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🌊Yuri🌊 · 2022年12月13日

问

  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）。

为什么p等于0.05

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1 个答案

星星_品职助教 · 2022年12月13日

同学你好，

p=0.05代表t表中，右侧单尾的（概率）面积为0.05。

本题给出条件为“ a 90% confidence interval”。数量科目中，置信区间全部都是位于分布中间的情况，所以此时两边尾部的总面积为1-90%=10%。

根据对称性可以得到单侧尾部面积为10%/2=5%，也就是0.05.

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