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1、Chapter 13Analysis of Variance andExperimental DesignLearning Objectives1.Understand how the analysis of variance procedure can be used to determine if the means of more than two populations are equal.2.Know the assumptions necessary to use the analysis of variance procedure.3.Understand the use of
2、the F distribution in performing the analysis of variance procedure.4.Know how to set up an ANOVA table and interpret the entries in the table.5.Be able to use output from computer software packages to solve analysis of variance problems.6.Know how to use Fishers least significant difference (LSD) p
3、rocedure and Fishers LSD with the Bonferroni adjustment to conduct statistical comparisons between pairs of populations means.7.Understand the difference between a completely randomized design, a randomized block design, and factorial experiments.8.Know the definition of the following terms:comparis
4、onwise Type I error ratepartitioningexperimentwise Type I error rateblockingfactormain effectlevelinteractiontreatmentreplication13 - 35Analysis of Variance and Experimental DesignSolutions:1.a. = (30 + 45 + 36)/3 = 37= 5(30 - 37)2 + 5(45 - 37)2 + 5(36 - 37)2 = 570MSTR = SSTR /(k - 1) = 570/2 = 285b
5、. = 4(6) + 4(4) + 4(6.5) = 66MSE = SSE /(nT - k) = 66/(15 - 3) = 5.5c.F = MSTR /MSE = 285/5.5 = 51.82 F.05 = 3.89 (2 degrees of freedom numerator and 12 denominator) Since F = 51.82 F.05 = 3.89, we reject the null hypothesis that the means of the three populations are equal. d.Source of VariationSum
6、 of SquaresDegrees of FreedomMean SquareFTreatments570228551.82Error66125.5Total636142.a.= (153 + 169 + 158)/3 = 160= 4(153 - 160)2 + 4(169 - 160) 2 + 4(158 - 160) 2 = 536MSTR= SSTR /(k - 1) = 536/2 = 268b. = 3(96.67) + 3(97.33) +3(82.00) = 828.00MSE = SSE /(nT - k) = 828.00 /(12 - 3) = 92.00c.F = M
7、STR /MSE = 268/92 = 2.91 F.05 = 4.26 (2 degrees of freedom numerator and 9 denominator) Since F = 2.91 F.05 = 3.89 we reject the null hypothesis that the means of the three populations are equal.d. Source of VariationSum of SquaresDegrees of FreedomMean SquareFTreatments1020251013.36Error4581238.17T
8、otal1478144.a.Source of VariationSum of SquaresDegrees of FreedomMean SquareFTreatments1200340080Error300605Total150063b.F.05 = 2.76 (3 degrees of freedom numerator and 60 denominator)Since F = 80 F.05 = 2.76 we reject the null hypothesis that the means of the 4 populations are equal.5.a. Source of
9、VariationSum of SquaresDegrees of FreedomMean SquareFTreatments12026020Error216723Total33674b.F.05 = 3.15 (2 numerator degrees of freedom and 60 denominator)F.05 = 3.07 (2 numerator degrees of freedom and 120 denominator)The critical value is between 3.07 and 3.15Since F = 20 must exceed the critica
10、l value, no matter what its actual value, we reject the null hypothesis that the 3 population means are equal.6. Manufacturer 1Manufacturer 2Manufacturer 3Sample Mean232821Sample Variance6.674.673.33= (23 + 28 + 21)/3 = 24 = 4(23 - 24) 2 + 4(28 - 24) 2 + 4(21 - 24) 2 = 104MSTR = SSTR /(k - 1) = 104/
11、2 = 52 = 3(6.67) + 3(4.67) + 3(3.33) = 44.01MSE = SSE /(nT - k) = 44.01/(12 - 3) = 4.89F = MSTR /MSE = 52/4.89 = 10.63 F.05 = 4.26 (2 degrees of freedom numerator and 9 denominator)Since F = 10.63 F.05 = 4.26 we reject the null hypothesis that the mean time needed to mix a batch of material is the s
