(12) Suppose N – 60,000, n =15,000, and s =50
(a) compute the standard error of mean (x) using the finite population correction factor
(b) Repeat part a assuming n =30,000
(c)Repeat part a assuming n = 60,000
(d)Compare parts (a), (b), and (c), and describe what happens to the standard error of the mean (x) as n is increased.
1 answer
a. SE = 50/sqrt(15000) * sqrt((60000-15000)/60000-1))