Question

In November 2018, 50,000 people living in the United States were asked whether they had a computer at home. 43,128 answered yes, they had a computer at home. There were 326,766,748 people living in the United States at the time of the survey. Determine an interval with 99% confidence for the number (how many) of people living in the United States who have a computer at home.

Answer 1

sample proportion, = 0.8626
sample size, n = 50000
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.8626 * (1 - 0.8626)/50000) = 0.0015

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 * 0.0015
ME = 0.0039

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.8626 - 2.58 * 0.0015 , 0.8626 + 2.58 * 0.0015)
CI = (0.8587 , 0.8665)