Stat 1-Online Fall 2017-Prof. Barbee tymor abolson 11/22/17 3:09 PM Homework: 9.

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Stat 1-Online Fall 2017-Prof. Barbee tymor abolson 11/22/17 3:09 PM Homework: 9.2 Homework Score: 0 of 1 pt 9.2.9-T Save 40f 8 (3 complete) Hw Score: 15.25%, 1 .22 of 8 pts Question Help A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2955 occupants not wearing seat beits, 30 were killed. Among 7636 occupants wearing seat belts, 13 were killed. Use a 0.01 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 altemative hypotheses for the hypothesis test? H1 : p1=p2 H1: p1 ,p2 Identify the test statistic. O B. Ho:P1=P2 H1 : p1 p2 E. Hoi P P2 H1 P1 P2 Round to two decimal places as needed.) Enter your answer in the answer bowand then click Check Answer Check Answer 5%taining - Clear All

Explanation / Answer

9.2.9.

Given that,
sample one, x1 =30, n1 =2955, p1= x1/n1=0.01
sample two, x2 =13, n2 =7636, p2= x2/n2=0
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.01-0)/sqrt((0.004*1(1/2955+1/7636))
zo =6.13
| zo | =6.13
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =6.134 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 6.1335 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 6.13
critical value: 2.33
decision: reject Ho
p-value: 0

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