Question

Out of 500 people sampled, 460 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places.

< p <

Answer 1

Solution :

Given that,

n = 500

x = 460

Point estimate = sample proportion = = x / n = 460/500=0.92

1 -   = 1-0.92 =0.08

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.92*0.08) /500 )

= 0.024

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.92-0.024 < p < 0.92+ 0.024

0.896< p < 0944