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维克多周 · 2020年06月13日

问一道题：NO.PZ2017092702000113 [ CFA I ]

问题如下：

For a sample size of 17, 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.

13.23.

B.

13.27.

C.

13.68.

解释：

B is correct.

The confidence interval is calculated using the following equation: $\overline X\pm t_{\alpha/2}\frac s{\sqrt n}$ Sample standard deviation (s) = $\sqrt{245.55}$ = 15.670. For a sample size of 17, degrees of freedom equal 16, so t_0.05= 1.746. The confidence interval is calculated as

$116.23\text{ }\pm\text{ }1.746\frac{15.67}{\sqrt{17}}\text{ }=\text{ }116.23\text{ }\pm\text{ }6.6357$ Therefore, the interval spans 109.5943 to 122.8656, meaning its width is equal to approximately 13.271. (This interval can be alternatively calculated as 6.6357 × 2).

看完了其他同学的提问和解答，还是没看明白为什么要用自由度来查表，而不是用17 0. 05来查，再有老师讲课的时候好像没怎么讲置信区间的计算公式吧

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星星_品职助教 · 2020年06月14日

同学你好，

自由度是t检验的参数。根据自由度和significance level α查表是t检验查表的规则。t表就是这么设定的，用的并不是sample size n。

置信区间的计算公式上课讲过，如果对于其中某个部分有问题可以继续提问。

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