Question

In a survey on supernatural experiences, 600 of 3000 randomly selected adult Americans surveyed reported that they had seen a ghost. Construct and interpret a 99% confidence interval for the proportion of all adult Americans who have seen a ghost. (Be sure to check all conditions necessary to your construction and interpret your interval in a sentence!)

Answer 1

SOLUTION:

Point estimate = sample proportion = = x / n = 600/3000=0.2

1 -   = 1-0.2=0.8

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* (((0.2*0.8) / 3000)

E = 0.019

A 99% confidence interval is,

- E < p < + E

0.2-0.019 < p < 0.2+0.019

0.181< p < 0.219

(0.181 , 0.219)