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Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 288 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county.

Express your answer in tri-inequality form. Give your answers as decimals, to three places.

Answer 1

You may use the formula CI = [ p^' - z * √p(1-p)/n , p^' + z * √p(1-p)/n ].

p^' = 288/400 = 0.72

The z-score for the 95% confidence interval is 2.57, and p' can be used in the place of p.

Therefore:

CI = [ 0.72 - 2.57 * √(0.72*0.28/400) , 0.72 + 2.57 * √(0.72*0.28/400) ] = [0.66, 0.78].