(1)(2018104)设总体 sim N(mu ,(sigma )^2) ,x1,x2,···,xn是取自总体X的简单随机样本,据-|||-此样本检验假设 _(0):mu =(mu )_(0) , _(1):mu neq (mu )_(0) ,若显著性水平为α,则 ()-|||-A.如果在 =0.05 下拒绝H0,那么在 alpha =0.01 下必拒绝H0-|||-B.如果在 =0.05 下拒绝H0,那么在 =0.01 下必接受H0-|||-C.如果在 alpha =0.05 下接受H0,那么在 alpha =0.01 下必拒绝H0-|||-D.如果在 =0.05 下接受H0,那么在 =0.01 下必接受H0

统计学上将()的事件称为小概率事件，表示该事件发生的可能性很小。A. P≤0.50B. P≤0.05C. P≤0.005D. P≤0.05或P≤0.01

常模样本对于所测群体应当具有较好的( )。A. 跨时空性B. 同一性C. 非限制性D. 代表性

设 sim N(mu ,(sigma )^2) ,σ^2未知,X为样本均值,S^2为样本方差,则μ的置信度为-|||-95%的置信区间为 () .-|||-(A) (overline (X)-dfrac (sigma )(sqrt {n)}Z0.05,overline (X)+dfrac (sigma )(sqrt {n)}Z0.05) (B) (overline (X)-dfrac (sigma )(sqrt {n)}Z0.025,overline (X)+dfrac (sigma )(sqrt {n)}Z0.025)-|||-(C) (overline (X)-dfrac (S)(sqrt {n)}(t)_(0.05),overline (X)+dfrac (S)(sqrt {n)}(t)_(0.05)) (D) (overline (X)-dfrac (S)(sqrt {n)}(t)_(0.025),overline (X)+dfrac (S)(sqrt {n)}(t)_(0.025))

【单选题】描述正态分布资料的变异程度,用下列哪个指标表示较好A. 全距B. 标准差C. 平均数D. 变异系数E. 四分位数间距

3.(2021,Ⅲ)设(X1 Y1),(X2,Y2),···,(Xn,Yn)为来自总体N(μ1,-|||-μ2;σ1^2,σ2^2,ρ)的简单随机样本,令 theta =(mu )_(1)-(mu )_(2), overline (X)=dfrac (1)(n)sum _(i=1)^nX overline (Y)=dfrac (1)(n)sum _(i=1)^n(Y)_(i),-|||-hat (theta )=overline (X)-overline (Y), 则 () .-|||-(A) (hat (theta ))=theta ,D(hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2}(n)-|||-(B) (hat (theta ))=theta ,D(hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2-2(rho )_(2)(Q)_(2)}(n)-|||-(C) (theta )neq theta , (hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2}(n)-|||-(D) (theta )neq theta , (hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2-2rho (sigma )_(1)(sigma )_(2)}}(n)

进行四个样本率比较的χ 2 检验,如χ 2>A. 各总体率不同或不全相同B. 有三个样本率不相同C. 各样本率均不相同D. 各样本率不同或不全相同E. 各总体率均不相同

设总体X的分布律为 X=1 =theta ， X=1 =theta ，现检验 X=1 =theta ，已知 X=1 =theta 是来自总体的简单随机样本，且拒绝域为 X=1 =theta ，则犯第二类错误的概率为_______.A.0.007B.0.016C.0.984D.0.968

5. 仿样本X1,X2,X是来自总体X,E(X)存在,问 C= ()-|||-时, (X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (1)(6)(X)_(3) 是E(X)的无偏估计量?-|||-(A) -dfrac (13)(30) (B) dfrac (7)(30) (C) dfrac (13)(30) (D) -dfrac (7)(30)

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)=(1)/(n)sum_(i=1)^nX_(i)，则theta的极大似然估计为：（）A. overline(X)B. overline(X)-1C. overline(X)-2D. overline(X)-3