Homework 11
Imtiaz, Jahir
2022-11-13
Required Libraries:
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':
##
## lengthslibrary(tidyr)## Warning: package 'tidyr' was built under R version 4.1.3Answer to question no 6.8
Entering the data:
time <- c(rep(-1,12), rep(1,12))
culture <- rep(c(rep(-1,2), rep(1,2)), 6)
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)Now, we analyze the model:
model eqn yijk = ai+bj+abij +eijk
where ai = factor 1 effect
bj = factor 2 effect
abij = 2 way interaction effect
eijk = error
Null hypothesis, Ho: ai = o
bj = 0
abij = 0
Alterantive hypothesis: Ha: ai≠0
bj ≠0
abij ≠0
Our ANOVA analysis is shown as follows:
anova6.8 <- aov(observation~time+culture+time*culture)
(lm((observation~time+culture+time*culture)))##
## Call:
## lm.default(formula = (observation ~ time + culture + time * culture))
##
## Coefficients:
## (Intercept) time culture time:culture
## 29.625 4.958 -0.625 -1.958summary(anova6.8)## Df Sum Sq Mean Sq F value Pr(>F)
## time 1 590.0 590.0 115.506 9.29e-10 ***
## culture 1 9.4 9.4 1.835 0.190617
## time:culture 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(anova6.8)
## hat values (leverages) are all = 0.1666667
## and there are no factor predictors; no plot no. 5
halfnormal(anova6.8)## Warning in halfnormal.lm(anova6.8): halfnormal not recommended for models with
## more residual df than model df##
## Significant effects (alpha=0.05, Lenth method):## [1] time time:culture e8
From our analysis of variance, we observe that the main effect Time and the interaction between time and culture has a significant effect for an alpha level of 0.05. Also from the half normal plot we do see that they are the outlier points indicating that they are significant parameters for the model.
From the residual plots, we see that the residuals have equal spread indicating the consideration of constant variance good enough. Also the residuals seem to have a fairly normal distribution. Hence, we consider that our model is sufficiently adequate.
The co-efficient of the estimates are also shown in the result boxes.
Answer to question no 6.12
Entering the data:
A <- c(-1,1,-1,1)
B <- c(-1,-1,1,1)
Ax <- as.factor(A)
Bx <- as.factor(B)
Ax <- rep(Ax, rep(4,4))
Bx <- rep(Bx, rep(4,4))
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)Part(a)
The effects are determined using the following formula:
effect = (contrast)*2/2^2(n)
Where, k = number of factors, n = replicates for each settings
The results are summarized in the table below:
summary(lm(obs~Ax*Bx))##
## Call:
## lm.default(formula = obs ~ Ax * Bx)
##
## 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 ***
## Ax1 -0.5988 0.3994 -1.499 0.160
## Bx1 0.3045 0.3994 0.762 0.461
## Ax1:Bx1 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: 0.1425The estimates of A,B, and interaction terms are -0.59, 0.30, and 0.56.
Part(b)
model eqn yijk = ai+bj+abij +eijk
where ai = factor 1 effect
bj = factor 2 effect
abij = 2 way interaction effect
eijk = error
Null hypothesis, Ho: ai = o
bj = 0
abij = 0
Alterantive hypothesis: Ha: ai≠0
bj ≠0
abij ≠0
The analysis of variance is performed as follows:
anova6.12 <- aov(obs~Ax*Bx)
summary(anova6.12)## Df Sum Sq Mean Sq F value Pr(>F)
## Ax 1 0.403 0.4026 1.262 0.2833
## Bx 1 1.374 1.3736 4.305 0.0602 .
## Ax:Bx 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 ' ' 1From the ANOVA table, we observe that none of the main effects or the two-way interaction effect has any significant effect on the response variable. So, we fail to reject the null hypothesis for all the main effects and the interaction term.
