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# Solucionario Walpole 8 ED

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\u3b2 = P (6 \u2264 X \u2264 12 | p = 0.5) = 0.9963\u2212 0.1509 = 0.8454. \u3b2 = P (6 \u2264 X \u2264 12 | p = 0.7) = 0.8732\u2212 0.0037 = 0.8695. (c) This test procedure is not good for detecting differences of 0.1 in p. 10.5 (a) \u3b1 = P (X < 110 | p = 0.6) + P (X > 130 | p = 0.6) = P (Z < \u22121.52) + P (Z > 1.52) = 2(0.0643) = 0.1286. (b) \u3b2 = P (110 < X < 130 | p = 0.5) = P (1.34 < Z < 4.31) = 0.0901. \u3b2 = P (110 < X < 130 | p = 0.7) = P (\u22124.71 < Z < \u22121.47) = 0.0708. (c) The probability of a Type I error is somewhat high for this procedure, although Type II errors are reduced dramatically. 10.6 (a) \u3b1 = P (X \u2264 3 | p = 0.6) = 0.0548. (b) \u3b2 = P (X > 3 | p = 0.3) = 1\u2212 0.6496 = 0.3504. \u3b2 = P (X > 3 | p = 0.4) = 1\u2212 0.3823 = 0.6177. \u3b2 = P (X > 3 | p = 0.5) = 1\u2212 0.1719 = 0.8281. 121 122 Chapter 10 One- and Two-Sample Tests of Hypotheses 10.7 (a) \u3b1 = P (X \u2264 24 | p = 0.6) = P (Z < \u22121.59) = 0.0559. (b) \u3b2 = P (X > 24 | p = 0.3) = P (Z > 2.93) = 1\u2212 0.9983 = 0.0017. \u3b2 = P (X > 24 | p = 0.4) = P (Z > 1.30) = 1\u2212 0.9032 = 0.0968. \u3b2 = P (X > 24 | p = 0.5) = P (Z > \u22120.14) = 1\u2212 0.4443 = 0.5557. 10.8 (a) n = 12, p = 0.7, and \u3b1 = P (X > 11) = 0.0712 + 0.0138 = 0.0850. (b) n = 12, p = 0.9, and \u3b2 = P (X \u2264 10) = 0.3410. 10.9 (a) n = 100, p = 0.7, µ = np = 70, and \u3c3 = \u221a npq = \u221a (100)(0.7)(0.3) = 4.583. Hence z = 82.5\u221270 4.583 = 0.3410. Therefore, \u3b1 = P (X > 82) = P (Z > 2.73) = 1\u2212 0.9968 = 0.0032. (b) n = 100, p = 0.9, µ = np = 90, and \u3c3 = \u221a npq = \u221a (100)(0.9)(0.1) = 3. Hence z = 82.5\u221290 3 = \u22122.5. So, \u3b2 = P (X \u2264 82) = P (X < \u22122.5) = 0.0062. 10.10 (a) n = 7, p = 0.4, \u3b1 = P (X \u2264 2) = 0.4199. (b) n = 7, p = 0.3, \u3b2 = P (X \u2265 3) = 1\u2212 P (X \u2264 2) = 1\u2212 0.6471 = 0.3529. 10.11 (a) n = 70, p = 0.4, µ = np = 28, and \u3c3 = \u221a npq = 4.099, with z = 23.5\u221228 4.099 = \u22121.10. Then \u3b1 = P (X < 24) = P (Z < \u22121.10) = 0.1357. (b) n = 70, p = 0.3, µ = np = 21, and \u3c3 = \u221a npq = 3.834, with z = 23.5\u221221 3.834 = 0.65 Then \u3b2 = P (X \u2265 24) = P (Z > 0.65) = 0.2578. 10.12 (a) n = 400, p = 0.6, µ = np = 240, and \u3c3 = \u221a npq = 9.798, with z1 = 259.5\u2212 240 9.978 = 1.990, and z2 = 220.5\u2212 240 9.978 = \u22121.990. Hence, \u3b1 = 2P (Z < \u22121.990) = (2)(0.0233) = 0.0466. (b) When p = 0.48, then µ = 192 and \u3c3 = 9.992, with z1 = 220.5\u2212 192 9.992 = 2.852, and z2 = 259.5\u2212 192 9.992 = 6.755. Therefore, \u3b2 = P (2.852 < Z < 6.755) = 1\u2212 0.9978 = 0.0022. 