In: Statistics and Probability
A U.S. Travel Data Center’s survey of 1500 adults found that 46% of respondents stated that they favor historical sites as vacations. Find the 99% confidence interval of the true proportion of all adults who favor visiting historical sites as vacations.
Solution :
Given that,
n = 1500
Point estimate = sample proportion =
= 46%=0.46
1 -
= 1-0.46 =0.54
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* (((0.46*0.54)
/1500 )
E = 0.033
A 99% confidence interval for proportion p is ,
- E < p <
+ E
0.46-0.033 < p < 0.46+0.033
0.427< p < 0.493
The 99% confidence interval for the proportion p is : (0.427 , 0.493)