part(c)
summary(lm(obs~Ax*Bx))##
## Call:
## lm.default(formula = obs ~ Ax * Bx)
##
## 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 ***
## Ax1 -0.5988 0.3994 -1.499 0.160
## Bx1 0.3045 0.3994 0.762 0.461
## Ax1:Bx1 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: 0.1425We can write our regression eqn as follows:
y = 14.52 - 0.59a + 0.304b + 0.563ab
part(d)
plot(anova6.12)
From our observation of the residuals plot, we see that the residuals do not show equal variance width in the fitted vs residual plots. Thus, the consideration of equal variance is violated. We can conclude that the model is inadequate.
y = 14.52 - 0.59a + 0.304b + 0.563ab
part(e)
Since the interaction effects are not significant, we can choose any main effect settings as per our convenience to maximize our yield of the response vairable which is the epitaxal layer thickness.
Answer to question no 6.21
Entering the data:
putt <- read.csv(file.choose())
View(putt)
putt$Length <- as.factor(putt$Length)
putt$Type <- as.factor(putt$Type)
putt$Break <- as.factor(putt$Break)
putt$Slope <- as.factor(putt$Slope)
pl <-rep(putt$Length, rep(7,16))
pt <- rep(putt$Type, rep(7,16))
pb <- rep(putt$Break, rep(7,16))
ps <- rep(putt$Slope, rep(7,16))
puttobs <- as.vector(t(putt[,c(5:11)]))
str(puttobs)## num [1:112] 10 18 14 12.5 19 16 18.5 0 16.5 4.5 ...Part(a)
We perform ANOVA for the 2^4 design with replication as follows:
anova6.21 <- aov(puttobs~pl*pt*pb*ps)
summary(anova6.21)## Df Sum Sq Mean Sq F value Pr(>F)
## pl 1 917 917.1 10.588 0.00157 **
## pt 1 388 388.1 4.481 0.03686 *
## pb 1 145 145.1 1.676 0.19862
## ps 1 1 1.4 0.016 0.89928
## pl:pt 1 219 218.7 2.525 0.11538
## pl:pb 1 12 11.9 0.137 0.71178
## pt:pb 1 115 115.0 1.328 0.25205
## pl:ps 1 94 93.8 1.083 0.30066
## pt:ps 1 56 56.4 0.651 0.42159
## pb:ps 1 2 1.6 0.019 0.89127
## pl:pt:pb 1 7 7.3 0.084 0.77294
## pl:pt:ps 1 113 113.0 1.305 0.25623
## pl:pb:ps 1 39 39.5 0.456 0.50121
## pt:pb:ps 1 34 33.8 0.390 0.53386
## pl:pt:pb:ps 1 96 95.6 1.104 0.29599
## Residuals 96 8316 86.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1From the results of our ANOVA analysis, we see that only the effect of Putt Length and Put type have significant main effects. Rest of all the factors or interactions does not seem to have any significant effect on the response variable.
Part(b)
plot(anova6.21)
From the residual vs fitted plots we can see that the quartile ranges have fluctuating values/ widths as we move along the fitted values. We can say the model is inadequate since the consideration of constant variance-which is a fundamental assumption ANOVA is violated.
Hence, the model is inadequate.
Answer to question no- 6.36
Entering data:
resA <- rep(c(-1,1),8)
resB <- rep(c(-1,-1,1,1),4)
resC <- rep(c(-1,-1,-1,-1,1,1,1,1),2)
resD <- c(rep(-1,8), rep(1,8))
resob <- 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)Part(a)
The factor effects are visualized in the half normal plot as follows:
anova6.35 <- aov(resob~resA*resB*resC*resD)
summary(lm(resob~resA*resB*resC*resD))##
## Call:
## lm.default(formula = resob ~ resA * resB * resC * resD)
##
## 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
## resA 3.16062 NaN NaN NaN
## resB -1.50187 NaN NaN NaN
## resC -0.22062 NaN NaN NaN
## resD -0.07937 NaN NaN NaN
## resA:resB -1.06938 NaN NaN NaN
## resA:resC -0.29812 NaN NaN NaN
## resB:resC 0.22937 NaN NaN NaN
## resA:resD -0.05687 NaN NaN NaN
## resB:resD -0.04688 NaN NaN NaN
## resC:resD 0.02937 NaN NaN NaN
## resA:resB:resC 0.34437 NaN NaN NaN
## resA:resB:resD -0.09688 NaN NaN NaN
## resA:resC:resD -0.01063 NaN NaN NaN
## resB:resC:resD 0.09438 NaN NaN NaN
## resA:resB:resC:resD 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: NAhalfnormal(anova6.35)##
## Significant effects (alpha=0.05, Lenth method):## [1] resA resB resA:resB resA:resB:resC
The factor effect estimates are shown in the table above.
