Question

Given 78 successes out of 124 trials, find the 90% confidence interval for the true population proportion.

Answer 1

The sample proportion, p, would be

p = 78 ÷ 124 = 0.63

To find the true population proportion, P, the equation would be


`p+z( \sqrt{ (p(1-p))/(n) } ) \ \ \textless \ P\ \textless \ p-z( \sqrt{ (p(1-p))/(n) } )`

where n = 124, and z = 1.645 for a 90% confidence interval

Substituting,


`0.63+1.645( \sqrt{ (0.63(1-0.63))/(124) } ) \ \ \textless \ P\ \textless \ 0.63-1.645( \sqrt{ (0.63(1-0.63))/(124) } )`


`0.7 \ \textless \ P\ \textless \ 0.56`