Question

Question 5 4 pts A random sample of size 200 was selected from a pool of university students and they were asked whether they use Facebook on a regular basis. Eighty replied yes. What would be your 90% confidence interval estimate of the population proportion of regular Facebook users based upon the given information? (Hint: use 1.65 as the critical z-value) O [46.5%, 53.5%] O [42.5%, 57.5%] O [48.5%, 51.5%] [34.3%, 45.7%)

Answer 1

sample proportion, = 0.4
sample size, n = 200
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.4 * (1 - 0.4)/200) = 0.0346

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.4 - 1.65 * 0.0346 , 0.4 + 1.65 * 0.0346)
CI = (0.343 , 0.457)

Option D