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# Solucionario Walpole 8 ED

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f Temperatures Error 1268.5333 112.8333 4 25 317.1333 4.5133 70.27 Total 1381.3667 29 with P -value< 0.0001. The results from Tukey\u2019s procedure can be obtained as follows: y¯0 y¯25 y¯100 y¯75 y¯50 55.167 60.167 64.167 70.500 72.833 194 Chapter 13 One-Factor Experiments: General The batteries activated at temperature 50 and 75 have significantly longer activated life. 13.24 The Duncan\u2019s procedure shows the following results: y¯E y¯A y¯C 0.3300 0.9422 1.0063 Hence, the sorption rate using the Esters is significantly lower than the sorption rate using the Aromatics or the Chloroalkanes. 13.25 Based on the definition, we have the following. SSB = k b\u2211 j=1 (y¯.j \u2212 y¯..)2 = k b\u2211 j=1 ( T.j k \u2212 T.. bk )2 = b\u2211 j=1 T 2.j k \u2212 2T 2 .. bk + T 2.. bk = b\u2211 j=1 T 2.j k \u2212 T 2 .. bk . 13.26 From the model yij = µ+ \u3b1i + \u3b2j + \u1ebij, and the constraints k\u2211 i=1 \u3b1i = 0 and b\u2211 j=1 \u3b2j = 0, we obtain y¯.j = µ+ \u3b2j + \u1eb¯.j and y¯.. = µ+ \u1eb¯... Hence SSB = k b\u2211 j=1 (y¯.j \u2212 y¯..)2 = k b\u2211 j=1 (\u3b2j + \u1eb¯.j \u2212 \u1eb¯..)2. Since E(\u1eb¯.j) = 0 and E(\u1eb¯..) = 0, we obtain E(SSB) = k b\u2211 j=1 \u3b22j + k b\u2211 j=1 E(\u1eb¯2.j)\u2212 bkE(\u1eb¯2..). We know that E(\u1eb¯2.j) = \u3c32 k and E(\u1eb¯2..) = \u3c32 bk . Then E(SSB) = k b\u2211 j=1 \u3b22j + b\u3c3 2 \u2212 \u3c32 = (b\u2212 1)\u3c32 + k b\u2211 j=1 \u3b22j . 13.27 (a) The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0, fertilizer effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.05. Critical region: f > 4.76. Computation: Solutions for Exercises in Chapter 13 195 Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Fertilizers Blocks Error 218.1933 197.6317 71.4017 3 2 6 72.7311 98.8158 11.9003 6.11 Total 487.2267 11 P -value= 0.0296. Decision: Reject H0. The means are not all equal. (b) The results of testing the contrasts are shown as: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f (f1, f3) vs (f2. f4) f1 vs f3 Error 206.6700 11.4817 71.4017 1 1 6 206.6700 11.4817 11.9003 17.37 0.96 The corresponding P -values for the above contrast tests are 0.0059 and 0.3639, respectively. Hence, for the first contrast, the test is significant and the for the second contrast, the test is insignificant. 13.28 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, no differences in the varieties H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.05. Critical region: f > 5.14. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Treatments Blocks Error 24.500 171.333 42.167 2 3 6 12.250 57.111 7.028 1.74 Total 238.000 11 P -value=0.2535. Decision: Do not reject H0; could not show that the varieties of potatoes differ in field. 13.29 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, brand effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.05. Critical region: f > 3.84. Computation: 196 Chapter 13 One-Factor Experiments: General Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Treatments Blocks Error 27.797 16.536 18.556 2 4 8 13.899 4.134 2.320 5.99 Total 62.889 14 P -value=0.0257. Decision: Reject H0; mean percent of foreign additives is not the same for all three brand of jam. The means are: Jam A: 2.36, Jam B: 3.48, Jam C: 5.64. Based on the means, Jam A appears to have the smallest amount of foreign additives. 13.30 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0, courses are equal difficulty 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 Subjects Students Error 42.150 1618.700 1112.100 3 4 12 14.050 404.675 92.675 0.15 Total 2772.950 19 P -value=0.9267. Decision: Fail to rejectH0; there is no significant evidence to conclude that courses are of different difficulty. 