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# Solucionario Walpole 8 ED

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AF , and BD, that have largest SS values and pool the other 2-way interactions to the error term. An ANOVA can be obtained. 250 Chapter 15 2k Factorial Experiments and Fractions Source of Degrees of Mean Computed Variation Freedom Square f P -value A B C D E F AC AF BD Error 1 1 1 1 1 1 1 1 1 6 81.54 166.54 5.64 4.41 40.20 1678.54 978.75 625.00 429.53 219.18 0.37 0.76 0.03 0.02 0.18 7.66 4.47 2.85 1.96 0.5643 0.4169 0.8778 0.8918 0.6834 0.0325 0.0790 0.1423 0.2111 Total 15 Main effect F , the location of detection, appears to be the only significant effect. The AC interaction, which is aliased with BE, has a P -value closed to 0.05. 15.31 To get all main effects and two-way interactions in the model, this is a saturated design, with no degrees of freedom left for error. Hence, we first get all SS of these effects and pick the 2-way interactions with large SS values, which are AD, AE, BD and BE. An ANOVA table is obtained. Source of Degrees of Mean Computed Variation Freedom Square f P -value A B C D E AD AE BD BE Error 1 1 1 1 1 1 1 1 1 6 388, 129.00 277, 202.25 4, 692.25 9, 702.25 1, 806.25 1, 406.25 462.25 1, 156.25 961.00 108.25 3, 585.49 2, 560.76 43.35 89.63 16.69 12.99 4.27 10.68 8.88 < 0.0001 < 0.0001 0.0006 < 0.0001 0.0065 0.0113 0.0843 0.0171 0.0247 Total 15 All main effects, plus AD, BD and BE two-way interactions, are significant at 0.05 level. 15.32 Consider a 24 design with letters A, B, C, and D, with design points {(1), a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd} . Using E = ABCD, we have the following design: {e, a, b, c, d, abe, ace, ade, bce, bde, cde, abc, abd, acd, bcd, abcde}. Solutions for Exercises in Chapter 15 251 15.33 Begin with a 23 with design points {(1), a, b, c, ab, ac, bc, abc}. Now, use the generator D = AB, E = AC, and F = BC. We have the following result: {def, af, be, cd, abd, ace, bcf, abcdef}. 15.34 We can use the D = AB, E = \u2212AC and F = BC as generators and obtain the result: {df, aef, b, cde, abde, ac, bcef, abcdf}. 15.35 Here are all the aliases A\u2261BD\u2261CE\u2261CDF\u2261BEF\u2261 \u2261ABCF \u2261ADEF \u2261ABCDE; B\u2261AD\u2261CF\u2261CDE\u2261AEF \u2261 \u2261ABCE \u2261BDEF \u2261ABCDF ; C\u2261AE\u2261BF\u2261BDE\u2261ADF\u2261 \u2261CDEF \u2261ABCD \u2261ABCEF ; D\u2261AB\u2261EF\u2261BCE\u2261ACF \u2261 \u2261BCDF \u2261ACDE \u2261ABDEF ; E\u2261AC\u2261DF\u2261ABF \u2261BCD\u2261 \u2261ABDE \u2261BCEF \u2261ACDEF ; F\u2261BC\u2261DE\u2261ACD\u2261ABE\u2261 \u2261ACEF \u2261ABDF \u2261BCDEF. 15.36 (a) The defining relation is ABC = \u2212I. (b) A = \u2212BC, B = \u2212AC, and C = \u2212AB. (c) The mean squares for A, B, and C are 1.50, 0.34, and 5.07, respectively. So, factor C, the amount of grain refiner, appears to be most important. (d) Low level of C. (e) All at the \u201clow\u201d level. (f) A hazard here is that the interactions may play significant roles. The following are two interaction plots. 1 1 0. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 AB Interaction A y 2 2 \u22121 1 B 1 2 \u22121 1 1 1 0. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 BC Interaction B y 2 2 \u22121 1 C 1 2 \u22121 1 252 Chapter 15 2k Factorial Experiments and Fractions 15.37 When the variables are centered and scaled, the fitted model is y\u2c6 = 12.7519 + 4.7194x1 + 0.8656x2 \u2212 1.4156x3. The lack-of-fit test results in an f -value of 81.58 with P -value < 0.0001. Hence, higher- order terms are needed in the model. 15.38 The ANOVA table for the regression model looks like the following. Coefficients Degrees of Freedom Intercept \u3b21 \u3b22 \u3b23 \u3b24 \u3b25 Two-factor interactions Lack of fit Pure error 1 1 1 1 1 1 10 16 32 Total 63 15.39 The defining contrasts are AFG, CEFG, ACDF, BEG, BDFG, CDG, BCDE, ABCDEFG, DEF, ADEG. 15.40 Begin with the basic line for N = 24; permute as described in Section 15.12 until 18 columns are formed. 