12、ame for each manufacturer.7. SuperiorPeerSubordinateSample Mean5.755.55.25Sample Variance1.642.001.93 = (5.75 + 5.5 + 5.25)/3 = 5.5= 8(5.75 - 5.5) 2 + 8(5.5 - 5.5) 2 + 8(5.25 - 5.5) 2 = 1MSTR = SSTR /(k - 1) = 1/2 = .5 = 7(1.64) + 7(2.00) + 7(1.93) = 38.99MSE = SSE /(nT - k) = 38.99/21 = 1.86F = MST
13、R /MSE = 0.5/1.86 = 0.27 F.05 = 3.47 (2 degrees of freedom numerator and 21 denominator)Since F = 0.27 F.05 = 3.68, we reject the null hypothesis that the mean perception score is the same for the three groups of specialists.9. Real Estate AgentArchitectStockbrokerSample Mean67.7361.1365.80Sample Va
14、riance117.72180.10137.12= (67.73 + 61.13 + 65.80)/3 = 64.89= 15(67.73 - 64.89) 2 + 15(61.13 - 64.89) 2 + 15(65.80 - 64.89) 2 = 345.47MSTR = SSTR /(k - 1) = 345.47/2 = 172.74 = 14(117.72) + 14(180.10) + 14(137.12) = 6089.16MSE = SSE /(nT - k) = 6089.16/(45-3) = 144.98F = MSTR /MSE = 172.74/144.98 = 1
15、.19 F.05 = 3.22 (2 degrees of freedom numerator and 42 denominator)Note: Table 4 does not show a value for 2 degrees of freedom numerator and 42 denominator. However, the value of 3.23 corresponding to 2 degrees of freedom numerator and 40 denominator can be used as an approximation.Since F = 1.19 a
16、 = 0.05, we cannot reject the null hypothesis that that the mean price/earnings ratio is the same for these three groups of firms.11.aLSD; significant difference LSD; significant difference LSD; significant differenceb. -15 3.23 = -18.23 to -11.7712.a. Sample 1Sample 2Sample 3Sample Mean517758Sample
17、 Variance96.6797.3481.99 = (51 + 77 + 58)/3 = 62= 4(51 - 62) 2 +4(77 - 62) 2 + 4(58 - 62) 2 = 1,448MSTR = SSTR /(k - 1) = 1,448/2 = 724 = 3(96.67) + 3(97.34) + 3(81.99) = 828 MSE = SSE /(nT - k) = 828/(12 - 3) = 92F = MSTR /MSE = 724/92 = 7.87 F.05 = 4.26 (2 degrees of freedom numerator and 9 denomi
18、nator)Since F = 7.87 F.05 = 4.26, we reject the null hypothesis that the means of the three populations are equal.b.LSD; significant differenceLSD; no significant differenceLSD; significant difference13.Since there does not appear to be any significant difference between the means of population 1 an
19、d population 3.14.23 - 28 3.54-5 3.54 = -8.54 to -1.4615.Since there are only 3 possible pairwise comparisons we will use the Bonferroni adjustment.a = .05/3 = .017t.017/2 = t.0085 which is approximately t.01 = 2.6021.06; no significant difference1.06; no significant difference1.06; significant diff
20、erence16.a.Machine 1Machine 2Machine 3Machine 4Sample Mean7.19.19.911.4Sample Variance1.21.93.701.02= (7.1 + 9.1 + 9.9 + 11.4)/4 = 9.38= 6(7.1 - 9.38) 2 + 6(9.1 - 9.38) 2 + 6(9.9 - 9.38) 2 + 6(11.4 - 9.38) 2 = 57.77MSTR = SSTR /(k - 1) = 57.77/3 = 19.26 = 5(1.21) + 5(.93) + 5(.70) + 5(1.02) = 19.30M
21、SE = SSE /(nT - k) = 19.30/(24 - 4) = .97F = MSTR /MSE = 19.26/.97 = 19.86 F.05 = 3.10 (3 degrees of freedom numerator and 20 denominator)Since F = 19.86 F.05 = 3.10, we reject the null hypothesis that the mean time between breakdowns is the same for the four machines.b.Note: ta/2 is based upon 20 d
22、egrees of freedomLSD; significant difference17.C = 6 (1,2), (1,3), (1,4), (2,3), (2,4), (3,4)a = .05/6 = .008 and a /2 = .004Since the smallest value for a /2 in the t table is .005, we will use t.005 = 2.845 as an approximation for t.004 (20 degrees of freedom)Thus, if the absolute value of the dif
