Question

the Kansas unemployment commission want to estimate the proportion of the labor force that was unemployed during last year in a certain depressed region. The commission wants to be 95% confident that their estimate is within 5% points (0.05) of the population proportion. If the population proportion has been estimated to be 0.15, how large a sample is required

Answer 1

Confidence level required = 95 %

z value = NORMSINV(1-(1-.95)/2) = 1.96

Proportion, p = 0.15

Margin of error required, e = 0.05

Sample size required, n = z²*p(1-p)/e²

= 1.96²*0.15*(1-0.15)/0.05²

= 196