问题如下图：
选项：
A. 
B. 
C. 
解释：
NO.PZ2015120604000140 问题如下 Here is a table shows a set of tPeter sets up a null hypothesis--- H0: μ≥8.5. Whis the calculatetest statisti A.1.7877 B.-1.7877 C.2.3656 A is correct.The test statisti z= (8.9−8.5) (1.79/ 64 ) =1.7877. 不是单边检验么 备择假设是小于号 应该是选择左边的统计量 不应该是负数么
NO.PZ2015120604000140问题如下Here is a table shows a set of tPeter sets up a null hypothesis--- H0: μ≥8.5. Whis the calculatetest statistic?A.1.7877B.-1.7877C.2.3656A is correct.The test statisti z= (8.9−8.5) (1.79/ 64 ) =1.7877.分子为什么要用样本的均值减去假设的数字而不是用总体的均值减假设的数
NO.PZ2015120604000140 -1.7877 2.3656 A is correct. The test statistiz= (8.9−8.5) (1.79/ 64 ) =1.7877.样本标准差和总体的标准差什么时候可以混用呢
NO.PZ2015120604000140不应该是样本均值减去假设数么 分子也就是8.9-8.5，所以应该是正的啊？