13.31 The hypotheses are H0 : \u3b11 = \u3b12 = · · · = \u3b16 = 0, station effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.01. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Stations Dates Error 230.127 3.259 44.018 5 5 25 46.025 0.652 1.761 26.14 Total 277.405 35 Solutions for Exercises in Chapter 13 197 P -value< 0.0001. Decision: Reject H0; the mean concentration is different at the different stations. 13.32 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, station 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 Stations Months Error 10.115 537.030 744.416 2 11 22 5.057 48.821 33.837 0.15 Total 1291.561 35 P -value= 0.8620. Decision: Do not reject H0; the treatment means do not differ significantly. 13.33 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, diet effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.01. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Diets Subjects Error 4297.000 6033.333 1811.667 2 5 10 2148.500 1206.667 181.167 11.86 Total 12142.000 17 P -value= 0.0023. Decision: Reject H0; differences among the diets are significant. 13.34 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, analyst effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.05. Computation: 198 Chapter 13 One-Factor Experiments: General Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Analysts Individuals Error 0.001400 0.021225 0.001400 2 3 6 0.000700 0.007075 0.000233 3.00 Total 0.024025 11 P -value= 0.1250. Decision: Do not reject H0; cannot show that the analysts differ significantly. 13.35 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = \u3b15 = 0, treatment effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.01. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Treatments Locations Error 79630.133 634334.667 689106.667 4 5 20 19907.533 126866.933 34455.333 0.58 Total 1403071.467 29 P -value= 0.6821. Decision: Do not reject H0; the treatment means do not differ significantly. 13.36 The hypotheses are H0 : \u3b11 = \u3b12 = \u3b13 = 0, treatment effects are zero H1 : At least one of the \u3b1i\u2019s is not equal to zero. \u3b1 = 0.01. Computation: Source of Sum of Degrees of Mean Computed Variation Squares Freedom Square f Treatments Subjects Error 203.2792 188.2271 212.8042 2 9 18 101.6396 20.9141 11.8225 8.60 Total 604.3104 29 P -value= 0.0024. Decision: RejectH0; the mean weight losses are different for different treatments and the therapists had the greatest effect on the weight loss. Solutions for Exercises in Chapter 13 199 13.37 The total sums of squares can be written as\u2211 i \u2211 j \u2211 k (yijk \u2212 y¯...)2 = \u2211 i \u2211 j \u2211 k [(y¯i.. \u2212 y¯...) + (y¯.j. \u2212 y¯...) + (y¯..k \u2212 y¯...) + (yijk \u2212 y¯i.. \u2212 y¯.j. \u2212 y¯..k + 2y¯...)]2 =r \u2211 i (y¯i.. \u2212 y¯...)2 + r \u2211 j (y¯.j. \u2212 y¯...)2 + r \u2211 k (y¯..k \u2212 y¯...)2 + \u2211 i \u2211 j \u2211 k (yijk \u2212 y¯i.. \u2212 y¯.j. \u2212 y¯..k + 2y¯...)2 + 6 cross-product terms, and all cross-product terms are equal to zeroes. 13.38 For the model yijk = µ+ \u3b1i + \u3b2j + \u3c4k + \u1ebijk, we have y¯..k = µ+ \u3c4k + \u1eb¯..k, and y¯... = µ+ \u1eb¯.... Hence SSTr = r \u2211 k (y¯..k \u2212 y¯...)2 = r \u2211 k (\u3c4k + \u1eb¯..k \u2212 \u1eb¯...)2, and E(SSTr) = r \u2211 k \u3c4 2k + r \u2211 k E(\u1eb¯2..k)\u2212 r2 \u2211 k E(\u1eb¯2...) = r \u2211 k \u3c4 2k + r \u2211 k \u3c32 r \u2212 r2\u3c3 2 r2 = r \u2211 k \u3c4 2k + r\u3c3 2 \u2212 \u3c32 =