From the half normal plot we see that the main effects A and B are significant. There is a significant 2-way interaction between factor A and B. Also a significant 3-way interaction between factor A,B, and C.
Based on the half-normal plot, our tentative model is:
y = A+B+A*B+A*B*C+e
Part(b)
Fitting the model identified in part(a) as follows:
anova6.35new <- aov(resob~resA+resB+resA*resB+resA*resB*resC)
summary(anova6.35new)## Df Sum Sq Mean Sq F value Pr(>F)
## resA 1 159.83 159.83 1563.061 1.84e-10 ***
## resB 1 36.09 36.09 352.937 6.66e-08 ***
## resC 1 0.78 0.78 7.616 0.02468 *
## resA:resB 1 18.30 18.30 178.933 9.33e-07 ***
## resA:resC 1 1.42 1.42 13.907 0.00579 **
## resB:resC 1 0.84 0.84 8.232 0.02085 *
## resA:resB:resC 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(anova6.35new)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
From the analysis of the residuals, we see that the consideration of constant variance has been violated as seen in the residuals vs fitted plot. So, we can say that our strong assumption of ANOVA has been violated. Hence, the model is inadequate.
Part(c)
We perform log transform on the response data and check the plots and summary again:
resoblog <- log(resob)
anova6.35new1 <- aov(resoblog~resA*resB*resC*resD)
halfnormal(anova6.35new1)##
## Significant effects (alpha=0.05, Lenth method):## [1] resA resB resA:resB:resC
After log-transformation, we see that the two-way interaction between A and B has turned out to be insignificant.
anova6.35new2 <- aov(resoblog~resA+resB+resA*resB*resC)
summary(anova6.35new2)## Df Sum Sq Mean Sq F value Pr(>F)
## resA 1 10.572 10.572 1994.556 6.98e-11 ***
## resB 1 1.580 1.580 298.147 1.29e-07 ***
## resC 1 0.001 0.001 0.124 0.73386
## resA:resB 1 0.010 0.010 1.839 0.21207
## resA:resC 1 0.025 0.025 4.763 0.06063 .
## resB:resC 1 0.000 0.000 0.054 0.82223
## resA:resB:resC 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(anova6.35new2)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
Apparently, the transformation has not changed the residuals vs fitted plots to make the model seem any more adequate than the last model. Hence, the log transformation did not help much.
part(d)
The fitted model is obtained from the estimates of the linear regression model as follows:
anova6.35new2 <- lm(resoblog~resA+resB+resA*resB+resA*resB*resC)
summary(anova6.35new2)##
## Call:
## lm.default(formula = resoblog ~ resA + resB + resA * resB + resA *
## resB * resC)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.1030 -0.0203 0.0000 0.0203 0.1030
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.185417 0.018201 65.129 3.44e-12 ***