15.41 The fitted model is y\u2c6 =190, 056.67 + 181, 343.33x1 + 40, 395.00x2 + 16, 133.67x3 + 45, 593.67x4 \u2212 29, 412.33x5 + 8, 405.00x6. The t-tests are given as Variable t P -value Intercept x1 x2 x3 x4 x5 x6 4.48 4.27 0.95 0.38 1.07 \u22120.69 0.20 0.0065 0.0079 0.3852 0.7196 0.3321 0.5194 0.8509 Only x1 and x2 are significant. Solutions for Exercises in Chapter 15 253 15.42 An ANOVA table is obtained. Source of Degrees of Mean Computed Variation Freedom Square f P -value Polymer 1 Polymer 2 Polymer 1*Polymer 2 Error 1 1 1 4 172.98 180.50 1.62 0.17 1048.36 1093.94 9.82 < 0.0001 < 0.0001 0.0351 Total 7 All main effects and interactions are significant. 15.43 An ANOVA table is obtained. Source of Degrees of Mean Computed Variation Freedom Square f P -value Mode Type Mode*Type Error 1 1 1 12 2, 054.36 4, 805.96 482.90 27.67 74.25 173.71 17.45 < 0.0001 < 0.0001 0.0013 Total 15 All main effects and interactions are significant. 15.44 Two factors at two levels each can be used with three replications of the experiment, giving 12 observations. The requirement that there must be tests on main effects and the interactions suggests that partial confounding be used The following design is indicated: Block Block Block 1 2 1 2 1 2 (1) ab a b a ab (1) b (1) a ab a Rep 1 Rep 2 Rep 3 15.45 Using the contrast method and compute sums of squares, we have Source of Variation d.f. MS f A B C D E F Error 1 1 1 1 1 1 8 0.0248 0.0322 0.0234 0.0676 0.0028 0.0006 0.0201 1.24 1.61 1.17 3.37 0.14 0.03 254 Chapter 15 2k Factorial Experiments and Fractions 15.46 With the defining contrasts ABCD, CDEFG, and BDF , we have L1 = \u3b31 + \u3b32 + \u3b33 + \u3b34, L2 = \u3b33 + \u3b34 + \u3b35 + \u3b36 + \u3b37. L3 = \u3b32 + \u3b34 + \u3b36. The principal block and the remaining 7 blocks are given by Block 1 Block 2 Block 3 Block 4 (1), eg abcd, bdg adf, bcf cdef, abcdeg bde, adefg bcefg, cdfg acg, abef ace, abfg a, aeg bcd, abdg df, abcf acdef, bcdeg abde, defg abcefg, acdfg cg, bef ce, bfg b, beg acd, dg abdf, cf bcdef, acdeg de, abdefg cefg, bcdfg abcg, aef abce, afg c, ceg abd, bcdg acdf, bf def, abdeg bcde, acdefg befg, dfg ag, abcef ae, abcfg Block 5 Block 6 Block 7 Block 8 d, deg abc, bg af, bcdf cef, abceg be, aefg bcdefg, cfg acdg, abdef acde, abdfg e, g abcde, bdeg adef, bcef cdf, abcdg bd, adfg bcfg, cdefg aceg, abf ac, abefg f, efg abcdf, bdfg ad, bc cde, abcdefg bdef, adeg bceg, cdg acfg, abe acef, abg ab, abeg cd, adg bdf, acf abcdef, cdeg ade, bdefg acefg, abcdfg bcg, ef bce, fg The two-way interactions AB \u2261 CD, AC \u2261 BD, AD \u2261 BC, BD \u2261 F , BF \u2261 D and DF \u2261 B. 15.47 A design (where L1 = L2 = L3 = L4 = 0 (mod 2) are used) is: {(1), abcg, abdh, abef, acdf, aceh, adeg, afgh, bcde, bcfh, bdfg, cdgh, cefg, defh, degh, abcdefgh} 15.48 In the four defining contrasts, BCDE, ACDF , ABCG, and ABDH , the length of interactions are all 4. Hence, it must be a resolution IV design. 15.49 Assuming three factors the design is a 23 design with 4 center runs. 15.50 (a) Consider a 23\u22121III design with ABC \u2261 I as defining contrast. Then the design points are Solutions for Exercises in Chapter 15 255 x1 x2 x3 \u22121 1 \u22121 1 0 0 \u22121 \u22121 1 1 0 0 \u22121 1 1 \u22121 0 0 For the noncentral design points, x¯1 = x¯2 = x¯3 = 0 and x¯ 2 1 = x¯ 2 2 = x¯ 2 3 = 1. Hence E(y¯f\u2212 y¯0) = \u3b20+\u3b21x¯1+\u3b22x¯2+\u3b23x¯3+\u3b211x¯21+\u3b222x¯22+\u3b233x¯23\u2212\u3b20 = \u3b211+\u3b222+\u3b233. (b) It is learned that the test for curvature that involves y¯f \u2212 y¯0 actually is testing the hypothesis \u3b211 + \u3b222 + \u3b233 = 0. Chapter 16 Nonparametric Statistics 16.1 The hypotheses H0 : µ\u2dc = 20 minutes H1 : µ\u2dc > 20 minutes. \u3b1 = 0.05. Test statistic: binomial variable X with p = 1/2.