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# Solucionario Walpole 8 ED

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for associate professors in research institutions is $2000 higher than for those in other institutions. 10.33 The hypotheses are H0 : µ1 \u2212 µ2 = 0.5 micromoles per 30 minutes, H1 : µ1 \u2212 µ2 > 0.5 micromoles per 30 minutes. \u3b1 = 0.01. Critical region: t > 2.485 with 25 degrees of freedom. Computation: s2p = (14)(1.5)2+(11)(1.2)2 25 = 1.8936, and t = (8.8\u22127.5)\u22120.5\u221a 1.8936 \u221a 1/15+1/12 = 1.50. Do not reject H0. 10.34 The hypotheses are H0 : µ1 \u2212 µ2 = 8, H1 : µ1 \u2212 µ2 < 8. Computation: s2p = (10)(4.7)2+(16)(6.1)2 26 = 31.395, and t = (85\u221279)\u22128\u221a 31.395 \u221a 1/11+1/17 = \u22120.92. Using 28 degrees of freedom and Table A.4, we obtain that 0.15 < P -value < 0.20. Decision: Do not reject H0. 10.35 The hypotheses are H0 : µ1 \u2212 µ2 = 0, H1 : µ1 \u2212 µ2 < 0. \u3b1 = 0.05 Critical region: t < \u22121.895 with 7 degrees of freedom. Computation: sp = \u221a (3)(1.363)+(4)(3.883) 7 = 1.674, and t = 2.075\u22122.860 1.674 \u221a 1/4+1/5 = \u22120.70. Decision: Do not reject H0. 10.36 The hypotheses are H0 : µ1 = µ2, H1 : µ1 6= µ2. Computation: sp = \u221a 51002+59002 2 = 5515, and t = 37,900\u221239,800 5515 \u221a 1/12+1/12 = \u22120.84. Using 22 degrees of freedom and since 0.20 < P (T < \u22120.84) < 0.3, we obtain 0.4 < P -value < 0.6. Decision: Do not reject H0. 128 Chapter 10 One- and Two-Sample Tests of Hypotheses 10.37 The hypotheses are H0 : µ1 \u2212 µ2 = 4 kilometers, H1 : µ1 \u2212 µ2 6= 4 kilometers. \u3b1 = 0.10 and the critical regions are t < \u22121.725 or t > 1.725 with 20 degrees of freedom. Computation: t = 5\u22124 (0.915) \u221a 1/12+1/10 = 2.55. Decision: Reject H0. 10.38 The hypotheses are H0 : µ1 \u2212 µ2 = 8, H1 : µ1 \u2212 µ2 < 8. \u3b1 = 0.05 and the critical region is t < \u22121.714 with 23 degrees of freedom. Computation: sp = \u221a (9)(3.2)2+(14)(2.8)2 23 = 2.963, and t = 5.5\u22128 2.963 \u221a 1/10+1/15 = \u22122.07. Decision: Reject H0 and conclude that µ1 \u2212 µ2 < 8 months. 10.39 The hypotheses are H0 : µII \u2212 µI = 10, H1 : µII \u2212 µI > 10. \u3b1 = 0.1. Degrees of freedom is calculated as v = (78.8/5 + 913.333/7)2 (78.8/5)2/4 + (913/333/7)2/6 = 7.38, hence we use 7 degrees of freedom with the critical region t > 2.998. Computation: t = (110\u221297.4)\u221210\u221a 78.800/5+913.333/7 = 0.22. Decision: Fail to reject H0. 