Question

Out of 500 people sampled, 380 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

Given that,

n = 500

x = 380

Point estimate = sample proportion = = x / n = 380/500=0.76

1 - = 1-0.76=0.24

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 ( Using z table )

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

= 2.576 (((0.76*0.24) / 500)

E = 0.049

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.76 -0.049 < p < 0.76+0.049

0.711< p < 0.809

The 99% confidence interval for the population proportion p is : 0.711, 0.809