Question

Your karate instructor has 1033 friends in Facebook. You randomly selected 87 of his friends and you found out that 43 of them have small kids. Construct a confidence interval for the proportion of your karate instructor's friends that have small children with a confidence level of 92% and interpret the results. (You must show the complete procedure even if the conditions are not met)

Answer #1

sample proportion, = 0.4943

sample size, n = 87

Standard error, SE = sqrt(pcap * (1 - pcap)/n)

SE = sqrt(0.4943 * (1 - 0.4943)/87) = 0.0536

Given CI level is 92%, hence α = 1 - 0.92 = 0.08

α/2 = 0.08/2 = 0.04, Zc = Z(α/2) = 1.75

Margin of Error, ME = zc * SE

ME = 1.75 * 0.0536

ME = 0.0938

CI = (pcap - z*SE, pcap + z*SE)

CI = (0.4943 - 1.75 * 0.0536 , 0.4943 + 1.75 * 0.0536)

CI = (0.4005 , 0.5881)

we are 92% confident that the proportion of your karate instructor's friends that have small children is between 0.4005 and 0.5881

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