Homework Week 11
Muneeb
11/10/2021
Question 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.
This is a 2^2 Factorial Design, where low value of a is ao = 12 and high value of a is a1 = 18. While low value of b is bo = 1 and high value of b is b1 = 2.
Model Equation:
\(Y_{i,j,k}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{i,j}\) + \(\epsilon_{i,j,k}\)
Loading Data
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)
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)
Obs <- c(21,22,23,28,20,26,25,26,24,25,29,27,37,39,38,38,35,36,31,34,29,33,30,35)
A <- as.factor(A)
B <- as.factor(B)
Data <- data.frame(A,B,Obs)
Model <- aov(Obs~A*B,data = Data)
summary(Model)## Df Sum Sq Mean Sq F value Pr(>F)
## A 1 590.0 590.0 115.506 9.29e-10 ***
## B 1 9.4 9.4 1.835 0.190617
## A:B 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 ' ' 1Answer: From the ANOVA analysis, we can conclude that interaction of Factor A & B is significant. Therefore, we’ll stop here and we will look at the interaction plot.
Interaction Plot:
interaction.plot(A,B,Obs)
Residual Analysis:
plot(Model)
Looking at the “Normal Q-Q” & “Residuals vs Fitted” plots we can see that the data is normally distributed. Evaluating constant variance assumption, we can see that varaince spread is slightly off and thus we’ll consider the model inadequate.
Question 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.
Model Equation:
\(Y_{i,j,k}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\alpha\beta_{i,j}\) + \(\epsilon_{i,j,k}\)
Loading 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)
Obs <- c(14.037,13.880,14.821,14.888,16.165,13.860,14.757,14.921,13.972,14.032,14.843,14.415,13.907,13.914,14.878,14.932)
A <- as.factor(A)
B <- as.factor(B)
Data <- data.frame(A,B,Obs)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))Factor Effects:
EffectA## [1] -0.31725EffectB## [1] 0.586EffectAB## [1] -0.2815Effect of A: -0.31725
Effect of B: 0.586
Effect of AB: -0.2815
Part B
Conduct an analysis of variance. Which factors are important?
Model <- aov(Obs~A*B,data = Data)
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 ' ' 1From the results of ANOVA analysis, we can conclude that interaction between factors A & B is insignificant. Thus removing interaction effect and testing for main effects.
Model <- aov(Obs~A+B,data = Data)
summary(Model)## 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 ' ' 1As evident from the ANOVA analysis, 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.
Model <- lm(Obs~A*B,data = Data)
coef(Model)## (Intercept) A1 B1 A1:B1
## 14.52025 -0.59875 0.30450 0.56300summary(Model)##
## Call:
## lm.default(formula = Obs ~ A * B, data = Data)
##
## 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) 14.5202 0.2824 51.414 1.93e-15 ***
## A1 -0.5987 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: 0.1425\(Y_{i,j,k}\) = 14.52025 - 0.59875\(\alpha_i\) + 0.30450\(\beta_j\) + 0.56300\(\alpha\beta\gamma_{i,j}\) + \(\epsilon_{i,j,k}\)
Part D
Analyze the residuals. Are there any residuals that should cause concern?
plot(Model)
Plotting the residuals, we can visualize that the model is inadequate and plots display that the data point 5 is an outlier which may be a point of concern for our analysis.
Part E
Discuss how you might deal with the potential outlier found in part (d).
We can perform a BoxCox transformation on the data and find out the appropriate value of lambda and then perform the ANOVA analysis on the transformed data.
Question 6.21
Loading Data
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,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,-1,-1,-1,-1,-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,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-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,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
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,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
Obs <- c(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)
Data <- data.frame(A,B,C,D,Obs)Part A
Analyze the data from this experiment. Which factors significantly affect putting performance?
Model <- aov(Obs~A*B*C*D,data = Data)
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 ' ' 1At 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(Model)
Looking at the “Normal Q-Q” & “Residuals vs Fitted” plots we can see that the model is inadequate. Though the data satisfies the normality assumption, there’s a wide spread in variance of data.
Question 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.
Loading 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)
Data <- data.frame(A,B,C,D,Obs)Part A
Estimate the factor effects. Plot the effect estimates on a normal probability plot and select a tentative model.
Model <- lm(Obs~A*B*C*D,data = Data)
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
summary(Model)##
## Call:
## lm.default(formula = Obs ~ A * B * C * D, data = Data)
##
## 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: NAFactor 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.
\(Y_{i,j,k}\) = 4.680625 + 3.160625\(\alpha_i\) - 1.501875\(\beta_j\) - 1.069375\(\alpha\beta{i,j}\) + \(\epsilon_{i,j,k}\)
Part B
Fit the model identified in part (a) and analyze the residuals. Is there any indication of model inadequacy?
Model2 <- aov(Obs~A+B+C+A*B+A*B*C,data = Data)
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
Looking at the “Normal Q-Q” & “Residuals vs Fitted” plots we can see that the model is inadequate. We can visualize that the data is neither normally distributed nor the variance can be characterized constant.
Further the ANOVA analysis presents that all factors and interactions A,B,C,AB,AC,BC & ABC are significant at a significance level of 0.05.
Part 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?
logobs <- log(Obs)
Data2 <- data.frame(A,B,C,D,logobs)
Model3 <- lm(logobs~A*B*C*D,data = Data2)
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 = Data2)
##
## 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 = Data2)
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)
## hat values (leverages) are all = 0.5
## and there are no factor predictors; no plot no. 5
After the log transformation, we can observe that the halfnormal plot only characterizes factors A, B and ABC as significant. Interaction effect AB has been deemed insignificant as per halfnormal plot after transformation. ANOVA analysis also presents factors A, B and ABC as significant.
After the log transformation, we can observe that data is now almost falling in a straight line and spread has also became better. It hasn’t improved considerable but variance has been stabilized to some extent and thus we can conclude that transformation has been useful.
Part D
Fit a model in terms of the coded variables that can be used to predict the resistivity
Based on the transformed data:
\(Y_{i,j,k,l}\) = 1.185417 + 0.812870345\(\alpha_i\) - 0.314277554\(\beta_j\) + \(\epsilon_{i,j,k,l}\)
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.
Loading Data
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)
Data <- data.frame(A,B,C,D,E,Obs)Part A
Model <- lm(Obs~A*B*C*D*E,data = Data)
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 = Data)
##
## 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 = Data)
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 ' ' 1Half 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(Model2)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5
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).
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)
Data <- data.frame(A,B,D,E,Obs)
Model <- lm(Obs~A*B*D*E,data = Data)
coef(Model)## (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(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 e10
summary(Model)##
## Call:
## lm.default(formula = Obs ~ A * B * D * E, data = Data)
##
## 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-10Model2 <- aov(Obs~A+B+D+E+A*B+A*D+A*E+B*E+D*E+A*B*E+A*D*E,data = Data)
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 ' ' 1plot(Model2)
## hat values (leverages) are all = 0.375
## and there are no factor predictors; no plot no. 5