HW Week 11
Md Ariful Haque Miah, Ayodegi
11/13/2022
Problem 6.8
library(DoE.base)## Warning: package 'DoE.base' was built under R version 4.1.3## 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':
##
## lengthstime<-c(rep(-1, 12),rep(1, 12) )
cul_med<-rep(c(rep(-1, 2), rep(1, 2)),6)
obs <- 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.factor(time)
cul_med <- as.factor(cul_med)
dat <- data.frame(time,cul_med,obs)
datmodel <- aov(obs~time*cul_med,data=dat)
summary(model) ## Df Sum Sq Mean Sq F value Pr(>F)
## time 1 590.0 590.0 115.506 9.29e-10 ***
## cul_med 1 9.4 9.4 1.835 0.190617
## time:cul_med 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 ' ' 1plot(model)
Hypothesis Test:
\(H_0: \alpha_i = 0\)
\(H_a: \alpha_i \neq 0\)
\(H_0: \beta_j\ = 0\)
\(H_a: \beta_j \neq 0\)
\(H_0: \alpha\beta_{ij}=0\)
\(H_a: \alpha\beta_{ij}\neq0\)
Only the factors Time and the interaction effects (Time:Culture Medium) are significant as the p value is less than \(\alpha=0.001\) level of significance so for these factors we reject \(H_0\) and the factor Culture medium is not significant as the p value is larger than \(\alpha=0.05\) level of significance so for this factors we fail to reject \(H_0\). Also Residual vs Fitted plot shows that the variance is constant and the errors are normally distributed and thus the model is adequate.
Problem 6.12 (a) Estimate the factor effects
A <- c(rep(-1,4),rep(1,4))
B <- c(rep(-1,8),rep(1,8))
obs <- 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)
A <- as.factor(A)
B <- as.factor(B)
dat <- data.frame(A,B,obs)
datFinding Corner Points:
one <- sum(dat$obs[1:4])
one## [1] 58.081a <- sum(dat$obs[5:8])
a## [1] 55.686b <- sum(dat$obs[9:12])
b## [1] 59.299ab <- sum(dat$obs[13:16])
ab## [1] 59.156n <- c(4)
fact_A <- 2*(a+ab-b-one)/(4*n)
fact_A## [1] -0.31725fact_B <- 2*(b+ab-a-one)/(4*n)
fact_B## [1] 0.586fact_AB <- 2*(ab+one-a-b)/(4*n)
fact_AB## [1] 0.2815Therefore, The factor Effect A = -0.31725, B = 0.586, Interaction effect AB = 0.2815
- Analysis of variance
A <- c(rep(-1,4),rep(1,4))
B <- c(rep(-1,8),rep(1,8))
obs <- 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)
A <- as.factor(A)
B <- as.factor(B)
dat <- data.frame(A,B,obs)
datmodel <- aov(obs~A*B,data=dat)
summary(model)## 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 ' ' 1Hypothesis Test: \(H_0: \alpha_i = 0\)
\(H_a: \alpha_i \neq 0\)
\(H_0: \beta_j\ = 0\)
\(H_a: \beta_j \neq 0\)
\(H_0: \alpha\beta_{ij}=0\)
\(H_a: \alpha\beta_{ij}\neq0\)
p values of the main effect A and the interaction effect (A:B) is greater than \(\alpha = 0.05\) level of significance. Hence the main effect A and the interaction effect (A:B) are not significant as we fail to reject \(H_0\). On the other hand p value of the main effect B is less than \(\alpha = 0.1\) level of significance. Hence the main effect B is significant as we reject \(H_0\).
