Confidence Level Intervals

Question

In a sample of 560 adults, 375 had children. Construct a 99% confidence interval for the true population proportion or adults with children.

_____ < p < p ____

Answer 1

proportions, a nice concept and with exact and approximate methods of solution.

An exact answer based on the Laplace-Bayes unbiased estimator and beta distribution would be P= 376/562 or 0.6690 and the 99% CI would be [0.617, 0.719]

Using the Normal Approximation (what your problem probably expects) , rounded to 5 decimal places:

P(success) = .66964; Variance = p*q = .22122; n=560 α= .01 Z_α/2 = 2.57583

CI for proportions = [p`- z<sub>α/2</sub> *√(V/n) , p`+ z_α/2 *√(V/n)]

that would be [ .66964 - 2.57583*√(.22122/560) < p < .66964 + 2.57583*√(.22122/550)]

or [.61845 , .72084] or .61845 < p < .72084

Answer 2

375/560 + z_.005*sqrt(375/560*(560-375)/560/560) < p < 375/560 + z_.995*sqrt(375/560*(560-375)/560/560)

.6185< p < .7208