A bank randomly selected 247 checking account customers and

A bank randomly selected 247 checking account customers and found that 113 of them also had savings accounts at this same bank. Construct a 90% confidence interval for the true proportion of checking account customers who also have savings accounts.

Find the Upper and Lower Limits.

What equation and how do I solve this?

Solution

p=113/247 = 0.4574899

Given a=1-0.9 =0.1, Z(0.05) = 1.645 (from standard normal table)

So Lower Limits is

p - Z*sqrt(p*(1-p)/n) = 0.4574899 -1.645*sqrt(0.4574899*(1-0.4574899)/247) =0.405345

So upper Limits is

p + Z*sqrt(p*(1-p)/n) = 0.4574899 +1.645*sqrt(0.4574899*(1-0.4574899)/247) =0.5096348