In a surey of 3258 adults 1410 say they have started paying

In a surey of 3258 adults, 1410 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion, Interpret results

Solution

p - Z*sqrt(p*(1-p)/n) =0.4327808-2.58*sqrt(0.4327808*(1-0.4327808)/3258) =0.4103857

So the upper bound is

p + Z*sqrt(p*(1-p)/n) =0.4327808+2.58*sqrt(0.4327808*(1-0.4327808)/3258) =0.4551759

We have 99% confident that the population proportion will be within this interval.