Question

In a sample of 300 adults, 195 had children. Construct a 90% confidence interval for the true population proportion of adults with children.

Give your answers as decimals, to three places

_____< p < _____

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 195 / 300= 0.65

alpha = 0.1

Z_/2 = 1.645

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

= 1.645 * (((0.65* 0.35) / 300)

= 0.045

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.65 - 0.045 < p < 0.65 + 0.045

0.605 < p < 0.695

(0.605, 0.695)