252y032app 3/25/03 Solutions for first problem in Take-Home part of Second ExamVariations 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 TotalHigh 6 9 9 24Average 9 19 23 51Low 7 8 10 25Total 22 36 42 100 a) We are comparing ,82x362n, 2222.3682p and ,103x,423n2381.42103p.Our we are testing 231: ppH . So the null hypothesis is 230: ppH a) Let 32ppp . So 0159.2381.2222.32 ppp and our hypotheses become0:320 ppH and 0:321 ppH. or 0:0pH and 0:1pH. 0955.009120.004319.004801.427619.2381.367778.2222.333222nqpnqpsp, 2308.781842362381.422222.3642361083233220nnpnpnp ,05. ,645.105.zz .960.1025.2zz Note that pq 1 and that q and pare between zero and one. 0957.009158.05159.17753.7692.2308.421361110031nnpqpUse one of the following: Confidence interval: Since the alternate hypothesis is 0:1pH, the confidence interval will be 0955.645.10159.0 pszpp or 1412.0p. This does not contradict0:0pH since any value of p between 0 and .1412 satisfies both the null hypothesis and the confidence interval, so do not reject 0H. Test ratio: 166.00957.0159.0pppz. Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below 645.1-05. zz. Since the test ratio is not in this zone, do not reject 0H. Critical value: Because the alternate hypothesis is 0:1pH, we need a critical value below zero. Use .1574.0957.645.100pcvzpp Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below -.1574. Since 0159.p is not in this zone, do not reject 0H.b) The p-value for this problem is .4325.0675.5.17.00159. zPpPSince this is not below ,05. do not reject 0H.c) 1872.0159.0955.960.10159.02pszpp or -.2031 to .17131252y032app 3/25/03d) 100n. The proportions in rows, rp, 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 Eare the same as in Oexcept for a possible small rounding error. (Note that 2 is computed two different ways here - only one way is needed.) 00.110000.4200.3600.2225.51.24.25512450.1000.950.542.2136.1622.1108.1064.828.595411LAHyrE Row O E OE 2OE EOE2 EO2 1 6 5.28 -0.72 0.5184 0.098182 6.8182 2 9 11.22 2.22 4.9284 0.439251 7.2193 3 7 5.50 -1.50 2.2500 0.409091 8.9091 4 9 8.64 -0.36 0.1296 0.015000 9.3750 5 19 18.36 -0.64 0.4096 0.022309 19.6623 6 8 9.00 1.00 1.0000 0.111111 7.1111 7 9 10.08 1.08 1.1664 0.115714 8.0357 8 23 21.42 -1.58 2.4964 0.116545 24.6965 9 10 10.50 0.50 0.2500 0.023810 9.5238 Total 100 100 0.00 1.35101 101.3510tindependenratedefectandyearsH :0 42211 crDF 4877.94205. EEOnEO22351.11003510.101 Since this is less than 9.4877, do not reject H0.(Diagram!)2252y032app 3/25/03Variations 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 TotalHigh 6 9 8 23Average 9 19 23 51Low 7 8 10 25Total 22 36 41 99 a) We are comparing ,82x362n, 2222.3682p and ,103x,413n2439.41103p.Our we are testing 231: ppH . So the null hypothesis is 230: ppH a) Let 32ppp . So 0217.2439.2222.32 ppp and our hypotheses become0:320 ppH and 0:321 ppH. or 0:0pH and 0:1pH. 0964.009299.004498.004801.417561.2439.367778.2222.333222nqpnqpsp, 2338.771841362439.412222.3641361083233220nnpnpnp ,05. ,645.105.zz .960.1025.2zz Note that pq 1 and that q and pare between zero and one. 0967.09344.7662.2338.411361110031nnpqpUse one of the following: Confidence interval: Since the alternate hypothesis is 0:1pH, the confidence interval will be 0964.645.10217.0 pszpp or 1379.0p. This does not contradict0:0pH since any value of p between 0 and .1379 satisfies both the null hypothesis and the confidence interval, so do not reject 0H. Test ratio: 224.00967.0217.0pppz. Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below 645.1-05. zz. Since the test ratio is not in this zone, do not reject 0H. Critical value: Because the alternate hypothesis is 0:1pH, we need a critical value below zero. Use .1591.0967.645.100pcvzpp Make a diagram of a Normal curve with zero in the middle. The ‘reject’ zone is the area below -.1591. Since 0217.p is not in this zone, do not reject 0H.b) The p-value for this problem is .4129.0871.5.224.00217. zPpPSincethis is not below ,05. do not reject 0H.c) 1889.0159.0964.960.10159.02pszpp or -.2048 to .17303252y032app 3/25/03d) 99n The proportions in rows, rp, 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 Eare the same as in Oexcept for a possible small rounding error. (Note that 2 is computed two different ways here - only one way is needed.) 0000.199.9899.4000.3600.222525.5152.2323.00.2500.5199.2235.1009.956.512.2155.1833.1152.936.811.595411LAHyrE Note slight rounding error. Row O E OE 2OE EOE2 EO2 1 6 5.11 -0.89 0.7921 0.155010 7.0450 2 9 11.33 2.33 5.4289 0.479162 7.1492 3 7 5.56 -1.44 2.0736 0.372950 8.8129 4 9 8.36 -0.64 0.4096
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