A. $\hat{\mu}_1, \hat{\mu}_2$ 为无偏估计量,$\hat{\mu}_2$ 是最有效的估计量
B. $\hat{\mu}_1, \hat{\mu}_2$ 为无偏估计量,$\hat{\mu}_3$ 是最有效的估计量
C. $\hat{\mu}_1, \hat{\mu}_3$ 为无偏估计量,$\hat{\mu}_1$ 是最有效的估计量
D. $\hat{\mu}_2$ 为无偏估计量,$\hat{\mu}_3$ 是最有效的估计量
设 X_1, X_2, X_3 为来自总体 X 的简单随机样本,其中 hat(mu)_1 = 0.4X_1 + 0.2X_2 + 0.4X_3,hat(mu)_
设X_1,X_2,...,X_n为来自总体X的简单随机样本,E(X)=mu,D(X)=sigma^2,记hat(mu)_1=(1)/(5)X_1+(3)/(10
设 X_1, X_2, X_3 为来自总体 X sim N(mu, sigma^2) 的一个样本,hat(mu)_1 = aX_1 + aX_2 + aX_3,
7.已知X1,X2为来自总体X的样本, (mu )_(1)=(X)_(1)+(X)_(2) , (mu )_(2)=0.2(X)_(1)+0.8(X)_(2)
5、设X_(1),X_(2),X_(3),X_(4)为来自总体X的样本,且EX=mu,记hat(mu)_(1)=(1)/(2)(X_(1)+X_(2)+X_(3
x1,x2是取自总体N(μ,1)(μ未知)的样本。hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;
设 X_(1),X_(2) 是来自正态总体 N(mu,1) 的样本,则对统计量 hat(mu)_(1)=(2)/(3)X_(1)+(1)/(3)X_(2), h
10.设总体Xsim N(mu,sigma^2),X_(1),X_(2)是来自总体X的样本,在mu的无偏估计量hat(mu)_(1)=(2)/(3)X_(1)+
设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+
样本X1,X2,X3来自总体X,若 hat (mu )=dfrac (1)(3)(X)_(1)+-|||-(X)_(2)+dfrac (1)(2)(X)_(3)