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# Solucionario Walpole 8 ED

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Fundamental Sampling Distributions and Data Descriptions 8.63 The box-and-whisker plot is shown below. 5 10 15 20 25 Box\u2212and_Whisker Plot The sample mean is 2.7967 and the sample standard deviation is 2.2273. 8.64 P ( S21 S22 > 1.26 ) = P ( S21/\u3c3 2 1 S22/\u3c3 2 2 > (15)(1.26) 10 ) = P (F > 1.89) \u2248 0.05, where F has 24 and 30 degrees of freedom. 8.65 No outliers. 8.66 The value 32 is a possible outlier. 8.67 µ = 5,000 psi, \u3c3 = 400 psi, and n = 36. (a) Using approximate normal distribution (by CLT), P (4800 < X¯ < 5200) = P ( 4800\u2212 5000 400/ \u221a 36 < Z < 5200\u2212 5000 400/ \u221a 36 ) = P (\u22123 < Z < 3) = 0.9974. (b) To find a z such that P (\u2212z < Z < z) = 0.99, we have P (Z < z) = 0.995, which results in z = 2.575. Hence, by solving 2.575 = 5100\u22125000 400/ \u221a n we have n \u2265 107. Note that the value n can be affected by the z values picked (2.57 or 2.58). 8.68 x¯ = 54,100 and s = 5801.34. Hence t = 54100\u2212 53000 5801.34/ \u221a 10 = 0.60. So, P (X¯ \u2265 54, 100) = P (T \u2265 0.60) is a value between 0.20 and 0.30, which is not a rare event. 8.69 nA = nB = 20, x¯A = 20.50, x¯B = 24.50, and \u3c3A = \u3c3B = 5. (a) P (X¯A \u2212 X¯B \u2265 4.0 | µA = µB) = P (Z > (24.5\u2212 20.5)/ \u221a 52/20 + 52/20) = P (Z > 4.5/(5/ \u221a 10)) = P (Z > 2.85) = 0.0022. Solutions for Exercises in Chapter 8 101 (b) It is extremely unlikely that µA = µB. 8.70 (a) nA = 30, x¯A = 64.5% and \u3c3A = 5%. Hence, P (X¯A \u2264 64.5 | µA = 65) = P (Z < (64.5\u2212 65)/(5/ \u221a 30)) = P (Z < \u22120.55) = 0.2912. There is no evidence that the µA is less than 65%. (b) nB = 30, x¯B = 70% and \u3c3B = 5%. It turns out \u3c3X¯B\u2212X¯A = \u221a 52 30 + 5 2 30 = 1.29%. Hence, P (X¯B \u2212 X¯A \u2265 5.5 | µA = µB) = P ( Z \u2265 5.5 1.29 ) = P (Z \u2265 4.26) \u2248 0. It does strongly support that µB is greater than µA. (c) i) Since \u3c3X¯B = 5\u221a 30 = 0.9129, X¯B \u223c n(x; 65%, 0.9129%). ii) X¯A \u2212 X¯B \u223c n(x; 0, 1.29%). iii) X¯A\u2212X¯B \u3c3 \u221a 2/30 \u223c n(z; 0, 1). 8.71 P (X¯B \u2265 70) = P ( Z \u2265 70\u221265 0.9129 ) = P (Z \u2265 5.48) \u2248 0. 8.72 It is known, from Table A.3, that P (\u22121.96 < Z < 1.96) = 0.95. Given µ = 20 and \u3c3 = \u221a 9 = 3, we equate 1.96 = 20.1\u221220 3/ \u221a n to obtain n = ( (3)(1.96) 0.1 )2 = 3457.44 \u2248 3458. 8.73 It is known that P (\u22122.575 < Z < 2.575) = 0.99. Hence, by equating 2.575 = 0.05 1/ \u221a n , we obtain n = ( 2.575 0.05 )2 = 2652.25 \u2248 2653. 