Yes, it is a valid latin square. For a valid latin sqaure, we need orthoganality. Meaning none of the latin letters should repeat in each row or column, which is satisfied here.
Model equation:
\(X_{ijk}=\mu+\tau_i+\beta{j}+\alpha_k+\epsilon_{ijk}\)
where \(\mu\) is the grand mean, \(\tau_i\) is the factor (ingredient) effect, \(\beta{j}\) is the batch effect, \(\alpha_k\) is the day effect, and \(\epsilon_{ijk}\) is the random variation of the data.
\(H_0:\tau_i=0\)
\(H_1:\tau_i\neq0\)
\(i=A,B,C,D,E\)
Batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
day <- c(seq(1,5),seq(1,5),seq(1,5),seq(1,5),seq(1,5))
ingredient <- 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')
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)
Batch <- as.factor(Batch)
day <- as.factor(day)
ingredient <- as.factor(ingredient)
data <- data.frame(obs,Batch,day,ingredient)
model <- aov(obs~Batch+day+ingredient,data = data)
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
## ingredient 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 ' ' 1As the resulted p-value for our factor of interest (ingredient) is 0.000488 and lower than our specified \(\alpha\) (0.05), we reject the null hypothesis and conclude that the choice of ingredient has a significant effect on the reaction time.