The population is 6000 programmer which is supposed to be normally distributed. A sample with 100 size is drawn from the population. Based on z-statistic, 95% confidence interval of sample mean of annual salary is 32.5 (in thousands) dollars ranges from 22 (in thousands) dollars to 43 (in thousands) dollars .Calculate the standard error of mean annual salary:
C is correct.
At the 95% level of significance, the critical value is ±1.96.
So the confidence interval is ,
From the equation t 32.5 + 1.96 standard error= 43 or 32.5 - 1.96standard error = 22, we get =5.3571.
根据题干的sample mean=32.5，基于z-statistics的95%的置信区间（得到关键值±1.96），和置信区间的公式，可以得到 32.5 + 1.96 standard error = 43 或 32.5 - 1.96 standard error = 22。
得到 standard error=5.3571