10.13 From Exercise 10.12(a) we have µ = 240 and \u3c3 = 9.798. We then obtain z1 = 214.5\u2212 240 9.978 = \u22122.60, and z2 = 265.5\u2212 240 9.978 = 2.60. Solutions for Exercises in Chapter 10 123 So \u3b1 = 2P (Z < \u22122.60) = (2)(0.0047) = 0.0094. Also, from Exercise 10.12(b) we have µ = 192 and \u3c3 = 9.992, with z1 = 214.5\u2212 192 9.992 = 2.25, and z2 = 265.5\u2212 192 9.992 = 7.36. Therefore, \u3b2 = P (2.25 < Z < 7.36) = 1\u2212 0.9878 = 0.0122. 10.14 (a) n = 50, µ = 15, \u3c3 = 0.5, and \u3c3X¯ = 0.5\u221a 50 = 0.071, with z = 14.9\u221215 0.071 = \u22121.41. Hence, \u3b1 = P (Z < \u22121.41) = 0.0793. (b) If µ = 14.8, z = 14.9\u221214.8 0.071 = 1.41. So, \u3b2 = P (Z > 1.41) = 0.0793. If µ = 14.9, then z = 0 and \u3b2 = P (Z > 0) = 0.5. 10.15 (a) µ = 200, n = 9, \u3c3 = 15 and \u3c3X¯ = 15 3 = 5. So, z1 = 191\u2212 200 5 = \u22121.8, and z2 = 209\u2212 200 5 = 1.8, with \u3b1 = 2P (Z < \u22121.8) = (2)(0.0359) = 0.0718. (b) If µ = 215, then z \u2212 1 = 191\u2212215 5 = \u22124.8 and z2 = 209\u22122155 = \u22121.2, with \u3b2 = P (\u22124.8 < Z < \u22121.2) = 0.1151\u2212 0 = 0.1151. 10.16 (a) When n = 15, then \u3c3X¯ = 15 5 = 3, with µ = 200 and n = 25. Hence z1 = 191\u2212 200 3 = \u22123, and z2 = 209\u2212 200 3 = 3, with \u3b1 = 2P (Z < \u22123) = (2)(0.0013) = 0.0026. (b) When µ = 215, then z \u2212 1 = 191\u2212215 3 = \u22128 and z2 = 209\u22122153 = \u22122, with \u3b2 = P (\u22128 < Z < \u22122) = 0.0228\u2212 0 = 0.0228. 10.17 (a) n = 50, µ = 5000, \u3c3 = 120, and \u3c3X¯ = 120\u221a 50 = 16.971, with z = 4970\u22125000 16.971 = \u22121.77 and \u3b1 = P (Z < \u22121.77) = 0.0384. (b) If µ = 4970, then z = 0 and hence \u3b2 = P (Z > 0) = 0.5. If µ = 4960, then z = 4970\u22124960 16.971 = 0.59 and \u3b2 = P (Z > 0.59) = 0.2776. 10.18 The OC curve is shown next. 124 Chapter 10 One- and Two-Sample Tests of Hypotheses 180 190 200 210 220 0. 0 0. 2 0. 4 0. 6 0. 8 OC curve µ Pr ob ab ilit y of a cc ep tin g th e nu ll h yp ot he sis 10.19 The hypotheses are H0 : µ = 800, H1 : µ 6= 800. Now, z = 788\u2212800 40/ \u221a 30 = \u22121.64, and P -value= 2P (Z < \u22121.64) = (2)(0.0505) = 0.1010. Hence, the mean is not significantly different from 800 for \u3b1 < 0.101. 10.20 The hypotheses are H0 : µ = 5.5, H1 : µ < 5.5. Now, z = 5.23\u22125.5 0.24/ \u221a 64 = \u22129.0, and P -value= P (Z < \u22129.0) \u2248 0. The White Cheddar Popcorn, on average, weighs less than 5.5oz. 10.21 The hypotheses are H0 : µ = 40 months, H1 : µ < 40 months. Now, z = 38\u221240 5.8/ \u221a 64 = \u22122.76, and P -value= P (Z < \u22122.76) = 0.0029. Decision: reject H0. 