8.74 µ = 9 and \u3c3 = 1. Then P (9\u2212 1.5 < X < 9 + 1.5) = P (7.5 < X < 10.5) = P (\u22121.5 < Z < 1.5) = 0.9322\u2212 0.0668 = 0.8654. Thus the proportion of defective is 1 \u2212 0.8654 = 0.1346. To meet the specifications 99% of the time, we need to equate 2.575 = 1.5 \u3c3 , since P (\u22122.575 < Z < 2.575) = 0.99. Therefore, \u3c3 = 1.5 2.575 = 0.5825. 8.75 With the 39 degrees of freedom, P (S2 \u2264 0.188 | \u3c32 = 1.0) = P (\u3c72 \u2264 (39)(0.188)) = P (\u3c72 \u2264 7.332) \u2248 0, which means that it is impossible to observe s2 = 0.188 with n = 40 for \u3c32 = 1. Note that Table A.5 does not provide any values for the degrees of freedom to be larger than 30. However, one can deduce the conclusion based on the values in the last line of the table. Also, computer software gives the value of 0. Chapter 9 One- and Two-Sample Estimation Problems 9.1 From Example 9.1 on page 271, we know that E(S2) = \u3c32. Therefore, E(S \u20322) = E [ n\u2212 1 n S2 ] = n\u2212 1 n E(S2) = n\u2212 1 n \u3c32. 9.2 (a) E(X) = np; E(P\u2c6 ) = E(X/n) = E(X)/n = np/n = p. (b) E(P \u2032) = E(X)+ \u221a n/2 n+ \u221a n = np+ \u221a n/2 n+ \u221a n 6= p. 9.3 lim n\u2192\u221e np+ \u221a n/2 n+ \u221a n = lim n\u2192\u221e p+1/2 \u221a n 1+1/ \u221a n = p. 9.4 n = 30, x¯ = 780, and \u3c3 = 40. Also, z0.02 = 2.054. So, a 96% confidence interval for the population mean can be calculated as 780\u2212 (2.054)(40/ \u221a 30) < µ < 780 + (2.054)(40/ \u221a 30), or 765 < µ < 795. 9.5 n = 75, x¯ = 0.310, \u3c3 = 0.0015, and z0.025 = 1.96. A 95% confidence interval for the population mean is 0.310\u2212 (1.96)(0.0015/ \u221a 75) < µ < 0.310 + (1.96)(0.0015/ \u221a 75), or 0.3097 < µ < 0.3103. 9.6 n = 50, x¯ = 174.5, \u3c3 = 6.9, and z0.01 = 2.33. (a) A 98% confidence interval for the population mean is 174.5\u2212 (2.33)(6.9/\u221a50) < µ < 174.5 + (2.33)(6.9/\u221a50), or 172.23 < µ < 176.77. (b) e < (2.33)(6.9)/ \u221a 50 = 2.27. 103 104 Chapter 9 One- and Two-Sample Estimation Problems 9.7 n = 100, x¯ = 23, 500, \u3c3 = 3900, and z0.005 = 2.575. (a) A 99% confidence interval for the population mean is 23, 500\u2212 (2.575)(3900/10) < µ < 23, 500 + (2.575)(3900/10), or 22, 496 < µ < 24, 504. (b) e < (2.575)(3900/10) = 1004. 9.8 n = [(2.05)(40)/10]2 = 68 when rounded up. 9.9 n = [(1.96)(0.0015)/0.0005]2 = 35 when rounded up. 9.10 n = [(1.96)(40)/15]2 = 28 when rounded up. 9.11 n = [(2.575)(5.8)/2]2 = 56 when rounded up. 9.12 n = 20, x¯ = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95% confidence interval for the population mean is 11.3\u2212 (2.093)(2.45/ \u221a 20) < µ < 11.3 + (2.093)(2.45/ \u221a 20), or 10.15 < µ < 12.45. 9.13 n = 9, x¯ = 1.0056, s = 0.0245, and t0.005 = 3.355 with 8 degrees of freedom. A 99% confidence interval for the population mean is 1.0056\u2212 (3.355)(0.0245/3) < µ < 1.0056 + (3.355)(0.0245/3), or 0.978 < µ < 1.033. 9.14 n = 10, x¯ = 230, s = 15, and t0.005 = 3.25 with 9 degrees of freedom. A 99% confidence interval for the population mean is 230\u2212 (3.25)(15/ \u221a 10) < µ < 230 + (3.25)(15/ \u221a 10), or 214.58 < µ < 245.42. 