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Shawnthesheep · 2020年10月22日

问一道题：NO.PZ2015120604000125

  1. CFA I
  2. Quantitative

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.

可以具体讲一下什么情况的standard error要除以根号n，什么时候不用吗？不是特别懂

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星星_品职助教 · 2020年10月22日

同学你好，

除以根号n的只有一种情况，就是当题干同时给出sample standard deviation（s），和sample size（n），要求计算standard error的时候。

这个时候standard error=s/√n。所以要观察题干有没有给出这两个条件。

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但是当题干给的条件本身就已经是standard error的时候，就不要再去除以根号n了，例如这道题考察置信区间的公式，置信区间的公式里就是X bar±critical value×standard error，这个时候代入公式求出来的就已经是standard error，自然就不用再去除以根号n了。

77 · 2021年02月10日

A sample with 100 size， 这个不是指n=100吗

星星_品职助教 · 2021年02月10日

@77 是指n=100，这个条件在这道题里没用。

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NO.PZ2015120604000125问题如下 The population is 6000 programmer whiis supposeto normally stribute A sample with 100 size is awn from the population. Baseon z-statisti95% confinintervof sample meof annusalary is 32.5 (in thousan) llars ranges from 22 (in thousan) llars to 43 (in thousan) llars .Calculate the stanrerror of meannusalary: A.1.96.B.3.99.C.5.36. C is correct.the 95% level of significance, the criticvalue is ±1.96.So the confinintervis 32.5±1.96 σ x ¯ , From the equation t 32.5 + 1.96 stanrerror = 43 or 32.5 - 1.96 stanrerror x ¯ = 22, we get σ x ¯ =5.36.=5.3571. 根据题干的sample mean=32.5，基于z-statistics的95%的置信区间（得到关键值±1.96），和置信区间的公式，可以得到 32.5 + 1.96 stanrerror = 43 或 32.5 - 1.96 stanrerror = 22。得到 stanrerror=5.3571 我可能公式有点搞混了，请问置信区间的计算什么时候使用 平均值+z×标准差，什么时候使用平均值+z×标准误

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NO.PZ2015120604000125 问题如下 The population is 6000 programmer whiis supposeto normally stribute A sample with 100 size is awn from the population. Baseon z-statisti95% confinintervof sample meof annusalary is 32.5 (in thousan) llars ranges from 22 (in thousan) llars to 43 (in thousan) llars .Calculate the stanrerror of meannusalary: A.1.96. B.3.99. C.5.36. C is correct.the 95% level of significance, the criticvalue is ±1.96.So the confinintervis 32.5±1.96 σ x ¯ , From the equation t 32.5 + 1.96 stanrerror = 43 or 32.5 - 1.96 stanrerror x ¯ = 22, we get σ x ¯ =5.36.=5.3571. 根据题干的sample mean=32.5，基于z-statistics的95%的置信区间（得到关键值±1.96），和置信区间的公式，可以得到 32.5 + 1.96 stanrerror = 43 或 32.5 - 1.96 stanrerror = 22。得到 stanrerror=5.3571 没看懂这道题啥意思，老师可以再讲一下吗

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