Assignment_14
Rezwan
2026-10-28
Problem 1
(a)
The fixed-effects model is
\[Y_{ijkl}=\mu+\alpha_i+\beta_j+\gamma_k+(\alpha\beta)_{ij}+(\alpha\gamma)_{ik}+(\beta\gamma)_{jk}+(\alpha\beta\gamma)_{ijk}+\varepsilon_{ijkl},\]
where the errors are independent and normally distributed with mean zero and constant variance \(\sigma^2\).
(b) Full factorial ANOVA
powder <- data.frame(Ammonium = factor(c(2,2,30,30,2,2,30,30,2,2,30,30,2,2,30,30)),StirRate = factor(c(rep(100,4),rep(150,4),rep(100,4),rep(150,4))),Temperature = factor(c(rep(8,8),rep(40,8))),Density = c(14.68,15.18,15.12,17.48,7.54,6.66,12.46,12.62,10.95,17.68,12.65,15.96,8.03,8.84,14.96,14.96))
powder## Ammonium StirRate Temperature Density
## 1 2 100 8 14.68
## 2 2 100 8 15.18
## 3 30 100 8 15.12
## 4 30 100 8 17.48
## 5 2 150 8 7.54
## 6 2 150 8 6.66
## 7 30 150 8 12.46
## 8 30 150 8 12.62
## 9 2 100 40 10.95
## 10 2 100 40 17.68
## 11 30 100 40 12.65
## 12 30 100 40 15.96
## 13 2 150 40 8.03
## 14 2 150 40 8.84
## 15 30 150 40 14.96
## 16 30 150 40 14.96Since all factors are fixed, the standard ANOVA F-tests are appropriate.
powder_model <- aov(Density ~ Ammonium * StirRate * Temperature,data = powder)
summary(powder_model)## Df Sum Sq Mean Sq F value Pr(>F)
## Ammonium 1 44.39 44.39 11.180 0.01018 *
## StirRate 1 70.69 70.69 17.804 0.00292 **
## Temperature 1 0.33 0.33 0.083 0.78117
## Ammonium:StirRate 1 28.12 28.12 7.082 0.02875 *
## Ammonium:Temperature 1 0.02 0.02 0.005 0.94281
## StirRate:Temperature 1 10.13 10.13 2.551 0.14889
## Ammonium:StirRate:Temperature 1 1.52 1.52 0.383 0.55341
## Residuals 8 31.76 3.97
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1At \(\alpha=0.05\), the significant terms are Ammonium (\(p=0.0102\)), Stir Rate (\(p=0.0029\)), and Ammonium:Stir Rate (\(p=0.0288\)). Temperature (\(p=0.7812\)), Ammonium:Temperature (\(p=0.9428\)), StirRate:Temperature (\(p=0.1489\)), and the three-way interaction (\(p=0.5534\)) are not significant.
Because the Ammonium:Stir Rate interaction is significant, the effect of changing stir rate depends on ammonium concentration. The following interaction plot averages over the two temperature level
with(powder,interaction.plot(x.factor=StirRate,trace.factor=Ammonium,response=Density,fun=mean, type="b", pch=c(16,17), lty=1:2,xlab="Stir Rate (RPM)",ylab="Mean Powder Density",trace.label="Ammonium (%)",main="Ammonium x Stir Rate Interaction"))
Conclusion: At the 5% significance level, ammonium concentration, stir rate, and their interaction significantly affect powder density.
Problem 2
Position has two levels, Temperature has three levels (800, 825, and 850 degrees C), and each combination has three observations.
The crossed factorial model is
\[Y_{ijk}=\mu+\alpha_i+\beta_j+(\alpha\beta)_{ij}+\varepsilon_{ijk}.\]
anode <- data.frame(Replicate = rep(1:3, times = 6),Temperature = factor(rep(c(800, 825, 850), each = 3, times = 2)),Position = factor(rep(c(1, 2), each = 9)),Density = c(570, 565, 583,1063, 1080, 1043,565, 510, 590,528, 547, 521,988, 1026, 1004,526, 538, 532))
anode## Replicate Temperature Position Density
## 1 1 800 1 570
## 2 2 800 1 565
## 3 3 800 1 583
## 4 1 825 1 1063
## 5 2 825 1 1080
## 6 3 825 1 1043
## 7 1 850 1 565
## 8 2 850 1 510
## 9 3 850 1 590
## 10 1 800 2 528
## 11 2 800 2 547
## 12 3 800 2 521
## 13 1 825 2 988
## 14 2 825 2 1026
## 15 3 825 2 1004
## 16 1 850 2 526
## 17 2 850 2 538
## 18 3 850 2 532(a) Both Temperature and Position Are Fixed Effects
library(GAD)
Position <- as.fixed(anode$Position)
Temperature <- as.fixed(anode$Temperature)
model_fixed <- aov(anode$Density ~ Position * Temperature)
gad(model_fixed)## $anova
## Analysis of Variance Table
##
## Response: anode$Density
## Df Sum Sq Mean Sq F value Pr(>F)
## Position 1 7160 7160 15.998 0.001762 **
## Temperature 2 945342 472671 1056.117 3.25e-14 ***
## Position:Temperature 2 818 409 0.914 0.427110
## Residuals 12 5371 448
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1At \(\alpha = 0.05\), Position and Temperature are significant, while their interaction is not significant.
