\bar{X}}} = D(\bar{X}_{1} - \bar{X}_{2}) = \sqrt{E(\bar{X}_{1} - \bar{X}_{2} - (\mu_{\bar{X}_{1}} - \mu_{\bar{X})2}))^{2}} \\ & = \sqrt{E((\bar{X}_{1} - \mu_{\bar{X}_{1}}) - (\bar{X}_{2} - \mu_{\bar{X}_{2}}))^{2}} \\ & = \sqrt{E(\bar{X}_{1} - \mu_{\bar{X}_{1}})^{2} + E(\bar{X}_{2} - \mu_{\bar{X}_{2}})^{2} - 2 \cdot E((\bar{X}_{1} - \mu_{\bar{X}_{1}})(\bar{X}_{2} - \mu_{\bar{X}_{2}}))}\\ & = \sqrt{Var(\bar{X}_{1})^{2} + Var(\bar{X}_{2})^{2} - 2\cdot Cov(\bar{X}_{1}, \bar{X}_{2})} \\ \because & Cov(\bar{X}, \bar{Y}) = Cov(\frac{1}{n}\sum_{i=1}^{n}X_{i}, \frac{1}{n}\sum_{j=1}^{n}Y_{j}) = \frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}Cov(X_{i}, Y_{j})\\ \because & \text{When } i \ne j \text{ } X_{i} \text{ and } Y_{j} \text{ are independent.} \\ \therefore & Cov(\bar{X},\bar{Y}) = \frac{1}{n^{2}} \sum_{i=i}^{n} Cov(X_{i}, Y_{i}) \\ \therefore & Cov(\bar{X}, \bar{Y}) = \frac{Cov(X,Y)}{n} \\ \because & r_{\bar{X},\bar{Y}} = \frac{Cov(X,Y)/n}{\sqrt{(Var(X)/n)\cdot (Var(Y)/n)}} = \frac{Cov(X,Y)}{\sqrt{Var(X)\cdot Var(Y)}} = r_{X,Y} \\ \therefore & Cov(\bar{X},\bar{Y}) = r_{X,Y}\cdot \sqrt{Var(\bar{X})\cdot Var(\bar{Y})} \\ \therefore & SE_{D_{\bar{X}}} = D(\bar{X}_{1} - \bar{X}_{2}) = \sqrt{Var(\bar{X}_{1}) + Var(\bar{X}_{2}) - 2r_{X_{1}, X_{2}}\sqrt{Var(\bar{X}_{1})Var(\bar{X}_{2})}} \\ & = \sqrt{\frac{\sigma^{2}_{1}}{n_{1}} + \frac{\sigma^{2}_{2}}{n_{2}} - 2r_{X_{1}, X_{2}} \cdot \frac{\sigma_{1}}{\sqrt{n_{1}}} \cdot \frac{\sigma_{2}}{\sqrt{n}_{2}}} \\ \because & n_{1} = n_{2} = n \\ \therefore & SE_{D_{\bar{X}}} = \sqrt{\frac{\sigma_{1}^{2}}{n} + \frac{\sigma^{2}_{2}}{n} - 2r_{X_{1}, X_{2}}\cdot \frac{\sigma_{1}}{\sqrt{n}} \cdot \frac{\sigma_{2}}{\sqrt{n}}} \\ \end{array}$$

• 使用Z检验 $Z = \frac{D_{\bar{X}} - \mu_{D_{\bar{X}}}}{SE_{D_{\bar{X}}}}$

两总体都是正态，两总体方差未知

• 两个样本平均数之间差异记为$D_{\bar{X}}$，分布为t分布，$\mu_{D_{\bar{X}}} = \mu_{1} - \mu_{2}$
• 对两个样本平均数差异$(\bar{X}_{1} - \bar{X}_{2})$的显著性检验实际上就是对$D_{\bar{X}}$与$\mu_{D_{\bar{X}}} = 0$差异的检验。

