Untitled

QUESTION 1:

Yes, this is a valid latin square because we do not have recurring values for each row and each column within the square.

QUESTION 2:

Model Equation: \(Y_{ijk} = \mu + \alpha_i + \beta_j + \tau_k + \epsilon_{ijk}\)

Hypothesis:

\(H_0: \tau_i = 0\) For every i

\(H_a : \tau_i \neq 0\) For same i

QUESTION 3:

#Data Entry

df<-expand.grid(seq(1,5),seq(1,5))
colnames(df)<-c("Day","Batch")
df$Day<-as.factor(df$Day)        
df$Batch<-as.factor(df$Batch)
str(df)

## 'data.frame':    25 obs. of  2 variables:
##  $ Day  : Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5 1 2 3 4 5 ...
##  $ Batch: Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 1 2 2 2 2 2 ...
##  - attr(*, "out.attrs")=List of 2
##   ..$ dim     : int [1:2] 5 5
##   ..$ dimnames:List of 2
##   .. ..$ Var1: chr [1:5] "Var1=1" "Var1=2" "Var1=3" "Var1=4" ...
##   .. ..$ Var2: chr [1:5] "Var2=1" "Var2=2" "Var2=3" "Var2=4" ...

df$Ingredient<-c("A","B","C","D","E",
                 "C","E","A","D","B",
                 "B","A","C","E","D",
                 "D","C","E","B","A",
                 "E","D","B","A","C")
df$Response<-c(8,7,1,7,3,
               11,2,7,3,8,
               4,9,10,1,5,
               6,8,6,6,10,
               4,2,3,8,8)
df$Ingredient<-as.factor(df$Ingredient)

#Analyze the problem using ANOVA

aov.model<-aov(Response~Ingredient, data=df)
summary(aov.model)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## Ingredient   4  91.04   22.76   3.938 0.0162 *
## Residuals   20 115.60    5.78                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Comment: After running the data through Anova, we found out that the P-value (0.0162) was less than the alpha(0.05), hence we reject \(H_0\).

df<-expand.grid(seq(1,5),seq(1,5))
colnames(df)<-c("Day","Batch")
df$Day<-as.factor(df$Day)        
df$Batch<-as.factor(df$Batch)
str(df)
df$Ingredient<-c("A","B","C","D","E",
                 "C","E","A","D","B",
                 "B","A","C","E","D",
                 "D","C","E","B","A",
                 "E","D","B","A","C")
df$Response<-c(8,7,1,7,3,
               11,2,7,3,8,
               4,9,10,1,5,
               6,8,6,6,10,
               4,2,3,8,8)
df$Ingredient<-as.factor(df$Ingredient)
aov.model<-aov(Response~Ingredient, data=df)
summary(aov.model)