(b) Both Temperature and Position Are Random Effects
The random-effects model is:
\[Y_{ijk} = \mu + A_i + B_j + (AB)_{ij} + \varepsilon_{ijk}\]
Position <- as.random(anode$Position)
Temperature <- as.random(anode$Temperature)
model_random <- aov(anode$Density ~ Position * Temperature)
gad(model_random)## $anova
## Analysis of Variance Table
##
## Response: anode$Density
## Df Sum Sq Mean Sq F value Pr(>F)
## Position 1 7160 7160 17.504 0.0526583 .
## Temperature 2 945342 472671 1155.518 0.0008647 ***
## Position:Temperature 2 818 409 0.914 0.4271101
## Residuals 12 5371 448
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1At \(\alpha = 0.05\), Temperature is significant, while Position and the interaction are not significant.
(c) Position Fixed and Temperature Random
In this mixed-effects model, Position is fixed and Temperature is random.
The model is:
\[Y_{ijk} = \mu + \alpha_i + B_j + (\alpha B)_{ij} + \varepsilon_{ijk}\]
Position <- as.fixed(anode$Position)
Temperature <- as.random(anode$Temperature)
model_mixed <- aov(anode$Density ~ Position * Temperature)
gad(model_mixed)## $anova
## Analysis of Variance Table
##
## Response: anode$Density
## Df Sum Sq Mean Sq F value Pr(>F)
## Position 1 7160 7160 17.504 0.05266 .
## Temperature 2 945342 472671 1056.117 3.25e-14 ***
## Position:Temperature 2 818 409 0.914 0.42711
## Residuals 12 5371 448
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1At \(\alpha = 0.05\), Temperature is significant, while Position and the interaction are not significant.
(d) Comparison of the Three Models
The p-values for the random-effects and mixed-effects models (b and c) are identical, while those for the fixed-effects model (a) differ because different error terms are used in the F-tests. Temperature is significant in all three models, whereas Position is significant only in the fixed-effects model. The interaction between Temperature and Position is not significant in any model at α = 0.05.
Complete R Code
library(GAD)
# Problem 1
Ammonium <- factor(c(2,2,30,30,2,2,30,30, 2,2,30,30,2,2,30,30))
StirRate <- factor(rep(c(100,150,100,150), each = 4))
Temperature <- factor(rep(c(8,40), each = 8))
Density <- c(14.68,15.18,15.12,17.48,7.54,6.66,12.46,12.62,10.95,17.68,12.65,15.96,8.03,8.84,14.96,14.96)
data1 <- data.frame(Ammonium, StirRate, Temperature, Density)
data1
# Question 1(b)
powder <- data.frame(Ammonium = factor(c(2,2,30,30,2,2,30,30,2,2,30,30,2,2,30,30)),StirRate = factor(c(rep(100,4),rep(150,4),rep(100,4),rep(150,4))),Temperature = factor(c(rep(8,8),rep(40,8))),Density = c(14.68,15.18,15.12,17.48,7.54,6.66,12.46,12.62,10.95,17.68,12.65,15.96,8.03,8.84,14.96,14.96))
powder
model1 <- aov(Density ~ Ammonium * StirRate * Temperature)
summary(model1)
interaction.plot(StirRate, Ammonium, Density,fun = mean,type = "b",pch = c(16,17),lty = 1:2,xlab = "Stir Rate (RPM)",ylab = "Mean Powder Density",trace.label = "Ammonium (%)")
# Problem 2
Position <- factor(rep(c(1,2), each = 9))
Temperature <- factor(rep(c(800,825,850),each = 3, times = 2))
Density <- c(570,565,583,1063,1080,1043, 565,510,590,528,547,521,988,1026,1004,526,538,532)
data2 <- data.frame(Position, Temperature, Density)
data2
# Question 2(a)
Position <- as.fixed(data2$Position)
Temperature <- as.fixed(data2$Temperature)
model2a <- aov(Density ~ Position * Temperature)
gad(model2a)
# Question 2(b)
Position <- as.random(data2$Position)
Temperature <- as.random(data2$Temperature)
model2b <- aov(Density ~ Position * Temperature)
gad(model2b)
# Question 2(c)
Position <- as.fixed(data2$Position)
Temperature <- as.random(data2$Temperature)
model2c <- aov(Density ~ Position * Temperature)
gad(model2c)