1.

It is a valid Latin square because it has 2 sources of nuisance variability which are number of days and the batch effect, which are known and controllable. Also, we can conclude that the observations are not repeated in each rows and column.

That Proves Orthognoality to the experiment.

2. The general linear model equation for the Latin Square is:

\[Y_{ijk} = \mu + \alpha_i + \beta_j + \tau_k + \epsilon_{ijk}\] where

\(Y_{ijk}\) = observed reaction time \(\mu\) = overall mean \(\alpha_i\) = batch effect \(\beta_j\) = day effect \(\tau_k\) = ingredient effect \(\epsilon_{ijk}\) = random error

3. R code and Analysis

reaction <- 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)
batch <- factor(rep(1:5, each=5))
day <- factor(rep(1:5, times=5))
Ingredient <- c(1,2,4,3,5,3,5,1,4,2,2,1,3,5,4,4,3,5,2,1,5,4,2,1,3)
Batch <- as.factor(batch)
Day <- as.factor(day)
Ingredient <- as.factor(Ingredient)
Data <- data.frame(reaction , Batch , Day, Ingredient)
str(Data)
## 'data.frame':    25 obs. of  4 variables:
##  $ reaction  : num  8 7 1 7 3 11 2 7 3 8 ...
##  $ Batch     : Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 1 2 2 2 2 2 ...
##  $ Day       : Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5 1 2 3 4 5 ...
##  $ Ingredient: Factor w/ 5 levels "1","2","3","4",..: 1 2 4 3 5 3 5 1 4 2 ...

The Hypothesis:

Null: \[H_0: \gamma = 0\\ H_a: \gamma \ne 0\]

Running the Annova Model:

aov.model <- aov(reaction~Ingredient+Batch+Day,data=Data)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Ingredient   4 141.44   35.36  11.309 0.000488 ***
## Batch        4  15.44    3.86   1.235 0.347618    
## Day          4  12.24    3.06   0.979 0.455014    
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Conclusion:

Since our Ingredients P-value is 0.000488, very small compared to the sig level of 0.05, therefore we will reject the Null Hypothesis, and conclude that different ingredients have certain effect on reaction time of a chemical process.

Rcode

reaction <- 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)
batch <- factor(rep(1:5, each=5))
day <- factor(rep(1:5, times=5))
Ingredient <- c(1,2,4,3,5,3,5,1,4,2,2,1,3,5,4,4,3,5,2,1,5,4,2,1,3)
Batch <- as.factor(batch)
Day <- as.factor(day)
Ingredient <- as.factor(Ingredient)
Data <- data.frame(reaction , Batch , Day, Ingredient)
str(Data)
## 'data.frame':    25 obs. of  4 variables:
##  $ reaction  : num  8 7 1 7 3 11 2 7 3 8 ...
##  $ Batch     : Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 1 2 2 2 2 2 ...
##  $ Day       : Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5 1 2 3 4 5 ...
##  $ Ingredient: Factor w/ 5 levels "1","2","3","4",..: 1 2 4 3 5 3 5 1 4 2 ...
aov.model <- aov(reaction~Ingredient+Batch+Day,data=Data)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Ingredient   4 141.44   35.36  11.309 0.000488 ***
## Batch        4  15.44    3.86   1.235 0.347618    
## Day          4  12.24    3.06   0.979 0.455014    
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1