Question

Suppose a random sample of size 58 is selected from a population with δ = 12. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). The population size is infinite (to 2 decimals). The population size is N = 50,000 (to 2 decimals). The population size is N = 5,000 (to 2 decimals). The population size is N = 500 (to 2 decimals).

Answer 1

1) The population size is infinite

SE = [ (standard deviation)/sqrt(n) ] = 12/sqrt(58) = 1.58

Here is the formula for the finite population correction, when the random variable is a mean score or proportion.

fpc = sqrt [ (N - n) / (N - 1) ]

Here is how the finite population correction is used to compute the standard error of a mean score.

SE = [ (standard deviation)/sqrt(n) ] * fpc

where, sample size (n) , population size (N)

2) The population size is N = 50,000

fpc = sqrt [ (50000 - 58) / (50000 - 1) ] = 0.9994

SE = (12/sqrt(58))*Fpc

= 1.57

3) The population size is N = 5000

fpc = sqrt [ (5000 - 58) / (5000 - 1) ] = 0.9942

SE = (12/sqrt(58))*Fpc

= 1.57

4) The population size is N = 500

fpc = sqrt [ (500 - 58) / (500 - 1) ] = 0.9411

SE = (12/sqrt(58))*Fpc

= 1.48