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

Question

A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2791 occupants not wearing seat belts, 30 were killed. Among 7816 occupants wearing seat belts, 16 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts (a) through (c) below.

a. Test the claim using a hypothesis test.

Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis test?

Explanation / Answer

Given that,
sample one, x1 =30, n1 =2791, p1= x1/n1=0.011
sample two, x2 =16, n2 =7816, p2= x2/n2=0.002
null, Ho: p1 = p2
alternate, seat belts are effective in reducing fatalities, H1: p1 > p2
level of significance, = 0.05
from standard normal table,right tailed z /2 =
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.011-0.002)/sqrt((0.004*0.996(1/2791+1/7816))
zo =6.005
| zo | =6.005
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =6.005 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 6.0054 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 < p2
alternate, H1: p1 > p2
test statistic: 6.005
critical value: 1.645
decision: reject Ho
p-value: 0

we have evidence to support seat belts are effective in reducing fatalities

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