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安在 · 2022年10月13日

求标准误不是有个公式么

  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


求标准误的公式这里不用吗

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星星_品职助教 · 2022年10月14日

同学你好，

本题并未给出总体标准差，无法使用中心极限定理对应的公式 σ/根号n

四喜 · 2022年10月14日

方差没有，怎么用公式，凑够十个字

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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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