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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)

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Answer:

CI 90 %  =  [ 0,05  ;  0,16 ]

Step-by-step 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 ]

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