Homework Week 11 - Fred Gersdorff
IE 5342 - Dr. Timothy I. Matis
Fred Gersdorff
11/14/2021
This report captures work done for the individual homework for Week 11. R code along with the results are provided. The required homework problems were taken from “Design and Analysis of Experiments 8th Edition”:
1) 6.8
2) 6.12
3) 6.21
3) 6.36
3) 6.39
Problem 4 (6.36)
Resistivity on a silicon wafer is influenced by several factors. The results of a 2^4 factorial experiment performed during a critical processing step is shown in the Table shown in the book.
- Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model.
# setup Libraries
library(knitr)
library(dplyr)
library(tidyr)
library(GAD)
library(tinytex)
library(ggplot2)
library(ggfortify)
library(car)
library(DoE.base)
# READ IN FILE
setwd("D:/R Files/")
dat <- read.csv("D:/R Files/6-8.csv",header=TRUE)
# SET UP DATA TYPES
dat$Time <- as.fixed(dat$Time)
dat$CultureMedium <- as.fixed(dat$CultureMedium)
Response <- dat$GrowthThe Linear Effects for this Model are:
\(\quad y_{ijk}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha \beta_{ij}\) \(\epsilon_{ijk}\)
Where:
\(\quad \alpha_i\) is the main effect of the \(\ i^{th}\) treatment of Time Rate used
\(\quad \beta_j\) is the main effect of the \(\ j^{th}\) treatment of Culture Medium
\(\quad \alpha \beta_{ij}\) is the interaction effect of the \(\ ij^{th}\) treatment of Time * Culture Medium, and
\(\quad \epsilon_{ijk}\) is the random error term.
The Hypotheses we will test are:
Time Main Effect:
\(\quad H_0\) : \(\alpha_i\) = 0 \(\forall\) i
\(\quad H_a\) : \(\alpha_i \neq\) 0 for at least one \(\ i\)
Culture Medium Main Effect:
\(\quad H_0\) : \(\beta_j\) = 0 \(\forall\) j
\(\quad H_a\) : \(\beta_j \neq\) 0 for at least one \(\ j\)
Time * Culture Medium Interaction Effect:
\(\quad H_0\) : \(\alpha \beta_{ij}\) = 0 \(\forall\) ij
\(\quad H_a\) : \(\alpha \beta_{ij} \neq\) 0 for at least one \(\ ij\)
\(\quad\)at a significance level of \(\alpha\) = 0.05
# SET UP model
model<-aov(Response~dat$Time*dat$CultureMedium)
#run model through GAD
GAD::gad(model)## Analysis of Variance Table
##
## Response: Response
## Df Sum Sq Mean Sq F value Pr(>F)
## dat$Time 1 590.04 590.04 115.5057 9.291e-10 ***
## dat$CultureMedium 1 9.38 9.38 1.8352 0.1906172
## dat$Time:dat$CultureMedium 1 92.04 92.04 18.0179 0.0003969 ***
## Residual 20 102.17 5.11
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We evaluate the hypothesis for the interaction effect first and find that it’s p-value < 0.000 which is significant to an \(\alpha\) of 0.05, so we reject the null hypothesis. Because of the Hierarchy Principle, we keep both main effects in the model even though the p-value for the Culture Medium is = 0.191 and we would have failed to reject the null hypothesis. We reject the null hypothesis for Time because its p-value is below our level of significance, with a p-value of <0.000.
Analyzing the residuals, we get the following:
autoplot(model)
We can see from the plots below that the residual are normal-ish and show constant variation so we find the model to be adequate.
