In a sample of 400 Americans over age 21, 75% said that they drank alcoholic beverages....

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?

Solutions

Expert Solution

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.

orchestra answered 1 year ago

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