Hugo(Xie Lanzhi) · 2020年05月16日
问题如下：
When sampling size is 50 with mean is 78, and population variance is unknown, which of the following distribution is taken if sampling forms a normal distribution?
选项：
A.z-statistic only
B.both t or z statistic
C.t-statistic only
解释：
B is correct, when sampling form a normal distribution with unknown variance, if sample size is larger than 30, both t or z-statistic is choose.
方差已知用z 方差未知用t，还有最后一句话是两个检验方法都不适用，是怎么说来着？有点忘了
NO.PZ2018062016000113 both t or z statistic t-statistic only B is correct, when sampling form a normstribution with unknown variance, if sample size is larger th30, both t or z-statistic is choose. 当样本形成方差未知的正态分布时，若样本量大于 30，则选择 t 或 z 统计量。 你好老师，上课的时候反复说方差已知用z，方差未知用t，这道题怎么？
NO.PZ2018062016000113 both t or z statistic t-statistic only B is correct, when sampling form a normstribution with unknown variance, if sample size is larger th30, both t or z-statistic is choose. 当样本形成方差未知的正态分布时，若样本量大于 30，则选择 t 或 z 统计量。 解析中，怎么断定样本量大于30呢？题目中给的是sample size，意思是取样时的数量，n的意思不是取样的次数吗？也就是sample1，sample2…sample50
老师，不是unknown variance都是T分布吗
知道符合正态分布的情况下，样本数量大于30，应该用Z，所以应该A是正确吧？假若不知道fu'he符合正态分布，样本数量大约30，才选择B。我的理解是否正确？