- Regression Equation:
A <- c(rep(-1,4),rep(1,4))
B <- c(rep(-1,8),rep(1,8))
obs <- 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)
A <- as.factor(A)
B <- as.factor(B)
dat <- data.frame(A,B,obs)
datmodel <- lm(obs~A+B+A*B,data=dat)
summary(model)##
## Call:
## lm.default(formula = obs ~ A + B + A * B, data = dat)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.61325 -0.14431 -0.00563 0.10188 1.64475
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 14.5203 0.2824 51.414 1.93e-15 ***
## A1 -0.5988 0.3994 -1.499 0.160
## B1 0.3045 0.3994 0.762 0.461
## A1:B1 0.5630 0.5648 0.997 0.339
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5648 on 12 degrees of freedom
## Multiple R-squared: 0.3535, Adjusted R-squared: 0.1918
## F-statistic: 2.187 on 3 and 12 DF, p-value: 0.1425y = \(\beta_0 + \beta_2 *x_2 + \epsilon\)
Hence, y = \(14.5203 + 0.3045 * x_2 + \epsilon\)
where \(\beta_0\) = the intercept = the grand mean of all 16 observations and the regression coefficient \(\beta_2\) is one-half the corresponding factor effect B estimate.
- The Residuals Analysis
A <- c(rep(-1,4),rep(1,4))
B <- c(rep(-1,8),rep(1,8))
obs <- 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)
A <- as.factor(A)
B <- as.factor(B)
dat <- data.frame(A,B,obs)
datmodel <- aov(obs~A*B,data=dat)
summary(model)## 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 ' ' 1plot(model)
There is a significant concern in the Residuals vs Fitted Plot, which shows that the variance is not constant. Hence the model is not adequate. Though the normality assumptions are fairly satisfied in the Normal Q-Q plot.
- The potential Outlier: Only the observation no. 2 might be the potential outlier according to the \(\sqrt\)(standardized residual) plot in part (d). Since There is 15 more observations left so only one outlier won’t cause any effect to grow an epitaxial layer on polished silicon wafers. Also,the interaction effects are not significant, we can choose any main effect to maximize the yield of the response vairable (epitaxal layer thickness).
Problem 6.21 (a)
A <- c(rep(-1,7),rep(1,7))
B <- c(rep(-1,14),rep(1,14))
C <- c(rep(-1,28),rep(1,28))
D <- c(rep(-1,56),rep(1,56))
obs <- 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)
A <- as.factor(A)
B <- as.factor(B)
C <- as.factor(C)
D <- as.factor(D)
dat <- data.frame(A,B,C,D,obs)
datmodel <- aov(obs~A*B*C*D,data=dat)
summary(model)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 917 917.1 10.588 0.00157 **
## B 1 388 388.1 4.481 0.03686 *
## C 1 145 145.1 1.676 0.19862
## D 1 1 1.4 0.016 0.89928
## A:B 1 219 218.7 2.525 0.11538
## A:C 1 12 11.9 0.137 0.71178
## B:C 1 115 115.0 1.328 0.25205
## A:D 1 94 93.8 1.083 0.30066
## B:D 1 56 56.4 0.651 0.42159
## C:D 1 2 1.6 0.019 0.89127
## A:B:C 1 7 7.3 0.084 0.77294
## A:B:D 1 113 113.0 1.305 0.25623
## A:C:D 1 39 39.5 0.456 0.50121
## B:C:D 1 34 33.8 0.390 0.53386
## A:B:C:D 1 96 95.6 1.104 0.29599
## Residuals 96 8316 86.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Hypothesis Test:
\(H_0: \alpha_i = 0\)
\(H_a: \alpha_i \neq 0\)
\(H_0: \beta_j\ = 0\)
\(H_a: \beta_j \neq 0\)
\(H_0: \gamma_k\ = 0\)
\(H_a: \gamma_k \neq 0\)
\(H_0: \delta_l\ = 0\)
\(H_a: \delta_l \neq 0\)
\(H_0: \alpha\beta_{ij}=0\)
\(H_a: \alpha\beta_{ij}\neq0\)
\(H_0: \alpha\gamma_{ik}=0\)
\(H_a: \alpha\gamma_{ik}\neq0\)
\(H_0: \alpha\delta_{il}=0\)
\(H_a: \alpha\delta_{il}\neq0\)
…..
