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1..1) A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the P-value. One-way ANOVA 2) Source DF SS MS F P Factor 4 987.71 246.93 33.09 < 0.0001 Error 25 186.53 7.46 Total 29 1174.24 A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the P-value. One-way ANOVA 2. (a) Fill in the blanks. Source DF SS MS F P Factor 3 36.15 12.05 1.21 0.3395 Error 16 159.89 9.99 Total 19 196.04 You may give bounds on the P-value. Completed table is: Source DF SS MS F P Treatment 4 1020.56 255.14 30.14 < 0.00001 Block 5 323.82 64.765 7.65 <0.001 Error 20 169.33 8.466 Total 29 1513.71 (b) How many blocks were used in this experiment? Six blocks were used. (c) What conclusions can you draw? The treatment effect is significant; the means of the five treatments are not all equal. 3. a) The assumption of normality is necessary to test the claim. According to the normal probability plots, the assumption of normality does not appear to be violated. This is evident from the fact that the data appear to fall along a straight line. Probability Plot of EX5-81V1 Normal - 95% CI 99 Mean StDev N AD P-Value 95 90 99.58 1.529 25 0.315 0.522 Percent 80 70 60 50 40 30 20 10 5 1 95 96 97 98 99 100 101 EX5-81V1 102 103 104 Probability Plot of EX5-81V2 Normal - 95% CI 99 Mean StDev N AD P-Value 95 90 Percent 80 70 60 50 40 30 20 10 5 1 100 102 104 106 EX5-81V2 108 110 112 b) 1) the parameters of interest are the variances of resistance of products, 12 , 22 2) H0: 12 22 3) H1: 12 22 4) = 0.05 5) The test statistic is f0 s12 s22 6) Reject H0 if f0 < f0.975,24,34 where f0.975,24,34 = 1 1 0.459 f0.025,34 ,24 2.18 or f0 > f0.025,24,34 where f0.025,24,34 =2.07 7) s1 = 1.53 n1 = 25 s2 =1.96 n2 = 35 104.9 1.930 33 0.328 0.506 f0 (153 . )2 (196 . )2 0.609 8) Since 0.601 > 0.459, do not reject H0 and conclude the variances are not significantly different at = 0.05. c) Two-Sample T-Test and CI: vendor1, vendor2 Two-sample T for vendor1 vs vendor2 N Mean StDev vendor1 25 99.58 1.53 vendor2 33 104.93 1.93 SE Mean 0.31 0.34 Difference = mu (vendor1) - mu (vendor2) Estimate for difference: -5.351 95% upper bound for difference: -4.567 T-Test of difference = -5 (vs <): T-Value = -0.75 Both use Pooled StDev = 1.7695 P-Value = 0.229 DF = 56 Since p-value > 0.05, we cannot reject the null hypothesis . That is we cannot prove that the mean resistance of vendor 2 is at least 5 higher than that of vendor 1. 4. a), d = 0.667 b) sd = 2.964, n = 12 95% confidence interval: s s d t / 2,n 1 d d d t / 2,n 1 d n n 2.964 2.964 0.667 2.201 d 0.667 2.201 12 12 1.216 d 2.55 indication that c) According to the normal probability plots, the assumption of normality does not appear to be since the data fall approximately along a straight line. Normal Probability Plot .999 .99 .95 Probability violated Since zero is contained within this interval, we are 95% confident there is no significant one design language is preferable. .80 .50 .20 .05 .01 .001 -5 0 5 diff A vera ge : 0 .66 66 67 S tDe v: 2 .9 644 4 N : 12 A nde rso n-Darlin g N orm al ity Te st A -Sq ua red : 0. 31 5 P -Va lue : 0.5 02 5. The effective life of insulating fluids at an accelerated load of 35 kV is being studied. Test data have been obtained for four types of fluids. The results from a completely randomized experiment were as follows: Fluid Type 1 2 3 4 17.6 16.9 21.4 19.3 18.9 15.3 23.6 21.1 (a) Is there any indication that the fluids differ? Life (in h) at 35 kV Load 16.3 17.4 18.6 17.1 19.4 18.5 16.9 17.5 20.1 19.5 20.5 18.3 21.6 20.3 22.3 19.8 Use = 0.05. Since the P-value = 0.05, there is a difference in means at = 0.05. Minitab Output One-way ANOVA: Life(in h) versus Fluid Type Source DF SS MS F P Fluid Type 3 29.92 9.97 3.10 0.050 Error 20 64.29 3.21 Total 23 94.21 (b) Which fluid would you select, given that the objective is long life? Descriptive Statistics: Life Variable Life Fluid Type 1 2 3 4 Mean 18.650 17.950 20.950 18.900 StDev 1.952 1.854 1.879 1.441 Minitab Output Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Fluid Type Individual confidence level = 98.89% Fluid Type = 1 subtracted from: Fluid Type Lower Center Upper 2 -3.598 -0.700 2.198 3 -0.598 2.300 5.198 4 -2.648 0.250 3.148 +---------+---------+---------+--------(---------*--------) (---------*--------) (---------*--------) +---------+---------+---------+---------6.0 -3.0 0.0 3.0 Fluid Type = 2 subtracted from: Fluid Type Lower Center Upper 3 0.102 3.000 5.898 4 -1.948 0.950 3.848 +---------+---------+---------+--------(---------*---------) (--------*---------) +---------+---------+---------+---------6.0 -3.0 0.0 3.0 Fluid Type = 3 subtracted from: Fluid Type Lower Center Upper 4 -4.948 -2.050 0.848 +---------+---------+---------+--------(--------*---------) +---------+---------+---------+---------6.0 -3.0 0.0 3.0 Fluid Type 3. Descriptive Statistics shows that Fluid Type 3 has the longest life. From the result of Tukey’s method, we found that, although there is no significant difference between Fluid Types 1, 3 and 4, there is a significant difference between Fluid Types 2 and 3, with Fluid Type 3 having a longer lifetime. (c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? Residual Plots for Life N orm a l P roba bil ity P l ot V ers us Fits 99 3.0 R e sidu a l Pe rc e n t 90 50 10 1 -5.0 1.5 0.0 -1.5 -3.0 -2.5 0.0 R e sidu a l 2.5 5.0 18 19 20 F itte d Va lu e His togra m V ers us Order 3.0 3.6 R e sidu a l F re qu e n c y 4.8 2.4 1.5 0.0 -1.5 1.2 0.0 21 -3.0 -2.4 -1.2 0.0 1.2 R e sidu a l 2.4 2 4 6 8 10 12 14 16 18 20 22 24 O bse rva tion O rde r There is nothing unusual in the residual plots. 6.. An article in the Fire Safety Journal (“The Effect of Nozzle Design on the Stability and Performance of Turbulent Water Jets,” Vol. 4, August 1981) describes an experiment in which a shape factor was determined for several different nozzle designs at six levels of jet efflux velocity. Interest focused on potential differences between nozzle designs, with velocity considered as a nuisance variable. The data are shown below: Jet Efflux Velocity (m/s) Nozzle Design 1 2 3 4 5 11.73 0.78 0.85 0.93 1.14 0.97 14.37 0.80 0.85 0.92 0.97 0.86 16.59 0.81 0.92 0.95 0.98 0.78 20.43 0.75 0.86 0.89 0.88 0.76 23.46 0.77 0.81 0.89 0.86 0.76 28.74 0.78 0.83 0.83 0.83 0.75 (a) Does nozzle design affect the shape factor? variance, using = 0.05. Compare nozzles with a scatter plot and with an analysis of Two-way ANOVA: Shape versus Nozzle Design, block Source DF SS Nozzle Design 4 0.102180 block 5 0.062867 Error 20 0.057300 Total 29 0.222347 S = 0.05353 R-Sq = 74.23% Since p-value< 0.05, MS F P 0.0255450 8.92 0.000 0.0125733 4.39 0.007 0.0028650 R-Sq(adj) = 62.63% nozzle design has a significant effect on shape factor. (b) Analyze the residual from this experiment. The plots shown below do not give any indication of serious problems. Three is some indication of a mild outlier on the normal probability plot and on the plot of residuals versus the predicted velocity. Residual Plots for Shape N orm a l P roba bility P l ot V ers us Fits 99 0.10 R e sidu a l Pe rc e n t 90 50 0.00 -0.05 10 1 0.05 -0.10 -0.10 -0.05 0.00 0.05 R e sidu a l 0.10 0.7 0.8 His togra m 0.9 F itte d Va lu e 1.0 V ers us Order 8 R e sidu a l F re qu e n c y 0.10 6 4 2 0 0.05 0.00 -0.05 -0.10 -0.05 0.00 0.05 R e sidu a l 0.10 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 O bse rva tion O rde r Individual Value Plot of RESI1 0.15 0.10 R ES I1 0.05 0.00 -0.05 -0.10 1 2 3 Noz z le De s ign 4 (c) Which nozzle designs are different with respect to shape factor Descriptive Statistics: Shape Nozzle Design Mean StDev 1 0.78167 0.02137 2 0.8533 0.0372 3 0.9017 0.0422 4 0.9433 0.1136 5 0.8133 0.0866 Tukey 95.0% Simultaneous Confidence Intervals Response Variable Shape All Pairwise Comparisons among Levels of Nozzle Design Nozzle Design = 1 subtracted from: Variable Shape Nozzle Design 2 3 4 5 Lower -0.02077 0.02757 0.06923 -0.06077 Center Upper 0.07167 0.1641 0.12000 0.2124 0.16167 0.2541 0.03167 0.1241 -----+---------+---------+---------+(-----*-----) (-----*-----) (-----*-----) (-----*-----) -----+---------+---------+---------+-0.15 0.00 0.15 0.30 Nozzle Design = 2 subtracted from: Nozzle Design 3 4 5 Lower Center Upper -----+---------+---------+---------+-0.0441 0.04833 0.14077 (-----*-----) -0.0024 0.09000 0.18243 (-----*-----) -0.1324 -0.04000 0.05243 (-----*-----) 5 -----+---------+---------+---------+-0.15 0.00 0.15 0.30 Nozzle Design = 3 subtracted from: Nozzle Design 4 5 Lower Center Upper -0.0508 0.04167 0.134100 -0.1808 -0.08833 0.004100 -----+---------+---------+---------+(-----*-----) (-----*-----) -----+---------+---------+---------+-0.15 0.00 0.15 0.30 Nozzle Design = 4 subtracted from: Nozzle Design 5 Lower Center -0.2224 -0.1300 Upper -----+---------+---------+---------+-0.03757 (-----*-----) -----+---------+---------+---------+-0.15 0.00 0.15 0.30 위의 결과로부터, Nozzle 1과 3, 1과4, 4와 5 가 서로 다르다.