3. 方差比值的区间估计 (已知期望)

由于

• $\frac{X_i-{\mu }_1}{{\sigma }_1}\sim N\left(0,1\right)$, $i=1,2,\cdots ,m$.
• $\frac{Y_j-{\mu }_2}{{\sigma }_2}\sim N\left(0,1\right)$, $j=1,2,\cdots ,n$,

所以

• $\frac{\sum _{i=1}^m{\left(X_i-{\mu }_1\right)}^2}{{\sigma }_1^2}\sim {\chi }^2\left(m\right)$,
• $\frac{\sum _{j=1}^n{\left(Y_j-{\mu }_2\right)}^2}{{\sigma }_2^2}\sim {\chi }^2\left(n\right)$,

从而由 $F$ 分布的定义有

$$\frac{n}{m}\frac{\sum _{i=1}^m{\left(X_i-{\mu }_1\right)}^2}{\sum _{j=1}^n{\left(Y_j-{\mu }_2\right)}^2}\frac{{\sigma }_2^2}{{\sigma }_1^2}\sim F\left(m,n\right),$$

再令

$$P\left(F_{\frac{\alpha }{2}}\left(m,n\right)\le \frac{n}{m}\frac{\sum _{i=1}^m{\left(X_i-{\mu }_1\right)}^2}{\sum _{j=1}^n{\left(Y_j-{\mu }_2\right)}^2}\frac{{\sigma }_2^2}{{\sigma }_1^2}\le F_{1-\frac{\alpha }{2}}\left(m,n\right)\right)=1-\alpha ,$$

求出分位点 $F_{\frac{\alpha }{2}}\left(m,n\right)$ 和 $F_{1-\frac{\alpha }{2}}\left(m,n\right)$ 并作等式变形立得 ${\sigma }_1^2/{\sigma }_2^2$ 的区间估计为

$$\left[\frac{1}{F_{1-\frac{\alpha }{2}}\left(m,n\right)}\frac{n\sum _{i=1}^m{\left(X_i-{\mu }_1\right)}^2}{m\sum _{j=1}^n{\left(Y_j-{\mu }_2\right)}^2},\frac{1}{F_{\frac{\alpha }{2}}\left(m,n\right)}\frac{n\sum _{i=1}^m{\left(X_i-{\mu }_1\right)}^2}{m\sum _{j=1}^n{\left(Y_j-{\mu }_2\right)}^2}\right].\text{\hspace{1em}(7.3.11)}$$