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🌊Yuri🌊 · 2022年12月13日

问

  1. CFA I
  2. Quantitative

NO.PZ2017092702000114

问题如下：

For a sample size of 65 with a mean of 31 taken from a normally distributed population with a variance of 529, a 99% confidence interval for the population mean will have a lower limit closest to:

选项：

A.

23.64.

B.

25.41.

C.

30.09.

解释：

A is correct.

To solve, use the structure of Confidence interval = Point estimate ± Reliability factor × Standard error, which, for a normally distributed population with known variance, is represented by the following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}

For a 99% confidence interval, use z0.005 = 2.58. Also, σ = 529\sqrt{529} = 23.

Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.6398

我们需要用到【置信区间结构】的计算公式解决本题：

the structure of Confidence interval = Point estimate ± Reliability factor × Standard error

即：X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}

当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529} = 23.

所以，下限为： =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398

请问老师，我有两个问题，第一个是如果题干没有告诉我应该用到什么分布的话，我怎么判断是应该用z分布还是t分布呢？

第二个问题是，我现在很混乱z分布和t分布的计算有什么区别呢》唯一的区别就是t分布的计算是自由度-1是吗，z分布就直接用sample zise的数量？

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1 个答案

星星_品职助教 · 2022年12月13日

同学你好，

1）根据原则“总体方差已知用z，总体方差未知用t ”来判断。也就是看题干中给的已知条件是总体标准差σ（用z分布），还是样本标准差s（用t分布）；

2）两个公式的总体形式都是一样的（见截图文字部分）。区别在于：①根据上一条原则，可知z分布计算用σ，t分布计算用s；②不同分布的关键值不同，分别是t α/2和z α/2。

3）t α/2和z α/2都是查表求得，t分布查表需要自由度n-1；z分布查表不需要自由度。绝大多数情况下，z分布的关键值可以直接使用背诵的一组关键值1.65/1.96/2.58，不用专门再查一遍。

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2023-06-06 23:49 1 · 回答

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2023-04-12 16:32 1 · 回答

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2022-11-06 16:45 1 · 回答

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2022-07-07 15:19 1 · 回答