experimentos fatoriais do tipo 2 k
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Experimentos Fatoriais do tipo 2 k. Capítulo 6. 6.5 Unreplicated 2 k Factorial Designs. These are 2 k factorial designs with one observation at each corner of the “cube” An unreplicated 2 k factorial design is also sometimes called a “ single replicate ” of the 2 k - PowerPoint PPT PresentationTRANSCRIPT
Experimentos Fatoriais do tipo 2k
Capítulo 6
6.5 Unreplicated 2k Factorial Designs
• These are 2k factorial designs with one observation at each corner of the “cube”
• An unreplicated 2k factorial design is also sometimes called a “single replicate” of the 2k
• These designs are very widely used• Risks…if there is only one observation at
each corner, is there a chance of unusual response observations spoiling the results?
• Modeling “noise”?
Spacing of Factor Levels in the Unreplicated 2k Factorial
Designs
If the factors are spaced too closely, it increases the chances that the noise will overwhelm the signal in the data
More aggressive spacing is usually best
Unreplicated 2k Factorial Designs
• Lack of replication causes potential problems in statistical testing– Replication admits an estimate of “pure error” (a
better phrase is an internal estimate of error)– With no replication, fitting the full model results in
zero degrees of freedom for error
• Potential solutions to this problem– Pooling high-order interactions to estimate error– Normal probability plotting of effects (Daniels,
1959)– Other methods…see text
Example of an Unreplicated 2k Design
• A 24 factorial was used to investigate the effects of four factors on the filtration rate of a resin
• The factors are A = temperature, B = pressure, C = mole ratio, D= stirring rate
• Experiment was performed in a pilot plant
The Resin Plant Experiment
The Resin Plant Experiment
dados=read.table("e:\\dox\\pfat2a4sr.txt",header=T) A=as.factor(dados$A) B=as.factor(dados$B) C=as.factor(dados$C) D=as.factor(dados$D)modeloC=dados$y~A+B+C+D+A:B+A:C+A:D+B:C+B:D+C:D+A:B:C+A:B:D+A:C:D+B:C:D+A:B:C:D
fitC=aov(modeloC)summary(fitC)
Df Sum Sq Mean SqA 1 1870.56 1870.56B 1 39.06 39.06C 1 390.06 390.06D 1 855.56 855.56A:B 1 0.06 0.06A:C 1 1314.06 1314.06A:D 1 1105.56 1105.56B:C 1 22.56 22.56B:D 1 0.56 0.56C:D 1 5.06 5.06A:B:C 1 14.06 14.06A:B:D 1 68.06 68.06A:C:D 1 10.56 10.56B:C:D 1 27.56 27.56A:B:C:D 1 7.56 7.56
Estimates of the Effects
The Half-Normal Probability Plot of Effects
Design Projection: ANOVA Summary for the Model as a 23 in Factors A, C, and D
modeloP=dados$y~A+C+D+A:C+A:D+C:D+A:C:D fitP=aov(modeloP)summary(fitP)
Df Sum Sq Mean Sq F value Pr(>F) A 1 1870.56 1870.56 83.3677 1.667e-05 ***C 1 390.06 390.06 17.3844 0.0031244 ** D 1 855.56 855.56 38.1309 0.0002666 ***A:C 1 1314.06 1314.06 58.5655 6.001e-05 ***A:D 1 1105.56 1105.56 49.2730 0.0001105 ***C:D 1 5.06 5.06 0.2256 0.6474830 A:C:D 1 10.56 10.56 0.4708 0.5120321 Residuals 8 179.50 22.44 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The Regression Model
regP=lm(dados$y~dados$A+dados$C+dados$D+dados$A*dados$C+dados$A*dados$D)summary(regP)
Residuals: Min 1Q Median 3Q Max -6.3750 -1.5000 0.0625 2.9063 5.7500
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 70.062 1.104 63.444 2.30e-14 ***dados$A 10.812 1.104 9.791 1.93e-06 ***dados$C 4.938 1.104 4.471 0.00120 ** dados$D 7.313 1.104 6.622 5.92e-05 ***dados$A:dados$C -9.062 1.104 -8.206 9.41e-06 ***dados$A:dados$D 8.312 1.104 7.527 2.00e-05 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.417 on 10 degrees of freedomMultiple R-squared: 0.966, Adjusted R-squared: 0.9489 F-statistic: 56.74 on 5 and 10 DF, p-value: 5.14e-07
Model Residuals are Satisfactory
Model Interpretation – Main Effects and Interactions
Outliers: suppose that cd = 375 (instead of 75)
Dealing with Outliers
• Replace with an estimate
• Make the highest-order interaction zero
• In this case, estimate cd such that ABCD = 0
• Analyze only the data you have
• Now the design isn’t orthogonal
• Consequences?
