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Jestin Job Sibychan
14/11/2021
Q no 6.8
Model Equation for two factor interaction:
\(Y_{ijk} = \mu_i + \alpha_i + \beta_j + \alpha\beta_{ij} + \epsilon_{ijk}\)
Where
\(\alpha_i\) is Main Effects of Factor A (Time)
\(\beta_j\) is Main Effects of Factor B (Culture medium)
\(\alpha\beta_{ij}\) is Interaction effects of Factors A and Factors B
library(DoE.base)## 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':
##
## lengthslibrary(GAD)## Loading required package: matrixStats## Loading required package: R.methodsS3## R.methodsS3 v1.8.1 (2020-08-26 16:20:06 UTC) successfully loaded. See ?R.methodsS3 for help.time<-c(rep(12,12),rep(18,12))
culture<-c(rep(1,2),rep(2,2),rep(1,2),rep(2,2),rep(1,2),rep(2,2),rep(1,2),rep(2,2),rep(1,2),rep(2,2),rep(1,2),rep(2,2))
observation<-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)
time<-as.fixed(time)
culture<-as.fixed(culture)
dat<-data.frame(observation,time,culture)
model<-aov(observation~time+culture+time*culture)
GAD::gad(model)## Analysis of Variance Table
##
## Response: observation
## Df Sum Sq Mean Sq F value Pr(>F)
## time 1 590.04 590.04 115.5057 9.291e-10 ***
## culture 1 9.38 9.38 1.8352 0.1906172
## time:culture 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 ' ' 1Here we can see that the p value of interaction between time and culture is less that alpha i.e 0.05 hence it is significant and we reject the null hypothesis
Interaction and Residual plots
interaction.plot(time,culture,observation)
plot(model)
From the above graphs we can see that it is normally distributed and the variance also looks a bit similar hence we can say that the model is adequate
Q 6.12
Factor Effects
Effect of a = 2*(a+ab-b-(1))/16
= -0.3185
Similarly effect at b = 2*(b+ab-a-(1))/16
=0.5872
Effect at ab = 2*(a+b-ab-(1))/16
= -0.28025
factorlevel<-c(rep(55,8),rep(58,8))
times<-c(rep(10,4),rep(15,4),rep(10,4),rep(15,4))
observation1<-c(14.037,16.165,13.972,13.907,14.821,14.757,14.843,14.878,13.880,13.860,14.032,13.914,14.888,14.921,14.415,14.932)
factorlevel<-as.fixed(factorlevel)
times<-as.fixed(times)
dat2<-data.frame(observation1,factorlevel,times)
model2<-aov(observation1~factorlevel+times+factorlevel*times)
GAD::gad(model2)## Analysis of Variance Table
##
## Response: observation1
## Df Sum Sq Mean Sq F value Pr(>F)
## factorlevel 1 0.4026 0.40259 1.2619 0.28327
## times 1 1.3736 1.37358 4.3054 0.06016 .
## factorlevel:times 1 0.3170 0.31697 0.9935 0.33856
## Residual 12 3.8285 0.31904
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1B)Here we can see that all p values for main effects as well as interaction the p values are greater than 0.05 hence we can say that the effects are not significant
model3<-lm(observation1~factorlevel+times+factorlevel*times)
coef(model3)## (Intercept) factorlevel58 times15
## 14.52025 -0.59875 0.30450
## factorlevel58:times15
## 0.56300C)Hence the regression equation for thickness is given above
plot(model2)
### D)From the above graphs we can see that the normality plot is not up to the mark hence the model is not adequate. ### hence the residuals causes concerns
6.21
FactorA<-rep(rep(c(-1,1),8),7)
FactorB<-rep(rep(c(1,1,-1,-1),4),7)
FactorC<-rep(rep(c(rep(1,4),rep(-1,4)),2),7)
FactorD<-rep(c(rep(1,8),rep(-1,8)),7)
Observation3<-c(10.0,0.0,4.0,0.0,0.0,5.0,6.5,16.5,4.5,19.5,15.0,41.5,8.0,21.5,0.0,18.0,18.0,16.5,6.0,10.0,0.0,20.5,18.5,4.5,18.0,18.0,16.0,39.0,4.5,10.5,0.0,5.0,14.0,4.5,1.0,34.0,18.5,18.0,7.5,0.0,14.5,16.0,8.5,6.5,6.5,6.5,0.0,7.0,12.5,17.5,14.5,11.0,19.5,20.0,6.0,23.5,10.0,5.5,0.0,3.5,10.0,0.0,4.5,10.0,19.0,20.5,12.0,25.5,16.0,29.5,0.0,8.0,0.0,10.0,0.5,7.0,13.0,15.5,1.0,32.5,16.0,17.5,14.0,21.5,15.0,19.0,10.0,8.0,17.5,7.0,9.0,8.5,41.0,24.0,4.0,18.5,18.5,33.0,5.0,0.0,11.0,10.0,0.0,8.0,6.0,36.0,3.0,36.0,14.0,16.0,6.5,8.0)
