Question 1

A)Full Factorial Model Equation

\(y_{ijkl} = \mu + \tau_i + \beta_j + \gamma_k + (\tau\beta)_{ij} + (\tau\gamma)_{ik} + (\beta\gamma)_{jk} + (\tau\beta\gamma)_{ijk} + \varepsilon_{ijkl}\)

Where,  

\(\tau_i = \text{Main Effect for Ammonium %}\)

\(\beta_j = \text{Main Effect for Stir Rate}\)

\(\gamma_k = \text{Main Effect for Temperature}\)

B)

To identify which factors are significant, we use a step-by-step testing method. We start by checking if the three factors together have any significant interaction, and then move on to test the two-factor interactions.

library(GAD)

df<-read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/main/PowderProduction.csv")
print(df)
##    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   14.96
df$Ammonium<-as.fixed(df$Ammonium)
df$StirRate<-as.fixed(df$StirRate)
df$Temperature<-as.fixed(df$Temperature)

str(df)
## 'data.frame':    16 obs. of  4 variables:
##  $ Ammonium   : Factor w/ 2 levels "2","30": 1 1 2 2 1 1 2 2 1 1 ...
##  $ StirRate   : Factor w/ 2 levels "100","150": 1 1 1 1 2 2 2 2 1 1 ...
##  $ Temperature: Factor w/ 2 levels "8","40": 1 1 1 1 1 1 1 1 2 2 ...
##  $ Density    : num  14.68 15.18 15.12 17.48 7.54 ...

Starting with 3 Factor interaction:

Null Hypothesis:

\[ H_0 : (\tau \beta \gamma)_{ijk} = 0 \quad \forall \, i,j,k \]

Alternate Hypothesis:

\[ H_a : (\tau \beta \gamma)_{ijk} \neq 0 \quad \exists \, i,j,k \]

model<-aov(Density~Ammonium*StirRate*Temperature,data=df)
gad(model)
## $anova
## Analysis of Variance Table
## 
## Response: Density
##                               Df Sum Sq Mean Sq F value   Pr(>F)   
## Ammonium                       1 44.389  44.389 11.1803 0.010175 * 
## StirRate                       1 70.686  70.686 17.8037 0.002918 **
## Temperature                    1  0.328   0.328  0.0826 0.781170   
## Ammonium:StirRate              1 28.117  28.117  7.0817 0.028754 * 
## Ammonium:Temperature           1  0.022   0.022  0.0055 0.942808   
## StirRate:Temperature           1 10.128  10.128  2.5510 0.148890   
## Ammonium:StirRate:Temperature  1  1.519   1.519  0.3826 0.553412   
## Residuals                      8 31.762   3.970                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
summary(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 ' ' 1

The p value is greater than .o5, so we fail to reject null hypothesis and their is no interaction between 3 factors.

Now 2 factor interaction between Ammonium and Temperature

Null Hypothesis:

\[ H_0 : (\tau \gamma)_{ik} = 0 \quad \forall \, i,k \]

Alternate Hypothesis:

\[ H_a : (\tau \gamma)_{ik} \neq 0 \quad \exists \, i,k \]

model<-aov(Density~Ammonium*Temperature,data=df)
gad(model)
## $anova
## Analysis of Variance Table
## 
## Response: Density
##                      Df  Sum Sq Mean Sq F value  Pr(>F)  
## Ammonium              1  44.389  44.389  3.7456 0.07686 .
## Temperature           1   0.328   0.328  0.0277 0.87069  
## Ammonium:Temperature  1   0.022   0.022  0.0018 0.96653  
## Residuals            12 142.212  11.851                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
summary(model)
##                      Df Sum Sq Mean Sq F value Pr(>F)  
## Ammonium              1  44.39   44.39   3.746 0.0769 .
## Temperature           1   0.33    0.33   0.028 0.8707  
## Ammonium:Temperature  1   0.02    0.02   0.002 0.9665  
## Residuals            12 142.21   11.85                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p value is greater than .05, hense we fail to reject null hypothesis.

Now 2 factor interaction between StirRate and Temperature

Null Hypothesis:

\[ H_0 : (\beta \gamma)_{jk} = 0 \quad \forall \, j,k \]

Alternate Hypothesis:

\[ H_a : (\beta \gamma)_{jk} \neq 0 \quad \exists \, j,k \]

model<-aov(Density~StirRate*Temperature,data=df)
gad(model)
## $anova
## Analysis of Variance Table
## 
## Response: Density
##                      Df  Sum Sq Mean Sq F value  Pr(>F)  
## StirRate              1  70.686  70.686  8.0167 0.01514 *
## Temperature           1   0.328   0.328  0.0372 0.85034  
## StirRate:Temperature  1  10.128  10.128  1.1487 0.30491  
## Residuals            12 105.809   8.817                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
summary(model)
##                      Df Sum Sq Mean Sq F value Pr(>F)  
## StirRate              1  70.69   70.69   8.017 0.0151 *
## Temperature           1   0.33    0.33   0.037 0.8503  
## StirRate:Temperature  1  10.13   10.13   1.149 0.3049  
## Residuals            12 105.81    8.82                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p valus is greater than .05, we fail to reject the null hypothesis.

