ISE_526_SolutionHW3 - ISE426/526 Solutions - Assignment 3...

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Unformatted text preview: ISE426/526 Solutions - Assignment 3 Page 1 of 18 Document Name: Assignment 3 Solutions Release Date: June 27, 2007 (Wednesday) 1 – a: The proposed problem has two nuisance factors to be averaged out in the design. This is best done, using Latin Square Designs. Plastics 1 2 3 4 5 yi.. P1 A = 3.0 B = 2.4 C = 1.9 D = 2.2 E = 1.7 11.2 P2 B = 2.1 C = 2.7 D = 2.3 E = 2.5 A = 3.1 12.7 P3 C = 2.1 D = 2.6 E = 2.5 A = 2.9 B = 2.1 12.2 P4 D = 2.0 E = 2.5 A = 3.2 B = 2.5 C = 2.2 12.4 P5 E = 2.1 A = 3.6 B = 2.4 C = 2.4 D = 2.1 12.6 y..k 11.3 13.8 12.3 12.5 11.2 y…=61.1 The theoretical model for a Latin Square Design is provided as: ijk k j i ijk y Where, = overall mean i i row effect (block) j j treatment effect k k column effect (block) ijk Random error The two nuisance factors under consideration in this problem are: operators and plastics with five methods of bonding. We will use MINITAB to solve this problem. The hypotheses for this design are set as: 5 4 3 2 1 : H j H : 1 , for at least one treatment For Block 1: 5 4 3 2 1 : H k H : 1 , for at least one block For Block 2: 5 4 3 2 1 : H i H : 1 , for at least one block ISE426/526 Solutions - Assignment 3 Page 2 of 18 MINITAB Output:------------------------------------------------------------------------------------------------------------ General Linear Model: Pounds Force versus Plastics, Operators, Method Factor Type Levels Values Plastics fixed 5 1, 2, 3, 4, 5 Operators fixed 5 1, 2, 3, 4, 5 Method fixed 5 1, 2, 3, 4, 5 Analysis of Variance for Pounds Force, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Plastics 4 0.28960 0.28960 0.07240 2.52 0.096 Operators 4 0.89360 0.89360 0.22340 7.77 0.002 Method 4 3.21360 3.21360 0.80340 27.96 0.000 Error 12 0.34480 0.34480 0.02873 Total 24 4.74160 S = 0.169509 R-Sq = 92.73% R-Sq(adj) = 85.46% Unusual Observations for Pounds Force Pounds Obs Force Fit SE Fit Residual St Resid 14 2.90000 3.21200 0.12223 -0.31200 -2.66 R R denotes an observation with a large standardized residual.------------------------------------------------------------------------------------------------------------ As can be seen from the ANOVA table, the F-value for Operators (Blocks) and methods (treatments) is greater than 4, and the p-value is less than 0.05. This means that the methods differ significantly from each other, as regards force required to break the bond. Blocking on operators is therefore useful, as there is difference between blocks (operators)....
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This note was uploaded on 11/22/2009 for the course ISE 526 taught by Professor Farrington during the Summer '07 term at University of Alabama - Huntsville.

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ISE_526_SolutionHW3 - ISE426/526 Solutions - Assignment 3...

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