设 $X_{1}, X_{2}, \cdots, X_{n}$ 为来自总体 $N\left(\mu_{1}, \sigma^{2}\right)$ 的简单随机样本, $Y_{1}, Y_{2}, \cdots, Y_{m}$ 为来自
总体 $N\left(\mu_{2}, 2 \sigma^{2}\right)$ 的简单随机样本, 且两样本相互独立, 记 $\bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$, $\bar{Y}=\frac{1}{m} \sum_{i=1}^{m} Y_{i}, \quad S_{1}^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}, \quad S_{2}^{2}=\frac{1}{m-1} \sum_{i=1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}$, 则( )
(A) $\frac{S_{1}^{2}}{S_{2}^{2}} \sim F(n, m)$
(B) $\frac{S_{1}^{2}}{S_{2}^{2}} \sim F(n-1, m-1)$
(C) $\frac{2 S_{1}^{2}}{S_{2}^{2}} \sim F(n, m)$
(D) $\frac{2 S_{1}^{2}}{S_{2}^{2}} \sim F(n-1, m-1)$
$X_{1}, X_{2}, \ldots, X_{n}$ 的样本方差 $S_{1}^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}$, $Y_{1}, Y_{2}, \ldots, Y_{n}$ 的样本方差 $S_{2}^{2}=\frac{1}{m-1} \sum_{i=1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}$, 则 $\frac{(n-1) S_{1}^{2}}{\sigma^{2}} \sim \chi^{2}(n-1)$, $\frac{(m-1) S_{2}^{2}}{2 \sigma^{2}} \sim \chi^{2}(m-1)$, 两个样本相互独立,

所以 $\frac{\frac{(n-1) S_{1}^{2}}{\sigma^{2}} /(n-1)}{\frac{(m-1) S_{2}^{2}}{2 \sigma^{2}} /(m-1)}=\frac{S_{1}^{2} / \sigma^{2}}{S_{2}^{2} / 2 \sigma^{2}}=\frac{2 S_{1}^{2}}{S_{2}^{2}} \sim F(n-1, m-1)$,

故选(D).