Homework 11
Ketan Beradia, Rahul Lamba
2022-11-13
Problem 6.8
A bacteriologist is interested in the effects of two different culture media and two different times on the growth of a particular virus. He or she performs six replicates of a 22 design, making the runs in random order. Analyze the bacterial growth data that follow and draw appropriate conclusions. Analyze the residuals and comment on the model’s adequacy.
Entering the data
Time<-c(rep(12,12),rep(18,12))
Medium<-c(rep(c(1,1,2,2),6))
Observations<-c(21,22,25,26,23,28,24,25,20,26,29,27,37,39,31,34,38,38,29,33,35,36,30,35)
dat1<-cbind(Time,Medium,Observations)
dat1<-as.data.frame(dat1)
dat1$Time<-as.factor(dat1$Time)
dat1$Medium<-as.factor(dat1$Medium)Analyzing
model1<-aov(Observations~Medium*Time,data = dat1)
summary(model1)## Df Sum Sq Mean Sq F value Pr(>F)
## Medium 1 9.4 9.4 1.835 0.190617
## Time 1 590.0 590.0 115.506 9.29e-10 ***
## Medium:Time 1 92.0 92.0 18.018 0.000397 ***
## Residuals 20 102.2 5.1
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Model Equation \(Yi,j,k = μ + αi + βj + αβi,j + ϵi,j,k\)
Let us consider alpha of 0.05
From the results we clearly see that the p value of the interaction between medium and time is less than alpha so they are significant.
plot(model1)
From the residuals vs fitted graph we also see that our data does not have constant variance, but our q q plot shows that the data appears to be in a straight line. Hence our model is inadequate.
Problem 6.12
An article in the AT&T Technical Journal (March/April 1986,Vol. 65, pp. 39–50) describes the application of two-level factorial designs to integrated circuit manufacturing. A basic processing step is to grow an epitaxial layer on polished silicon wafers. The wafers mounted on a susceptor are positioned inside a bell jar, and chemical vapors are introduced. The susceptor is rotated, and heat is applied until the epitaxial layer is thick enough. An experiment was run using two factors: arsenic flow rate (A) and deposition time (B). Four replicates were run, and the epitaxial layer thickness was measured ( m). The data are shown in Table P6.1.
Entering the data
A<-c(rep(-1,4),rep(1,4),rep(-1,4),rep(1,4))
B<-c(rep(-1,8),rep(1,8))
Observations<-c(14.037,16.165,13.972,13.907,13.880,13.860,14.032,13.914,14.821,14.757,14.843,14.878,14.888,14.921,14.415,14.932)
dat2<-cbind(A,B,Observations)
dat2<-as.data.frame(dat2)
dat2$A<-as.factor(dat2$A)
dat2$B<-as.factor(dat2$B)Analyzing
Model Equation \(Yi,j,k = μ + αi + βj + αβi,j + ϵi,j,k\)
model2<-aov(Observations~A*B)PART A: Estimate the factor effects.
One<-c(14.037,16.165,13.972,13.907)
A<-c(13.88,13.86,14.032,13.914)
B<-c(14.821,14.757,14.843,14.878)
AB<-c(14.888,14.921,14.415,14.932)
S1<-sum(One)
SA<-sum(A)
SB<-sum(B)
SAB<-sum(AB)
EffectA<-(2*(SA+SAB-S1-SB)/(4*4))
EffectB<-(2*(SB+SAB-S1-SA)/(4*4))
EffectAB<-(2*(SA+SB-S1-SAB)/(4*4))
EffectA## [1] -0.31725EffectB## [1] 0.586EffectAB## [1] -0.2815Effect A = -0.31725
Effect B = 0.586
Effect AB = -0.2815
PART B: Conduct an analysis of variance. Which factors are important?
summary(model2)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 0.403 0.4026 1.262 0.2833
## B 1 1.374 1.3736 4.305 0.0602 .
## A:B 1 0.317 0.3170 0.994 0.3386
## Residuals 12 3.828 0.3190
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1For an alpha of 0.05 we see that none of the factors are important because the p value is greater than alpha assumed.
