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Benny™ · 2022年11月04日

课上好像没怎么说这块

  1. CFA I
  2. Quantitative

NO.PZ2017092702000113

问题如下：

For a sample size of 37, with a mean of 116.23 and a variance of 245.55, the width of a 90% confidence interval using the appropriate t-distribution is closest to:

选项：

A.

8.5480.

B.

8.6970

C.

8.8456.

解释：

B is correct.

The confidence interval is calculated using the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frac s{\sqrt n}

Sample standard deviation (s) = 245.55\sqrt{245.55} = 15.670.

For a sample size of 37, degrees of freedom equal 36, so t0.05 = 1.688.

The confidence interval is calculated as：


Therefore, the interval spans 120.5785 to 111.8815, meaning its width is equal to approximately 8.6970. (This interval can be alternatively calculated as 4.3485 × 2).

样本标准差的计算如下：

245.55\sqrt{245.55} = 15.670.

当样本=37，自由度=36，那么 t0.05 = 1.688.

置信区间计算如下：


因此，置信区间为：111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。

tα/2是什么呢？课上好像没详细说这个公式

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1 个答案

星星_品职助教 · 2022年11月04日

同学你好，

本题的α为1-90%=10%。

由于数量科目中，所有的置信区间都对应双尾的情况，所以此时两侧尾部面积之和为10%。即单侧尾部面积为5%。

教材的t表是根据单侧尾部的面积去查表的，也就是根据5%来得到关键值。所以这个关键值就是t（5%），也就是t（10%/2）。推广一下就是t（α/2）

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