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曾兰芝🍒 · 2021年02月17日

老师，为什么题干这种检验师t分布，5年师60个月，n足够大，不是可以认为是z分布吗。。。。

  1. CFA I
  2. Quantitative

NO.PZ2015120604000125

问题如下：

The population is 6000 programmer which is supposed to be normally distributed. A sample with 100 size is drawn from the population. Based on z-statistic, 95% confidence interval of sample mean of annual salary is 32.5 (in thousands) dollars ranges from 22 (in thousands) dollars to 43 (in thousands) dollars .Calculate the standard error of mean annual salary:

选项：

A.

1.96.

B.

3.99.

C.

5.36.

解释：

C is correct.

At the 95% level of significance, the critical value is z α/2 =1.96 .

So the confidence interval is 32.5±1.96 σ x ¯ ,

From the equation t 32.5 + 1.96 σ x ¯ = 43 or 32.5 - 1.96 σ x ¯ = 22, we get σ x ¯ =5.36.

我不知道是我哪里的知识缺了一块吗。为什么根据z表，95%的置信区间求出来的critical value 是1.96.。。。不应该是1.65吗。。。。。

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星星_品职助教 · 2021年02月17日

同学你好，

正态分布下，95%对应的置信区间的关键值是±1.96；

±1.65对应的是90%的置信区间。

这一组值是要求背诵的，不用再去查表了。但如果要查表自己验证的话，需要注意z表是累积概率，即关键值对应的是其左侧的所有面积。所以要查95%置信区间的关键值，不能对应95%的概率面积，而要对应的是97.5%的。

附上讲义和z表查表结果作参考。


曾兰芝🍒 · 2021年02月17日

懂了，谢谢老师！！！

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