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# Solucionario Walpole 8 ED

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for a future observation is 11.3± (2.093)(2.45) \u221a 1 + 1/20 = 11.3± 5.25, which yields (6.05, 16.55). 106 Chapter 9 One- and Two-Sample Estimation Problems 9.26 n = 12, x¯ = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95% prediction interval for a future observation is 79.3± (2.201)(7.8) \u221a 1 + 1/12 = 79.3± 17.87, which yields (61.43, 97.17). 9.27 n = 15, x¯ = 3.7867, s = 0.9709, and t0.025 = 2.145 with 14 degrees of freedom. A 95% prediction interval for a new observation is 3.7867± (2.145)(0.9709) \u221a 1 + 1/15 = 3.7867± 2.1509, which yields (1.6358, 5.9376). 9.28 n = 9, x¯ = 1.0056, s = 0.0245, 1 \u2212 \u3b1 = 0.95, and \u3b3 = 0.05, with k = 3.532. The tolerance interval is 1.0056± (3.532)(0.0245) which yields (0.919, 1.092). 9.29 n = 15, x¯ = 3.84, and s = 3.07. To calculate an upper 95% prediction limit, we obtain t0.05 = 1.761 with 14 degrees of freedom. So, the upper limit is 3.84 + (1.761)(3.07) \u221a 1 + 1/15 = 3.84 + 5.58 = 9.42. This means that a new observation will have a chance of 95% to fall into the interval (\u2212\u221e, 9.42). To obtain an up- per 95% tolerance limit, using 1 \u2212 \u3b1 = 0.95 and \u3b3 = 0.05, with k = 2.566, we get 3.84 + (2.566)(3.07) = 11.72. Hence, we are 95% confident that (\u2212\u221e, 11.72) will contain 95% of the orthophosphorous measurements in the river. 9.30 n = 50, x¯ = 78.3, and s = 5.6. Since t0.05 = 1.677 with 49 degrees of freedom, the bound of a lower 95% prediction interval for a single new observation is 78.3 \u2212 (1.677)(5.6) \u221a 1 + 1/50 = 68.91. So, the interval is (68.91,\u221e). On the other hand, with 1 \u2212 \u3b1 = 95% and \u3b3 = 0.01, the k value for a one-sided tolerance limit is 2.269 and the bound is 78.3\u2212 (2.269)(5.6) = 65.59. So, the tolerance interval is (65.59,\u221e). 9.31 Since the manufacturer would be more interested in the mean tensile strength for future products, it is conceivable that prediction interval and tolerance interval may be more interesting than just a confidence interval. 9.32 This time 1 \u2212 \u3b1 = 0.99 and \u3b3 = 0.05 with k = 3.126. So, the tolerance limit is 78.3\u2212 (3.126)(5.6) = 60.79. Since 62 exceeds the lower bound of the interval, yes, this is a cause of concern. 9.33 In Exercise 9.27, a 95% prediction interval for a new observation is calculated as (1.6358, 5.9377). Since 6.9 is in the outside range of the prediction interval, this new observation is likely to be an outlier. 