10.40 The hypotheses are H0 : µS = µN , H1 : µS 6= µN . Degrees of freedom is calculated as v = (0.3914782/8 + 0.2144142/24)2 (0.3914782/8)2/7 + (0.2144142/24)2/23 = 8. Computation: t = 0.97625\u22120.91583\u221a 0.3914782/8+0.2144142/24 = \u22120.42. Since 0.3 < P (T < \u22120.42) < 0.4, we obtain 0.6 < P -value < 0.8. Decision: Fail to reject H0. Solutions for Exercises in Chapter 10 129 10.41 The hypotheses are H0 : µ1 = µ2, H1 : µ1 6= µ2. \u3b1 = 0.05. Degrees of freedom is calculated as v = (7874.3292/16 + 2479/5032/12)2 (7874.3292/16)2/15 + (2479.5032/12)2/11 = 19 degrees of freedom. Critical regions t < \u22122.093 or t > 2.093. Computation: t = 9897.500\u22124120.833\u221a 7874.3292/16+2479.5032/12 = 2.76. Decision: Reject H0 and conclude that µ1 > µ2. 10.42 The hypotheses are H0 : µ1 = µ2, H1 : µ1 6= µ2. \u3b1 = 0.05. Critical regions t < \u22122.776 or t > 2.776, with 4 degrees of freedom. Computation: d¯ = \u22120.1, sd = 0.1414, so t = \u22120.10.1414/\u221a5 = \u22121.58. Decision: Do not reject H0 and conclude that the two methods are not significantly different. 10.43 The hypotheses are H0 : µ1 = µ2, H1 : µ1 > µ2. Computation: d¯ = 0.1417, sd = 0.198, t = 0.1417 0.198/ \u221a 12 = 2.48 and 0.015 < P -value < 0.02 with 11 degrees of freedom. Decision: Reject H0 when a significance level is above 0.02. 10.44 The hypotheses are H0 : µ1 \u2212 µ2 = 4.5 kilograms, H1 : µ1 \u2212 µ2 < 4.5 kilograms. Computation: d¯ = 3.557, sd = 2.776, t = 3.557\u22124.5 2.778/ \u221a 7 = \u22120.896, and 0.2 < P -value < 0.3 with 6 degrees of freedom. Decision: Do not reject H0. 130 Chapter 10 One- and Two-Sample Tests of Hypotheses 10.45 The hypotheses are H0 : µ1 = µ2, H1 : µ1 < µ2. Computation: d¯ = \u221254.13, sd = 83.002, t = \u221254.1383.002/\u221a15 = \u22122.53, and 0.01 < P -value < 0.015 with 14 degrees of freedom. Decision: Reject H0. 10.46 The hypotheses are H0 : µ1 = µ2, H1 : µ1 6= µ2. \u3b1 = 0.05. Critical regions are t < \u22122.365 or t > 2.365 with 7 degrees of freedom. Computation: d¯ = 198.625, sd = 210.165, t = 198.625 210.165/ \u221a 8 = 2.67. Decision: Reject H0; length of storage influences sorbic acid residual concentrations. 10.47 n = (1.645+1.282) 2(0.24)2 0.32 = 5.48. The sample size needed is 6. 10.48 \u3b2 = 0.1, \u3c3 = 5.8, \u3b4 = 35.9 \u2212 40 = \u22124.1. Assume \u3b1 = 0.05 then z0.05 = 1.645, z0.10 = 1.28. Therefore, n = (1.645 + 1.28)2(5.8)2 (\u22124.1)2 = 17.12 \u2248 18 due to round up. 10.49 1\u2212 \u3b2 = 0.95 so \u3b2 = 0.05, \u3b4 = 3.1 and z0.01 = 2.33. Therefore, n = (1.645 + 2.33)2(6.9)2 3.12 = 78.28 \u2248 79 due to round up. 10.50 \u3b2 = 0.05, \u3b4 = 8, \u3b1 = 0.05, z0.05 = 1.645, \u3c31 = 6.28 and \u3c32 = 5.61. Therefore, n = (1.645 + 1.645)2(6.282 + 5.612) 82 = 11.99 \u2248 12 due to round up. 10.51 n = 1.645+0.842) 2(2.25)2 [(1.2)(2.25)]2 = 4.29. The sample size would be 5. 