We will try removing interaction effect and testing for main effects.
model21<-aov(Observations~A+B,data = dat2)
summary(model21)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 0.403 0.4026 1.263 0.2815
## B 1 1.374 1.3736 4.308 0.0584 .
## Residuals 13 4.145 0.3189
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We clearly see that our main effects are also insignificant.
PART 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.
model23<-lm(Observations~A*B,data = dat2)
coef(model23)## (Intercept) A1 B1 A1:B1
## 14.52025 -0.59875 0.30450 0.56300\(Yi,j,k = 14.52025 - 0.59875αi + 0.30450βj + 0.56300αβγi,j + ϵi,j,k\)
PART D: Analyze the residuals. Are there any residuals that should cause concern?
plot(model2)
## hat values (leverages) are all = 0.25
## and there are no factor predictors; no plot no. 5
From the residuals vs fitted graph we can see that there is an outlier.
PART E: Discuss how you might deal with the potential outlier found in part (d).
We can just ignore the outlier and do the analysis again.
Problem 6.21
Data Entry:
Dist<-c(10,18,14,12.5,19,16,18.5,0,16.5,4.5,17.5,20.5,17.5,33,4,6,1,14.5,12,14,5,0,10,34,11,25.5,21.5,0,0,0,18.5,19.5,16,15,11,5,20.5,18,20,29.5,19,10,6.5,18.5,7.5,6,0,10,0,16.5,4.5,0,23.5,8,8,8,4.5,18,14.5,10,0,17.5,6,19.5,18,16,5.5,10,7,36,15,16,8.5,0,0.5,9,3,41.5,39,6.5,3.5,7,8.5,36,8,4.5,6.5,10,13,41,14,21.5,10.5,6.5,0,15.5,24,16,0,0,0,4.5,1,4,6.5,18,5,7,10,32.5,18.5,8)
FA <- c(-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1)
FB <- c(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
FC <- c(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
FD <- c(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
FA <- as.factor(FA)
FB <- as.factor(FB)
FC <- as.factor(FC)
FD <- as.factor(FD)
dat621 <- data.frame(FA,FB,FC,FD,Dist)Part A
Analyze the data from this experiment. Which factors significantly affect putting performance?
Model621<- aov(Dist~FA*FB*FC*FD,data = dat621)
summary(Model621)## Df Sum Sq Mean Sq F value Pr(>F)
## FA 1 917 917.1 10.588 0.00157 **
## FB 1 388 388.1 4.481 0.03686 *
## FC 1 145 145.1 1.676 0.19862
## FD 1 1 1.4 0.016 0.89928
## FA:FB 1 219 218.7 2.525 0.11538
## FA:FC 1 12 11.9 0.137 0.71178
## FB:FC 1 115 115.0 1.328 0.25205
## FA:FD 1 94 93.8 1.083 0.30066
## FB:FD 1 56 56.4 0.651 0.42159
## FC:FD 1 2 1.6 0.019 0.89127
## FA:FB:FC 1 7 7.3 0.084 0.77294
## FA:FB:FD 1 113 113.0 1.305 0.25623
## FA:FC:FD 1 39 39.5 0.456 0.50121
## FB:FC:FD 1 34 33.8 0.390 0.53386
## FA:FB:FC:FD 1 96 95.6 1.104 0.29599
## Residuals 96 8316 86.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Comment: At a significance level 0f 0.05, only main effects of factors A & B, aka length of putt and type of putter significantly affect putting performance.
Part B
Analyze the residuals from this experiment. Are there any indications of model inadequacy?
plot(Model621)
From the residual vs fitted graph we see that the data does not have constant variance and from the q q plot we also see that the data is not normal.
Problem 6.36
Resistivity on a silicon wafer is influenced by several factors. The results of a 24 factorial experiment performed during a critical processing step is shown in Table P6.10.
Entering the data:
library(DoE.base)## Warning: package 'DoE.base' was built under R version 4.2.2## Loading required package: grid## Loading required package: conf.design## Registered S3 method overwritten by 'DoE.base':
## method from
## factorize.factor conf.design##
## Attaching package: 'DoE.base'## The following objects are masked from 'package:stats':
##
## aov, lm## The following object is masked from 'package:graphics':
##
## plot.design## The following object is masked from 'package:base':
##
## lengthsF_A <- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
F_B <- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
F_C <- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
F_D <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
Obs <- c(1.92,11.28,1.09,5.75,2.13,9.53,1.03,5.35,1.6,11.73,1.16,4.68,2.16,9.11,1.07,5.3)
dat636 <- data.frame(F_A,F_B,F_C,F_D,Obs)Part A
Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model.
