A simple random sample of fornt seat occupants involved in car crashes is obtain

Question

A simple random sample of fornt seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts 31 killed. Among 7765 occupants wearing seat belts 16 were killed Constract a 90% confidence interval estimate of the difference between the fatality rate is higer for those not wearing seat belts and those wearing seat belts. What does the result suggest about the effectiveness of seat belts.
Then use 0.05 significance level to test the claim that the fatality rate is higher for those not wearing seat belts.

Explanation / Answer

n1=2823, p1=31/2823= 0.011
n2=7765 , p2=16/7765= 0.002

a=0.1, |Z(0.05)|=1.645 (check standard normal table)
So 90% CI is
(p1-p2)± Z*[p1*(1-p1)/n1 +p2*(1-p2)/n2]

--> (0.011-0.002)±1.645*sqrt(0.011*(1-0.011)/2823 +0.002*(1-0.002)/7765)

--> ( 0.005664764, 0.01233524)

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