Yes, it is a valid Latin Square since all letters appears in each row and each column.
Linear effect equation:
\[ Y_{ijk}=\mu+\tau_{i}+\beta_{j}+\alpha_{k}+\epsilon_{ijk} \]
Where,
µ = Grand Mean
τi = Treatment effect
βj = Block-1 effect
αk = Block-2 effect
εijk = Random error
i = number of treatments
j = number of block-1
k = number of block-2
Null hypotheses = \(H_{0}:\mu_{ 1}=\mu_{ 2}=\mu_{ 3}=\mu_{ 4}=\mu_{ 5}\)
Alternative Hypotheses = At least one mu differs
Batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
Day <- c(rep(seq(1,5),5))
Obs <- 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)
Ingredients <- 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")
Batch <- as.factor(Batch)
Day <- as.factor(Day)
Ingredients <- as.factor(Ingredients)
dat<- data.frame(Batch, Day, Obs, Ingredients)
model <- aov(Obs~Batch+Day+Ingredients)
summary(model)
## 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
## Ingredients 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
P-value is larger than alpha 0.05, there is no evidence to reject Ho.
Batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
Day <- c(rep(seq(1,5),5))
Obs <- 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)
Ingredients <- 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")
Batch <- as.factor(Batch)
Day <- as.factor(Day)
Ingredients <- as.factor(Ingredients)
dat<- data.frame(Batch, Day, Obs, Ingredients)
model <- aov(Obs~Batch+Day+Ingredients)
summary(model)