Problem 2 (6.12)
The full explanation of this problem can be found in the text book. a) Estimate the factor effects.
dat2 <- read.csv("D:/R Files/6-12.csv",header=TRUE)
FactorA <- dat2$A
FactorB <- dat2$B
Result <- dat2$Result
mod <- lm(Result~FactorA*FactorB, data = dat2)
coef(mod)## (Intercept) FactorA FactorB FactorA:FactorB
## 14.513875 -0.158625 0.293000 0.140750
b) Conduct an analysis of variance. Which factors are important?
anova(mod)## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## FactorA 1 0.4026 0.40259 1.2619 0.28327
## FactorB 1 1.3736 1.37358 4.3054 0.06016 .
## FactorA:FactorB 1 0.3170 0.31697 0.9935 0.33856
## Residuals 12 3.8285 0.31904
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1This first round shows that none of the factors are significant. I’ll remove the interaction term and run the ANOVA again.
mod <- lm(Result~FactorA+FactorB, data = dat2)
anova(mod)## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## FactorA 1 0.4026 0.40259 1.2625 0.28150
## FactorB 1 1.3736 1.37358 4.3075 0.05835 .
## Residuals 13 4.1454 0.31888
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We now see that both main effects are still not significant. If I remove Factor A because it is least significant and run again, I get the following for FactorB:
mod <- lm(Result~FactorB, data = dat2)
anova(mod)## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## FactorB 1 1.3736 1.37358 4.2282 0.0589 .
## Residuals 14 4.5480 0.32486
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Which is still not significant. Therefore none of the factors or interactions were found to be important / significant.
c) Write down a regression equation that could be used to predict epitaxial layer thickness over the region of arsenic flow rate and deposition time used in this experiment. The equation is:
Result = 14.514 -0.159A + 0.293B + 0.141 AB
d)Analyze the residuals. Are there any residuals that should cause concern? Analyzing the residuals, we get the following:
mod <- lm(Result~FactorA*FactorB, data = dat2)
autoplot(mod)
The residual for first observation of Replicate II is an outlier with the original value of the observation at 16.165.
- Discuss how you might deal with the potential outlier found in part (d).
Ways to deal with the outlier are varied we could, first see if it was due to a transcription error and then correct accordingly. We could replace it while maintaining orthogonality or under the right circumstances, we could disregard the value and make note of it. —-
Problem 3 (6.21)
The full explanation of this problem can be found in the textbook. a) Analyze the data from this experiment. Which factors significantly affect putting performance?
dat3 <- read.csv("D:/R Files/6-21.csv",header=TRUE)
PuttLength <- dat3$PuttLength
PutterType <- dat3$PutterType
PuttBreak <- dat3$PuttBreak
PuttSlope <- dat3$PuttSlope
DistanceFromCup <- dat3$DistanceFromCup
PuttLength <- as.fixed(PuttLength)
PutterType <- as.fixed(PutterType)
PuttBreak <- as.fixed(PuttBreak)
PuttSlope <- as.fixed(PuttSlope)
mod3 <- lm(DistanceFromCup~PuttLength+PutterType+PuttBreak+PuttSlope+
PuttLength*PutterType+
PuttLength*PuttBreak+
PutterType*PuttBreak+
PuttLength*PuttSlope+
PutterType*PuttSlope+
PuttBreak*PuttSlope+
PuttLength*PutterType*PuttBreak+
PuttLength*PutterType*PuttSlope+
PuttLength*PuttBreak*PuttSlope+
PuttLength*PuttBreak*PuttSlope+
PutterType*PuttBreak*PuttSlope+
PuttLength*PutterType*PuttBreak*PuttSlope)
GAD::gad(mod3)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.5878 0.001572 **
## PutterType 1 388.1 388.15 4.4809 0.036862 *
## PuttBreak 1 145.1 145.15 1.6756 0.198615
## PuttSlope 1 1.4 1.40 0.0161 0.899280
## PuttLength:PutterType 1 218.7 218.68 2.5245 0.115377
## PuttLength:PuttBreak 1 11.9 11.90 0.1373 0.711776
## PutterType:PuttBreak 1 115.0 115.02 1.3278 0.252054
## PuttLength:PuttSlope 1 93.8 93.81 1.0829 0.300658
## PutterType:PuttSlope 1 56.4 56.43 0.6515 0.421588
## PuttBreak:PuttSlope 1 1.6 1.63 0.0188 0.891271
## PuttLength:PutterType:PuttBreak 1 7.3 7.25 0.0837 0.772939
## PuttLength:PutterType:PuttSlope 1 113.0 113.00 1.3045 0.256228
## PuttLength:PuttBreak:PuttSlope 1 39.5 39.48 0.4558 0.501207
## PutterType:PuttBreak:PuttSlope 1 33.8 33.77 0.3899 0.533858
## PuttLength:PutterType:PuttBreak:PuttSlope 1 95.6 95.65 1.1042 0.295994
## Residual 96 8315.8 86.62
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
I’ll remove the 4th level interaction effect and run again because it isn’t significant to an alpha of 0.05.
