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：

13.23.

13.27.

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).

老师，这道题的SAMPLE SIZE是17，应该是大于30，才服从T分布吧？

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1 个答案

星星_品职助教 · 2020年03月27日

同学你好，

是否服从t分布和sample size无关。N大于30只是t分布趋近于正态分布的条件。对于本题而言，题干中已经说明了要求用t分布，直接用t分布即可。

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问一道题：NO.PZ2017092702000113

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