Question

Out of 100 people sampled, 12 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 four places ] <p<

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 12 / 100 = 0.12

1 - $\hat p$ = 1 - 0.12 = 0.88

Z_$\alpha$/2 = Z0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576 * ($\sqrt$((0.12 * 0.88) / 100)

= 0.0837

A 99% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.12 - 0.0837 < p < 0.12 + 0.0837

0.0363 < p < 0.2037

The 99% confidence interval for the population proportion p is : 0.0363 < p < 0.2037