mod31 <- lm(DistanceFromCup~PuttLength+PutterType+PuttBreak+PuttSlope+
PuttLength*PutterType+
PuttLength*PuttBreak+
PutterType*PuttBreak+
PuttLength*PuttSlope+
PutterType*PuttSlope+
PuttBreak*PuttSlope+
PuttLength*PutterType*PuttBreak+
PuttLength*PutterType*PuttSlope+
PuttLength*PuttBreak*PuttSlope+
PuttLength*PuttBreak*PuttSlope+
PutterType*PuttBreak*PuttSlope)
GAD::gad(mod31)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.5764 0.001576 **
## PutterType 1 388.1 388.15 4.4761 0.036934 *
## PuttBreak 1 145.1 145.15 1.6738 0.198822
## PuttSlope 1 1.4 1.40 0.0161 0.899331
## PuttLength:PutterType 1 218.7 218.68 2.5218 0.115536
## PuttLength:PuttBreak 1 11.9 11.90 0.1372 0.711915
## PutterType:PuttBreak 1 115.0 115.02 1.3264 0.252277
## PuttLength:PuttSlope 1 93.8 93.81 1.0818 0.300889
## PutterType:PuttSlope 1 56.4 56.43 0.6508 0.421816
## PuttBreak:PuttSlope 1 1.6 1.63 0.0188 0.891326
## PuttLength:PutterType:PuttBreak 1 7.3 7.25 0.0836 0.773051
## PuttLength:PutterType:PuttSlope 1 113.0 113.00 1.3031 0.256452
## PuttLength:PuttBreak:PuttSlope 1 39.5 39.48 0.4553 0.501419
## PutterType:PuttBreak:PuttSlope 1 33.8 33.77 0.3894 0.534062
## Residual 97 8411.4 86.72
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We can now see that none of the 3rd level interaction effects are close to being significant so we will remove them and re-run the model.
mod32 <- lm(DistanceFromCup~PuttLength+PutterType+PuttBreak+PuttSlope+
PuttLength*PutterType+
PuttLength*PuttBreak+
PutterType*PuttBreak+
PuttLength*PuttSlope+
PutterType*PuttSlope+
PuttBreak*PuttSlope)
GAD::gad(mod32)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.7649 0.001421 **
## PutterType 1 388.1 388.15 4.5558 0.035227 *
## PuttBreak 1 145.1 145.15 1.7036 0.194779
## PuttSlope 1 1.4 1.40 0.0164 0.898432
## PuttLength:PutterType 1 218.7 218.68 2.5668 0.112255
## PuttLength:PuttBreak 1 11.9 11.90 0.1396 0.709444
## PutterType:PuttBreak 1 115.0 115.02 1.3500 0.248009
## PuttLength:PuttSlope 1 93.8 93.81 1.1010 0.296542
## PutterType:PuttSlope 1 56.4 56.43 0.6624 0.417645
## PuttBreak:PuttSlope 1 1.6 1.63 0.0191 0.890357
## Residual 101 8604.9 85.20
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We then remove each secondary interaction in order of highest p-value and discover that as we remove each term, none of the 2nd order interactions are significant. The step-wise process hasn’t been shown here for brevity. Running the model with main effects, we get the following:
mod33 <- lm(DistanceFromCup~PuttLength+PutterType+PuttBreak+PuttSlope)
GAD::gad(mod33)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.7812 0.001386 **
## PutterType 1 388.1 388.15 4.5627 0.034956 *
## PuttBreak 1 145.1 145.15 1.7062 0.194280
## PuttSlope 1 1.4 1.40 0.0164 0.898342
## Residual 107 9102.4 85.07
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We now remove PuttSlope becasue it isn’t significant and re-run:
mod34 <- lm(DistanceFromCup~PuttLength+PutterType+PuttBreak)
GAD::gad(mod34)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.8803 0.001317 **
## PutterType 1 388.1 388.15 4.6046 0.034125 *
## PuttBreak 1 145.1 145.15 1.7219 0.192233
## Residual 108 9103.8 84.29
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We now remove PuttBreak becasue it isn’t significant and re-run:
mod35 <- lm(DistanceFromCup~PuttLength+PutterType)
GAD::gad(mod35)## Analysis of Variance Table
##
## Response: DistanceFromCup
## Df Sum Sq Mean Sq F value Pr(>F)
## PuttLength 1 917.1 917.15 10.8087 0.00136 **
## PutterType 1 388.1 388.15 4.5743 0.03469 *
## Residual 109 9248.9 84.85
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1And we now find the significant factors, we see PuttLength at a p-value of 0.00136 and PutterType at a p-value of 0.03469.
