Question

Out of 500 people sampled, 470 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

< p <

Answer 1

Solution :

Given that,

n = 500

x = 470

Point estimate = sample proportion = = x / n = 470/500=0.940

1 - = 1-0.940=0.06

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.94*0.06) /500 )

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.94-0.017 < p <0.94+0.017

0.923< p < 0.957