## resA 0.812870 0.018201 44.660 6.98e-11 ***
## resB -0.314278 0.018201 -17.267 1.29e-07 ***
## resC -0.006409 0.018201 -0.352 0.73386
## resA:resB -0.024685 0.018201 -1.356 0.21207
## resA:resC -0.039724 0.018201 -2.182 0.06063 .
## resB:resC -0.004226 0.018201 -0.232 0.82223
## resA:resB:resC 0.063434 0.018201 3.485 0.00826 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.0728 on 8 degrees of freedom
## Multiple R-squared: 0.9966, Adjusted R-squared: 0.9935
## F-statistic: 330.2 on 7 and 8 DF, p-value: 3.296e-09The fitted equation then:
y = 1.19 + 0.81A - 0.31B - 0.006C - 0.024AB - 0.039AC - 0.004BC + 0.06 ABC
Answer to question no- 6.39
Entering the data:
relA <- rep(c(-1,1),16)
relB <- rep(c(-1,-1,1,1),8)
relC <- rep(c(-1,-1,-1,-1,1,1,1,1),4)
relD <- rep(c(rep(-1,8), rep(1,8)),2)
relE <- c(rep(-1,16), rep(1,16))
relobs <- 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)Part(a)
The ANOVA results and the significant main and interaction effects are identified in the following sections:
anova6.39 <- aov(relobs~relA*relB*relC*relD*relE)
halfnormal(anova6.39)##
## Significant effects (alpha=0.05, Lenth method):## [1] relD relE relA:relD relA relD:relE
##
## [6] relB:relE relA:relB relA:relB:relE relA:relE relA:relD:relE
From the half-normal plot, we see that we have 3 significant main effects A,D,E. We have five 2-way interactions as follows: A*D, D*E, B*E, and A*B, A*E. We have two significant 3-way interaction terms - A*D*E and A*B*E.
Part(b)
The residuals in the new model (containing only the significant terms) are tested as follows:
anova6.39new <- aov(relobs~relD+relE+relA*relD+relA+relD*relE+
relB*relE+relA*relB+relA*relB*relE+relA*relE+
relA*relD*relE)
summary(anova6.39new <- lm(relobs~relD+relE+relA*relD+relA+relD*relE+
relB*relE+relA*relB+relA*relB*relE+relA*relE+
relA*relD*relE))##
## Call:
## lm.default(formula = relobs ~ relD + relE + relA * relD + relA +
## relD * relE + relB * relE + relA * relB + relA * relB * relE +
## relA * relE + relA * relD * relE)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.04875 -0.67375 -0.00687 0.65281 2.25375
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.18031 0.22548 45.150 < 2e-16 ***
## relD 2.98844 0.22548 13.254 2.30e-11 ***
## relE 2.18781 0.22548 9.703 5.24e-09 ***
## relA 1.61594 0.22548 7.167 6.10e-07 ***
## relB 0.04344 0.22548 0.193 0.849178
## relD:relA 1.66656 0.22548 7.391 3.87e-07 ***
## relD:relE 1.38969 0.22548 6.163 5.07e-06 ***
## relE:relB 1.28344 0.22548 5.692 1.43e-05 ***
## relA:relB 1.23656 0.22548 5.484 2.28e-05 ***
## relE:relA 1.02719 0.22548 4.556 0.000192 ***
## relE:relA:relB 1.18531 0.22548 5.257 3.82e-05 ***
## relD:relE:relA 0.90156 0.22548 3.998 0.000706 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.275 on 20 degrees of freedom
## Multiple R-squared: 0.9643, Adjusted R-squared: 0.9447
## F-statistic: 49.15 on 11 and 20 DF, p-value: 5.069e-12summary(anova6.39new)##
## Call:
## lm.default(formula = relobs ~ relD + relE + relA * relD + relA +
## relD * relE + relB * relE + relA * relB + relA * relB * relE +
## relA * relE + relA * relD * relE)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.04875 -0.67375 -0.00687 0.65281 2.25375
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.18031 0.22548 45.150 < 2e-16 ***
## relD 2.98844 0.22548 13.254 2.30e-11 ***
## relE 2.18781 0.22548 9.703 5.24e-09 ***
## relA 1.61594 0.22548 7.167 6.10e-07 ***
## relB 0.04344 0.22548 0.193 0.849178
## relD:relA 1.66656 0.22548 7.391 3.87e-07 ***
## relD:relE 1.38969 0.22548 6.163 5.07e-06 ***
## relE:relB 1.28344 0.22548 5.692 1.43e-05 ***
## relA:relB 1.23656 0.22548 5.484 2.28e-05 ***
## relE:relA 1.02719 0.22548 4.556 0.000192 ***
## relE:relA:relB 1.18531 0.22548 5.257 3.82e-05 ***
## relD:relE:relA 0.90156 0.22548 3.998 0.000706 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.275 on 20 degrees of freedom
## Multiple R-squared: 0.9643, Adjusted R-squared: 0.9447
## F-statistic: 49.15 on 11 and 20 DF, p-value: 5.069e-12plot(anova6.39new)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
From the residual analysis we can conclude that the model is inadequate as the assumption of constant variance seems to be violated as per the residuals vs fitted value plot.
