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# Solucionario Walpole 8 ED

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9.57 n = 1600, p\u2c6 = 2 3 , q\u2c6 = 1 3 , and z0.025 = 1.96. (a) 2 3 ± (1.96) \u221a (2/3)(1/3) 1600 = 2 3 ± 0.023, which yields 0.644 < p < 0.690. (b) Error \u2264 (1.96) \u221a (2/3)(1/3) 1600 = 0.023. 9.58 n = (1.96) 2(0.32)(0.68) (0.02)2 = 2090 when round up. 9.59 n = (2.05) 2(0.57)(0.43) (0.02)2 = 2576 when round up. 9.60 n = (2.575) 2(0.228)(0.772) (0.05)2 = 467 when round up. 9.61 n = (2.33) 2(0.08)(0.92) (0.05)2 = 160 when round up. 9.62 n = (1.96) 2 (4)(0.01)2 = 9604 when round up. 9.63 n = (2.575) 2 (4)(0.01)2 = 16577 when round up. 9.64 n = (1.96) 2 (4)(0.04)2 = 601 when round up. 9.65 nM = nF = 1000, p\u2c6M = 0.250, q\u2c6M = 0.750, p\u2c6F = 0.275, q\u2c6F = 0.725, and z0.025 = 1.96. So (0.275\u2212 0.250)± (1.96) \u221a (0.250)(0.750) 1000 + (0.275)(0.725) 1000 = 0.025± 0.039, which yields \u22120.0136 < pF \u2212 pM < 0.0636. Solutions for Exercises in Chapter 9 111 9.66 n1 = 250, n2 = 175, p\u2c61 = 80 250 = 0.32, p\u2c62 = 40 175 = 0.2286, and z0.05 = 1.645. So, (0.32\u2212 0.2286)± (1.645) \u221a (0.32)(0.68) 250 + (0.2286)(0.7714) 175 = 0.0914± 0.0713, which yields 0.0201 < p1 \u2212 p2 < 0.1627. From this study we conclude that there is a significantly higher proportion of women in electrical engineering than there is in chemical engineering. 9.67 n1 = n2 = 500, p\u2c61 = 120 500 = 0.24, p\u2c62 = 98 500 = 0.196, and z0.05 = 1.645. So, (0.24\u2212 0.196)± (1.645) \u221a (0.24)(0.76) 500 + (0.196)(0.804) 500 = 0.044± 0.0429, which yields 0.0011 < p1 \u2212 p2 < 0.0869. Since 0 is not in this confidence interval, we conclude, at the level of 90% confidence, that inoculation has an effect on the incidence of the disease. 9.68 n5\u25e6C = n15\u25e6C = 20, p\u2c65\u25e6C = 0.50, p\u2c615\u25e6C = 0.75, and z0.025 = 1.96. So, (0.5\u2212 0.75)± (1.96) \u221a (0.50)(0.50) 20 + (0.75)(0.25) 20 = \u22120.25± 0.2899, which yields \u22120.5399 < p5\u25e6C \u2212 p15\u25e6C < 0.0399. Since this interval includes 0, the significance of the difference cannot be shown at the confidence level of 95%. 9.69 nnow = 1000, p\u2c6now = 0.2740, n91 = 760, p\u2c691 = 0.3158, and z0.025 = 1.96. So, (0.2740\u2212 0.3158)± (1.96) \u221a (0.2740)(0.7260) 1000 + (0.3158)(0.6842) 760 = \u22120.0418± 0.0431, which yields \u22120.0849 < pnow \u2212 p91 < 0.0013. Hence, at the confidence level of 95%, the significance cannot be shown. 9.70 n90 = n94 = 20, p\u2c690 = 0.337, and 0\u2c694 = 0.362 (a) n90p\u2c690 = (20)(0.337) \u2248 7 and n94p\u2c694 = (20)(0.362) \u2248 7. (b) Since z0.025 = 1.96, (0.337\u22120.362)± (1.96) \u221a (0.337)(0.663) 20 + (0.362)(0.638) 20 = \u22120.025± 0.295, which yields \u22120.320 < p90 \u2212 p94 < 0.270. Hence there is no evidence, at the confidence level of 95%, that there is a change in the proportions. 