b) Analyze the residuals from this experiment. Are there any indications of model inadequacy?
The resiudal plots are below:
autoplot(mod35)
The residuals appear to be significantly not normal at the higher end of the theoretical quantiles so the model is inadequate.
Problem 4 (6.36)
Resistivity on a silicon wafer is influenced by several factors. The results of a 2^4 factorial experiment performed during a critical processing step is shown in the Table in the book.
- Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model.
dat4 <- read.csv("D:/R Files/6-36.csv",header=TRUE)
A <- dat4$A
B <- dat4$B
C <- dat4$C
D <- dat4$D
Resistivity <- dat4$Resistivity
A <- as.numeric(A)
B <- as.numeric(B)
C <- as.numeric(C)
D <- as.numeric(D)
dat636 <- data.frame(A,B,C,D,Resistivity)
mod5 <- lm(Resistivity~A*B*C*D, data = dat636)
anova(mod5)## Warning in anova.lm(mod5): ANOVA F-tests on an essentially perfect fit are
## unreliable## Analysis of Variance Table
##
## Response: Resistivity
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 159.833 159.833 NaN NaN
## B 1 36.090 36.090 NaN NaN
## C 1 0.779 0.779 NaN NaN
## D 1 0.101 0.101 NaN NaN
## A:B 1 18.297 18.297 NaN NaN
## A:C 1 1.422 1.422 NaN NaN
## B:C 1 0.842 0.842 NaN NaN
## A:D 1 0.052 0.052 NaN NaN
## B:D 1 0.035 0.035 NaN NaN
## C:D 1 0.014 0.014 NaN NaN
## A:B:C 1 1.898 1.898 NaN NaN
## A:B:D 1 0.150 0.150 NaN NaN
## A:C:D 1 0.002 0.002 NaN NaN
## B:C:D 1 0.143 0.143 NaN NaN
## A:B:C:D 1 0.322 0.322 NaN NaN
## Residuals 0 0.000 NaNhalfnormal(mod5)
The half-normal plot shows that Factors A, B and AB are significant from the plot.
b) Fit the model identified in part (a) and analyze the residuals, Is there any indication of model inadequacy?
dat6362 <- data.frame(A,B,C,D,Resistivity)
mod52 <- lm(Resistivity~A+B+A*B, data= dat6362)
anova(mod52)## Analysis of Variance Table
##
## Response: Resistivity
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 159.833 159.833 333.088 4.049e-10 ***
## B 1 36.090 36.090 75.211 1.630e-06 ***
## A:B 1 18.297 18.297 38.130 4.763e-05 ***
## Residuals 12 5.758 0.480
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1autoplot(mod52)
After plotting the residuals, the model appears to be inadequate because the residuals aren’t normally distributed.
c) Repeat the analysis from parts (a) and (b) using ln(y) as the response variable. Is there an indication that the transformation has been useful?
