In: Statistics and Probability
In a sample of 400 Americans over age 21, 75% said that they drank alcoholic beverages. What is 98% confidence interval of the true proportion of all Americans who drink alcoholic beverages?
Solution: Here, the given information is
n=400, p=75% = 0.75
98% confidence interval
First find
= 1- = 1-0.02/2
=1-0.01
=0.99 ( see this value in z table)
=2.33
Margin of error
E= *
=2.33*
=2.33*0.02165
=0.05044
*
( 0.75-0.05044 , 0.75+0.05044)
(0.699, 0.800)
0.699 to 0.800 is the 98% confidence interval of the true proportion of all Americans who drink alcoholic beverages.