Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 132 had kids. Based on this, construct a 95% confidence interval for the proportion, p, of adult residents who are parents in this county.

Give your answers as decimals, to three places.

Answer 1

Solution :

Given that,

n = 200

x = 132

Point estimate = sample proportion = = x / n =132/200=0.66

1 -   = 1- 0.66 =0.34

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.66*0.34) /200 )

= 0.066

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.66-0.066< p < 0.66+0.066

0.594< p < 0.726

The 95% confidence interval for the population proportion p is : 0.594, 0.726