1. Is this a valid Latin Square?

# Data from the assignment table
Day   <- factor(rep(1:5, each = 5))
Batch <- factor(rep(1:5, times = 5))
Trt   <- 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"
))
y <- 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
)
dat <- data.frame(Day, Batch, Trt, y)

# Check Latin-square property
xtabs(~ Day + Trt, dat)
##    Trt
## Day A B C D E
##   1 1 1 1 1 1
##   2 1 1 1 1 1
##   3 1 1 1 1 1
##   4 1 1 1 1 1
##   5 1 1 1 1 1
xtabs(~ Batch + Trt, dat)
##      Trt
## Batch A B C D E
##     1 1 1 1 1 1
##     2 1 1 1 1 1
##     3 1 1 1 1 1
##     4 1 1 1 1 1
##     5 1 1 1 1 1

Answer:
Each treatment appears once per Day and once per Batch, so it is a valid Latin Square.

2. Model Equation

\[ y_{ijk} = \mu + \tau_i + \rho_j + \kappa_k + \varepsilon_{ijk} \]

where
- \(\tau_i\): Treatment (ingredient) effect
- \(\rho_j\): Day effect (row block)
- \(\kappa_k\): Batch effect (column block)
- \(\varepsilon_{ijk}\): random error

3. Analyze the Data and Draw Conclusions (\(\alpha = 0.05\))

# Latin Square ANOVA
fit <- aov(y ~ Trt + Day + Batch, data = dat)
summary(fit)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Trt          4 141.44   35.36  11.309 0.000488 ***
## Day          4  15.44    3.86   1.235 0.347618    
## Batch        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

Answer:
The Treatment (Trt) line tests if ingredient means differ after adjusting for Day and Batch.
Since the p-value for Trt is less than 0.05, we reject H₀ and conclude that ingredient means are not all equal.

Treatment Means

tapply(y, Trt, mean)
##   A   B   C   D   E 
## 8.4 5.6 8.8 3.4 3.2

Answer:
Average reaction times by ingredient:
A = 8.4, B = 5.6, C = 8.8, D = 3.4, E = 3.2.
Ingredients A and C are higher, while D and E are lower.

Tukey Test

TukeyHSD(fit, "Trt")
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = y ~ Trt + Day + Batch, data = dat)
## 
## $Trt
##     diff        lwr        upr     p adj
## B-A -2.8 -6.3646078  0.7646078 0.1539433
## C-A  0.4 -3.1646078  3.9646078 0.9960012
## D-A -5.0 -8.5646078 -1.4353922 0.0055862
## E-A -5.2 -8.7646078 -1.6353922 0.0041431
## C-B  3.2 -0.3646078  6.7646078 0.0864353
## D-B -2.2 -5.7646078  1.3646078 0.3365811
## E-B -2.4 -5.9646078  1.1646078 0.2631551
## D-C -5.4 -8.9646078 -1.8353922 0.0030822
## E-C -5.6 -9.1646078 -2.0353922 0.0023007
## E-D -0.2 -3.7646078  3.3646078 0.9997349

Answer:
From the Tukey test, ingredients D and E are significantly lower than A and C.
B is not significantly different from the others.

Conclusion