问题如下:
For a sample size of 65 with a mean of 31 taken from a normally distributed population with a variance of 529, a 99% confidence interval for the population mean will have a lower limit closest to:
选项:
A.23.64.
B.25.41.
C.30.09.
解释:
A is correct.
To solve, use the structure of Confidence interval = Point estimate ± Reliability factor × Standard error, which, for a normally distributed population with known variance, is represented by the following formula:
For a 99% confidence interval, use z0.005 = 2.58. Also, σ = = 23. Therefore, the lower limit = .
请问,为什么不用自由度64,还有单尾0.005查表t分布,而是直接用z分布的呢?是因为题目里面说了正态分布吗?