9.34 n = 12, x¯ = 48.50, s = 1.5, 1 \u2212 \u3b1 = 0.95, and \u3b3 = 0.05, with k = 2.815. The lower bound of the one-sided tolerance interval is 48.50\u2212 (2.815)(1.5) = 44.275. Their claim is not necessarily correct. Solutions for Exercises in Chapter 9 107 9.35 n1 = 25, n2 = 36, x¯1 = 80, x¯2 = 75, \u3c31 = 5, \u3c32 = 3, and z0.03 = 1.88. So, a 94% confidence interval for µ1 \u2212 µ2 is (80\u2212 75)\u2212 (1.88) \u221a 25/25 + 9/36 < µ1 \u2212 µ2 < (80\u2212 75) + (1.88) \u221a 25/25 + 9/36, which yields 2.9 < µ1 \u2212 µ2 < 7.1. 9.36 nA = 50, nB = 50, x¯A = 78.3, x¯B = 87.2, \u3c3A = 5.6, and \u3c3B = 6.3. It is known that z0.025 = 1.96. So, a 95% confidence interval for the difference of the population means is (87.2\u2212 78.3)± 1.96 \u221a 5.62/50 + 6.32/50 = 8.9± 2.34, or 6.56 < µA \u2212 µB < 11.24. 9.37 n1 = 100, n2 = 200, x¯1 = 12.2, x¯2 = 9.1, s1 = 1.1, and s2 = 0.9. It is known that z0.01 = 2.327. So (12.2\u2212 9.1)± 2.327 \u221a 1.12/100 + 0.92/200 = 3.1± 0.30, or 2.80 < µ1\u2212µ2 < 3.40. The treatment appears to reduce the mean amount of metal removed. 9.38 n1 = 12, n2 = 10, x¯1 = 85, x¯2 = 81, s1 = 4, s2 = 5, and sp = 4.478 with t0.05 = 1.725 with 20 degrees of freedom. So (85\u2212 81)± (1.725)(4.478) \u221a 1/12 + 1/10 = 4± 3.31, which yields 0.69 < µ1 \u2212 µ2 < 7.31. 9.39 n1 = 12, n2 = 18, x¯1 = 84, x¯2 = 77, s1 = 4, s2 = 6, and sp = 5.305 with t0.005 = 2.763 with 28 degrees of freedom. So, (84\u2212 77)± (2.763)(5.305) \u221a 1/12 + 1/18 = 7± 5.46, which yields 1.54 < µ1 \u2212 µ2 < 12.46. 9.40 n1 = 10, n2 = 10, x¯1 = 0.399, x¯2 = 0.565, s1 = 0.07279, s2 = 0.18674, and sp = 0.14172 with t0.025 = 2.101 with 18 degrees of freedom. So, (0.565\u2212 0.399)± (2.101)(0.14172) \u221a 1/10 + 1/10 = 0.166± 0.133, which yields 0.033 < µ1 \u2212 µ2 < 0.299. 9.41 n1 = 14, n2 = 16, x¯1 = 17, x¯2 = 19, s 2 1 = 1.5, s 2 2 = 1.8, and sp = 1.289 with t0.005 = 2.763 with 28 degrees of freedom. So, (19\u2212 17)± (2.763)(1.289) \u221a 1/16 + 1/14 = 2± 1.30, which yields 0.70 < µ1 \u2212 µ2 < 3.30. 108 Chapter 9 One- and Two-Sample Estimation Problems 9.42 n1 = 12, n2 = 10, x¯1 = 16, x¯2 = 11, s1 = 1.0, s2 = 0.8, and sp = 0.915 with t0.05 = 1.725 with 20 degrees of freedom. So, (16\u2212 11)± (1.725)(0.915) \u221a 1/12 + 1/10 = 5± 0.68, which yields 4.3 < µ1 \u2212 µ2 < 5.7. 9.43 nA = nB = 12, x¯A = 36, 300, x¯B = 38, 100, sA = 5, 000, sB = 6, 100, and v = 50002/12 + 61002/12 (50002/12)2 11 + (6100 2/12)2 11 = 21, with t0.025 = 2.080 with 21 degrees of freedom. So, (36, 300\u2212 38, 100)± (2.080) \u221a 50002 12 + 61002 12 = \u22121, 800± 4, 736, which yields \u22126, 536 < µA \u2212 µB < 2, 936. 9.44 n = 8, d¯ = \u22121112.5, sd = 1454, with t0.005 = 3.499 with 7 degrees of freedom. So, \u22121112.5± (3.499)1454\u221a 8 = \u22121112.5± 1798.7, which yields \u22122911.2 < µD < 686.2. 