ResistivityLog <- log(Resistivity)
dat6363 <- data.frame(A,B,C,D,ResistivityLog)
mod6 <- lm(ResistivityLog~A*B*C*D, data = dat6363)
anova(mod6)## Warning in anova.lm(mod6): ANOVA F-tests on an essentially perfect fit are
## unreliable## Analysis of Variance Table
##
## Response: ResistivityLog
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 10.5721 10.5721 NaN NaN
## B 1 1.5803 1.5803 NaN NaN
## C 1 0.0007 0.0007 NaN NaN
## D 1 0.0052 0.0052 NaN NaN
## A:B 1 0.0097 0.0097 NaN NaN
## A:C 1 0.0252 0.0252 NaN NaN
## B:C 1 0.0003 0.0003 NaN NaN
## A:D 1 0.0015 0.0015 NaN NaN
## B:D 1 0.0002 0.0002 NaN NaN
## C:D 1 0.0051 0.0051 NaN NaN
## A:B:C 1 0.0644 0.0644 NaN NaN
## A:B:D 1 0.0143 0.0143 NaN NaN
## A:C:D 1 0.0002 0.0002 NaN NaN
## B:C:D 1 0.0002 0.0002 NaN NaN
## A:B:C:D 1 0.0157 0.0157 NaN NaN
## Residuals 0 0.0000 NaNhalfnormal(mod6)
mod53 <- lm(ResistivityLog~A+B+C+(A*B*C), data= dat6363)
anova(mod53)## Analysis of Variance Table
##
## Response: ResistivityLog
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 10.5721 10.5721 1994.5559 6.975e-11 ***
## B 1 1.5803 1.5803 298.1470 1.289e-07 ***
## C 1 0.0007 0.0007 0.1240 0.733861
## A:B 1 0.0097 0.0097 1.8393 0.212066
## A:C 1 0.0252 0.0252 4.7632 0.060627 .
## B:C 1 0.0003 0.0003 0.0539 0.822233
## A:B:C 1 0.0644 0.0644 12.1466 0.008256 **
## Residuals 8 0.0424 0.0053
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1autoplot(mod53)
The transformation on Resistivity corrected the non-normal residuals. The model is now adeuqate.
- Fit a model in terms of the coded variables that can be used predict the resistivity.
coef(mod53)## (Intercept) A B C A:B A:C
## 1.185417116 0.812870345 -0.314277554 -0.006408558 -0.024684570 -0.039723700
## B:C A:B:C
## -0.004225796 0.063434408The model is: 1.1854 + 0.8129A - 0.3143B - 0.0064C - 0.0247AB - 0.0397AC - 0.0042BC + 0.0634ABC
Problem 5 (6.39)
An article in Quality and Reliability Engineering…
a) Analyze the data from this experiment. Identify the significant factors and interactions.
Running the anova and half normality plot, we see that only the interaction ADE is significant.
