Question

Out of 300 people sampled, 99 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 four places

< p

Answer 1

Solution :

Given that,

n = 300

x = 99

Point estimate = sample proportion = = x / n = 99/300=0.33

1 -   = 1- 0.33 =0.67

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z 0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.33*0.67) /300 )

E = 0.053

A 95% confidence interval p is ,

- E < p < + E

0.33 - 0.053 < p < 0.33+0.053

0.277< p