9.71 s2 = 0.815 with v = 4 degrees of freedom. Also, \u3c720.025 = 11.143 and \u3c7 2 0.975 = 0.484. So, (4)(0.815) 11.143 < \u3c32 < (4)(0.815) 0.484 , which yields 0.293 < \u3c32 < 6.736. Since this interval contains 1, the claim that \u3c32 seems valid. 112 Chapter 9 One- and Two-Sample Estimation Problems 9.72 s2 = 16 with v = 19 degrees of freedom. It is known \u3c720.01 = 36.191 and \u3c7 2 0.99 = 7.633. Hence (19)(16) 36.191 < \u3c32 < (19)(16) 7.633 , or 8.400 < \u3c32 < 39.827. 9.73 s2 = 6.0025 with v = 19 degrees of freedom. Also, \u3c720.025 = 32.852 and \u3c7 2 0.975 = 8.907. Hence, (19)(6.0025) 32.852 < \u3c32 < (19)(6.0025) 8.907 , or 3.472 < \u3c32 < 12.804, 9.74 s2 = 0.0006 with v = 8 degrees of freedom. Also, \u3c720.005 = 21.955 and \u3c7 2 0.995 = 1.344. Hence, (8)(0.0006) 21.955 < \u3c32 < (8)(0.0006) 1.344 , or 0.00022 < \u3c32 < 0.00357. 9.75 s2 = 225 with v = 9 degrees of freedom. Also, \u3c720.005 = 23.589 and \u3c7 2 0.995 = 1.735. Hence, (9)(225) 23.589 < \u3c32 < (9)(225) 1.735 , or 85.845 < \u3c32 < 1167.147, which yields 9.27 < \u3c3 < 34.16. 9.76 s2 = 2.25 with v = 11 degrees of freedom. Also, \u3c720.05 = 19.675 and \u3c7 2 0.95 = 4.575. Hence, (11)(2.25) 19.675 < \u3c32 < (11)(2.25) 4.575 , or 1.258 < \u3c32 < 5.410. 9.77 s21 = 1.00, s 2 2 = 0.64, f0.01(11, 9) = 5.19, and f0.01(9, 11) = 4.63. So, 1.00/0.64 5.19 < \u3c321 \u3c322 < (1.00/0.64)(4.63), or 0.301 < \u3c321 \u3c322 < 7.234, which yields 0.549 < \u3c31 \u3c32 < 2.690. 9.78 s21 = 5000 2, s22 = 6100 2, and f0.05(11, 11) = 2.82. (Note: this value can be found by using \u201c=finv(0.05,11,11)\u201d in Microsoft Excel.) So,( 5000 6100 )2 1 2.82 < \u3c321 \u3c322 < ( 5000 6100 )2 (2.82), or 0.238 < \u3c321 \u3c322 < 1.895. Since the interval contains 1, it is reasonable to assume that \u3c321 = \u3c3 2 2 . 9.79 s2I = 76.3, s 2 II = 1035.905, f0.05(4, 6) = 4.53, and f0.05(6, 4) = 6.16. So,( 76.3 1035.905 )( 1 4.53 ) < \u3c32I \u3c32II < ( 76.3 1035.905 ) (6.16), or 0.016 < \u3c32I \u3c32II < 0.454. Hence, we may assume that \u3c32I 6= \u3c32II . Solutions for Exercises in Chapter 9 113 9.80 sA = 0.7794, sB = 0.7538, and f0.025(14, 14) = 2.98 (Note: this value can be found by using \u201c=finv(0.025,14,14)\u201d in Microsoft Excel.) So,( 0.7794 0.7538 )2( 1 2.98 ) < \u3c32A \u3c32B < ( 0.7794 0.7538 )2 (2.98), or 0.623 < \u3c32A \u3c32B < 3.186. Hence, it is reasonable to assume the equality of the variances. 9.81 The likelihood function is L(x1, . . . , xn) = n\u220f i=1 f(xi; p) = n\u220f i=1 pxi(1\u2212 p)1\u2212xi = pnx¯(1\u2212 p)n(1\u2212x¯). Hence, lnL = n[x¯ ln(p) + (1 \u2212 x¯) ln(1 \u2212 p)]. Taking derivative with respect to p and setting the derivative to zero, we obtain \u2202 ln(L) \u2202p = n ( x¯ p \u2212 1\u2212x¯ 1\u2212p ) = 0, which yields x¯ p \u2212 1\u2212x¯ 1\u2212p = 0. Therefore, p\u2c6 = x¯. 