Math 132B Fall 2017 Sec 02 David Nay | 12/9/17 3:55 PM Homework: Homework 23 (Se
ID: 3319928 • Letter: M
Question
Math 132B Fall 2017 Sec 02 David Nay | 12/9/17 3:55 PM Homework: Homework 23 (Section 9.1) Score: 0 of 1 pt 9.1.13-T Save 4 or 10 (3 complete) HW Score: 23.33%, 233 of 10 pts Question Help A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2707 occupants not wearing seat belts, 31 were killed. Among 7617 occupants wearing seat belts, 13 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 altemative hypotheses for the hypothesis test? OA, Ho:P1=P2 B. Ho:p1=p2 H, p1 = p2 F. Ho P1 P2 H1 P1 P2 H1 : p1 #p2 Identify the test statistic. Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer Clear Al Check Answer remainingExplanation / Answer
9.1.13
Given that,
sample one, x1 =31, n1 =2707, p1= x1/n1=0.01
sample two, x2 =13, n2 =7617, p2= x2/n2=0
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.011-0)/sqrt((0.004*1(1/2707+1/7617))
zo =6.69
| zo | =6.69
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =6.685 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 6.6853 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 6.69
critical value: 1.64
decision: reject Ho
p-value: 0