In: Statistics and Probability
A poll conducted in 2013 surveyed a random sample of 2003 American adults, and 18% of them said that they have seen a ghost. Construct a 95% confidence interval for the proportion of American adults who have seen a ghost.
Solution :
Given that,
n = 2003
Point estimate = sample proportion =
=18%=0.18
1 -
= 0.82
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2
= Z0.025 = 1.96 ( Using z table )
Margin of error = E = Z/2 *
(((
* (1 -
)) / n)
= 1.96 (((0.18*0.82)
/2003 )
E = 0.017
A 95% confidence interval for population proportion p is ,
- E < p <
+ E
0.18 - 0.017 < p < 0.18+0.017
(0.163 , 0.197)