9.82 (a) The likelihood function is L(x1, . . . , xn) = n\u220f i=1 f(xi;\u3b1, \u3b2) = (\u3b1\u3b2) n n\u220f i=1 x\u3b2\u22121i e \u2212\u3b1x\u3b2i = (\u3b1\u3b2)ne \u2212\u3b1 nP i=1 x\u3b2i ( n\u220f i=1 xi )\u3b2\u22121 . (b) So, the log-likelihood can be expressed as lnL = n[ln(\u3b1) + ln(\u3b2)]\u2212 \u3b1 n\u2211 i=1 x\u3b2i + (\u3b2 \u2212 1) n\u2211 i=1 ln(xi). To solve for the maximum likelihood estimate, we need to solve the following two equations \u2202 lnL \u2202\u3b1 = 0, and \u2202 lnL \u2202\u3b2 = 0. 9.83 (a) The likelihood function is L(x1, . . . , xn) = n\u220f i=1 f(xi;µ, \u3c3) = n\u220f i=1 { 1\u221a 2\u3c0\u3c3xi e\u2212 [ln(xi)\u2212µ] 2 2\u3c32 } = 1 (2\u3c0)n/2\u3c3n n\u220f i=1 xi exp { \u2212 1 2\u3c32 n\u2211 i=1 [ln(xi)\u2212 µ]2 } . 114 Chapter 9 One- and Two-Sample Estimation Problems (b) It is easy to obtain lnL = \u2212n 2 ln(2\u3c0)\u2212 n 2 ln(\u3c32)\u2212 n\u2211 i=1 ln(xi)\u2212 1 2\u3c32 n\u2211 i=1 [ln(xi)\u2212 µ]2. So, setting 0 = \u2202 lnL \u2202µ = 1 \u3c32 n\u2211 i=1 [ln(xi) \u2212 µ], we obtain µ\u2c6 = 1n n\u2211 i=1 ln(xi), and setting 0 = \u2202 lnL \u2202\u3c32 = \u2212 n 2\u3c32 + 1 2\u3c34 n\u2211 i=1 [ln(xi)\u2212 µ]2, we get \u3c3\u2c62 = 1n n\u2211 i=1 [ln(xi)\u2212 µ\u2c6]2. 9.84 (a) The likelihood function is L(x1, . . . , xn) = 1 \u3b2n\u3b1\u393(\u3b1)n n\u220f i=1 ( x\u3b1\u22121i e \u2212xi/\u3b2) = 1 \u3b2n\u3b1\u393(\u3b1)n ( n\u220f i=1 xi )\u3b1\u22121 e \u2212 nP i=1 (xi/\u3b2) . (b) Hence lnL = \u2212n\u3b1 ln(\u3b2)\u2212 n ln(\u393(\u3b1)) + (\u3b1\u2212 1) n\u2211 i=1 ln(xi)\u2212 1 \u3b2 n\u2211 i=1 xi. Taking derivatives of lnL with respect to \u3b1 and \u3b2, respectively and setting both as zeros. Then solve them to obtain the maximum likelihood estimates. 9.85 L(x) = px(1\u2212 p)1\u2212x, and lnL = x ln(p) + (1\u2212 x) ln(1\u2212 p), with \u2202 lnL \u2202p = x p \u2212 1\u2212x 1\u2212p = 0, we obtain p\u2c6 = x = 1. 9.86 From the density function b\u2217(x; p) = ( x\u22121 k\u22121 ) pk(1\u2212 p)x\u2212k, we obtain lnL = ln ( x\u2212 1 k \u2212 1 ) + k ln p+ (n\u2212 k) ln(1\u2212 p). Setting \u2202 lnL \u2202p = k p \u2212 n\u2212k 1\u2212p = 0, we obtain p\u2c6 = k n . 9.87 For the estimator S2, V ar(S2) = 1 (n\u2212 1)2V ar [ n\u2211 i=1 (xi \u2212 x¯)2 ] = 1 (n\u2212 1)2V ar(\u3c3 2\u3c72n\u22121) = 1 (n\u2212 1)2\u3c3 4[2(n\u2212 1)] = 2\u3c3 4 n\u2212 1 . For the estimator \u3c3\u3022, we have V ar(\u3c3\u3022) = 2\u3c34(n\u2212 1) n2 . Solutions for Exercises in Chapter 9 115 9.88 n = 7, d¯ = 3.557, sd = 2.776, and t0.025 = 2.447 with 6 degrees of freedom. So, 3.557± (2.447)2.776\u221a 7 = 3.557± 2.567, which