Construct a confidence interval of the population proportion

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

x=120, n=300, 90% confidence

The 90% confidence interval is ___, ____

Use ascending Order, Round to three decimal places as needed

Solution

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

p= 120/300 =0.4

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.4 -1.645*sqrt (0.4*0.6/300) =0.353

So the upper bound is

p + Z*sqrt(p*(1-p)/n)= 0.4 +1.645*sqrt (0.4*0.6/300) =0.447