…..
…..
\(H_0: \alpha\beta\gamma\delta_{ijkl}=0\)
\(H_a: \alpha\beta\gamma\delta_{ijkl}\neq0\)
From the anova model summary and analyzing the p value we see that None of the interaction terms are significant but the main effect factors “length of putt” and “the type of putter” are significant and these factors significantly affect putting performance.
Problem 6.21 (b)
plot(model)
Residual vs fitted plot shows that the variance is not constant and hence the model is not adequate though the errors are assumed to be normally distributed from the normal Q-Q plot.
QUESTION 6.36
Getting in data
library(DoE.base)
A<- c(-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1)
B<- c(-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1)
C<- c(-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,1,1,1,1)
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)
dat<- data.frame(A,B,C,D,obs)6.36) A
model<-lm(obs~A*B*C*D,data=dat)
coef(model)## (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(model)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B A:B:C
From the Half-normal plot,we can see that Factors A,B, A:B , and A:B:C are significant at alpha=0.5
summary(model)##
## Call:
## lm.default(formula = obs ~ A * B * C * D, data = dat)
##
## 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
## A 3.16062 NaN NaN NaN
## B -1.50187 NaN NaN NaN
## C -0.22062 NaN NaN NaN
## D -0.07937 NaN NaN NaN
## A:B -1.06938 NaN NaN NaN
## A:C -0.29812 NaN NaN NaN
## B:C 0.22937 NaN NaN NaN
## A:D -0.05687 NaN NaN NaN
## B:D -0.04688 NaN NaN NaN
## C:D 0.02937 NaN NaN NaN
## A:B:C 0.34437 NaN NaN NaN
## A:B:D -0.09688 NaN NaN NaN
## A:C:D -0.01063 NaN NaN NaN
## B:C:D 0.09438 NaN NaN NaN
## A:B:C: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: NATo select our tentative model, we would consider the significant factors from our half-normal plots,which are factors A,B,A:B,A:B:C.
So therefore our tentative model can be written as
\(Y_{i,j,k,l}=4.68062+3.16062\alpha _{i}-1.50187\beta_{j}-1.06938\alpha\beta_{ij}+0.34437\alpha\beta\gamma+\epsilon_{ijkl}\)
6.36) B
model2<- aov(obs~A+B+C+A*B+A*B*C,data =dat)
summary(model2)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 159.83 159.83 1563.061 1.84e-10 ***
## B 1 36.09 36.09 352.937 6.66e-08 ***
## C 1 0.78 0.78 7.616 0.02468 *
## A:B 1 18.30 18.30 178.933 9.33e-07 ***
## A:C 1 1.42 1.42 13.907 0.00579 **
## B:C 1 0.84 0.84 8.232 0.02085 *
## A:B: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(model2)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
From the Normal Q-Q plot, we can’t assume Normality in the data since the data points in the plot doesn’t appears to fall on a straight line
Also, from the Residuals vs fitted plot, we can’t assume constant variance since all the residuals have varying spread, depleting that the variances are not equal.
6.36) C
Using a Log transformation on our data below
logobs <- log(obs)
dat2 <- data.frame(A,B,C,D,logobs)
model3<- lm(logobs~A*B*C*D,data = dat2)
