Question

Out of 500 people sampled, 325 had kids. Based on this, construct a 99% 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 = 500

x = 325

Point estimate = sample proportion = = x / n = 325 / 500 = 0.650

1 - = 1 - 0.650 = 0.35

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 * (((0.650 * 0.35) /500 )

= 0.055

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.650 - 0.055 < p < 0.650 + 0.055

0.595 < p < 0.705

The 99% confidence interval for the population proportion p is : 0.595 , 0.705