9.45 n = 9, d¯ = 2.778, sd = 4.5765, with t0.025 = 2.306 with 8 degrees of freedom. So, 2.778± (2.306)4.5765\u221a 9 = 2.778± 3.518, which yields \u22120.74 < µD < 6.30. 9.46 nI = 5, nII = 7, x¯I = 98.4, x¯II = 110.7, sI = 8.375, and sII = 32.185, with v = (8.7352/5 + 32.1852/7)2 (8.7352/5)2 4 + (32.185 2/7)2 6 = 7 So, t0.05 = 1.895 with 7 degrees of freedom. (110.7\u2212 98.4)± 1.895 \u221a 8.7352/5 + 32.1852/7 = 12.3± 24.2, which yields \u221211.9 < µII \u2212 µI < 36.5. 9.47 n = 10, d¯ = 14.89%, and sd = 30.4868, with t0.025 = 2.262 with 9 degrees of freedom. So, 14.89± (2.262)30.4868\u221a 10 = 14.89± 21.81, which yields \u22126.92 < µD < 36.70. Solutions for Exercises in Chapter 9 109 9.48 nA = nB = 20, x¯A = 32.91, x¯B = 30.47, sA = 1.57, sB = 1.74, and Sp = 1.657. (a) t0.025 \u2248 2.042 with 38 degrees of freedom. So, (32.91\u2212 30.47)± (2.042)(1.657) \u221a 1/20 + 1/20 = 2.44± 1.07, which yields 1.37 < µA \u2212 µB < 3.51. (b) Since it is apparent that type A battery has longer life, it should be adopted. 9.49 nA = nB = 15, x¯A = 3.82, x¯B = 4.94, sA = 0.7794, sB = 0.7538, and sp = 0.7667 with t0.025 = 2.048 with 28 degrees of freedom. So, (4.94\u2212 3.82)± (2.048)(0.7667) \u221a 1/15 + 1/15 = 1.12± 0.57, which yields 0.55 < µB \u2212 µA < 1.69. 9.50 n1 = 8, n2 = 13, x¯1 = 1.98, x¯2 = 1.30, s1 = 0.51, s2 = 0.35, and sp = 0.416. t0.025 = 2.093 with 19 degrees of freedom. So, (1.98\u2212 1.30)± (2.093)(0.416) \u221a 1/8 + 1/13 = 0.68± 0.39, which yields 0.29 < µ1 \u2212 µ2 < 1.07. 9.51 (a) n = 200, p\u2c6 = 0.57, q\u2c6 = 0.43, and z0.02 = 2.05. So, 0.57± (2.05) \u221a (0.57)(0.43) 200 = 0.57± 0.072, which yields 0.498 < p < 0.642. (b) Error \u2264 (2.05) \u221a (0.57)(0.43) 200 = 0.072. 9.52 n = 500.p\u2c6 = 485 500 = 0.97, q\u2c6 = 0.03, and z0.05 = 1.645. So, 0.97± (1.645) \u221a (0.97)(0.03) 500 = 0.97± 0.013, which yields 0.957 < p < 0.983. 9.53 n = 1000, p\u2c6 = 228 1000 = 0.228, q\u2c6 = 0.772, and z0.005 = 2.575. So, 0.228± (2.575) \u221a (0.228)(0.772) 1000 = 0.228± 0.034, which yields 0.194 < p < 0.262. 9.54 n = 100, p\u2c6 = 8 100 = 0.08, q\u2c6 = 0.92, and z0.01 = 2.33. So, 0.08± (2.33) \u221a (0.08)(0.92) 100 = 0.08± 0.063, which yields 0.017 < p < 0.143. 110 Chapter 9 One- and Two-Sample Estimation Problems 9.55 (a) n = 40, p\u2c6 = 34 40 = 0.85, q\u2c6 = 0.15, and z0.025 = 1.96. So, 0.85± (1.96) \u221a (0.85)(0.15) 40 = 0.85± 0.111, which yields 0.739 < p < 0.961. (b) Since p = 0.8 falls in the confidence interval, we can not conclude that the new system is better. 9.56 n = 100, p\u2c6 = 24 100 = 0.24, q\u2c6 = 0.76, and z0.005 = 2.575. (a) 0.24± (2.575) \u221a (0.24)(0.76) 100 = 0.24± 0.110, which yields 0.130 < p < 0.350. (b) Error \u2264 (2.575) \u221a (0.24)(0.76) 100 = 0.110.