252y032app 3 25 03 Solutions for first problem in Take Home part of Second Exam Variations on Solution If your Social Security Number ended in 9 To summarize the information in the problem 05 Defect Rate 1 yr 1 4 yr 5 9yr Total 6 9 9 24 High 9 19 23 51 Average 7 8 10 25 Low 22 36 42 100 Total a We are comparing x 2 8 n 2 36 p 3 10 42 p2 8 36 2222 and x 3 10 n 3 42 2381 Our we are testing H 1 p 3 p 2 So the null hypothesis is H 0 p 3 p 2 a Let p p 2 p 3 So p p 2 p 3 2222 2381 0159 and our hypotheses become H 0 p 2 p 3 0 and H 1 p 2 p 3 0 or H 0 p 0 and H 1 p 0 s p p0 p q p2 q2 2222 7778 2381 7619 3 3 004801 004319 009120 0955 n2 n3 36 42 n p n 3 p 3 36 2222 42 2381 18 8 10 2 2 2308 36 42 n 2 n3 36 42 78 05 z z 05 1 645 are between zero and one p p 0 q 0 1 1 n1 n3 z 2 z 025 1 960 2308 7692 1 1 36 42 Note that q 1 p and that q and p 17753 05159 009158 0957 Use one of the following Confidence interval Since the alternate hypothesis is H 1 p 0 the confidence interval will be p p z s p 0 0159 1 645 0955 or p 0 1412 This does not contradict H 0 p 0 since any value of p between 0 and 1412 satisfies both the null hypothesis and the confidence interval so do not reject H 0 Test ratio z p p 0 0159 0 166 Make a diagram of a Normal curve with zero in p 0957 the middle The reject zone is the area below z z 05 1 645 Since the test ratio is not in this zone do not reject H 0 Critical value Because the alternate hypothesis is H 1 p 0 we need a critical value below zero Use p cv p 0 z p 0 1 645 0957 1574 Make a diagram of a Normal curve with zero in the middle The reject zone is the area below 1574 Since p 0159 is not in this zone do not reject H 0 b The p value for this problem is P p 0159 P z 0 17 5 0675 4325 Since this is not below 05 do not reject H 0 c p p z 2 s p 0 0159 1 960 0955 0159 1872 or 2031 to 1713 1 252y032app 3 25 03 d n 100 The proportions in rows p r are used with column totals to get the items in E O is at the top of the page Note that row and column sums in E are the same as in O except for a possible small rounding error Note that 2 is computed two different ways here only one way is needed 1 yr 1 4 5 9 5 28 8 64 10 08 E H A 11 22 L 24 24 51 51 16 36 21 42 5 50 9 00 10 50 25 25 22 00 36 00 42 00 100 1 00 O Row 1 2 3 4 5 6 7 8 9 Total 6 9 7 9 19 8 9 23 10 100 E 5 28 11 22 5 50 8 64 18 36 9 00 10 08 21 42 10 50 100 E O 0 72 2 22 1 50 0 36 0 64 1 00 1 08 1 58 0 50 0 00 E 0 5184 4 9284 2 2500 0 1296 0 4096 1 0000 1 1664 2 4964 0 2500 H 0 years and defect rate independent 205 4 9 4877 O2 n E O E 2 E O 2 E O 2 E O2 E 0 098182 6 8182 0 439251 7 2193 0 409091 8 9091 0 015000 9 3750 0 022309 19 6623 0 111111 7 1111 0 115714 8 0357 0 116545 24 6965 0 023810 9 5238 1 35101 101 3510 DF r 1 c 1 2 2 4 101 3510 100 1 351 Since this is less than 9 4877 do not reject H 0 Diagram 2 252y032app 3 25 03 Variations on Solution If your Social Security Number ended in 8 To summarize the information in the problem 05 Defect Rate 1 yr 1 4 yr 5 9yr Total 6 9 8 23 High 9 19 23 51 Average 7 8 10 25 Low 22 36 41 99 Total a We are comparing x 2 8 n 2 36 p 3 10 41 p2 8 36 2222 and x 3 10 n3 41 2439 Our we are testing H 1 p 3 p 2 So the null hypothesis is H 0 p 3 p 2 a Let p p 2 p 3 So p p 2 p3 2222 2439 0217 and our hypotheses become H 0 p 2 p 3 0 and H 1 p 2 p 3 0 or H 0 p 0 and H 1 p 0 s p p0 p q p2 q2 2222 7778 2439 7561 3 3 004801 004498 009299 0964 n2 n3 36 41 n p n 3 p 3 36 2222 41 2439 18 8 10 2 2 2338 36 41 n 2 n3 36 41 77 05 z z 05 1 645 are between zero and one p p 0 q 0 1 1 n1 n3 z 2 z 025 1 960 2338 7662 1 Note that q 1 p and that q and p 1 36 41 09344 0967 Use one of the following Confidence interval Since the alternate hypothesis is H 1 p 0 the confidence interval will be p p z s p 0 0217 1 645 0964 or p 0 1379 This does not contradict H 0 p 0 since any value of p between 0 and 1379 satisfies both the null hypothesis and the confidence interval so do not reject H 0 Test ratio z p p 0 0217 0 224 Make a diagram of a Normal curve with zero in p 0967 the middle The reject zone is the area below z z 05 1 645 Since the test ratio is not in this zone do not reject H 0 Critical value Because the alternate hypothesis is H 1 p 0 we need a critical value below zero Use p cv p 0 z p 0 1 645 0967 1591 Make a diagram of a Normal curve with zero in the middle The reject zone is the area below 1591 Since p 0217 is not in this zone do not reject H 0 b The p value for this problem is P p 0217 P z 0 224 5 0871 4129 Since this is not below 05 do not reject H 0 …
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