10.52 \u3c3 = 1.25, \u3b1 = 0.05, \u3b2 = 0.1, \u3b4 = 0.5, so \u2206 = 0.5 1.25 = 0.4. Using Table A.8 we find n = 68. 10.53 (a) The hypotheses are H0 : Mhot \u2212Mcold = 0, H1 : Mhot \u2212Mcold 6= 0. Solutions for Exercises in Chapter 10 131 (b) Use paired T -test and find out t = 0.99 with 0.3 < P -value < 0.4. Hence, fail to reject H0. 10.54 Using paired T -test, we find out t = 2.4 with 8 degrees of freedom. So, 0.02 < P -value < 0.025. Reject H0; breathing frequency significantly higher in the presence of CO. 10.55 The hypotheses are H0 : p = 0.40, H1 : p > 0.40. Denote by X for those who choose lasagna. P -value = P (X \u2265 9 | p = 0.40) = 0.4044. The claim that p = 0.40 is not refuted. 10.56 The hypotheses are H0 : p = 0.40, H1 : p > 0.40. \u3b1 = 0.05. Test statistic: binomial variable X with p = 0.4 and n = 15. Computation: x = 8 and np0 = (15)(0.4) = 6. Therefore, from Table A.1, P -value = P (X \u2265 8 | p = 0.4) = 1\u2212 P (X \u2264 7 | p = 0.4) = 0.2131, which is larger than 0.05. Decision: Do not reject H0. 10.57 The hypotheses are H0 : p = 0.5, H1 : p < 0.5. P -value = P (X \u2264 5 | p = 0.05) = 0.0207. Decision: Reject H0. 10.58 The hypotheses are H0 : p = 0.6, H1 : p < 0.6. So P -value \u2248 P ( Z < 110\u2212 (200)(0.6)\u221a (200)(0.6)(0.4) ) = P (Z < \u22121.44) = 0.0749. Decision: Fail to reject H0. 132 Chapter 10 One- and Two-Sample Tests of Hypotheses 10.59 The hypotheses are H0 : p = 0.2, H1 : p < 0.2. Then P -value \u2248 P ( Z < 136\u2212 (1000)(0.2)\u221a (1000)(0.2)(0.8) ) = P (Z < \u22125.06) \u2248 0. Decision: Reject H0; less than 1/5 of the homes in the city are heated by oil. 10.60 The hypotheses are H0 : p = 0.25, H1 : p > 0.25. \u3b1 = 0.05. Computation: P -value \u2248 P ( Z > 28\u2212 (90)(0.25)\u221a (90)(0.25)(0.75) ) = P (Z > 1, 34) = 0.091. Decision: Fail to reject H0; No sufficient evidence to conclude that p > 0.25. 10.61 The hypotheses are H0 : p = 0.8, H1 : p > 0.8. \u3b1 = 0.04. Critical region: z > 1.75. Computation: z = 250\u2212(300)(0.8)\u221a (300)(0.8)(0.2) = 1.44. Decision: Fail to reject H0; it cannot conclude that the new missile system is more accurate. 10.62 The hypotheses are H0 : p = 0.25, H1 : p > 0.25. \u3b1 = 0.05. Critical region: z > 1.645. Computation: z = 16\u2212(48)(0.25)\u221a (48)(0.25)(0.75) = 1.333. Decision: Fail to reject H0. On the other hand, we can calculate P -value = P (Z > 1.33) = 0.0918. Solutions for Exercises in Chapter 10 133 10.63 The hypotheses are H0 : p1 = p2, H1 : p1 6= p2. Computation: p\u2c6 = 63+59 100+125 = 0.5422, z = (63/100)\u2212(59/125)\u221a (0.5422)(0.4578)(1/100+1/125) = 2.36, with P -value = 2P (Z > 2.36) = 0.0182. Decision: Reject H0 at level 0.0182. The proportion of urban residents who favor