例 6.3.1 (分位数求解)

查表或用 R 软件,求

(1) ${\chi }_{0.99}^2\left(10\right),{\chi }_{0.05}^2\left(20\right)$ 和 ${\chi }_{0.95}^2\left(60\right)$.

(2) $t_{0.95}\left(10\right),t_{0.10}\left(20\right)$ 和 $t_{0.90}\left(50\right)$.

(3) $F_{0.99}\left(5,4\right),F_{0.05}\left(3,7\right)$.

解

(1)

$${\chi }_{0.99}^2\left(10\right)=\mathrm{qchisq}\left(0.99,10\right)=23.20925,$$ $${\chi }_{0.05}^2\left(20\right)=\mathrm{qchisq}\left(0.05,20\right)=10.85081,$$ $${\chi }_{0.95}^2\left(60\right)=\text{ qchisq }\left(0.95,60\right)=79.08194.$$

查表时,一般查不到 ${\chi }_{0.95}^2\left(60\right)$,而是用正态分布近似,即 ${\chi }_{\alpha }^2\left(n\right)\approx u_{\alpha }\sqrt{2n}+n$

$${\chi }_{0.95}^2\left(60\right)\approx \text{ qnorm }\left(0.95\right)\sqrt{2\times 60}+60=78.01847.$$

(2)

$$t_{0.95}\left(10\right)=\mathrm{qt}\left(0.95,10\right)=1.812461,$$ $$t_{0.10}\left(20\right)=\mathrm{qt}\left(0.10,20\right)=-1.325341,$$ $$t_{0.90}\left(50\right)=\mathrm{qt}\left(0.90,50\right)=1.298714.$$

查表时,一般查不到 $t_{0.90}\left(50\right)$,而是用正态分布近似,即 $t_{\alpha }\left(n\right)\approx u_{\alpha }$.

$$t_{0.90}\left(50\right)\approx u_{0.90}=\text{ qnorm }\left(0.90\right)=1.281552.$$

(3)

$$F_{0.99}\left(5,4\right)=\mathrm{qf}\left(0.99,5,4\right)=15.52186,$$ $$F_{0.05}\left(3,7\right)=\mathrm{qf}\left(0.05,3,7\right)=0.1125272.$$

查表时,一般查不到 $F_{0.05}\left(3,7\right)$,而是用 $F_{\alpha }\left(m,n\right)=1/F_{1-\alpha }\left(n,m\right)$,

$$F_{0.05}\left(3,7\right)=\frac{1}{F_{0.95}\left(7,3\right)}=\frac{1}{\mathrm{qf}\left(0.95,7,3\right)}=0.1125272.$$