Construct a confidence interval of the population proportion

Construct a confidence interval of the population proportion at the given level of confidence

X=80, n=200, 90% confidence

The 90% confidnece Level is ( _ and _ )

(use ascending order. Round to three decimal places as needed.)

Solution

Given a=0.1, Z(0.05) = 1.645 (from standard normal tabe)

p=80/200 =0.4

So the lower bound is p - Z*sqrt(p*(1-p)/n)

=0.4 - 1.645*sqrt(0.4*0.6/200)

=0.343

So the upper bound is p + Z*sqrt(p*(1-p)/n)

=0.4 + 1.645*sqrt(0.4*0.6/200)

=0.457