Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 96 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.

< p < Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p = ±±

Answer 1

$\hat{p}$ = 96 / 400 = 0.24, 1 - $\hat{p}$ = 0.76, n = 400

The Zcritical (2 tail) for $\alpha$ = 0.01, is 2.576

The Confidence Interval is given by $\hat{p}$ $\pm$ ME, where

ME = Zcritical * \sqrt{\frac{\hat{p}(1-\hat{p})}{n} }= 2.576 * \sqrt{\frac{0.24*0.76}{400}} = 0.055

The Lower Limit = 0.24 - 0.055 = 0.185

The Upper Limit = 0.24 + 0.055 = 0.295

The 99% Confidence Interval is 0.185 < p < 0.295

In $\hat{p}$ $\pm$ ME. The 99% CI is 0.24 $\pm$ 0.055