Construct a confidence interval of the population proportion

Construct a confidence interval of the population proportion at the given level of confidence. x=75,n- 150, 99% confidence The 99% confidence interval is ([], []). (Use ascending order. Round to three decimal places as needed.)

Solution

p=75/150=0.5

Given a=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) =0.5-2.58*sqrt(0.5*0.5/150)=0.395

So the upper bound is

p + Z*sqrt(p*(1-p)/n) =0.5+2.58*sqrt(0.5*0.5/150)=0.605