dat3<-data.frame(FactorA,FactorB,FactorC,FactorD,Observation3)
model4<-lm(Observation3~FactorA*FactorB*FactorC*FactorD,data=dat3)
halfnormal(model4)## Warning in halfnormal.lm(model4): halfnormal not recommended for models with
## more residual df than model df##
## Significant effects (alpha=0.05, Lenth method):## [1] FactorA e74 e92 FactorB e53
### A)From the above graph we can see that factorA and Factor B are significant. Hence FactorA and FactorB affects the performance(i.e Length of putt and type of putt)
model5<-aov(Observation3~FactorA*FactorB,data=dat3)
summary(model5)## Df Sum Sq Mean Sq F value Pr(>F)
## FactorA 1 917 917.1 10.969 0.00126 **
## FactorB 1 388 388.1 4.642 0.03342 *
## FactorA:FactorB 1 219 218.7 2.615 0.10875
## Residuals 108 9030 83.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model5)
## hat values (leverages) are all = 0.03571429
## and there are no factor predictors; no plot no. 5
### B)From the above summary we can say that the Factor A and Factor B has p value less than 0.05 hence they are significant ### And from the graph we can say that that the normality plot looks good and hence the model is adequate
6.34
a<-c(-1,1)
A<-rep(a,8)
b<-c(-1,-1,1,1)
B<-rep(b,4)
c<-c(rep(-1,4),rep(1,4))
C<-rep(c,2)
D<-c(rep(-1,8),rep(1,8))
Observation4<-c(1.92,11.28,1.09,5.75,2.13,9.53,1.03,5.35,1.60,11.73,1.16,4.68,2.16,9.11,1.07,5.30)
dat4<-data.frame(Observation4,A,B,C,D)
model6<-lm(Observation4~A*B*C*D,data = dat4)
coef(model6)## (Intercept) A B C D A:B
## 4.680625 3.160625 -1.501875 -0.220625 -0.079375 -1.069375
## A:C B:C A:D B:D C:D A:B:C
## -0.298125 0.229375 -0.056875 -0.046875 0.029375 0.344375
## A:B:D A:C:D B:C:D A:B:C:D
## -0.096875 -0.010625 0.094375 0.141875halfnormal(model6)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B A:B:C
### From Above graph we can see that A,B,AB are signicicant ABC is given significant but by looking at the diagram it seems unsignificant hence cosidring it as not significant
AA<-as.fixed(dat4$A)
BB<-as.fixed(dat4$B)
dat5<-data.frame(Observation4,AA,BB)
model7<-aov(Observation4~AA*BB,data = dat5)
GAD::gad(model7)## Analysis of Variance Table
##
## Response: Observation4
## Df Sum Sq Mean Sq F value Pr(>F)
## AA 1 159.833 159.833 333.088 4.049e-10 ***
## BB 1 36.090 36.090 75.211 1.630e-06 ***
## AA:BB 1 18.297 18.297 38.130 4.763e-05 ***
## Residual 12 5.758 0.480
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model7)
From the above graph of normal probability plot the data does not look like like its normally distributed hence the model is not adequate
Doing log transform
LogObservation<-log(Observation4)
dat6<-data.frame(LogObservation,A,B,C,D)
model8<-lm(LogObservation~A*B*C*D,data = dat6)
coef(model8)## (Intercept) A B C D A:B
## 1.185417116 0.812870345 -0.314277554 -0.006408558 -0.018077390 -0.024684570
## A:C B:C A:D B:D C:D A:B:C
## -0.039723700 -0.004225796 -0.009578245 0.003708723 0.017780432 0.063434408
## A:B:D A:C:D B:C:D A:B:C:D
## -0.029875960 -0.003740235 0.003765760 0.031322043halfnormal(model8)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B:C
### From the above graph we can see that A and B are Significant
dat7<-data.frame(LogObservation,AA,BB)
model9<-aov(LogObservation~AA+BB,data = dat7)
GAD::gad(model9)## Analysis of Variance Table
