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# Solucionario Walpole 8 ED

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an ANOVA analysis, we obtain the P -value as 0.0121. Hence, the loom variance component is significantly different from 0 at level 0.05. (c) The suspicion is supported by the data. 13.54 The hypotheses are H0 : µ1 = µ2 = µ3 = µ4, H1 : At least two of the µi\u2019s are not equal. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Garlon levels Error 3.7289 9.5213 3 12 1.2430 0.7934 1.57 Total 13.2502 15 P -value= 0.2487. Decision: Do not reject H0; there is insufficient evidence to claim that the concentration levels of Garlon would impact the heights of shoots. 206 Chapter 13 One-Factor Experiments: General 13.55 Bartlett\u2019s statistic is b = 0.8254. Conclusion: do not reject homogeneous variance assumption. 13.56 The hypotheses are H0 : \u3c4A = \u3c4B = \u3c4C = \u3c4D = \u3c4E = 0, ration effects are zero H1 : At least one of the \u3c4i\u2019s is not zero. \u3b1 = 0.01. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Lactation Periods Cows Rations Error 245.8224 353.1224 89.2624 21.3392 4 4 4 12 61.4556 88.2806 22.3156 1.7783 12.55 Total 709.5464 24 P -value= 0.0003. Decision: Reject H0; different rations have an effect on the daily milk production. 13.57 It can be shown that y¯C = 76.16, y¯1 = 81.20, y¯2 = 81.54 and y¯3 = 80.98. Since this is a one-sided test, we find d0.01(3, 16) = 3.05 and\u221a 2s2 n = \u221a (2)(3.52575) 5 = 1.18756. Hence d1 = 81.20\u2212 76.16 1.18756 = 4.244, d2 = 81.54\u2212 76.16 1.18756 = 4.532, d3 = 80.98\u2212 76.16 1.18756 = 4.059, which are all larger than the critical value. Hence, significantly higher yields are obtained with the catalysts than with no catalyst. 13.58 (a) The hypotheses for the Bartlett\u2019s test are H0 : \u3c3 2 A = \u3c3 2 B = \u3c3 2 C = \u3c3 2 D, H1 : The variances are not all equal. \u3b1 = 0.05. Critical region: We have n1 = n2 = n3 = n4 = 5, N = 20, and k = 4. Therefore, we reject H0 when b < b4(0.05, 5) = 0.5850. Computation: sA = 1.40819, sB = 2.16056, sC = 1.16276, sD = 0.76942 and hence sp = 1.46586. From these, we can obtain that b = 0.7678. Decision: Do not reject H0; there is no sufficient evidence to conclude that the variances are not equal. Solutions for Exercises in Chapter 13 207 (b) The hypotheses are H0 : µA = µB = µC = µD, H1 : At least two of the µi\u2019s are not equal. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Laboratories Error 85.9255 34.3800 3 16 28.6418 2.1488 13.33 Total 120.3055 19 P -value= 0.0001. Decision: Reject H0; the laboratory means are significantly different. (c) The normal probability plot is given as follows: \u22122 \u22121 0 1 2 \u2212 2 \u2212 1 0 1 2 Theoretical Quantiles Sa m pl e Qu an tile s 13.59 The hypotheses for the Bartlett\u2019s test are H0 : \u3c3 2 1 = \u3c3 2 2 = \u3c3 2 3 = \u3c3 2 4 , H1 : The variances are not all equal. \u3b1 = 0.01. Critical region: We have n1 = n2 = n3 = 4, n4 = 9 N = 21, and k = 4. Therefore, we reject H0 when b < b4(0.01, 4, 4, 4, 9) = (4)(0.3475) + (4)(0.3475) + (4)(0.3475) + (9)(0.6892) 21 = 0.4939. Computation: s21 = 0.41709, s 2 2 = 0.93857, s 2 3 = 0.25673, s 2 4 = 1.72451 and hence s2p = 1.0962. Therefore, b = [(0.41709)3(0.93857)3(0.25763)3(1.72451)8]1/17 1.0962 = 0.79. 208 Chapter 13 One-Factor Experiments: General Decision: Do not reject H0; the variances are not significantly different. 13.60 The hypotheses for the Cochran\u2019s test are H0 : \u3c3 2 A = \u3c3 2 B = \u3c3 2 C , H1 : The variances are not all equal. \u3b1 = 0.01. Critical region: g > 0.6912. Computation: s2A = 29.5667, s 2 B = 10.8889, s 2 C = 3.2111, and hence \u2211 s2i = 43.6667. Now, g = 29.5667 43.6667 = 0.6771. Decision: Do not reject H0; the variances are not significantly different. 