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，查表 Degree of Freedom = 16, T0.05(two-tailed)= 2.12 不是 1.746呀（ T0.1)

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星星_品职助教 · 2019年11月01日

同学你好，

这道题需要用5%的单尾概率，而不是双尾。这么理解，90%的置信区间相当于剩余的两边面积一共是10%，那么单边就是5%了。所以是单尾的表，就要查df=16，单尾5%，原版书附录里t分布的表就是单尾的表，查出来恰好就是1.746。；双尾的表就要查df=16,双尾（一共）10%。

所以这就是自己查表的意义，因为考试时不一定会给单尾，还是双尾的表，还有可能是给出一堆描述：例如one tail t(16,5%）=1.746，t(17,5%）=1.740，或者two tail t(16,10%)=1.746这种关键值的形式，要求考生去选择。所以自己查表可以暴露出来平常之前没想到的问题，考前就可以给搞定~加油

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NO.PZ2017092702000113 问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480. B.8.6970 C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2ns Sample stanrviation (s) = 245.55\sqrt{245.55}245.55 = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}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）。为什么解答过程中提到了自由度36，是用来做什么的呢

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问一道题：NO.PZ2017092702000113

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