Question

Out of 100 people sampled, 50 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

Answer 1

Solution :

Given that,

n = 100

x = 50

Point estimate = sample proportion = = x / n = 50/100=0.5

1 -   = 1- 0.5 =0.5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.5*0.5) / 100)

E = 0.098

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.5-0.098 < p <0.5+0.098

0.402< p < 0.598

The 95% confidence interval for the population proportion p is : 0.402, 0.598