题目
3 (2023,9题,5分)设X_(1),X_(2),...,X_(n)为来自总体N(mu_(1),sigma^2)的简单随机样本，Y_(1),Y_(2),...,为来自总体N(mu_(2),2sigma^2)的简单随机样本，且两样本相互独立.记overline(X)=(1)/(n)sum_(i=1)^nX_(i),overline(Y)=(1)/(m)sum_(i=1)^mY_(i), S_(1)^2=(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2,S_(2)^2=(1)/(m-1)sum_(i=1)^m(Y_(i)-overline(Y))^2,则(A)(S_(1)^2)/(S_(2)^2)sim F(n,m). (B)(S_(1)^2)/(S_(2)^2)sim F(n-1,m-1).(C)(2S_(1)^2)/(S_(2)^2)sim F(n,m). (D)(2S_(1)^2)/(S_(2)^2)sim F(n-1,m-1).
3 (2023,9题,5分)设$X_{1},X_{2},\cdots,X_{n}$为来自总体$N(\mu_{1},\sigma^{2})$的简单随机样本，$Y_{1},Y_{2},\cdots,$为来自总体$N(\mu_{2},2\sigma^{2})$的简单随机样本，且两样本相互独立.记$\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i},\overline{Y}=\frac{1}{m}\sum_{i=1}^{m}Y_{i},$ $S_{1}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\overline{X})^{2},S_{2}^{2}=\frac{1}{m-1}\sum_{i=1}^{m}(Y_{i}-\overline{Y})^{2},$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)$\frac{2S_{1}^{2}}{S_{2}^{2}}\sim F(n,m).$ (D)$\frac{2S_{1}^{2}}{S_{2}^{2}}\sim F(n-1,m-1).$
题目解答
答案
设 $X_1, X_2, \cdots, X_n$ 来自 $N(\mu_1, \sigma^2)$，$Y_1, Y_2, \cdots, Y_m$ 来自 $N(\mu_2, 2\sigma^2)$，两样本独立。
样本方差 $S_1^2$ 和 $S_2^2$ 满足：
$\frac{(n-1)S_1^2}{\sigma^2} \sim \chi^2(n-1), \quad \frac{(m-1)S_2^2}{2\sigma^2} \sim \chi^2(m-1)$
由 F 分布定义，若 $U \sim \chi^2(k)$，$V \sim \chi^2(l)$ 独立，则 $\frac{U/k}{V/l} \sim F(k, l)$。
令 $U = \frac{(n-1)S_1^2}{\sigma^2}$，$V = \frac{(m-1)S_2^2}{2\sigma^2}$，则：
$\frac{U/(n-1)}{V/(m-1)} = \frac{2S_1^2}{S_2^2} \sim F(n-1, m-1)$
答案： $\boxed{D}$