##
## Response: LogObservation
## Df Sum Sq Mean Sq F value Pr(>F)
## AA 1 10.5721 10.5721 962.95 1.408e-13 ***
## BB 1 1.5803 1.5803 143.94 2.095e-08 ***
## Residual 13 0.1427 0.0110
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model9)
From the above graph we can see that the model is adequate
dat8<-data.frame(LogObservation,A,B)
model10<-lm(LogObservation~A+B,data = dat8)
coef(model10)## (Intercept) A B
## 1.1854171 0.8128703 -0.3142776The Model equation in terms of coded variables is log(resistivity) = 1.18+(0.81)A+(-0.314)B+epsilon ijk
6.39
a1<-c(-1,1)
b1<-c(-1,-1,1,1)
c1<-c(-1,-1,-1,-1,1,1,1,1)
d1<-c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1)
A1<-c(rep(a1,16))
B1<-c(rep(b1,8))
C1<-c(rep(c1,4))
D1<-c(rep(d1,2))
E1<-c(rep(-1,16),rep(1,16))
Observation5<-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)
dat9<-data.frame(A1,B1,C1,D1,E1,Observation5)
model11<-lm(Observation5~A1*B1*C1*D1*E1,data=dat9)
halfnormal(model11)##
## Significant effects (alpha=0.05, Lenth method):## [1] D1 E1 A1:D1 A1 D1:E1 B1:E1 A1:B1 A1:B1:E1
##
## [9] A1:E1 A1:D1:E1
coef(model11)## (Intercept) A1 B1 C1 D1
## 10.1803125 1.6159375 0.0434375 -0.0121875 2.9884375
## E1 A1:B1 A1:C1 B1:C1 A1:D1
## 2.1878125 1.2365625 -0.0015625 -0.1953125 1.6665625
## B1:D1 C1:D1 A1:E1 B1:E1 C1:E1
## -0.0134375 0.0034375 1.0271875 1.2834375 0.3015625
## D1:E1 A1:B1:C1 A1:B1:D1 A1:C1:D1 B1:C1:D1
## 1.3896875 0.2503125 -0.3453125 -0.0634375 0.3053125
## A1:B1:E1 A1:C1:E1 B1:C1:E1 A1:D1:E1 B1:D1:E1
## 1.1853125 -0.2590625 0.1709375 0.9015625 -0.0396875
## C1:D1:E1 A1:B1:C1:D1 A1:B1:C1:E1 A1:B1:D1:E1 A1:C1:D1:E1
## 0.3959375 -0.0740625 -0.1846875 0.4071875 0.1278125
## B1:C1:D1:E1 A1:B1:C1:D1:E1
## -0.0746875 -0.3553125from above plot we can see that apart from C many of the others are significant
dat10<-data.frame(A1,B1,D1,E1,Observation5)
model12<-aov(Observation5~A1+B1+D1+E1+A1*B1+B1*E1+D1*E1+A1*D1+A1*E1+A1*B1*E1+A1*D1*E1,data = dat10)
plot(model12)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
### From the above plot we can say that the model is adequate but because there is change in vriance our Assumption is wrong
C)
model13<-aov(Observation5~A1*B1*D1*E1,data = dat10)
halfnormal(model13)##
## Significant effects (alpha=0.05, Lenth method):## [1] D1 E1 A1:D1 A1 D1:E1 B1:E1 A1:B1 A1:B1:E1
##
## [9] A1:E1 A1:D1:E1 e10
coef(model13)## (Intercept) A1 B1 D1 E1 A1:B1
## 10.1803125 1.6159375 0.0434375 2.9884375 2.1878125 1.2365625
## A1:D1 B1:D1 A1:E1 B1:E1 D1:E1 A1:B1:D1
## 1.6665625 -0.0134375 1.0271875 1.2834375 1.3896875 -0.3453125
## A1:B1:E1 A1:D1:E1 B1:D1:E1 A1:B1:D1:E1
## 1.1853125 0.9015625 -0.0396875 0.4071875summary(model13)## Df Sum Sq Mean Sq F value Pr(>F)
## A1 1 83.56 83.56 57.233 1.14e-06 ***
## B1 1 0.06 0.06 0.041 0.841418
## D1 1 285.78 285.78 195.742 2.16e-10 ***
## E1 1 153.17 153.17 104.910 1.97e-08 ***
## A1:B1 1 48.93 48.93 33.514 2.77e-05 ***
## A1:D1 1 88.88 88.88 60.875 7.66e-07 ***
## B1:D1 1 0.01 0.01 0.004 0.950618
## A1:E1 1 33.76 33.76 23.126 0.000193 ***
## B1:E1 1 52.71 52.71 36.103 1.82e-05 ***
## D1:E1 1 61.80 61.80 42.328 7.24e-06 ***
## A1:B1:D1 1 3.82 3.82 2.613 0.125501
## A1:B1:E1 1 44.96 44.96 30.794 4.40e-05 ***
## A1:D1:E1 1 26.01 26.01 17.815 0.000650 ***
## B1:D1:E1 1 0.05 0.05 0.035 0.854935
## A1:B1:D1:E1 1 5.31 5.31 3.634 0.074735 .
## Residuals 16 23.36 1.46
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Factor C does not seem to be important
After removing factorc the design remains 2^4 design
We can see that there is no change in significant factors
model14<-lm(Observation5~A1+B1+D1+E1, data = dat10)
coef(model14)## (Intercept) A1 B1 D1 E1
## 10.1803125 1.6159375 0.0434375 2.9884375 2.1878125