Part(c)
From our initial sets of observations, we see that the factor C does not appear to by itself or in interaction with other main effects as a significant parameter. so we drop it out of the model and check:
anova6.39new1 <- aov(relobs~relA*relB*relD*relE)
plot(anova6.39new1)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
summary(anova6.39new1)## Df Sum Sq Mean Sq F value Pr(>F)
## relA 1 83.56 83.56 57.233 1.14e-06 ***
## relB 1 0.06 0.06 0.041 0.841418
## relD 1 285.78 285.78 195.742 2.16e-10 ***
## relE 1 153.17 153.17 104.910 1.97e-08 ***
## relA:relB 1 48.93 48.93 33.514 2.77e-05 ***
## relA:relD 1 88.88 88.88 60.875 7.66e-07 ***
## relB:relD 1 0.01 0.01 0.004 0.950618
## relA:relE 1 33.76 33.76 23.126 0.000193 ***
## relB:relE 1 52.71 52.71 36.103 1.82e-05 ***
## relD:relE 1 61.80 61.80 42.328 7.24e-06 ***
## relA:relB:relD 1 3.82 3.82 2.613 0.125501
## relA:relB:relE 1 44.96 44.96 30.794 4.40e-05 ***
## relA:relD:relE 1 26.01 26.01 17.815 0.000650 ***
## relB:relD:relE 1 0.05 0.05 0.035 0.854935
## relA:relB:relD:relE 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 ' ' 1We do not observe any change in the model adequacy from the residual plots. Slight changes are observed in the F-statistic of all other main and interaction effects owing to the change in the DOF of the entire model for dropping out factor C.
Other than that, no other sigificant changes are seen when comparing to our initial results in part a.
part(d)
summary(lm(relobs~relD+relE+relA*relD+relA+relD*relE+
relB*relE+relA*relB+relA*relB*relE+relA*relE+
relA*relD*relE))##
## Call:
## lm.default(formula = relobs ~ relD + relE + relA * relD + relA +
## relD * relE + relB * relE + relA * relB + relA * relB * relE +
## relA * relE + relA * relD * relE)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.04875 -0.67375 -0.00687 0.65281 2.25375
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.18031 0.22548 45.150 < 2e-16 ***
## relD 2.98844 0.22548 13.254 2.30e-11 ***
## relE 2.18781 0.22548 9.703 5.24e-09 ***
## relA 1.61594 0.22548 7.167 6.10e-07 ***
## relB 0.04344 0.22548 0.193 0.849178
## relD:relA 1.66656 0.22548 7.391 3.87e-07 ***
## relD:relE 1.38969 0.22548 6.163 5.07e-06 ***
## relE:relB 1.28344 0.22548 5.692 1.43e-05 ***
## relA:relB 1.23656 0.22548 5.484 2.28e-05 ***
## relE:relA 1.02719 0.22548 4.556 0.000192 ***
## relE:relA:relB 1.18531 0.22548 5.257 3.82e-05 ***
## relD:relE:relA 0.90156 0.22548 3.998 0.000706 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.275 on 20 degrees of freedom
## Multiple R-squared: 0.9643, Adjusted R-squared: 0.9447
## F-statistic: 49.15 on 11 and 20 DF, p-value: 5.069e-12Now, the fitted model equation that can be used to maximize the yield is as follows:
Y = 10.18 + 2.98D + 2.18E + 1.61A + 0.04B + 1.67A*D + 1.38 D*E + 1.28B*E + 1.23A*B + 1.02A*E + 1.18A*B*E + 0.90A*D*E
Since all the terms are positive and have significant effect on the response var, we can use this to maximize the yield of the predicted response.
