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# FinalExamSolutions - IE 410 Fall 2003 Final Exam Solutions...

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Unformatted text preview: IE 410 Fall 2003 Final Exam Solutions Question 1. a) Examination of the residuals from the multifactor Anova clearly shows a problem. Could be an outlier, could be non-constant variance. There are a few things you can try: 1. Box-Cox transform: Lambda turns out to be about -2/3 2. Transform with addend: Lambda=0 addend of about -240 works ok 3. Try replacing the outlier with another value For this data, there does not seem to be any particular analysis that is completely satisfying. In any event, you should at the very least see if the basic conclusions change with the analysis . Some of the result do change, especially with regard to a couple of the interactions and factor C. Some of the results dont change, namely the significance of factors A and B. The main point here is that you had to at least try something and see what changes and what doesnt change. On the following pages, I tried a couple of things to see what changes. Original Data Analysis of Variance for Y - Type III Sums of Squares-------------------------------------------------------------------------------- Source Sum of Squares Df Mean Square F-Ratio P-Value-------------------------------------------------------------------------------- MAIN EFFECTS A:Feed 4.14567E6 2 2.07284E6 125.25 0.0000 B:Material 355278.0 1 355278.0 21.47 0.0001 C:Speed 152636.0 4 38158.9 2.31 0.0812 INTERACTIONS AB 231963.0 2 115982.0 7.01 0.0032 AC 260849.0 8 32606.1 1.97 0.0855 BC 84146.8 4 21036.7 1.27 0.3031 ABC 114658.0 8 14332.3 0.87 0.5551 RESIDUAL 496476.0 30 16549.2-------------------------------------------------------------------------------- TOTAL (CORRECTED) 5.84168E6 59 Residual Plot for Y-470-270-70 130 330 530 residual 400 800 1200 1600 2000 2400 predicted Y Normal Probability Plot-470-270-70 130 330 530 RESIDUALS 0.1 1 5 20 50 80 95 99 99.9 percentage Residual NPP from transform with Lambda=0.0001 and addend = -200 Normal Probability Plot-90-50-10 30 70 110 150 RES0.0001AM200 0.1 1 5 20 50 80 95 99 99.9 percentage Residual Plot for Y_0.0001AM200 residual predicted Y_0.0001AM200-140-90-40 10 60 110 160 1100 1300 1500 1700 1900 2100 2300 Analysis of Variance for Y_0.0001AM200 - Type III Sums of Squares-------------------------------------------------------------------------------- Source Sum of Squares Df Mean Square F-Ratio P-Value-------------------------------------------------------------------------------- MAIN EFFECTS A:FeedRate 3.07105E6 2 1.53553E6 473.30 0.0000 B:Material 138694.0 1 138694.0 42.75 0.0000 C:Speed 45460.4 4 11365.1 3.50 0.0184 INTERACTIONS AB 2379.89 2 1189.94 0.37 0.6960 AC 79181.9 8 9897.73 3.05 0.0124 BC 6548.19 4 1637.05 0.50 0.7326 ABC 6001.79 8 750.224 0.23 0.9819 RESIDUAL 97328.3 30 3244.28-------------------------------------------------------------------------------- TOTAL (CORRECTED) 3.60432E6 59 Analysis in which outlier is replaced by the value, then a log transform is taken Normal Probability Plot-60-30 30 60 90 RESOut0.0001 0.1 1 5 20 50 80 95 99 99.999....
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FinalExamSolutions - IE 410 Fall 2003 Final Exam Solutions...

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