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cui1130 · 2023年02月07日

没看懂这道题啥意思，老师可以再讲一下吗

  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 ±1.96.

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

From the equation t 32.5 + 1.96 standard error = 43 or 32.5 - 1.96 standard error 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 standard error = 43 或 32.5 - 1.96 standard error = 22。

得到 standard error=5.3571


没看懂这道题啥意思，老师可以再讲一下吗

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1 个答案

星星_品职助教 · 2023年02月07日

同学你好，

本题给出了置信区间上下限的数字，要计算样本均值的标准误（standard error）。

根据置信区间的公式，

可知置信区间=32.5±1.96*standard error。其中32.5 + 1.96 *standard error= 43 ；32.5 - 1.96standard error = 22。根据任意一个式子均可以解出standard error的值。

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此后提问需要具体标明具体是哪里不明白，和答案解析哪里没有看懂。

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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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