SOLUTION: In a survey of 3236 ​adults, 1495 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Inter

Algebra ->  Probability-and-statistics -> SOLUTION: In a survey of 3236 ​adults, 1495 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Inter      Log On


   

Question 1121478: In a survey of 3236 ​adults, 1495 say they have started paying bills online in the last year.
Construct a​ 99% confidence interval for the population proportion. Interpret the results.
A​ 99% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis
.
​(Round to three decimal places as​ needed.)
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
0.462 is p hat, the point estimate
SE is sqrt (p*(1-p)/n)
This is sqrt[ 0.462*0.538/3236]=0.0088
z=2.576 for CI 99% z0.995
the half interval is z*SE=0.0227 or 0.023
(0.439, 0.485)