Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 235 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

Solution :

Given that,

Point estimate = sample proportion = = x / n = 235 / 500 = 0.47

Z_/2 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.47 * 0.53) / 500)

= 0.057

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.47 - 0.057 < p < 0.47 + 0.057

0.413 < p < 0.527

0.47 0.057