Question

Out of 400 people sampled, 20 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

Answer 1

Solution :

Given that,

n = 400

x = 20

= x / n = 20 / 400 = 0.05

1 - = 1 - 0.05 = 0.95

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (((0.05 * 0.95) /400 )

= 0.028

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.05 - 0.028 < p < 0.05 + 0.028

0.022 < p < 0.078

(0.022,0.078)

The 99% confidence interval 0.022 to 0.078