coef(model3)## (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(model3)##
## Significant effects (alpha=0.05, Lenth method):## [1] A B A:B:C
summary(model3)##
## Call:
## lm.default(formula = logobs ~ A * B * C * D, data = dat2)
##
## Residuals:
## ALL 16 residuals are 0: no residual degrees of freedom!
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.185417 NaN NaN NaN
## A 0.812870 NaN NaN NaN
## B -0.314278 NaN NaN NaN
## C -0.006409 NaN NaN NaN
## D -0.018077 NaN NaN NaN
## A:B -0.024685 NaN NaN NaN
## A:C -0.039724 NaN NaN NaN
## B:C -0.004226 NaN NaN NaN
## A:D -0.009578 NaN NaN NaN
## B:D 0.003709 NaN NaN NaN
## C:D 0.017780 NaN NaN NaN
## A:B:C 0.063434 NaN NaN NaN
## A:B:D -0.029876 NaN NaN NaN
## A:C:D -0.003740 NaN NaN NaN
## B:C:D 0.003766 NaN NaN NaN
## A:B:C:D 0.031322 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: NAmodel4<-aov(logobs~A+B+C+A*B*C,data=dat2)
summary(model4)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 10.572 10.572 1994.556 6.98e-11 ***
## B 1 1.580 1.580 298.147 1.29e-07 ***
## C 1 0.001 0.001 0.124 0.73386
## A:B 1 0.010 0.010 1.839 0.21207
## A:C 1 0.025 0.025 4.763 0.06063 .
## B:C 1 0.000 0.000 0.054 0.82223
## A:B:C 1 0.064 0.064 12.147 0.00826 **
## Residuals 8 0.042 0.005
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(model4)
After the Log transformation, we can observe that from the half-normal plots that only factors A, B, and A:B:C are significant and interactions of A:B seems insignificant as compared to pre-log transformation.
The confirmation of the half normal plot was double checked using ANOVA analysis that showed factors A,B, and A:B:C to be significant as shown above.
Based on the residuals plots, after the log transformation was performed we observed that
From the Normal Q-Q plot, we can not assume Normality in the data since the data points in the plot does-not appear to fall on a straight line
Also, from the Residuals vs fitted plot, we cannot assume constant variance since all the residuals have varying spread, depleting that the variances are not equal.
6.36) D
Fitting a model of the coded variables we have
\(Y_{ijkl}=1.185417+0.812870\alpha _{i}-0.314278\beta_{j}+0.063434\alpha\beta\gamma+\epsilon_{{ijkl}}\)
QUESTION 6.39
Getting in the data we have
library(DoE.base)
A<-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)
B<-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)
C<-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)
D<-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)
E<- 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)
obs<- 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)
dat<- data.frame(A,B,C,D,E,obs)6.39) A
model<-lm(obs~A*B*C*D*E,data=dat)
coef(model)## (Intercept) A B C D E
## 10.1803125 1.6159375 0.0434375 -0.0121875 2.9884375 2.1878125
## A:B A:C B:C A:D B:D C:D
## 1.2365625 -0.0015625 -0.1953125 1.6665625 -0.0134375 0.0034375
## A:E B:E C:E D:E A:B:C A:B:D
## 1.0271875 1.2834375 0.3015625 1.3896875 0.2503125 -0.3453125
## A:C:D B:C:D A:B:E A:C:E B:C:E A:D:E
## -0.0634375 0.3053125 1.1853125 -0.2590625 0.1709375 0.9015625
## B:D:E C:D:E A:B:C:D A:B:C:E A:B:D:E A:C:D:E
## -0.0396875 0.3959375 -0.0740625 -0.1846875 0.4071875 0.1278125
## B:C:D:E A:B:C:D:E
## -0.0746875 -0.3553125halfnormal(model)##
## Significant effects (alpha=0.05, Lenth method):## [1] D E A:D A D:E B:E A:B A:B:E A:E A:D:E
summary(model)##
