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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）。

For a sample size of 37, degrees of freedom equal 36, so t0.05 = 1.688.请问这个1.688是怎么得来的？t分布只能查表吗

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2 个答案

星星_品职助教 · 2023年09月14日

@Katherine

1.65是正态分布下的关键值，本题为t分布，需要查表。

同样的问题提问一次即可。

星星_品职助教 · 2022年12月13日

同学你好，

1.688为关键值，需要通过查表求得。

90%的置信区间说明单尾面积为（1-90%）/2=5%，即对应t表中的p=0.05；自由度=37-1=36。查表结果如下：

Katherine · 2023年09月12日

为什么不能是1.65呢

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