Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 88 had kids. Based on this, construct a 99% confidence interval for the proportion pp of adult residents who are parents in this county.

Give your answers as decimals, to 4 places.

Answer = ____ ± ____

(Note this is NOT in interval form, so check your answer carefully)

Answer 1

Solution :

n = 200

x = 88

= x / n = 88 / 200 = 0.440

1 - = 1 - 0.440 = 0.560

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * (((0.440 * 0.560) / 200)

= 0.0904

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.440 - 0.0904 < p < 0.440 + 0.0904

0.3496 < p < 0.5304