Question

Suppose we take a poll (random sample) of 3556 students classified as Juniors and find that 3051 of them believe that they will find a job immediately after graduation.

What is the 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.

Answer 1

sample proportion, = 0.858
sample size, n = 3556
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.858 * (1 - 0.858)/3556) = 0.0059

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0059
ME = 0.0152

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.858 - 2.58 * 0.0059 , 0.858 + 2.58 * 0.0059)
CI = (0.8428 , 0.8732)