Model636 <- lm(Obs~F_A*F_B*F_C*F_D,data = dat636)
coef(Model636)## (Intercept) F_A F_B F_C F_D
## 4.680625 3.160625 -1.501875 -0.220625 -0.079375
## F_A:F_B F_A:F_C F_B:F_C F_A:F_D F_B:F_D
## -1.069375 -0.298125 0.229375 -0.056875 -0.046875
## F_C:F_D F_A:F_B:F_C F_A:F_B:F_D F_A:F_C:F_D F_B:F_C:F_D
## 0.029375 0.344375 -0.096875 -0.010625 0.094375
## F_A:F_B:F_C:F_D
## 0.141875halfnormal(Model636)##
## Significant effects (alpha=0.05, Lenth method):## [1] F_A F_B F_A:F_B F_A:F_B:F_C
summary(Model636)##
## Call:
## lm.default(formula = Obs ~ F_A * F_B * F_C * F_D, data = dat636)
##
## Residuals:
## ALL 16 residuals are 0: no residual degrees of freedom!
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.68062 NaN NaN NaN
## F_A 3.16062 NaN NaN NaN
## F_B -1.50187 NaN NaN NaN
## F_C -0.22062 NaN NaN NaN
## F_D -0.07937 NaN NaN NaN
## F_A:F_B -1.06938 NaN NaN NaN
## F_A:F_C -0.29812 NaN NaN NaN
## F_B:F_C 0.22937 NaN NaN NaN
## F_A:F_D -0.05687 NaN NaN NaN
## F_B:F_D -0.04688 NaN NaN NaN
## F_C:F_D 0.02937 NaN NaN NaN
## F_A:F_B:F_C 0.34437 NaN NaN NaN
## F_A:F_B:F_D -0.09688 NaN NaN NaN
## F_A:F_C:F_D -0.01063 NaN NaN NaN
## F_B:F_C:F_D 0.09438 NaN NaN NaN
## F_A:F_B:F_C:F_D 0.14188 NaN NaN NaN
##
## Residual standard error: NaN on 0 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: NaN
## F-statistic: NaN on 15 and 0 DF, p-value: NAComment: Factor effects are visible in summary of the model. From the half normal plot of the model we can observe that Factors A,B,AB are significant. We’ll ignore the interaction effect ABC since it doesn’t differ considerably from the distribution. Therefore, we’ll select a tentative model comprising of these main and interaction effects and we’ll drop the rest from our model.
Yi,j,k = 4.680625 + 3.160625αi - 1.501875βj - 1.069375αβi,j + ϵi,j,k
Part B
Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy?
Model636_2 <- aov(Obs~F_A+F_B+F_C+F_A*F_B+F_A*F_B*F_C,data = dat636)
summary(Model636_2)## Df Sum Sq Mean Sq F value Pr(>F)
## F_A 1 159.83 159.83 1563.061 1.84e-10 ***
## F_B 1 36.09 36.09 352.937 6.66e-08 ***
## F_C 1 0.78 0.78 7.616 0.02468 *
## F_A:F_B 1 18.30 18.30 178.933 9.33e-07 ***
## F_A:F_C 1 1.42 1.42 13.907 0.00579 **
## F_B:F_C 1 0.84 0.84 8.232 0.02085 *
## F_A:F_B:F_C 1 1.90 1.90 18.556 0.00259 **
## Residuals 8 0.82 0.10
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(Model636_2)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
From the Analysis we see that all the factors and the interactions are significant at alpha of 0.05.
From the residuals vs fitted graph we see that there is no constant variance and from the q q plot we see that the data is also not normally distributed.
Question 6.39
An article in Quality and Reliability Engineering International (2010, Vol. 26, pp. 223–233) presents a 25 factorial design. The experiment is shown in Table P6.12.
