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# Solucionario Walpole 8 ED

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ad bd d ac bc abd a b cd abcd L1 = 0 L1 = 1 L1 = 0 L1 = 1 L2 = 0 L2 = 0 L2 = 1 L2 = 1 Since (ABC)(ABD) = A2B2CD = CD (mod 2), then CD is the other effect con- founded. 15.16 (a) L1 = \u3b31+ \u3b32+ \u3b34+ \u3b35, L2 = \u3b31+ \u3b35. We find that the following treatment combi- nations are in the principal block (L1 = 0, L2 = 0): (1), c, ae, bd, ace, abde, abcde. The other blocks are constructed by multiplying the treatment combinations in the principal block modulo 2 by a, b, and ab, respectively, to give the following blocking arrangement: Block 1 Block 2 Block 3 Block 4 (1) c ae bd ace bcd abde abcde a ac e abd ce abcd bde bcde b bc abe d abce cd ade acde ab abc bce ad bce acd de cde (b) (ABDE)(AE) = BD (mod 2). Therefore BD is also confounded with days. (c) Yates\u2019 technique gives the following sums of squares for the main effects: SSA = 21.9453, SSB = 40.2753, SSC = 2.4753, SSD = 7.7028, SSE = 1.0878. 15.17 L1 = \u3b31 + \u3b32 + \u3b33, L2 = \u3b31 + \u3b32. Block Block Block 1 2 1 2 1 2 abc a b c ab ac bc (1) abc a b c ab ac bc (1) (1) c ab abc a b ac bc Rep 1 Rep 2 Rep 3 ABC Confounded ABC Confounded AB Confounded Solutions for Exercises in Chapter 15 245 For treatment combination (1) we find L1 (mod 2) = 0 and L2 (mod 2) = 0. For treatment combination a we find L1 (mod 2) = 1 and L2 (mod 2) = 1. Replicate 1 and Replicate 2 have L1 = 0 in one block and L1 = 1 in the other. Replicate 3 has L2 = 0 in one block and L2 = 1 in the other. Analysis of Variance Source of Variation Degrees of Freedom Blocks A B C AB AC BC ABC Error 5 1 1 1 1 \u2032 1 1 1 \u2032 11 Total 23 Relative information on ABC = 1 3 and relative information on AB = 2 3 . 15.18 (a) The ANOVA table is shown here. Source of Degrees of Mean Computed Variation Freedom Square f P -value Operators A B C D Error 1 1 1 1 1 10 0.1225 4.4100 3.6100 9.9225 2.2500 2.8423 0.04 1.55 1.27 3.49 0.79 0.2413 0.2861 0.0912 0.3945 Total 15 None of the main effects is significant at 0.05 level. (b) ABC is confounded with operators since all treatments with positive signs in the ABC contrast are in one block and those with negative signs are in the other block. 15.19 (a) One possible design would be: Machine 1 2 3 4 (1) a c d ab b abc abd ce ace e cde abce bce abe abcde acd cd ad ac bde abde bcde be ade de acde ae bcd abcd bd bc (b) ABD, CDE, and ABCE. 246 Chapter 15 2k Factorial Experiments and Fractions 15.20 (a) y\u2c6 = 43.9 + 1.625x1 \u2212 8.625x2 + 0.375x3 + 9.125x1x2 + 0.625x1x3 + 0.875x2x3. (b) The Lack-of-fit test results in a P -value of 0.0493. There are possible quadratic terms missing in the model. 15.21 (a) The P -values of the regression coefficients are: Parameter Intercept x1 x2 x3 x1x2 x1x3 x2x3 x1x2x3 P -value < 0.0001 0.5054 0.0772 0.0570 0.0125 0.0205 0.7984 0.6161 and s2 = 0.57487 with 4 degrees of freedom. So x2, x3, x1x2 and x1x3 are impor- tant in the model. (b) t = y¯f\u2212y¯C\u221a s2(1/nf+1/nC) = 52.075\u221249.275\u221a (0.57487)(1/8+1/4) = 6.0306. Hence the P -value = 0.0038 for testing quadratic curvature. It is significant. (c) Need one additional design point different from the original ones. 15.22 (a) No. (b) It could be as follows. Machine 1 2 3 4 (1) a b d ad e abe ade bc abc c bcd abce bce ace abcde acd cd abcd ac abd bd ad ab cde acde bcde ce bde abde de be ADE, BCD and ABCE are confounded with blocks. (c) Partial confounding. 15.23 To estimate the quadratic terms, it might be good to add points in the middle of the edges. Hence (\u22121, 0), (0,\u22121), (1, 0), and (0, 1) might be added. 