独立样本的平均数差异检验

1. 两个总体方差一致或相等
   • 标准误 $SE_{D_{\bar{X}}} = \sqrt{\frac{n_{1}s_{1}^{2} + n_{2}s_{2}^{2}}{n_{1} + n_{2} -2}\big(\frac{1}{n_{1}} + \frac{1}{n_{2}}\big)}$

     proof

     $$\begin{array}{ll} \because & \text{The variance of two samples are homogeneity.} \\ \therefore & SE_{\bar{D}_{\bar{X}}} = \sqrt{ \frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma^{2}_{2}}{n_{2}}} = \sqrt{\sigma_{0}^{2} \big(\frac{1}{n_{1}} + \frac{1}{n_{2}})} \\ \because & \sigma^{2}_{0} \text{ is the unbiased estimation of the population consist of } X_{1} \text{ and } X_{2} \\ \therefore & \sigma^{2}_{0} = \frac{\sum_{i=1}^{n_{1}}(X_{1i} - \bar{X}_{1})^{2} + \sum_{i=1}^{n_{2}}(X_{2i} - \bar{X}_{2})^{2}}{n_{1} + n_{2} - 2} = \frac{n1\cdot s_{1}^{2}+n_{2}\cdot s_{2}^{2}}{n_{1} + n_{2} - 2} \\ \therefore & SE_{D_{\bar{X}}} = \sqrt{\frac{n_{1} \cdot s_{1}^{2} + n_{2} \cdot s_{2}^{2}}{n_{1} + n_{2} - 2}\big(\frac{1}{n_{1}} + \frac{1}{n_{2}}\big)} \\ \end{array}$$

   • $t = \frac{\bar{X}_{1} - \bar{X}_{2}}{SE_{D_{\bar{X}}}} \sim t(n_{1} + n_{2} - 2)$
2. 两个总体方差不齐
   • 不考

相关样本的平均数差异检验

1. 相关系数未知
   • 标准误 $SE_{D_{\bar{X}}} = \sqrt{\frac{s_{d}^{2}}{n-1}} \quad \text{for } s_{d}^{2} \text{ is the variance of sample's } d.$
   • $t = \frac{\bar{X}_{1} - \bar{X}_{2}}{SE_{D_{\bar{X}}}} \sim t(n-1)$
2. 相关系数已知
   • 标准误 $SE_{D_{\bar{X}}} = \sqrt{\frac{s_{d}^{2}}{n-1}} = \sqrt{\frac{s_{1}^{2} + s_{2}^{2} - 2r\cdot s_{1}s_{2}}{n-1}}$
   • $t = \frac{\bar{X}_{1} - \bar{X}_{2}}{SE_{D_{\bar{X}}}} \sim t(n-1)$

两总体非正态，大样本$(n>30)$

1. 独立样本的的平均数差异检验
   • $D_{\bar{X}} \sim \mathcal{N}(0, \frac{\sigma^{2}_{1}}{n_{1}} + \frac{\sigma^{2}_{2}}{n_{2}})$ $$Z' = \frac{\bar{X}_{1} - \bar{X}_{2}}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}} + \frac{\sigma^{2}_{2}}{n_{2}}}} \quad \text{or} \quad Z' = \frac{\bar{X}_{1} - \bar{X}_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}} + \frac{s^{2}_{2}}{n_{2}}}}$$
   • 注 此处的$s_{1}、s_{2}$为样本方差而非总体的无偏估计
2. 相关样本的的平均数差异检验
   • $D_{\bar{X}} \sim \mathcal{N}(0, \frac{\sigma^{2}_{1} + \sigma^{2}_{2} - 2r\cdot\sigma_{1}\sigma_{2}}{n})$ $$Z' = \frac{\bar{X}_{1} - \bar{X}_{2}}{\sqrt{\frac{\sigma_{1}^{2} + \sigma_{2}^{2} - 2r\cdot\sigma_{1}\sigma_{2}}{n}}} \quad \text{or} \quad Z' = \frac{\bar{X}_{1} - \bar{X}_{2}}{\sqrt{\frac{s_{1}^{2} + s_{2}^{2} - 2r\cdot s_{1}s_{2}}{n}}}$$
   • 注 此处的$s_{1}、s_{2}$为样本方差而非总体的无偏估计