9.15 n = 12, x¯ = 48.50, s = 1.5, and t0.05 = 1.796 with 11 degrees of freedom. A 90% confidence interval for the population mean is 48.50\u2212 (1.796)(1.5/ \u221a 12) < µ < 48.50 + (1.796)(1.5/ \u221a 12), or 47.722 < µ < 49.278. 9.16 n = 12, x¯ = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95% confidence interval for the population mean is 79.3\u2212 (2.201)(7.8/ \u221a 12) < µ < 79.3 + (2.201)(7.8/ \u221a 12), or 74.34 < µ < 84.26. Solutions for Exercises in Chapter 9 105 9.17 n = 25, x¯ = 325.05, s = 0.5, \u3b3 = 5%, and 1\u2212 \u3b1 = 90%, with k = 2.208. So, 325.05± (2.208)(0.5) yields (323.946, 326.151). Thus, we are 95% confident that this tolerance interval will contain 90% of the aspirin contents for this brand of buffered aspirin. 9.18 n = 15, x¯ = 3.7867, s = 0.9709, \u3b3 = 1%, and 1 \u2212 \u3b1 = 95%, with k = 3.507. So, by calculating 3.7867± (3.507)(0.9709) we obtain (0.382, 7.192) which is a 99% tolerance interval that will contain 95% of the drying times. 9.19 n = 100, x¯ = 23,500, s = 3, 900, 1 \u2212 \u3b1 = 0.99, and \u3b3 = 0.01, with k = 3.096. The tolerance interval is 23,500± (3.096)(3,900) which yields 11,425 < µ < 35,574. 9.20 n = 12, x¯ = 48.50, s = 1.5, 1\u2212 \u3b1 = 0.90, and \u3b3 = 0.05, with k = 2.655. The tolerance interval is 48.50± (2.655)(1.5) which yields (44.52, 52.48). 9.21 By definition, MSE = E(\u398\u2c6\u2212 \u3b8)2 which can be expressed as MSE = E[\u398\u2c6\u2212E(\u398\u2c6) + E(\u398\u2c6)\u2212 \u3b8]2 = E[\u398\u2c6\u2212E(\u398\u2c6)]2 + E[E(\u398\u2c6)\u2212 \u3b8]2 + 2E[\u398\u2c6\u2212 E(\u398\u2c6)]E[E(\u398\u2c6)\u2212 \u3b8]. The third term on the right hand side is zero since E[\u398\u2c6\u2212E(\u398\u2c6)] = E[\u398\u2c6]\u2212E(\u398\u2c6) = 0. Hence the claim is valid. 9.22 (a) The bias is E(S \u20322)\u2212 \u3c32 = n\u22121 n \u3c32 \u2212 \u3c32 = \u3c32 n . (b) lim n\u2192\u221e Bias = lim n\u2192\u221e \u3c32 n = 0. 9.23 Using Theorem 8.4, we know that X2 = (n\u22121)S 2 \u3c32 follows a chi-squared distribution with n\u2212 1 degrees of freedom, whose variance is 2(n\u2212 1). So, V ar(S2) = V ar ( \u3c32 n\u22121X 2 ) = 2 n\u22121\u3c3 4, and V ar(S \u20322) = V ar ( n\u22121 n S2 ) = ( n\u22121 n )2 V ar(S2) = 2(n\u22121)\u3c3 4 n2 . Therefore, the variance of S \u20322 is smaller. 9.24 Using Exercises 9.21 and 9.23, MSE(S2) MSE(S \u20322) = V ar(S2) + [Bias(S2)]2 V ar(S \u20322) + [Bias(S \u20322)]2 = 2\u3c34/(n\u2212 1) 2(n\u2212 1)\u3c34/n2 + \u3c34/n2 = 1 + 3n\u2212 1 2n2 \u2212 3n+ 1 , which is always larger than 1 when n is larger than 1. Hence theMSE of S \u20322 is usually smaller. 9.25 n = 20, x¯ = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95% prediction interval