Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 120 had kids. Based on this, construct a 90% 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,

n = 600

x = 120

Point estimate = sample proportion = = x / n 120 / 600 = 0.2

1 - = 1 - 0.2 = 0.8

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.2 * 0.8) / 600)

= 0.027

A 90% confidence interval for population proportion p is ,

   - E < p < + E

0.2 - 0.027 < p < 0.2 + 0.027

0.173 < p < 0.227

± E   

P = 0.2 ± 0.027