Data Entry:
Yeild<-c(8.11,5.56,5.77,5.82,9.17,7.8,3.23,5.69,8.82,14.23,9.2,8.94,8.68,11.49,6.25,9.12,7.93,5,7.47,12,9.86,3.65,6.4,11.61,12.43,17.55,8.87,25.38,13.06,18.85,11.78,26.05)
FactorA<-rep(c(-1,1),16)
FactorB<-rep(c(-1,-1,1,1),8)
FactorC<-rep(c(-1,-1,-1,-1,1,1,1,1),4)
FactorD<-c(rep(-1,8),rep(1,8),rep(-1,8),rep(1,8))
FactorE<-c(rep(-1,16),rep(1,16))Part A:
library(DoE.base)
dat639<-data.frame(Yeild,FactorA,FactorB,FactorC,FactorD,FactorE)
model639<-lm(Yeild~FactorA*FactorB*FactorC*FactorD*FactorE,data=dat639)
coef(model639)## (Intercept) FactorA
## 10.1803125 1.6159375
## FactorB FactorC
## 0.0434375 -0.0121875
## FactorD FactorE
## 2.9884375 2.1878125
## FactorA:FactorB FactorA:FactorC
## 1.2365625 -0.0015625
## FactorB:FactorC FactorA:FactorD
## -0.1953125 1.6665625
## FactorB:FactorD FactorC:FactorD
## -0.0134375 0.0034375
## FactorA:FactorE FactorB:FactorE
## 1.0271875 1.2834375
## FactorC:FactorE FactorD:FactorE
## 0.3015625 1.3896875
## FactorA:FactorB:FactorC FactorA:FactorB:FactorD
## 0.2503125 -0.3453125
## FactorA:FactorC:FactorD FactorB:FactorC:FactorD
## -0.0634375 0.3053125
## FactorA:FactorB:FactorE FactorA:FactorC:FactorE
## 1.1853125 -0.2590625
## FactorB:FactorC:FactorE FactorA:FactorD:FactorE
## 0.1709375 0.9015625
## FactorB:FactorD:FactorE FactorC:FactorD:FactorE
## -0.0396875 0.3959375
## FactorA:FactorB:FactorC:FactorD FactorA:FactorB:FactorC:FactorE
## -0.0740625 -0.1846875
## FactorA:FactorB:FactorD:FactorE FactorA:FactorC:FactorD:FactorE
## 0.4071875 0.1278125
## FactorB:FactorC:FactorD:FactorE FactorA:FactorB:FactorC:FactorD:FactorE
## -0.0746875 -0.3553125halfnormal(model639)##
## Significant effects (alpha=0.05, Lenth method):## [1] FactorD FactorE FactorA:FactorD
##
## [4] FactorA FactorD:FactorE FactorB:FactorE
##
## [7] FactorA:FactorB FactorA:FactorB:FactorE FactorA:FactorE
##
## [10] FactorA:FactorD:FactorE
summary(model639)##
## Call:
## lm.default(formula = Yeild ~ FactorA * FactorB * FactorC * FactorD *
## FactorE, data = dat639)
##
## Residuals:
## ALL 32 residuals are 0: no residual degrees of freedom!
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.180312 NaN NaN NaN
## FactorA 1.615938 NaN NaN NaN
## FactorB 0.043438 NaN NaN NaN
## FactorC -0.012187 NaN NaN NaN
## FactorD 2.988437 NaN NaN NaN
## FactorE 2.187813 NaN NaN NaN
## FactorA:FactorB 1.236562 NaN NaN NaN
## FactorA:FactorC -0.001563 NaN NaN NaN
## FactorB:FactorC -0.195313 NaN NaN NaN
## FactorA:FactorD 1.666563 NaN NaN NaN
## FactorB:FactorD -0.013438 NaN NaN NaN
## FactorC:FactorD 0.003437 NaN NaN NaN
## FactorA:FactorE 1.027188 NaN NaN NaN
## FactorB:FactorE 1.283437 