If n=180 and ˆ p (p-hat) =0.2, find the margin of error at a 99% confidence level

Question

If n=180 and  (p-hat) =0.2, find the margin of error at a 99% confidence level

Answer 1

ME = critical z value * sqrt(phat * (1-phat)/n)

convert confidence level to decimal: 0.99

critical probability = halfway between 0.99 and 1 = 0.995

find z-score for 0.995 probability from table or using technology = 2.58

that is the critical z value so:

ME = 2.58 * sqrt(0.2 * 0.8/180) = 0.0769