Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 52 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county.

Provide the point estimate and margin of error. Give your answers as decimals, to three places.

Answer 1

Solution :

n = 400

x = 225

= x / n = 52 / 400 = 0.130

1 - = 1 - 0.130 = 0.870

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 * (((0.130 * 0.870) / 400)

= 0.043

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.130 - 0.043 < p < 0.130 + 0.043

0.087 < p < 0.173