Complete Code Chunk:
#problem 6.8
time <- c(rep(-1,12), rep(1,12))
culture <- rep(c(rep(-1,2), rep(1,2)), 6)
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)
anova6.8 <- aov(observation~time+culture+time*culture)
(lm((observation~time+culture+time*culture)))
summary(anova6.8)
plot(anova6.8)
#problem 6.12
library(dplyr)
A <- c(-1,1,-1,1)
B <- c(-1,-1,1,1)
Ax <- as.factor(A)
Bx <- as.factor(B)
Ax <- rep(Ax, rep(4,4))
Bx <- rep(Bx, rep(4,4))
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)
anova6.12 <- aov(obs~Ax*Bx)
summary(anova6.12)
halfnormal(anova6.12)
plot(anova6.12)
#Problem 6.21
library(tidyr)
putt <- read.csv(file.choose())
View(putt)
putt$Length <- as.factor(putt$Length)
putt$Type <- as.factor(putt$Type)
putt$Break <- as.factor(putt$Break)
putt$Slope <- as.factor(putt$Slope)
pl <-rep(putt$Length, rep(7,16))
pt <- rep(putt$Type, rep(7,16))
pb <- rep(putt$Break, rep(7,16))
ps <- rep(putt$Slope, rep(7,16))
puttobs <- as.vector(t(putt[,c(5:11)]))
str(puttobs)
anova6.21 <- aov(puttobs~pl*pt*pb*ps)
summary(anova6.21)
plot(anova6.21)
#problem 6.36
install.packages("DoE.base")
library(DoE.base)
resA <- rep(c(-1,1),8)
resB <- rep(c(-1,-1,1,1),4)
resC <- rep(c(-1,-1,-1,-1,1,1,1,1),2)
resD <- c(rep(-1,8), rep(1,8))
resob <- 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)
anova6.35 <- aov(resob~resA*resB*resC*resD)
halfnormal(anova6.35)
summary(anova6.35)
anova6.35new <- aov(resob~resA+resB+resA*resB+resA*resB*resC)
summary(anova6.35new)
plot(anova6.35new)
resoblog <- log(resob)
anova6.35new1 <- aov(resoblog~resA*resB*resC*resD)
halfnormal(anova6.35new1)
anova6.35new2 <- lm(resoblog~resA+resB+resA*resB*resC)
summary(anova6.35new2)
#problem 6.39
relA <- rep(c(-1,1),16)
relB <- rep(c(-1,-1,1,1),8)
relC <- rep(c(-1,-1,-1,-1,1,1,1,1),4)
relD <- rep(c(rep(-1,8), rep(1,8)),2)
relE <- c(rep(-1,16), rep(1,16))
relobs <- 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)
anova6.39 <- aov(relobs~relA*relB*relC*relD*relE)
halfnormal(anova6.39)
anova6.39new <- aov(relobs~relD+relE+relA*relD+relA+relD*relE+
relB*relE+relA*relB+relA*relB*relE+relA*relE+
relA*relD*relE)
summary(anova6.39new <- lm(relobs~relD+relE+relA*relD+relA+relD*relE+
relB*relE+relA*relB+relA*relB*relE+relA*relE+
relA*relD*relE))
summary(anova6.39new)
plot(anova6.39new)