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# Solucionario Walpole 8 ED

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(r \u2212 1)\u3c32 + r \u2211 k \u3c4 2k . 13.39 The hypotheses are H0 : \u3c41 = \u3c42 = \u3c43 = \u3c44 = 0, professor effects are zero H1 : At least one of the \u3c4i\u2019s is not equal to zero. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Time Periods Courses Professors Error 474.50 252.50 723.50 287.50 3 3 3 6 158.17 84.17 241.17 47.92 5.03 Total 1738.00 15 P -value= 0.0446. Decision: Reject H0; grades are affected by different professors. 200 Chapter 13 One-Factor Experiments: General 13.40 The hypotheses are H0 : \u3c4A = \u3c4B = \u3c4C = \u3c4D = \u3c4E = 0, color additive effects are zero H1 : At least one of the \u3c4i\u2019s is not equal to zero. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Workers Days Additives Error 12.4344 14.7944 3.9864 9.2712 4 4 4 12 3.1086 3.6986 0.9966 0.7726 1.29 Total 40.4864 24 P -value= 0.3280. Decision: Do not reject H0; color additives could not be shown to have an effect on setting time. 13.41 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, dye effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Amounts Plants Error 1238.8825 53.7004 101.2433 2 1 20 619.4413 53.7004 5.0622 122.37 Total 1393.8263 23 P -value< 0.0001. Decision: Reject H0; color densities of fabric differ significantly for three levels of dyes. 13.42 (a) After a transformation g(y) = \u221a y, we come up with an ANOVA table as follows. Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Materials Error 7.5123 6.2616 2 27 3.7561 0.2319 16.20 Total 13.7739 29 Solutions for Exercises in Chapter 13 201 (b) The P -value< 0.0001. Hence, there is significant difference in flaws among three materials. (c) A residual plot is given below and it does show some heterogeneity of the variance among three treatment levels. 1.0 1.5 2.0 2.5 3.0 \u2212 0. 5 0. 0 0. 5 1. 0 material re si du al (d) The purpose of the transformation is to stabilize the variances. (e) One could be the distribution assumption itself. Once the data is transformed, it is not necessary that the data would follow a normal distribution. (f) Here the normal probability plot on residuals is shown. \u22122 \u22121 0 1 2 \u2212 0. 5 0. 0 0. 5 1. 0 Theoretical Quantiles Sa m pl e Qu an tile s It appears to be close to a straight line. So, it is likely that the transformed data are normally distributed. 13.43 (a) The hypotheses are H0 : \u3c3 2 \u3b1 = 0, H1 : \u3c3 2 \u3b1 6= 0 \u3b1 = 0.05. Computation: 202 Chapter 13 One-Factor Experiments: General Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Operators Error 371.8719 99.7925 3 12 123.9573 8.3160 14.91 Total 471.6644 15 P -value= 0.0002. Decision: Reject H0; operators are different. (b) \u3c3\u2c62 = 8.316 and \u3c3\u2c62\u3b1 = 123.9573\u22128.3160 4 = 28.910. 13.44 The model is yij = µ+ Ai +Bj + \u1ebij . Hence y¯.j = µ+ A¯. +Bj + \u1eb¯.j , and y¯.. = µ+ A¯. + B¯. + \u1eb¯... Therefore, SSB = k b\u2211 j=1 (y¯.j \u2212 y¯..)2 = k b\u2211 j=1 [(Bj \u2212 B¯.) + (\u1eb¯.j \u2212 \u1eb¯..)]2, and E(SSB) = k b\u2211 j=1 E(B2j )\u2212 kbE(B¯2. ) + k b\u2211 j=1 E(\u1eb¯2.j)\u2212 kbE(\u1eb¯2..) = kb\u3c32\u3b2 \u2212 k\u3c32\u3b2 + b\u3c32 \u2212 \u3c32 = (b\u2212 1)\u3c32 + k(b\u2212 1)\u3c32\u3b2. 13.45 (a) The hypotheses are H0 : \u3c3 2 \u3b1 = 0, H1 : \u3c3 2 \u3b1 6= 0 \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Treatments Blocks Error 23.238 45.283 27.937 3 4 12 7.746 11.321 2.328 3.33 Total 96.458 19 P -value= 0.0565. Decision: Not able to show a significant difference in the ran- dom treatments at 0.05 level, although the P -value shows marginal significance. (b) \u3c32\u3b1 = 7.746\u22122.328 5 = 1.084, and \u3c32\u3b2 = 11.321\u22122.328 4 = 2.248. 