The Drilling Experiment Example 6.3
A = drill load, B = flow, C = speed, D = type of mud, y = advance rate of the drill
Normal Probability Plot of Effects –The Drilling Experiment
Residual Plots
DESIGN-EXPERT Plotadv._rate
Predicted
Re
sid
ua
ls
Residuals vs. Predicted
-1.96375
-0.82625
0.31125
1.44875
2.58625
1.69 4.70 7.70 10.71 13.71
• The residual plots indicate that there are problems with the equality of variance assumption
• The usual approach to this problem is to employ a transformation on the response
• Power family transformations are widely used
Transformations are typically performed to – Stabilize variance– Induce at least approximate normality– Simplify the model
Residual Plots
*y y
Selecting a Transformation
• Empirical selection of lambda• Prior (theoretical) knowledge or experience can
often suggest the form of a transformation• Analytical selection of lambda…the Box-Cox
(1964) method (simultaneously estimates the model parameters and the transformation parameter lambda)
• Box-Cox method implemented in Design-Expert
(15.1)
The Box-Cox MethodDESIGN-EXPERT Plotadv._rate
LambdaCurrent = 1Best = -0.23Low C.I. = -0.79High C.I. = 0.32
Recommend transform:Log (Lambda = 0)
Lambda
Ln
(Re
sid
ua
lSS
)
Box-Cox Plot for Power Transforms
1.05
2.50
3.95
5.40
6.85
-3 -2 -1 0 1 2 3
A log transformation is recommended
The procedure provides a confidence interval on the transformation parameter lambda
If unity is included in the confidence interval, no transformation would be needed
Effect Estimates Following the Log Transformation
Three main effects are large
No indication of large interaction effects
What happened to the interactions?
ANOVA Following the Log Transformation
Following the Log Transformation
The Log Advance Rate Model
• Is the log model “better”?
• We would generally prefer a simpler model in a transformed scale to a more complicated model in the original metric
• What happened to the interactions?
• Sometimes transformations provide insight into the underlying mechanism
Other Examples of Unreplicated 2k Designs
• The sidewall panel experiment (Example 6.4, pg. 245)– Two factors affect the mean number of defects– A third factor affects variability– Residual plots were useful in identifying the dispersion
effect
• The oxidation furnace experiment (Example 6.5, pg. 245)– Replicates versus repeat (or duplicate) observations?– Modeling within-run variability
Por que trabalha-se com as variáveis de planejamento codificadas?
I R A B V
4 1 -1 -1 3.802
4 1 -1 -1 4.013
6 1 1 -1 6.065
6 1 1 -1 5.992
4 2 -1 1 7.934
4 2 -1 1 8.159
6 2 1 1 11.865
6 2 1 1 12.138
EXEMPLO
• Na análise com as variáveis codificadas, as magnitudes dos coeficientes do modelo são diretamente comparáveis, isto é, elas são adimensionais, e medem os efeitos da variação de cada fator de planejamento sobre um intervalo unitário.
• Além disso, são todas estimadas com a mesma precisão.
• Variáveis codificadas são muito efetivas para determinar o tamanho relativo dos efeitos dos fatores.
• Em geral, os coeficientes obtidos usando-se as unidades originais não são diretamente comparáveis, mas eles podem ter significado físico.
• Em quase todas as situações, a análise codificada é preferível.