Question

Out of 300 people sampled, 249 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 = 300

x = 249

= x / n = 249 / 300 = 0.83

Point estimate = 0.83

1 - = 1 - 0.83 = 0.17

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 * (((0.83 * 0.17) / 300)

= 0.056

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.83 - 0.056 < p < 0.83 + 0.056

0.774 < p < 0.886

(0.774,0.886)

The 99% confidence interval for p is 0.774 to 0.886.