0.0.1 Answer (1)

From the data table, we found that each ingridents (A, B, C, D, E) are coming once in each row and column. Therefore, it is a valid Latin Square.

0.0.2 Answer (2)

0.0.3 Model equation:

\(X_{ijk}\) = \(\mu\) + \(\tau_i\) + \(\beta_j\) + \(\gamma_k\) + \(\epsilon_{ijk}\)

0.0.4 Hypothesis :

\(H_0: \tau_i = 0, \forall 0\)

\(H_a:\tau_i \neq 0\) for some i

0.0.5 Answer (3)

batches<- c(rep(1,5), rep(2,5), rep(3,5), rep(4,5),rep(5,5))
days<- 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)

batches<- as.factor(batches)
days<- as.factor(days)
ing<- as.factor(ing)
?aov
anova<- aov(obs~batches+days+ing)
summary(anova)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## batches      4  15.44    3.86   1.235 0.347618    
## days         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 ' ' 1

0.0.6 From the ANOVA, we got the P-value for the Ingridents is 0.000488 which is less than the alpha =0.05. Therefore, we reject the null hypothesis.