23、ference between any two sample means exceeds 1.62, there is sufficient evidence to reject the hypothesis that the corresponding population means are equal.Means(1,2)(1,3)(1,4)(2,3)(2,4)(3,4)| Difference |22.84.30.82.31.5Significant ?YesYesYesNoYesNo18.n1 = 12 n2 = 8 n3 = 10ta/2 is based upon 27 degr
24、ees of freedomComparing 1 and 2LSD; significant differenceComparing 1 and 3|9.95 - 13.5| = 3.55 LSD; significant differenceComparing 2 and 3|14.75 - 13.5| = 1.25 F.05 = 3.68, we reject the hypothesis that the means for the three treatments are equal.20.a. Source of VariationSum of SquaresDegrees of
25、FreedomMean SquareFTreatments148827445.50Error203015135.3Total351817b.|156-142| = 14 14.31; significant difference|142-134| = 8 2.69 we reject H023. Source of VariationSum of SquaresDegrees of FreedomMean SquareFTreatments1502754.80Error2501615.63Total40018F.05 = 3.63 (2 degrees of freedom numerator
26、 and 16 denominator) Since F = 4.80 F.05 = 3.63, we reject the null hypothesis that the means of the three treatments are equal.24. Source of VariationSum of SquaresDegrees of FreedomMean SquareFTreatments1200260043.99Error6004413.64Total180046F.05 = 3.23 (2 degrees of freedom numerator and 40 denom
27、inator)F.05 = 3.15 (2 degrees of freedom numerator and 60 denominator)The critical F value is between 3.15 and 3.23.Since F = 43.99 exceeds the critical value, we reject the hypothesis that the treatment means are equal.25. ABCSample Mean119107100Sample Variance146.8996.43173.78= 8(119 - 107.93) 2 +
28、 10(107 - 107.93) 2 + 10(100 - 107.93) 2 = 1617.9MSTR = SSTR /(k - 1) = 1617.9 /2 = 809.95 = 7(146.86) + 9(96.44) + 9(173.78) = 3,460MSE = SSE /(nT - k) = 3,460 /(28 - 3) = 138.4F = MSTR /MSE = 809.95 /138.4 = 5.85 F.05 = 3.39 (2 degrees of freedom numerator and 25 denominator)Since F = 5.85 F.05 =
29、3.39, we reject the null hypothesis that the means of the three treatments are equal.26.a. Source of VariationSum of SquaresDegrees of FreedomMean SquareFTreatments4560222809.87Error624027231.11Total1080029b.F.05 = 3.35 (2 degrees of freedom numerator and 27 denominator)Since F = 9.87 F.05 = 3.35, w
30、e reject the null hypothesis that the means of the three assembly methods are equal.27. Source of VariationSum of SquaresDegrees of FreedomMean SquareFBetween61.64320.5517.56Error23.41201.17Total85.0523F.05 = 3.10 (3 degrees of freedom numerator and 20 denominator)Since F = 17.56 F.05 = 3.10, we rej
31、ect the null hypothesis that the mean breaking strength of the four cables is the same.28.506070Sample Mean332928Sample Variance3217.59.5 = (33 + 29 + 28)/3 = 30= 5(33 - 30) 2 + 5(29 - 30) 2 + 5(28 - 30) 2 = 70MSTR = SSTR /(k - 1) = 70 /2 = 35 = 4(32) + 4(17.5) + 4(9.5) = 236MSE = SSE /(nT - k) = 23
32、6 /(15 - 3) = 19.67F = MSTR /MSE = 35 /19.67 = 1.78 F.05 = 3.89 (2 degrees of freedom numerator and 12 denominator)Since F = 1.78 F.05 = 3.89, we cannot reject the null hypothesis that the mean yields for the three temperatures are equal.29. Direct ExperienceIndirect ExperienceCombinationSample Mean17.020.425.0Sample Variance5.016.264.01= (17 + 20.4 + 25)/3 = 20.8= 7(17 - 20.8) 2 + 7(20.4 - 20.8) 2 + 7(25 - 20.8) 2 = 225.68MSTR = SSTR /(k - 1) = 225.68 /2 = 112.84 = 6(5.01) + 6(6.26) + 6(4.01) = 91.68MSE = SSE /(nT - k) = 91.68 /(21 - 3) = 5.09F = MSTR /MSE = 112.84 /5.09 = 22.17F