问题如下图：
选项：
A. 
B. 
C. 
解释：
请问一个基本的问题，样本方差计算时用的N-1,而标准误的时候用的N，而这题查表时用自由度又是N-1，这里面有什么规律可以记忆吗？
星星_品职助教 · 2019年09月27日
同学你好，
自由度是个很复杂的概念，证明超出了CFA的范畴，只能试着去关联理解一下。最方便的办法是对于每一个涉及到自由度的计算或者是类似概念，做一个表格分门别类的总结一下，也可以避免混淆。例如你提到的三个概念：
样本方差：有两种理解方式，第一是因为有个不能变的数字X bar，所以可以“自由”变动的就是n-1；或者说对于这个估计量，分母必须要是n-1才能做到无偏。
t统计量的自由度，一级最主要涉及到就是检验总体均值等于一个特定的数，这种检验下自由度为n-1。关于这个还可以去看一下讲义上types of hypothesis最后的那个表格，不同的检验对应不同的统计量和自由度。
最后要注意就是标准误用n是源自中心极限定理。
加油
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