Question

Out of 300 people sampled, 216 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places I am 90% confident that the proportion of people who have kids is between____and___.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 216 / 300 = 0.720

1 - = 1 - 0.720 = 0.28

Z/2 = 1.645

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

= 1.645 * (((0.720 * 0.28) / 300)

Margin of error = E = 0.043

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.720 - 0.043 < p < 0.720 + 0.043

0.677 < p < 0.763

90% confident that the proportion of people who have kids is between 0.677 and 0.763