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Yukikiki · 2022年09月27日

代入选项算的解题方法是否可行?

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%

老师,既然题目说是正态分布,而且是选择题。想问下能否利用正态分布的特性,代入选项解题?正太分布中轴是均值,题干给出的置信区间应该也是轴对称。

直接带入选项,分位点-均值:19.13-15=4.13

均值-分位点到均值的距离:15-4.13=10.87

然后选出最符合的的选项


1 个答案
已采纳答案

星星_品职助教 · 2022年09月27日

同学你好,

本题可以用这种方式解,但这不是常规解法。如果选项中多出一项9.88% to 20.13%,这种方法就无法区分了。

这类题型的常规解法是直接代入置信区间的公式,建议用常规思路解题。

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