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Grace Zhu · 2020年04月13日

问一道题:NO.PZ2017092702000113

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: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 17, degrees of freedom equal 16, so t0.05 = 1.746. The confidence interval is calculated as

116.23 ± 1.74615.6717 = 116.23 ± 6.6357116.23\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).

什么情况下z a/2, T a/2 的置信区间直接用,比如99%=2.58,什么时候要查表?

1 个答案

星星_品职助教 · 2020年04月14日

同学你好,

上课要求背诵的1.65,1.96,2.58这些值也都是查表得到的。适用于正态分布。

例如这道题,如果题干说求90%正态分布的置信区间,则直接应用1.65即可。但这道题说的是t分布,且题干没有说可以用正态分布的临界值替代t分布的临界值,这个时候就要根据自由度重新查表了。这道题N=17,属于小样本,所以查表值1.746和1.65有一定差距,如果是N很大,t分布是趋近于正态分布的,最后的结果和1.65也不会差很远的。

 

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