Question

PRINTER VERSION 4 BACK NEXT Reserve Problems Chapter 10 Section 6 Problem 1 Rafting down two different rivers took place. 313 bosts rafted down the first river, and accidents (capsizing, bost damage, etc.) happened to 32 of them. 97 boats rafted down the second river, and accidents happened to 16 of them. Use the -values rounded to two decimal places to obtain the answers. ) The second river is considered to be a more complicated route to raft. Is there evidence for this assumption? Find the value of the test. Use a 0.10. Round your answer to four decimal places (eg. 98.7654), P-value b) Construct a 90% one-sided confidence limit for the difference in proportions that can be used to answer the question in part (a). Round your answer to four decimal places (e.g. 98.7654). h - P1

Answer 1

Let p, be the proportion of accident for rafting boats in first river and P2 be the proportion of accident for rafting boats in second river. To test Ho: P, = P2 against H, :P, <P2. That is the second river is more probable for accidents. Given a sample of rafting boats from first river n, = 313 , out of these x, = 32 met with accident, and a sample of rafting boats from second river n2 = 97 out of these x, = 16 met with accident. From 32 these sample the proportion of accidents from first river, p and that of the second river is 313 16 32+16 np' +n,p2 48 P Assume P, = P2 = p , where p = This implies p = and 97 n+n2 313+97 410 q = 1-p x 32 16 n The test statistic used is z = n ра -0.0627 = -1.68 0.0374 ра + 313 48 362 х 410 410 313 97 48 362 X 410 410 97 n n2 + p-value is 0.0465 For a = 0.10, from std normal table za = 1.28. It can be verified that, z < -zą Then Reject H.. OR P-value is less than a = 0.10 . Then Reject H.. That is the second river is more probable for accidents. (1-a)100% Confidence interval for P-P2 is X Pilt P272 PA - + n + P272 n2 =, n n2 n n n n2 n n2 ā, =1-P1,72 =1-P2 For (1-a)=0.90, that is for, a = 0.10,2q =1.6449 2 281 81 81 Hence 90% confidence interval is, 32 16 - 1.6449 313 hox 313 97 313 97 32 16 313 97 281 +1.6449313 313 sẽ x} + 97 [(-0.0627) - 1.6449x(0.04 14), (-0.0627)+1.6449x(0.04 14) ] [(-0.0627)– 0.0681, (-0.0627)+0.0681 ] = [-0.1308, 0.0054)