FA12

Question 1

Yes, it is a valid Latin Square, because the observations are not repeated in each row and column. This provides orthogonality to the experiment.

Question 2

The model equation is

\[ Y_{i,j,k} = \mu + \alpha_{i} +\beta_{j} + \gamma_{k} + \epsilon_{i,j,k} \]

Question 3

df<-expand.grid(seq(1,5),seq(1,5))# create the data frame

colnames(df)<-c("Day","Batch") # rename the column names
df$Day<-as.factor(df$Day)      # convert the data into factor        
df$Batch<-as.factor(df$Batch)  # convert the data into factor

df$Factor <- c("A","B","D","C","E",
               "C","E","A","D","B",
               "B","A","C","E","D",
               "D","C","E","B","A",
               "E","D","B","A","C") # input the factors

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) # input the response

df$factor <- as.factor(df$Factor) # convert the data into factor

str(df) #confirm all the data is factor

## 'data.frame':    25 obs. of  5 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 ...
##  $ Factor  : chr  "A" "B" "D" "C" ...
##  $ Response: num  8 7 1 7 3 11 2 7 3 8 ...
##  $ factor  : Factor w/ 5 levels "A","B","C","D",..: 1 2 4 3 5 3 5 1 4 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" ...

model <- lm(Response~Batch+Day+Factor, data = df) # create the model 
aov <- aov(model) # anova test
summary(aov) #showing result

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Batch        4  15.44    3.86   1.235 0.347618    
## Day          4  12.24    3.06   0.979 0.455014    
## Factor       4 141.44   35.36  11.309 0.000488 ***
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Conclusion:

Since our ingrediants P-Value is 0.000488, very small compared to the sig level of 0.05, therefore we will reject the Null Hypothesis. So, we conclude that different ingredients have certain effect on reaction time of a chemical process. The Factor(Ingredient) has a significant effect on the reaction time.

Complete Code

df<-expand.grid(seq(1,5),seq(1,5))# create the data frame

colnames(df)<-c("Day","Batch") # rename the column names
df$Day<-as.factor(df$Day)      # convert the data into factor        
df$Batch<-as.factor(df$Batch)  # convert the data into factor

df$Factor <- c("A","B","D","C","E",
               "C","E","A","D","B",
               "B","A","C","E","D",
               "D","C","E","B","A",
               "E","D","B","A","C") # input the factors

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) # input the response

df$factor <- as.factor(df$Factor) # convert the data into factor

str(df) #confirm all the data is factor

model <- lm(Response~Batch+Day+Factor, data = df) # create the model 
aov <- aov(model) # anova test
summary(aov) #showing result