Question

Out of 300 people sampled, 264 had kids. Based on this, construct a 99% 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 = 300

x = 264

= x / n = 264 / 300 = 0.880

1 - = 1 - 0.880 = 0.120

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.880 * 0.120) / 300)

= 0.048

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.880 - 0.048 < p < 0.880 + 0.048

0.832 < p < 0.928

The 99% confidence interval for the population proportion p is : ( 0.832 , 0.928)