13.46 From the model yijk = µ+ Ai +Bj + Tk + \u1ebij , Solutions for Exercises in Chapter 13 203 we have y¯..k = µ+ A¯. + B¯. + Tk + \u1eb¯..k, and y¯... = µ+ A¯. + B¯. + T¯. + \u1eb¯.... Hence, SSTr = r \u2211 k (y¯..k \u2212 y¯...)2 = r \u2211 k [(Tj \u2212 T¯.) + (\u1eb¯..k \u2212 \u1eb¯...)]2, and E(SSTr) = r \u2211 k E(T 2k )\u2212 r2E(T¯ 2. ) + r \u2211 k E(\u1eb¯2..k)\u2212 r2E(\u1eb¯2...) = r2\u3c32\u3c4 \u2212 r\u3c32\u3c4 + r\u3c32 \u2212 \u3c32 = (r \u2212 1)(\u3c32 + r\u3c32\u3b2). 13.47 (a) The matrix is A = \uf8ee\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8f0 bk b b · · · b k k · · · k b b 0 · · · 0 1 1 · · · 1 b 0 b · · · 0 1 1 · · · 1 ... ... ... . . . ... ... ... . . . ... b 0 0 · · · b 1 1 · · · 1 k 1 1 · · · k 0 0 · · · 0 k 1 1 · · · 0 k 0 · · · 0 ... ... ... . . . ... ... ... . . . ... k 1 1 · · · 0 0 0 · · · k \uf8f9\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fb , where b = number of blocks and k = number of treatments. The vectors are b \u2032 = (µ, \u3b11, \u3b12, · · · , \u3b1k, \u3b21, \u3b22, · · · , \u3b2b)\u2032, and g \u2032 = (T.., T1., T2., · · · , Tk., T.1, T.2, · · · , T.b)\u2032 . (b) Solving the system Ab = g with the constraints k\u2211 i=1 \u3b1i = 0 and b\u2211 j=1 \u3b2j = 0, we have µ\u2c6 = y¯.., \u3b1\u2c6i = y¯i. \u2212 y¯.., for i = 1, 2, . . . , k, \u3b2\u2c6j = y¯.j \u2212 y¯.., for j = 1, 2, . . . , b. Therefore, R(\u3b11, \u3b12, . . . , \u3b1k, \u3b21, \u3b22, . . . , \u3b2b) = b \u2032 g \u2212 T 2 .. bk = k\u2211 i=1 T 2i. b + b\u2211 j=1 T 2.j k \u2212 2T 2 .. bk . 204 Chapter 13 One-Factor Experiments: General To find R(\u3b21, \u3b22, . . . , \u3b2b | \u3b11, \u3b12, . . . , \u3b1k) we first find R(\u3b11, \u3b12, . . . , \u3b1k). Setting \u3b2j = 0 in the model, we obtain the estimates (after applying the constraint k\u2211 i=1 \u3b1i = 0) µ\u2c6 = y¯.., and \u3b1\u2c6i = y¯i. \u2212 y¯.., for i = 1, 2, . . . , k. The g vector is the same as in part (a) with the exception that T.1, T.2, . . . , T.b do not appear. Thus one obtains R(\u3b11, \u3b12, . . . , \u3b1k) = k\u2211 i=1 T 2i. b \u2212 T 2 .. bk and thus R(\u3b21, \u3b22, . . . , \u3b2b | \u3b11, \u3b12, . . . , \u3b1k) = R(\u3b11, \u3b12, . . . , \u3b1k, \u3b21, \u3b22, . . . , \u3b2b) \u2212 R(\u3b11, \u3b12, . . . , \u3b1k) = b\u2211 j=1 T 2.j k \u2212 T 2 .. bk = SSB. 13.48 Since 1\u2212 \u3b2 = P [ F (3, 12) > 3.49 1 + (4)(1.5) ] = P [F (12, 3) < 2.006] < 0.95. Hence we do not have large enough samples. We then find, by trial and error, that n = 16 is sufficient since 1\u2212 \u3b2 = P [ F (3, 60) > 2.76 1 + (16)(1.5) ] = P [F (60, 3) < 9.07] > 0.95. 13.49 We know \u3c62 = b 4\u2211 i=1 \u3b12i 4\u3c32 = b 2 , when 4\u2211 i=1 \u3b12i \u3c32 = 2.0. If b = 10, \u3c6 = 2.24; v1 = 3 and v2 = 27 degrees of freedom. If b = 9, \u3c6 = 2.12; v1 = 3 and v2 = 24 degrees of freedom. If b = 8, \u3c6 = 2.00; v1 = 3 and v2 = 21 degrees of freedom. From Table A.16 we see that b = 9 gives the desired result. 13.50 For the randomized complete block design we have E(S21) = E ( SSA k \u2212 1 ) = \u3c32 + b k\u2211 i=1 \u3b12i k \u2212 1 . Therefore, \u3bb = v1[E(S 2 1)] 2\u3c32 \u2212 v1 2 = (k \u2212 1) [ \u3c32 + b k\u2211 i=1 \u3b12i /(k \u2212 1) ] 2\u3c32 \u2212 k \u2212 1 2 = b k\u2211 i=1 \u3b12i 2\u3c32 , Solutions for Exercises in Chapter 13 205 and then \u3c62 = E(S21)\u2212 \u3c32 \u3c32 · v1 v1 + 1 = [\u3c32 + b k\u2211 i=1 \u3b12i /(k \u2212 1)]\u2212 \u3c32 \u3c32 · k \u2212 1 k = b k\u2211 i=1 \u3b12i k\u3c32 . 13.51 (a) The model is yij = µ+ \u3b1i + \u1ebij , where \u3b1i \u223c n(0, \u3c32\u3b1). (b) Since s2 = 0.02056 and s21 = 0.01791, we have \u3c3\u2c6 2 = 0.02056 and s21\u2212s2 10 = 0.01791\u22120.02056 10 = \u22120.00027, which implies \u3c3\u2c62\u3b1 = 0. 13.52 (a) The P -value of the test result is 0.7830. Hence, the variance component of pour is not significantly different from 0. (b) We have the ANOVA table as follows: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Pours Error 0.08906 1.02788 4 20 0.02227 0.05139 0.43 Total 1.11694 24 Since s21\u2212s2 5 = 0.02227\u22120.05139 5 < 0, we have \u3c3\u2c62\u3b1 = 0. 13.53 (a) yij = µ+ \u3b1i + \u1ebij , where \u3b1i \u223c n(x; 0, \u3c32\u3b1). (b) Running