Conduct the Hypothesis Test:A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed. Use this data with a 0.05 significance level to test the claim that the fatality rate is higher for those not wearing seat belts.

Answer:There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.Step-by-step explanation: Among 2823 occupants not wearing seat belts, 31 were killed.[tex]y_1=31\\n_1=2823[/tex]Among 7765 occupants wearing seat belts, 16 were killed.[tex]y_2=16\\n_2=7765[/tex]Let [tex]p_1[/tex]and [tex]p_2[/tex] be the fatality rate for hose not wearing seat belts and wearing seat belts receptively .Claim :the fatality rate is higher for those not wearing seat beltsSo, [tex]H_0:p_1=p_2[/tex] [tex]H_a:p_1> p_2[/tex]We will use Comparing Two Proportions [tex]\widehat{p_1}=\frac{y_1}{n_1}[/tex][tex]\widehat{p_1}=\frac{31}{2823}[/tex][tex]\widehat{p_1}=0.01098[/tex][tex]\widehat{p_2}=\frac{y_2}{n_2}[/tex][tex]\widehat{p_2}=\frac{16}{7765}[/tex][tex]\widehat{p_2}=0.00206[/tex][tex]\widehat{p}=\frac{y_1+y_2}{n_1+n_2}=\frac{31+16}{2823+7765} =0.00443[/tex]Formula of test statistic : [tex]\frac{\widehat{p_1}-\widehat{p_2}}{\sqrt{\widehat{p}(1-\widehat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}[/tex]Test statistic : [tex]\frac{0.01098-0.00206}{\sqrt{0.00443(1-0.00443)(\frac{1}{2823}+\frac{1}{7765})}}[/tex]Test statistic : [tex]\frac{0.01098-0.00206}{\sqrt{0.00443(1-0.00443)(\frac{1}{2823}+\frac{1}{7765})}}[/tex]Test statistic : [tex]6.111[/tex]Significance level- 0.05So, z at 0.05 = 1.96z statistic > z critical So, We failed to accept null hypothesis Hence There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.