Question

Out of 300 people sampled, 207 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 = 207

Point estimate = sample proportion = = x / n = 207/300=0.69

1 -   = 1- 0.69 =0.31

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 (((0.69*0.31) /300 )

E = 0.0523

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.69-0.0523 < p <0.69+ 0.0523

0.6377< p < 0.7423