15.24 The alias for each effect is obtained by multiplying each effect by the defining contrast and reducing the exponents modulo 2. A \u2261CDE, AB \u2261BCDE, BD\u2261ABCE, B \u2261ABCDE, AC \u2261DE, BE\u2261ABCD, C \u2261ADE, AD\u2261CE, ABC\u2261BDE, D \u2261ACE, AE\u2261CD, ABD\u2261BCE, E \u2261ACD, BC \u2261ABDE, ABE\u2261BCD, 15.25 (a) With BCD as the defining contrast, we have L = \u3b32+\u3b33+\u3b34. The 1 2 fraction cor- responding to L = 0 (mod 2 is the principal block: {(1), a, bc, abc, bd, abd, cd, acd}. (b) To obtain 2 blocks for the 1 2 fraction the interaction ABC is confounded using L = \u3b31 + \u3b32 + \u3b33: Solutions for Exercises in Chapter 15 247 Block 1 Block 2 (1) bc abd acd a abc bd cd (c) Using BCD as the defining contrast we have the following aliases: A\u2261ABCD, AB\u2261ACD, B\u2261CD, AC\u2261ABD, C\u2261BD, AD\u2261ABC, D\u2261BC. Since AD andABC are confounded with blocks there are only 2 degrees of freedom for error from the unconfounded interactions. Analysis of Variance Source of Variation Degrees of Freedom Blocks A B C D Error 1 1 1 1 1 2 Total 7 15.26 With ABCD and BDEF as defining contrasts, we have L1 = \u3b31 + \u3b32 + \u3b33 + \u3b34, L2 = \u3b32 + \u3b34 + \u3b35 + \u3b36. The following treatment combinations give L1 = 0, L2 = 0 (mod 2) and thereby suffice as the 1 4 fraction: {(1), ac, bd, abcd, abe, bce, ade, abf, bcf, adf, cdf, ef, acef, bdef, abcdef}. The third defining contrast is given by (ABCD)(BDEF ) = AB2CD2EF = ACEF (mod 2). The effects that are aliased with the six main effects are: A\u2261BCD \u2261ABDEF\u2261CEF, B\u2261ACD \u2261DEF\u2261ABCEF, C\u2261ABD \u2261BCDEF\u2261AEF, D\u2261ABC \u2261BEF\u2261ACDEF, E\u2261ABCDE\u2261BDF \u2261ACF, F\u2261ABCDF\u2261BDE\u2261ACE. 248 Chapter 15 2k Factorial Experiments and Fractions 15.27 (a) With ABCE and ABDF , and hence (ABCE)(ABDF ) = CDEF as the defining contrasts, we have L1 = \u3b31 + \u3b32 + \u3b33 + \u3b35, L2 = \u3b31 + \u3b32 + \u3b34 + \u3b36. The principal block, for which L1 = 0, and L2 = 0, is as follows: {(1), ab, acd, bcd, ce, abce, ade, bde, acf, bcf, df, abdf, aef, bef, cdef, abcdef}. (b) The aliases for each effect are obtained by multiplying each effect by the three defining contrasts and reducing the exponents modulo 2. A \u2261BCE \u2261BDF \u2261ACDEF, B \u2261ACE \u2261ADF \u2261BCDEF, C \u2261ABE \u2261ABCDF\u2261DEF, D \u2261ABCDE\u2261ABF \u2261CEF, E \u2261ABC \u2261ABDEF\u2261CDF, F \u2261ABCEF\u2261ABD \u2261CDE, AB \u2261CE \u2261DF \u2261ABCDEF, AC \u2261BE \u2261BCDF\u2261ADEF, AD \u2261BCDE\u2261BF \u2261ACEF, AE \u2261BC \u2261BDEF\u2261ACDF, AF \u2261BCEF\u2261BD \u2261ACDE, CD \u2261ABDE \u2261ABCF \u2261EF, DE \u2261ABCD\u2261ABEF \u2261CF, BCD\u2261ADE \u2261ACF \u2261BEF, DCF\u2261AEF \u2261ACD \u2261BDE, . Since E and F do not interact and all three-factor and higher interactions are negligible, we obtain the following ANOVA table: Source of Variation Degrees of Freedom A B C D E F AB AC AD BC BD CD Error 1 1 1 1 1 1 1 1 1 1 1 1 3 Total 15 15.28 The ANOVA table is shown here and the error term is computed by pooling all the interaction effects. Factor E is the only significant effect, at level 0.05, although the decision on factor G is marginal. Solutions for Exercises in Chapter 15 249 Source of Degrees of Mean Computed Variation Freedom Square f P -value A B C D E F G Error 1 1 1 1 1 1 1 8 1.44 4.00 9.00 5.76 16.00 3.24 12.96 2.97 0.48 1.35 3.03 1.94 5.39 1.09 4.36 0.5060 0.2793 0.1199 0.2012 0.0488 0.3268 0.0701 Total 15 15.29 All two-factor interactions are aliased with each other. So, assuming that two-factor as well as higher order interactions are negligible, a test on the main effects is given in the ANOVA table. Source of Degrees of Mean Computed Variation Freedom Square f P -value A B C D Error 1 1 1 1 3 6.125 0.605 4.805 0.245 1.053 5.81 0.57 4.56 0.23 0.0949 0.5036 0.1223 0.6626 Total 7 Apparently no main effects is significant at level 0.05. Comparatively factors A and C are more significant than the other two. Note that the degrees of freedom on the error term is only 3, the test is not very powerful. 15.30 Two-factor interactions are aliased with each other. There are total 7 two-factor in- teractions that can be estimated. Among those 7, we picked the three, which are AC,