10.22 The hypotheses are H0 : µ = 162.5 centimeters, H1 : µ 6= 162.5 centimeters. Now, z = 165.2\u2212162.5 6.9/ \u221a 50 = 2.77, and P -value= 2P (Z > 2.77) = (2)(0.0028) = 0.0056. Decision: reject H0 and conclude that µ 6= 162.5. Solutions for Exercises in Chapter 10 125 10.23 The hypotheses are H0 : µ = 20, 000 kilometers, H1 : µ > 20, 000 kilometers. Now, z = 23,500\u221220,000 3900/ \u221a 100 = 8.97, and P -value= P (Z > 8.97) \u2248 0. Decision: reject H0 and conclude that µ 6= 20, 000 kilometers. 10.24 The hypotheses are H0 : µ = 8, H1 : µ > 8. Now, z = 8.5\u22128 2.25/ \u221a 225 = 3.33, and P -value= P (Z > 3.33) = 0.0004. Decision: Reject H0 and conclude that men who use TM, on average, mediate more than 8 hours per week. 10.25 The hypotheses are H0 : µ = 10, H1 : µ 6= 10. \u3b1 = 0.01 and df = 9. Critical region: t < \u22123.25 or t > 3.25. Computation: t = 10.06\u221210 0.246/ \u221a 10 = 0.77. Decision: Fail to reject H0. 10.26 The hypotheses are H0 : µ = 220 milligrams, H1 : µ > 220 milligrams. \u3b1 = 0.01 and df = 9. Critical region: t > 1.729. Computation: t = 224\u2212220 24.5/ \u221a 20 = 4.38. Decision: Reject H0 and claim µ > 220 milligrams. 10.27 The hypotheses are H0 : µ1 = µ2, H1 : µ1 > µ2. Since sp = \u221a (29)(10.5)2+(29)(10.2)2 58 = 10.35, then P [ T > 34.0 10.35 \u221a 1/30 + 1/30 ] = P (Z > 12.72) \u2248 0. Hence, the conclusion is that running increases the mean RMR in older women. 126 Chapter 10 One- and Two-Sample Tests of Hypotheses 10.28 The hypotheses are H0 : µC = µA, H1 : µC > µA, with sp = \u221a (24)(1.5)2+(24)(1.25)2 48 = 1.3807. We obtain t = 20.0\u221212.0 1.3807 \u221a 2/25 = 20.48. Since P (T > 20.48) \u2248 0, we conclude that the mean percent absorbency for the cotton fiber is significantly higher than the mean percent absorbency for acetate. 10.29 The hypotheses are H0 : µ = 35 minutes, H1 : µ < 35 minutes. \u3b1 = 0.05 and df = 19. Critical region: t < \u22121.729. Computation: t = 33.1\u221235 4.3/ \u221a 20 = \u22121.98. Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average, to take the test. 10.30 The hypotheses are H0 : µ1 = µ2, H1 : µ1 6= µ2. Since the variances are known, we obtain z = 81\u221276\u221a 5.22/25+3.52/36 = 4.22. So, P -value\u2248 0 and we conclude that µ1 > µ2. 10.31 The hypotheses are H0 : µA \u2212 µB = 12 kilograms, H1 : µA \u2212 µB > 12 kilograms. \u3b1 = 0.05. Critical region: z > 1.645. Computation: z = (86.7\u221277.8)\u221212\u221a (6.28)2/50+(5.61)2/50 = \u22122.60. So, fail to reject H0 and conclude that the average tensile strength of thread A does not exceed the average tensile strength of thread B by 12 kilograms. 10.32 The hypotheses are H0 : µ1 \u2212 µ2 = $2, 000, H1 : µ1 \u2212 µ2 > $2, 000. Solutions for Exercises in Chapter 10 127 \u3b1 = 0.01. Critical region: z > 2.33. Computation: z = (70750\u221265200)\u22122000\u221a (6000)2/200+(5000)2/200 = 6.43, with a P -value= P (Z > 6.43) \u2248 0. Reject H0 and conclude that the mean salary