Question

Out of 300 people sampled, 78 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

< p <

Answer 1

Solution :

Given that,

n = 300

x = 78

Point estimate = sample proportion = = x / n = 78/300=0.26

1 -   = 1- 0.26 =0.74

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 (((0.26*0.74) / 300)

E = 0.04963

=0.050 rounded

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.26 - 0.050 < p < 0.26 +0.050

(0.210 , 0.310)