## Call:
## lm.default(formula = obs ~ A * B * C * D * E, data = dat)
##
## 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
## A 1.615938 NaN NaN NaN
## B 0.043438 NaN NaN NaN
## C -0.012187 NaN NaN NaN
## D 2.988437 NaN NaN NaN
## E 2.187813 NaN NaN NaN
## A:B 1.236562 NaN NaN NaN
## A:C -0.001563 NaN NaN NaN
## B:C -0.195313 NaN NaN NaN
## A:D 1.666563 NaN NaN NaN
## B:D -0.013438 NaN NaN NaN
## C:D 0.003437 NaN NaN NaN
## A:E 1.027188 NaN NaN NaN
## B:E 1.283437 NaN NaN NaN
## C:E 0.301563 NaN NaN NaN
## D:E 1.389687 NaN NaN NaN
## A:B:C 0.250313 NaN NaN NaN
## A:B:D -0.345312 NaN NaN NaN
## A:C:D -0.063437 NaN NaN NaN
## B:C:D 0.305312 NaN NaN NaN
## A:B:E 1.185313 NaN NaN NaN
## A:C:E -0.259062 NaN NaN NaN
## B:C:E 0.170938 NaN NaN NaN
## A:D:E 0.901563 NaN NaN NaN
## B:D:E -0.039687 NaN NaN NaN
## C:D:E 0.395938 NaN NaN NaN
## A:B:C:D -0.074063 NaN NaN NaN
## A:B:C:E -0.184688 NaN NaN NaN
## A:B:D:E 0.407187 NaN NaN NaN
## A:C:D:E 0.127812 NaN NaN NaN
## B:C:D:E -0.074688 NaN NaN NaN
## A:B:C:D:E -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: NAsummary(model)##
## Call:
## lm.default(formula = obs ~ A * B * C * D * E, data = dat)
##
## 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
## A 1.615938 NaN NaN NaN
## B 0.043438 NaN NaN NaN
## C -0.012187 NaN NaN NaN
## D 2.988437 NaN NaN NaN
## E 2.187813 NaN NaN NaN
## A:B 1.236562 NaN NaN NaN
## A:C -0.001563 NaN NaN NaN
## B:C -0.195313 NaN NaN NaN
## A:D 1.666563 NaN NaN NaN
## B:D -0.013438 NaN NaN NaN
## C:D 0.003437 NaN NaN NaN
## A:E 1.027188 NaN NaN NaN
## B:E 1.283437 NaN NaN NaN
## C:E 0.301563 NaN NaN NaN
## D:E 1.389687 NaN NaN NaN
## A:B:C 0.250313 NaN NaN NaN
## A:B:D -0.345312 NaN NaN NaN
## A:C:D -0.063437 NaN NaN NaN
## B:C:D 0.305312 NaN NaN NaN
## A:B:E 1.185313 NaN NaN NaN
## A:C:E -0.259062 NaN NaN NaN
## B:C:E 0.170938 NaN NaN NaN
## A:D:E 0.901563 NaN NaN NaN
## B:D:E -0.039687 NaN NaN NaN
## C:D:E 0.395938 NaN NaN NaN
## A:B:C:D -0.074063 NaN NaN NaN
## A:B:C:E -0.184688 NaN NaN NaN
## A:B:D:E 0.407187 NaN NaN NaN
## A:C:D:E 0.127812 NaN NaN NaN
## B:C:D:E -0.074688 NaN NaN NaN
## A:B:C:D:E -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: NAmodel2<- aov(obs~A+B+D+E+A*B+A*D+A*E+B*E+D*E+A*B*E+A*D*E,data=dat)
summary(model2)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 83.56 83.56 51.362 6.10e-07 ***
## B 1 0.06 0.06 0.037 0.849178
## D 1 285.78 285.78 175.664 2.30e-11 ***
## E 1 153.17 153.17 94.149 5.24e-09 ***
## A:B 1 48.93 48.93 30.076 2.28e-05 ***
## A:D 1 88.88 88.88 54.631 3.87e-07 ***
## A:E 1 33.76 33.76 20.754 0.000192 ***
## B:E 1 52.71 52.71 32.400 1.43e-05 ***
## D:E 1 61.80 61.80 37.986 5.07e-06 ***
## A:B:E 1 44.96 44.96 27.635 3.82e-05 ***
## A:D:E 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 ' ' 1We can observe that from the half-normal plots that A,D,E,A:D,D:E,B:E,A:B,A:E,A:B:E,A:D:E are significant
The confirmation of the half normal plot was double checked using ANOVA analysis that showed that the above factors mentioned were significant
6.39) B
plot(model2)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
From the Normal Q-Q plot, we can’t assume Normality in the data since the data points in the plot doesn’t appears to fall on a straight line
Also, from the Residuals vs fitted plot, we can’t assume constant variance since all the residuals have varying spread, depleting that the variances are not equal.
