A random sample of 56 credit card holders showed that 41 reg

A random sample of 56 credit card holders showed that 41 regularly paid their credit card bills on time. Find a 95% confidence interval for p, the proportion of all credit card holders who pay their credit card bills on time.

Solution

p=41/56 = 0.7321429

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

SO 95% confidence interval is

p +/- Z*sqrt(p*(1-p)/n)

--> 0.7321429 +/- 1.96*sqrt(0.7321429*(1-0.7321429)/56)

--> (0.6161553, 0.8481305)