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：

- 10.88% to 19.13%.

10.88% to 19.13%.

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]=

$\bar{x} \pm z_{α / 2} \frac{σ}{\sqrt{n}} = 15 % \pm 1.65 \times \frac{25 %}{\sqrt{100}} = [10.88 %, 19.13 %]$

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

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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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代入选项算的解题方法是否可行？

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