Question

out of 200 people sampled 162 had kids. based on this construct a 99% confidence interval for the true population proportion of people with kids

Answer 1

Solution :

Given that,

n = 200

x = 162

Point estimate = sample proportion = = x / n = 162 / 200 = 0.81

1 - = 1 - 0.81 = 0.19

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 (((0.81 * 0.19) / 200 )

= 0.071

A 99% confidence interval for population proportion p is ,

± E   

= 0.81  ± 0.071

= ( 0.739, 0.881 )