Question

A national health organization warns that alcohol use among middle school students is on the rise. Concerned, a local health agency randomly and anonymously surveys 110 of the middle school students in its city. 21 of them report having been intoxicated. Create a 90% confidence interval. Find p-hat. (p -hat = x/n)

Answer 1

CI 90 % = [ 0,05 ; 0,16 ]

Explanation:

From sample we got

n = 110

x = 21

p = 21/110 = 0.19 then q = 1 - p q = 1 - 0.19 q = 0.81

To create a 90 % confidence Interval, the significance level α = 10 %

α = 0.1 α/2 = 0.05

t(c) for α/2 = 0.05 is from z-table z(c) = 1.64

CI 90 % = ( p ± z(c) * √(p*q)/n

CI 90 % = [ 0,11 ± 1.64 * √ (0.19*0.81)/110 ]

CI 90 % = [ 0.11 ± 1.64 * 0.037 ]

CI 90 % = [ 0.11 ± 0,06 ]

CI 90 % = [ 0,05 ; 0,16 ]