Question

Out of 100 people sampled, 60 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 = 100

x = 60

= x / n = 60 / 100= 0.6

1 - = 1 - 0. 6= 0.4

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576* (((0.6* 0.4) / 100) = 0.1262

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.6- 0.1262< p < 0.6+ 0.1262

0.4738< p < 0.7262

The 99% confidence interval for the population proportion p is : ( 0.4738, 0.7262)