NaN NaN NaN
## FactorC:FactorE 0.301563 NaN NaN NaN
## FactorD:FactorE 1.389687 NaN NaN NaN
## FactorA:FactorB:FactorC 0.250313 NaN NaN NaN
## FactorA:FactorB:FactorD -0.345312 NaN NaN NaN
## FactorA:FactorC:FactorD -0.063437 NaN NaN NaN
## FactorB:FactorC:FactorD 0.305312 NaN NaN NaN
## FactorA:FactorB:FactorE 1.185313 NaN NaN NaN
## FactorA:FactorC:FactorE -0.259062 NaN NaN NaN
## FactorB:FactorC:FactorE 0.170938 NaN NaN NaN
## FactorA:FactorD:FactorE 0.901563 NaN NaN NaN
## FactorB:FactorD:FactorE -0.039687 NaN NaN NaN
## FactorC:FactorD:FactorE 0.395938 NaN NaN NaN
## FactorA:FactorB:FactorC:FactorD -0.074063 NaN NaN NaN
## FactorA:FactorB:FactorC:FactorE -0.184688 NaN NaN NaN
## FactorA:FactorB:FactorD:FactorE 0.407187 NaN NaN NaN
## FactorA:FactorC:FactorD:FactorE 0.127812 NaN NaN NaN
## FactorB:FactorC:FactorD:FactorE -0.074688 NaN NaN NaN
## FactorA:FactorB:FactorC:FactorD:FactorE -0.355312 NaN NaN NaN
##
## Residual standard error: NaN on 0 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: NaN
## F-statistic: NaN on 31 and 0 DF, p-value: NAModel639_1 <- aov(Yeild~FactorA+FactorB+FactorD+FactorE+FactorA*FactorB+FactorA*FactorD+FactorA*FactorE+FactorB*FactorE+FactorD*FactorE+FactorA*FactorB*FactorE+FactorA*FactorD*FactorE,data = dat639)
summary(Model639_1)## Df Sum Sq Mean Sq F value Pr(>F)
## FactorA 1 83.56 83.56 51.362 6.10e-07 ***
## FactorB 1 0.06 0.06 0.037 0.849178
## FactorD 1 285.78 285.78 175.664 2.30e-11 ***
## FactorE 1 153.17 153.17 94.149 5.24e-09 ***
## FactorA:FactorB 1 48.93 48.93 30.076 2.28e-05 ***
## FactorA:FactorD 1 88.88 88.88 54.631 3.87e-07 ***
## FactorA:FactorE 1 33.76 33.76 20.754 0.000192 ***
## FactorB:FactorE 1 52.71 52.71 32.400 1.43e-05 ***
## FactorD:FactorE 1 61.80 61.80 37.986 5.07e-06 ***
## FactorA:FactorB:FactorE 1 44.96 44.96 27.635 3.82e-05 ***
## FactorA:FactorD:FactorE 1 26.01 26.01 15.988 0.000706 ***
## Residuals 20 32.54 1.63
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Comment: Half normal plot displays factors A,D,E,A:D,D:E,B:E,A:B,A:E,A:B:E,A:D:E as significant. ANOVA analysis also presents the factors A,D,E,AB,AD,AE,BE,DE,ABE,ADE as significant.
Part B
Analyze the residuals from this experiment. Are there any indications of model inadequacy or violations of the assumptions?
plot(Model639_1)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
Comment: Looking at the “Normal Q-Q” & “Residuals vs Fitted” plots we can conclude that the model is inadequate. Though the data satisfies the normality assumption, there’s a wide spread in variance of data.
Part 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).
