A random sample of 1000 people was taken 450 of the people i

A random sample of 1000 people was taken. 450 of the people in the sample favored candidate A. The 95% confidence interval for the true proportion f people who favors Candidate A is?

Solution

p=450/1000=0.45

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) =0.45 - 1.96*sqrt(0.45*(1-0.45)/1000) =0.419165

So the upper bound is

p + Z*sqrt(p*(1-p)/n) =0.45 + 1.96*sqrt(0.45*(1-0.45)/1000) =0.480835