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 2877 occupants not wearing seat belts, 28 were killed. Among 7645 occupants wearing seat belts, 14 were killed Use a 0.01 significance level to test the claim that seat belts are effective in reducing Complete parts (a) through (c) below. A. H_0: P_1 lessthanorequalto P_2 H_1: P_1 notequalto P_2 B. H_0: P_1 = P_2 H_1: P_1 notequalto P_2 C. H_0: P_1 notequalto P_2 H_1: P_1 = P_2 D. H_0: P_1 = P_2 H_1: P_1 P_2 F. H_0: P_1 greaterthanorequalto P_2 H_1: P_1 notequalto P_2 Identify the test statistic. z = 5.73 (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.)

Explanation / Answer

Solution:

Ho: p1 = p2
Ha: p1 > p2

p1-hat = 28/2877 = 0.009732

p2-hat = 14/7645 = 0.001831

pc = (28+14)/(2877+7645) = 42/10522 = 0.003992

Test Statistic z = (p1-hat-p2-hat)/[(pc)(1-pc)(1/n1+1/n2)]
= (0.009732 - 0.001831)/[ 0.003992*(1- 0.003992)((1/2877+1/2877)]

= 4.7524

p-value = 0.0001

Decision: we can't reject Ho since the p - value (0.0001) is less than 0.01

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