13.61 The hypotheses for the Bartlett\u2019s test are H0 : \u3c3 2 1 = \u3c3 2 2 = \u3c3 2 3, H1 : The variances are not all equal. \u3b1 = 0.05. Critical region: reject H0 when b < b4(0.05, 9, 8, 15) = (9)(0.7686) + (8)(0.7387) + (15)(0.8632) 32 = 0.8055. Computation: b = [(0.02832) 8(0.16077)7(0.04310)14 ]1/29 0.067426 = 0.7822. Decision: Reject H0; the variances are significantly different. 13.62 (a) The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0, H1 : At least one of the \u3b1i\u2019s is not zero. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Diets Blocks Error 822.1360 17.1038 106.3597 3 5 15 274.0453 3.4208 7.0906 38.65 Total 945.5995 23 with P -value< 0.0001. Decision: Reject H0; diets do have a significant effect on mean percent dry matter. Solutions for Exercises in Chapter 13 209 (b) We know that y¯C = 35.8483, y¯F = 36.4217, y¯T = 45.1833, y¯A = 49.6250, and\u221a 2s2 n = \u221a (2)(7.0906) 6 = 1.5374. Hence, dAmmonia = 49.6250\u2212 35.8483 1.5374 = 8.961, dUrea Feeding = 36.4217\u2212 35.8483 1.5374 = 0.3730, dUrea Treated = 45.1833\u2212 35.8483 1.5374 = 6.0719. Using the critical value d0.05(3, 15) = 2.61, we obtain that only \u201cUrea Feeding\u201d is not significantly different from the control, at level 0.05. (c) The normal probability plot for the residuals are given below. \u22122 \u22121 0 1 2 \u2212 4 \u2212 2 0 2 4 Theoretical Quantiles Sa m pl e Qu an tile s 13.63 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, H1 : At least one of the \u3b1i\u2019s is not zero. \u3b1 = 0.05. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Diet Error 0.32356 0.20808 2 12 0.16178 0.01734 9.33 Total 0.53164 14 210 Chapter 13 One-Factor Experiments: General with P -value= 0.0036. Decision: Reject H0; zinc is significantly different among the diets. 13.64 (a) The gasoline manufacturers would want to apply their results to more than one model of car. (b) Yes, there is a significant difference in the miles per gallon for the three brands of gasoline. (c) I would choose brand C for the best miles per gallon. 13.65 (a) The process would include more than one stamping machine and the results might differ with different machines. (b) The mean plot is shown below. 1 1 1 3. 2 3. 4 3. 6 3. 8 4. 0 4. 2 4. 4 Material N um be r o f G as ke ts 2 2 2 cork plastic rubber (c) Material 1 appears to be the best. (d) Yes, there is interaction. Materials 1 and 3 have better results with machine 1 but material 2 has better results with machine 2. 13.66 (a) The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, H1 : At least one of the \u3b1i\u2019s is not zero. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Paint Types Error 227875.11 336361.83 2 15 113937.57 22424.12 5.08 Total 564236.94 17 with P -value= 0.0207. Decision: Reject H0 at level 0.05; the average wearing quality differs significantly for three paints. (b) Using Tukey\u2019s test, it turns out the following. Solutions for Exercises in Chapter 13 211 y¯1. y¯3. y¯2. 197.83 419.50 450.50 Types 2 and 3 are not significantly different, while Type 1 is significantly different from Type 2. (c) We plot the residual plot and the normal probability plot for the residuals as follows. 1.0 1.5 2.0 2.5 3.0 \u2212 20 0 \u2212 10 0 0 10 0 20 0 Type R es id ua l \u22122 \u22121 0 1 2 \u2212 20 0 \u2212 10 0 0 10 0 20 0 Theoretical Quantiles Sa m pl e Qu an tile s It appears that the heterogeneity in variances may be violated, as is the normality assumption. (d) We do a log transformation of the data, i.e., y \u2032 = log(y). The ANOVA result