6.39) C
A<- 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)
B<- 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)
D<- 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)
E<- 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)
obs<- 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)
dat<- data.frame(A,B,D,E,obs)model4<- lm(obs~A*B*D*E,data =dat)
coef(model4)## (Intercept) A B D E A:B
## 10.1803125 1.6159375 0.0434375 2.9884375 2.1878125 1.2365625
## A:D B:D A:E B:E D:E A:B:D
## 1.6665625 -0.0134375 1.0271875 1.2834375 1.3896875 -0.3453125
## A:B:E A:D:E B:D:E A:B:D:E
## 1.1853125 0.9015625 -0.0396875 0.4071875halfnormal(model4)##
## Significant effects (alpha=0.05, Lenth method):## [1] D E A:D A D:E B:E A:B A:B:E A:E A:D:E e10
summary(model4)##
## Call:
## lm.default(formula = obs ~ A * B * D * E, data = dat)
##
## 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 ***
## A 1.61594 0.21360 7.565 1.14e-06 ***
## B 0.04344 0.21360 0.203 0.841418
## D 2.98844 0.21360 13.991 2.16e-10 ***
## E 2.18781 0.21360 10.243 1.97e-08 ***
## A:B 1.23656 0.21360 5.789 2.77e-05 ***
## A:D 1.66656 0.21360 7.802 7.66e-07 ***
## B:D -0.01344 0.21360 -0.063 0.950618
## A:E 1.02719 0.21360 4.809 0.000193 ***
## B:E 1.28344 0.21360 6.009 1.82e-05 ***
## D:E 1.38969 0.21360 6.506 7.24e-06 ***
## A:B:D -0.34531 0.21360 -1.617 0.125501
## A:B:E 1.18531 0.21360 5.549 4.40e-05 ***
## A:D:E 0.90156 0.21360 4.221 0.000650 ***
## B:D:E -0.03969 0.21360 -0.186 0.854935
## A:B:D:E 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-10model5<- aov(obs~A+B+D+E+A*B+A*D+A*E+B*E+D*E+A*B*E+A*D*E,data=dat)
summary(model5)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 83.56 83.56 51.362 6.10e-07 ***
## B 1 0.06 0.06 0.037 0.849178
## D 1 285.78 285.78 175.664 2.30e-11 ***
## E 1 153.17 153.17 94.149 5.24e-09 ***
## A:B 1 48.93 48.93 30.076 2.28e-05 ***
## A:D 1 88.88 88.88 54.631 3.87e-07 ***
## A:E 1 33.76 33.76 20.754 0.000192 ***
## B:E 1 52.71 52.71 32.400 1.43e-05 ***
## D:E 1 61.80 61.80 37.986 5.07e-06 ***
## A:B:E 1 44.96 44.96 27.635 3.82e-05 ***
## A:D:E 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(model5)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
Factor C was dropped totally from our model initially because it seemed insignificant. After dropping factor C we realized that from the half-normal plots that A,D,E,A:D,D:E,B:E,A:B,A:E,A:B:E,A:D:E are significant
The confirmation of the half normal plot was double checked using ANOVA analysis that showed that the above factors mentioned were significant
Which was similar to the results gotten when factor C was included.
6.39) D
Fitting our model in a linear equation we can see that
\(Y_{ijkl}=10.18031+1.61594\alpha _{i}+0.04344\beta_{j}+2.98844\gamma_{k}+1.66656\alpha\gamma+1.38969\gamma\delta+1.28344\beta\delta+1.23656\alpha\beta+1.02719\alpha\delta+1.18531\alpha\beta\delta+0.90156\alpha\gamma\delta\)
To maximize the predicted response, the above linear model should be follows
Note
Because all the factors are of positive coefficient, therefore they should be at a +1 level to produce maximum response.