1. Yes, this is a valid Latin square. Each ingredient appears exactly once in each batch and exactly once on each day.

  2. The model equation is

y_ijk = μ + τ_i + β_j + γ_k + ε_ijk

where μ is the overall mean, τ_i is the ingredient effect, β_j is the batch effect, γ_k is the day effect, and ε_ijk is the random error.

  1. H0: All five ingredients have the same mean reaction time.

H1: At least one ingredient has a different mean reaction time.

batch <- factor(rep(1:5, each = 5))
day <- factor(rep(1:5, times = 5))

ingredient <- 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"))

time <- 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)

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

Since the p-value for ingredient is 0.000488, which is less than 0.05, we reject H0. There is a significant difference in mean reaction time among the five ingredients.