Now 2 factor interaction between Ammonium and StirRate

Null Hypothesis:

\[ H_0 : (\tau \beta)_{ij} = 0 \quad \forall \, i,j \]

Alternate Hypothesis:

\[ H_a : (\tau \beta)_{ij} \neq 0 \quad \exists \, i,j \]

model<-aov(Density~Ammonium*StirRate,data=df)
gad(model)
## $anova
## Analysis of Variance Table
## 
## Response: Density
##                   Df Sum Sq Mean Sq F value    Pr(>F)    
## Ammonium           1 44.389  44.389 12.1727 0.0044721 ** 
## StirRate           1 70.686  70.686 19.3841 0.0008612 ***
## Ammonium:StirRate  1 28.117  28.117  7.7103 0.0167511 *  
## Residuals         12 43.759   3.647                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
summary(model)
##                   Df Sum Sq Mean Sq F value   Pr(>F)    
## Ammonium           1  44.39   44.39   12.17 0.004472 ** 
## StirRate           1  70.69   70.69   19.38 0.000861 ***
## Ammonium:StirRate  1  28.12   28.12    7.71 0.016751 *  
## Residuals         12  43.76    3.65                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p value is less than .05, hense we reject null hypothesis.

interaction.plot(df$StirRate, df$Ammonium, df$Density, main = "Ammonium vs Stir Rate", ylab="Density")

since the lines are not parallel , the interaction between ammonium and StirRate is significant.

Question 2

Hypothesis:

Null Hypothesis:

\[ H_0 : (\alpha \beta)_{ij} = 0 \quad \forall \, i,j \]

Alternate Hypothesis:

\[ H_a : (\alpha \beta)_{ij} \neq 0 \quad \exists \, i,j \] The data :

Position <- c(1, 2)
Temperature <- c(800, 825, 850)
df <- expand.grid(Position = Position, Temperature = Temperature)
df1 <- rbind(df, df, df)   

# response
response <- c(570,1063,565,565,1080,510,583,1043,590,528,988,526,547,1026,538,521,1004,532)
df1$response <- response

# Convert to factors for GAD
df1$Position <- as.fixed(df1$Position)
df1$Temperature <- as.fixed(df1$Temperature)

# Part A: Fixed effects
model1 <- aov(response ~ Position + Temperature + Position*Temperature, data = df1)
GAD::gad(model1)
## $anova
## Analysis of Variance Table
## 
## Response: response
##                      Df Sum Sq Mean Sq  F value    Pr(>F)    
## Position              1   1267    1267   1.7056     0.216    
## Temperature           2 229965  114983 154.8241 2.696e-09 ***
## Position:Temperature  2 718547  359273 483.7613 3.381e-12 ***
## Residuals            12   8912     743                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Part B: Random effects
df1$Position <- as.random(df1$Position)
df1$Temperature <- as.random(df1$Temperature)
model2 <- aov(response ~ Position + Temperature + Position*Temperature, data = df1)
GAD::gad(model2)
## $anova
## Analysis of Variance Table
## 
## Response: response
##                      Df Sum Sq Mean Sq  F value    Pr(>F)    
## Position              1   1267    1267   0.0035    0.9581    
## Temperature           2 229965  114983   0.3200    0.7576    
## Position:Temperature  2 718547  359273 483.7613 3.381e-12 ***
## Residuals            12   8912     743                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Part C: Mixed effects
df1$Position <- as.fixed(df1$Position)
df1$Temperature <- as.random(df1$Temperature)
model3 <- aov(response ~ Position + Temperature + Position*Temperature, data = df1)
GAD::gad(model3)
## $anova
## Analysis of Variance Table
## 
## Response: response
##                      Df Sum Sq Mean Sq  F value    Pr(>F)    
## Position              1   1267    1267   0.0035    0.9581    
## Temperature           2 229965  114983 154.8241 2.696e-09 ***
## Position:Temperature  2 718547  359273 483.7613 3.381e-12 ***
## Residuals            12   8912     743                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Comment:-

Across all three models, the interaction between Position and Temperature is highly significant, indicating a strong combined influence on the response. The main effect of Position is not significant in any case, suggesting it has little independent impact. When all factors are treated as fixed (Part A), Temperature shows a highly significant effect; however, when treated as random (Part B), its significance disappears because the variance is attributed to random variation. In the mixed model (Part C), Temperature regains significance, reflecting its consistent influence when properly tested against random variation.