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🌅 🌅多吉旺珍🌞🌞 · 2020年07月28日

问一道题：NO.PZ2015120604000125 [ CFA I ]

  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 K dollers ranges from 22 K dollars to 43 K 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.

标准差 标准误 傻傻分不清楚了 有什么好的方法吗
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1 个答案

星星_品职助教 · 2020年07月29日

同学你好，

标准误也是标准差的概念，只不过标准误是“样本统计量”的标准差。

例如现在有一个总体，那么就有总体的均值和标准差σ；有一个样本，样本就有样本的均值和样本的标准差s，而标准误是“样本均值”的标准差的意思。所以主要看的是最终求的是谁的标准差。

 

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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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