Question

Out of 300 people sampled, 78 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,

Point estimate = sample proportion = = x / n = 78 /300 = 0.260

1 - = 0.740

Z_/2 = 2.576

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

= 2.576 * (((0.260 * 0.740) / 300)

= 0.065

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.260 - 0.065 < p < 0.260 + 0.065

0.195 < p < 0.325