dat639_2<-data.frame(Yeild,FactorA,FactorB,FactorD,FactorE)
model639_2<-lm(Yeild~FactorA*FactorB*FactorD*FactorE,data=dat639_2)
coef(model639_2)## (Intercept) FactorA
## 10.1803125 1.6159375
## FactorB FactorD
## 0.0434375 2.9884375
## FactorE FactorA:FactorB
## 2.1878125 1.2365625
## FactorA:FactorD FactorB:FactorD
## 1.6665625 -0.0134375
## FactorA:FactorE FactorB:FactorE
## 1.0271875 1.2834375
## FactorD:FactorE FactorA:FactorB:FactorD
## 1.3896875 -0.3453125
## FactorA:FactorB:FactorE FactorA:FactorD:FactorE
## 1.1853125 0.9015625
## FactorB:FactorD:FactorE FactorA:FactorB:FactorD:FactorE
## -0.0396875 0.4071875halfnormal(model639_2)##
## Significant effects (alpha=0.05, Lenth method):## [1] FactorD FactorE FactorA:FactorD
##
## [4] FactorA FactorD:FactorE FactorB:FactorE
##
## [7] FactorA:FactorB FactorA:FactorB:FactorE FactorA:FactorE
##
## [10] FactorA:FactorD:FactorE e10
summary(model639_2)##
## Call:
## lm.default(formula = Yeild ~ FactorA * FactorB * FactorD * FactorE,
## data = dat639_2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.4750 -0.5637 0.0000 0.5637 1.4750
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.18031 0.21360 47.661 < 2e-16 ***
## FactorA 1.61594 0.21360 7.565 1.14e-06 ***
## FactorB 0.04344 0.21360 0.203 0.841418
## FactorD 2.98844 0.21360 13.991 2.16e-10 ***
## FactorE 2.18781 0.21360 10.243 1.97e-08 ***
## FactorA:FactorB 1.23656 0.21360 5.789 2.77e-05 ***
## FactorA:FactorD 1.66656 0.21360 7.802 7.66e-07 ***
## FactorB:FactorD -0.01344 0.21360 -0.063 0.950618
## FactorA:FactorE 1.02719 0.21360 4.809 0.000193 ***
## FactorB:FactorE 1.28344 0.21360 6.009 1.82e-05 ***
## FactorD:FactorE 1.38969 0.21360 6.506 7.24e-06 ***
## FactorA:FactorB:FactorD -0.34531 0.21360 -1.617 0.125501
## FactorA:FactorB:FactorE 1.18531 0.21360 5.549 4.40e-05 ***
## FactorA:FactorD:FactorE 0.90156 0.21360 4.221 0.000650 ***
## FactorB:FactorD:FactorE -0.03969 0.21360 -0.186 0.854935
## FactorA:FactorB:FactorD:FactorE 0.40719 0.21360 1.906 0.074735 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.208 on 16 degrees of freedom
## Multiple R-squared: 0.9744, Adjusted R-squared: 0.9504
## F-statistic: 40.58 on 15 and 16 DF, p-value: 7.07e-10Model_3<- aov(Yeild~FactorA+FactorB+FactorD+FactorE+FactorA*FactorB+FactorA*FactorD+FactorA*FactorE+FactorB*FactorE+FactorD*FactorE+FactorA*FactorB*FactorE+FactorA*FactorD*FactorE,data = dat639_2)
summary(Model_3)## Df Sum Sq Mean Sq F value Pr(>F)
## FactorA 1 83.56 83.56 51.362 6.10e-07 ***
## FactorB 1 0.06 0.06 0.037 0.849178
## FactorD 1 285.78 285.78 175.664 2.30e-11 ***
## FactorE 1 153.17 153.17 94.149 5.24e-09 ***
## FactorA:FactorB 1 48.93 48.93 30.076 2.28e-05 ***
## FactorA:FactorD 1 88.88 88.88 54.631 3.87e-07 ***
## FactorA:FactorE 1 33.76 33.76 20.754 0.000192 ***
## FactorB:FactorE 1 52.71 52.71 32.400 1.43e-05 ***
## FactorD:FactorE 1 61.80 61.80 37.986 5.07e-06 ***
## FactorA:FactorB:FactorE 1 44.96 44.96 27.635 3.82e-05 ***
## FactorA:FactorD:FactorE 1 26.01 26.01 15.988 0.000706 ***
## Residuals 20 32.54 1.63
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(Model_3)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
Comment: Since factor C was insignificant and thus didn’t seem important, so dropped it. Even after dropping factor C and analyzing the data with four active factors still the results are same as in part A. ANOVA analysis in this case too presents the factors A,D,E,AB,AD,AE,BE,DE,ABE,ADE as significant.
Part D
Find settings of the active factors that maximize the predicted response.
Based on the transformed data:
Denoting factors A,B,D,E with alpha,beta,gamma & delta.
Yi,j,k,l = 10.1803125 + 1.6159375αi + 0.0434375βj + 2.9884375γk+ 2.1878125δl + ϵi,j,k,l
Since in the equation all the factors have positive coefficients, thus they must be at +1 level (high) to produce max response.