A. $X^{2}(n-1)$
B. $N(0,1)$
C. $t(n-1)$
D. $t(R)$
1.6 总体X-N(mu,sigma^2),x_(1),x_(2),...,x_(n)为其样本,bar(x)=(1)/(n)sum_(i=1)^nx_(i),s
12.设x_(1),x_(2),...,x_(n),x_(n+1)是来自N(mu,sigma^2)的样本,overline(x)_(n)=(1)/(n)sum_
17.设x_(1),x_(2),...,x_(n),x_(n+1)是来自N(mu,sigma^2)的样本,又设overline(x)_(n)=(1)/(n)su
设X_(1),X_(2),...,X_(n)为总体X的简单样本,则样本均值overline(X)=(1)/(n)sum_(i=1)^nX_(i).A. 对B.
3.设n个随机变量X_(1),X_(2),...,X_(n)独立同分布,D(X_(1))=sigma^2,overline(X)=(1)/(n)sum_(i=1
设总体 X sim N(mu, sigma^2), X_(1), X_(2), ..., X_(n) 为来自总体X的简单随机样本,则 sum_(i=1)^n((
6.设总体Xsim N(mu,sigma^2),X_(1),X_(2),...,X_(20)为其样本,S^2=(1)/(19)sum_(i=1)^20(X_(i
5、设X_(1),X_(2),...,X_(n)是正态总体N(mu,sigma^2)的一个样本,S^2=(1)/(n-1)sum_(i=1)^n(X_(i)-o
设X_(1),X_(2)...,X_(n)是来自总体X的样本,则(1)/(n-1)sum_(i=1)^n(X_(i)-overline(X))^2为().A.
3.设总体Xsim N(theta+3,1),theta为未知参数,X_(1),X_(2)...X_(n)是来自该总体的样本,样本均值overline(X)=(