1 Question 1

This is a valid latin square because no repetitions of ingredients is observed in row or column, which ensure ortogonality.

2 Question 2

The model equation is given as: \[ x_{ijk}= \mu +\tau_i + \beta_j + \alpha_k + \epsilon_{ijk} \]

where \(x\) is an individual observation, \(\mu\) is the mean of all observations, \(\tau\) is the effect of treatments, \(\beta\) is the effect of first block factor, \(\alpha\) is the effect of second block factor, and \(\epsilon\) is the random error.

3 Question 3

We are interested in analyze the following hypotheses:

\[H_0:\tau_i=0 \\ H_a:\tau_i\neq 0 \]

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)

day <- as.factor(c(seq(1,5),seq(1,5),seq(1,5),seq(1,5),seq(1,5)))

batch <- as.factor(c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5)))


treatment <- as.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'))

aov.model <- aov(obs~treatment+batch+day)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## treatment    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

From the ANOVA test we can observe that there were significantly difference between each treatment (p-value=0.000488). In contrast no significantly differences were observed for batch and day, which means that these factors do not have an impact in observation values. This is what we expected since the blocks should have an additive effect to the linear model instead of interaction.