Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 345 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,

Point estimate = sample proportion = = x / n = 345/500=0.69

1 -   = 1-0.69 = 0.31

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* (((0.69*0.31) /500 )

E = 0.053

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.69-0.053 < p < 0.69+0.053

0.637< p < 0.743