dat5 <- read.csv("D:/R Files/6-39.csv",header=TRUE)
A <- dat5$A
B <- dat5$B
C <- dat5$C
D <- dat5$D
E <- dat5$E
Result <- dat5$y
A <- as.numeric(A)
B <- as.numeric(B)
C <- as.numeric(C)
D <- as.numeric(D)
E <- as.numeric(E)
dat639 <- data.frame(A,B,C,D,E,Result)
mod639 <- lm(Result~A*B*C*D*E, data = dat639)
anova(mod639)## Warning in anova.lm(mod639): ANOVA F-tests on an essentially perfect fit are
## unreliable## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 2.674 2.674 NaN NaN
## B 1 0.991 0.991 NaN NaN
## C 1 0.257 0.257 NaN NaN
## D 1 81.505 81.505 NaN NaN
## E 1 26.082 26.082 NaN NaN
## A:B 1 13.533 13.533 NaN NaN
## A:C 1 15.249 15.249 NaN NaN
## B:C 1 4.766 4.766 NaN NaN
## A:D 1 5.176 5.176 NaN NaN
## B:D 1 33.395 33.395 NaN NaN
## C:D 1 0.498 0.498 NaN NaN
## A:E 1 76.107 76.107 NaN NaN
## B:E 1 104.221 104.221 NaN NaN
## C:E 1 27.288 27.288 NaN NaN
## D:E 1 32.140 32.140 NaN NaN
## A:B:C 1 51.537 51.537 NaN NaN
## A:B:D 1 1.062 1.062 NaN NaN
## A:C:D 1 26.481 26.481 NaN NaN
## B:C:D 1 0.179 0.179 NaN NaN
## A:B:E 1 11.725 11.725 NaN NaN
## A:C:E 1 20.082 20.082 NaN NaN
## B:C:E 1 30.167 30.167 NaN NaN
## A:D:E 1 179.409 179.409 NaN NaN
## B:D:E 1 45.816 45.816 NaN NaN
## C:D:E 1 2.983 2.983 NaN NaN
## A:B:C:D 1 44.157 44.157 NaN NaN
## A:B:C:E 1 0.518 0.518 NaN NaN
## A:B:D:E 1 5.687 5.687 NaN NaN
## A:C:D:E 1 15.750 15.750 NaN NaN
## B:C:D:E 1 50.125 50.125 NaN NaN
## A:B:C:D:E 1 2.605 2.605 NaN NaN
## Residuals 0 0.000 NaNhalfnormal(mod639)
b) Analyze the residuals from this experiment. Are there any indications of model inadequacy or violations of the assumptions?
mod6391 <- lm(Result~A+D+E+(A*D)+(A*E)+(D*E)+(A*D*E), data = dat639)
anova(mod6391)## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 2.67 2.674 0.1261 0.725657
## D 1 81.50 81.505 3.8425 0.061677 .
## E 1 26.08 26.082 1.2296 0.278465
## A:D 1 5.18 5.176 0.2440 0.625802
## A:E 1 76.11 76.107 3.5881 0.070310 .
## D:E 1 32.14 32.140 1.5152 0.230268
## A:D:E 1 179.41 179.409 8.4582 0.007708 **
## Residuals 24 509.07 21.211
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1autoplot(mod6391)
Analyzing the residuals shows that they are normal and of constant variance, so the model is adequate.
c) One of the factors from this experiment does not seem to be important. If you drop this factor, what type of design remains? Analyze the data using the full factorial model for only the four active factors. Compare your results with those obtained in part (a).
Factor C is not significant in this experiment. If we remove Factor C, we have a non-hierarchical model. The half-normal plot below shows that the significant factors would now be ADE, BE, D and AE. This is much different from the results obtained in part a when we had ADE as significant. Adding C in with noise allowed us to see BE, D and AE as significant.
mod6392 <- lm(Result~A*B*D*E, data = dat639)
anova(mod6392)## Analysis of Variance Table
##
## Response: Result
## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 2.674 2.674 0.1462 0.707234
## B 1 0.991 0.991 0.0542 0.818933
## D 1 81.505 81.505 4.4562 0.050862 .
## E 1 26.082 26.082 1.4260 0.249816
## A:B 1 13.533 13.533 0.7399 0.402395
## A:D 1 5.176 5.176 0.2830 0.602048
## B:D 1 33.395 33.395 1.8259 0.195416
## A:E 1 76.107 76.107 4.1611 0.058227 .
## B:E 1 104.221 104.221 5.6982 0.029671 *
## D:E 1 32.140 32.140 1.7573 0.203584
## A:B:D 1 1.062 1.062 0.0581 0.812629
## A:B:E 1 11.725 11.725 0.6411 0.435057
## A:D:E 1 179.409 179.409 9.8091 0.006434 **
## B:D:E 1 45.816 45.816 2.5050 0.133050
## A:B:D:E 1 5.687 5.687 0.3109 0.584829
## Residuals 16 292.640 18.290
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1halfnormal(mod6392)
- Find settings of the active factors that maximize the predicted response.
Using another software package’s response optimizer, I found that the following solution optimizes the predicted response: A = -1; B = 1; C = -1; D = 1; E= -1. This generates the result of y = 22.19.