1. Is this a valid latin square?

Yes, this is a valid latin square as each ingredient appears exactly once in each row and column.

2. Model equation:

\[Y_{ij} = \mu + \beta_i + \delta_j + \tau_{k(i,j)} + \epsilon_{ij}\]

Where:

\(\mu\) is the overall mean

\(\beta_i\) is the batch effect

\(\delta_j\) is the day effect

\(\tau_{k(i,j)}\) is the ingredient effect

3. Data analysis

\(H_0\): All five ingredients have equal mean reaction times.

\(H_1\): At least one ingredient has a different mean reaction time.

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

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

ingrd <- 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
)

data <- data.frame(batch, day, ingrd, time)

model <- aov(time ~ batch + day + ingrd, 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    
## ingrd        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
Source df SS MS F p-value
Batch 4 15.44 3.86 1.2345 0.3476
Day 4 12.24 3.06 0.9787 0.4550
Ingredient 4 141.44 35.36 11.3092 0.000488
Error 12 37.52 3.1267

Here since p = 0.000488 < 0.05, we reject the null hypothesis. We find that the ingredient significantly affects the mean reaction time.