ISE_526_SolutionHW4

# ISE_526_SolutionHW4 - ISE426/526 Solutions Assignment 4...

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Unformatted text preview: ISE426/526 Solutions - Assignment 4 Page 1 of 20 Document Name: Assignment 4 Solutions Release Date: June 25, 2007 (Monday) 1-a: The given problem requires us to analyze an experiment studying the effect of several factors on the dyeing of cotton. The experiments involve three operators, three cycle times and two temperature levels. This is a three factor design with two factors having three levels and one factor having two levels. Theoretical model for this design is given by: ijkl ijk jk ik ij k j i ijkl y ) ( ) ( ) ( ) ( The hypotheses for this design are set as: 3 2 1 : H i H : 1 , for at least one treatment For factor 2: 3 2 1 : H j H : 1 , for at least one treatment For factor 3: 2 1 : H k H : 1 , for at least one treatment For interactions: ij H ) ( : ij H ) ( : 1 , for at least one combination ik H ) ( : ik H ) ( : 1 , for at least one combination jk H ) ( : jk H ) ( : 1 , for at least one combination ijk H ) ( : ijk H ) ( : 1 , for at least one combination ISE426/526 Solutions - Assignment 4 Page 2 of 20 1-b: MINITAB output for this design follows:------------------------------------------------------------------------------------------------------------ General Linear Model: Replicates versus A, B, C Factor Type Levels Values A fixed 3 1, 2, 3 B fixed 2 1, 2 C fixed 3 1, 2, 3 Analysis of Variance for Replicates, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P A 2 868.78 868.78 434.39 21.70 0.000 B 1 0.02 0.02 0.02 0.00 0.976 C 2 192.33 192.33 96.17 4.80 0.014 A*B 2 48.26 48.26 24.13 1.21 0.311 A*C 4 380.22 380.22 95.06 4.75 0.004 B*C 2 23.59 23.59 11.80 0.59 0.560 A*B*C 4 22.96 22.96 5.74 0.29 0.885 Error 36 720.67 720.67 20.02 Total 53 2256.83 S = 4.47421 R-Sq = 68.07% R-Sq(adj) = 52.99% Unusual Observations for Replicates Obs Replicates Fit SE Fit Residual St Resid 5 25.0000 33.3333 2.5832 -8.3333 -2.28 R 6 23.0000 34.6667 2.5832 -11.6667 -3.19 R R denotes an observation with a large standardized residual.------------------------------------------------------------------------------------------------------------ As can be seen in the results, the main factors of significance are A and C (p- value lesser than 0.05), along with two-way interaction AC. So it can be inferred that, the cycle time and operator and their two-way interaction are significant from the perspective of the model. ISE426/526 Solutions - Assignment 4 Page 3 of 20 1 – c: Analysis of residuals is performed in MINITAB. A four-in-one plot for the provided data-set is shown below: Residual Percent 10 5-5-10 99 90 50 10 1 N 54 AD 0.516 P-Value 0.183 Fitted Value Residual 40 35 30 25 5-5-10 Residual Frequency 4-4-8-12 16 12 8 4 Observation Order...
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ISE_526_SolutionHW4 - ISE426/526 Solutions Assignment 4...

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