For the Latin Square become valid, the Latin letters should not repeat in each row and column. So, since it is a two-source of nuisance variability that is exactly the case, the Latin square is valid.
Since it is a two-block Latin Square (Latin Square), letโs consider the following equation:
\[ y_{ijk}=\mu+\tau_i+\beta_j+\alpha_k+\epsilon_{ijk} \]
Where \(\tau_i\) is the treatment effect, \(\beta_j\) is the nuisance variability of the Batch (block one), \(\alpha_k\) is the nuisance variability of the Days (block two) and \(\epsilon_{ijk}\) is the random error.
The hypothesis of the statistics test is:
\[ H_o: \tau_i = 0 \;for \;all \;i \\ H_a: \tau_i \neq 0 \;for \;some \;i \;(i=1,2,3,4,5) \]
library(GAD)## Loading required package: matrixStats## Loading required package: R.methodsS3## R.methodsS3 v1.8.2 (2022-06-13 22:00:14 UTC) successfully loaded. See ?R.methodsS3 for help.batch <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
day <- c(rep(seq(1,5),5))
ing <- 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.fixed(batch)
day <- as.fixed(day)
ing <- as.fixed(ing)
data <- data.frame(batch,day,ing,obs)
model <- aov(obs~batch+day+ing, 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
## ing 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 ' ' 1Conclusion: Since the p-value of ingredients is lesser than alpha (\(p-value_{ing}=0.000488 < \alpha\)) we can reject \(H_o\).
Therefore, changing the ingredient can significantly change the reaction time of chemical process, because the treatment effect is different than zero.