问题如下图:
选项:
A.
B.
C.
解释:
请问老师coefficience = U (均值)吗?Upper bound=U+Critical Value*Standard Error; Lower bound=U-Critical Value*Standard Error. 我这样理解这个公式有误吗?
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?