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卓娅 · 2021年11月29日

置信区间

  1. CFA I
  2. Quantitative

NO.PZ2015120604000130

问题如下：

We want to use z-statistic to construct confidence interval for a normally distribution. Assume the sample size is 100, sample mean is 15% and the standard deviation of sample is 25%. The significance level is supposed to be 10% , the confidence interval is  :

选项：

A.

- 10.88% to 19.13%.

B.

10.88% to 19.13%.

C.

10.88% to 20.57%.

解释：

B is correct.

In this case, we can know that the Confidence Interval=[Point Estimate +/- (reliability factor)*Standard error]=

x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%

请老师说说这题的解题思路，谢谢诶

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2 个答案

星星_品职助教 · 2021年11月29日

同学你好，

根据“... use z-statistic”和“The significance level is supposed to be 10%”得到本题的critical value 或者叫 reliability factor为1.65.

由于要构建sample mean的置信区间，所以根据中心极限定理得到sample mean的标准差（即标准误）为25%/根号100=2.5%

所以最终的置信区间根据公式“Confidence Interval=Point Estimate +/- (reliability factor)*Standard error”，可得：

15%±1.65×2.5%=10.875% to 19.125%。选择B选项

星星_品职助教 · 2021年11月29日

同学你好，

根据“... use z-statistic”和“The significance level is supposed to be 10%”得到本题的critical value 或者叫 reliability factor为1.65.

由于要构建sample mean的置信区间，所以根据中心极限定理得到sample mean的标准差（即标准误）为25%/根号100=2.5%

所以最终的置信区间根据公式“Confidence Interval=Point Estimate +/- (reliability factor)*Standard error”，可得：

15%±1.65×2.5%=10.875% to 19.125%。选择B选项

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