问题如下图：
选项：
A. 
B. 
C. 
解释：
像这种题目题目中没给查表的数2.011，那考试中也不给要自己背吗
NO.PZ201709270100000107 95%的置信区间那个criticvalue为什么不是1.96
NO.PZ201709270100000107 请问此题在哪个条件中体现了95%的 “T” 统计量呢？
NO.PZ201709270100000107 老师，我记得前面有一个类似的题目求置信区间，用的是mean加上t*stanrerror，就是用0.3979+t*stanrerror，为什么这里就直接用系数的估计值？
−0.3947. 1.4528. B is correct. The calculation for the confinintervis −4.1589 ± (2.011 ×1.8718). The upper bounis −0.3947. The 2.011 is the critict-value for the 5% level of significan(2.5% in one tail) for 48 grees of freem.这道题的2.011是如何得到的？
−0.3947. 1.4528. B is correct. The calculation for the confinintervis −4.1589 ± (2.011 ×1.8718). The upper bounis −0.3947. The 2.011 is the critict-value for the 5% level of significan(2.5% in one tail) for 48 grees of freem.为啥不能用1.96？
