问题如下图：
选项：
A. 
B. 
C. 
解释：
我百度了一个表查出来 双尾 自由度16 t0.05怎么是2.12啊？单尾的 t0.05才是1.746
菲菲_品职助教 · 2019年04月17日
同学你好，你查错列了，双尾是要查0.1那一列的，因为这边是90%。
木水日 · 2019年04月22日
答案里写的不是t0.05么
菲菲_品职助教 · 2019年04月24日
不是说写t0.05就是查0.05。查表有查表的方法。这个t0.05代表着一个尾巴。而表上的双尾是一个整体。
木水日 · 2019年05月04日
那这种情况怎么区分，还是说有哪节课讲了的？可以详细说一下么，不然我下次遇见还是不知道该查哪个啊。。
菲菲_品职助教 · 2019年05月04日
其实不用想的这么复杂，也不用太关注公式里的下标这种，只要记住，双尾10%就查双尾10%，单尾10%就查单尾10%，就这么简单啦。
木水日 · 2019年05月05日
好
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
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.8.6970C.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）。 百分之90，为什么不能是1.65呢
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.8.6970C.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）。 是图里阴影部分吗？为啥
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）。 这个题目没有给出T分布的表格，怎么算出来呢
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p等于0.05