Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 46 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county.

Give your answers as decimals, to three places.

< p <

Answer 1

Solution :

Given that,

n = 200

x = 46

Point estimate = sample proportion = = x / n = 46/200=0.23

1 -   = 1-0.23 =0.77

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 ( Using z table )

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

= 2.576* (((0.23*0.77) /200 )

E = 0.077

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.23-0.077< p <0.23+ 0.077

0.153< p < 0.307

The 99% confidence interval for the population proportion p is : 0.153< p < 0.307