NO.PZ2022062760000022
问题如下:
For a sample of the past 30 monthly stock returns for McCreary, Inc., the mean return is 4% and the sample
standard deviation is 20%. The population variance is unknown but the standard error of the sample mean is
estimated to be:
The related t-table values are shown below (ti,j denotes the (100-j)
th percentile of t-distribution value with i
degrees of freedom):
What is the 95% confidence interval for the mean monthly return?
选项:
A.[-3.466%, 11.466%]
[-3.453%, 11.453%]
[-2.201%, 10.201%]
[-2.194%, 10.194%]
解释:
中文解析:
自由度=30-1=29
t检验是双尾检验,单尾是2.5%
查表得检测值=2.045
变量的变化范围=均值+/- 检测值*σ
[4% − 2.045(3.651%), 4% + 2.045(3.651%)] = [−3.466%, 11.466%]
选A
Here the t-reliability factor is used since the population variance is unknown. Since there are 30 observations, the degrees of freedom are 30 - 1 = 29. The t-test is a two-tailed test. So, the correct critical t-value is t29,2.5 = 2.045, thus the 95% confidence interval for the mean return is:
[4% − 2.045(3.651%), 4% + 2.045(3.651%)] = [−3.466%, 11.466%]
比如t 2.5 即是单尾2.5%的value?