Part 1: Is this a valid Latin Square?

yes, Each ingredient (A–E) appears exactly once in each Day (row) and once in each Batch (column).

Part 2: Model

Y(ijk) = mu + alpha(i) + beta(j) + tau(k) + error(ijk)

Part 3: Analysis at alpha = 0.05

# ---- Data entry ----
Day    <- rep(1:5, each = 5)
Batch  <- rep(1:5, times = 5)
Ingredient <- 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)

dat <- data.frame(Day = factor(Day),
                  Batch = factor(Batch),
                  Ingredient = factor(Ingredient),
                  Time = Time)

dat
##    Day Batch Ingredient Time
## 1    1     1          A    8
## 2    1     2          B    7
## 3    1     3          D    1
## 4    1     4          C    7
## 5    1     5          E    3
## 6    2     1          C   11
## 7    2     2          E    2
## 8    2     3          A    7
## 9    2     4          D    3
## 10   2     5          B    8
## 11   3     1          B    4
## 12   3     2          A    9
## 13   3     3          C   10
## 14   3     4          E    1
## 15   3     5          D    5
## 16   4     1          D    6
## 17   4     2          C    8
## 18   4     3          E    6
## 19   4     4          B    6
## 20   4     5          A   10
## 21   5     1          E    4
## 22   5     2          D    2
## 23   5     3          B    3
## 24   5     4          A    8
## 25   5     5          C    8
# ----  Latin Square ANOVA ----
dat_latin <- aov(Time ~ Ingredient + Day + Batch, data = dat)
summary(dat_latin)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## Ingredient   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
TukeyHSD(dat_latin, "Ingredient")
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = Time ~ Ingredient + Day + Batch, data = dat)
## 
## $Ingredient
##     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

The ANOVA shows that the Ingredient factor is significant (p = 0.000488), while Day and Batch are not significant (p = 0.347 and 0.455, respectively). This means that only the type of ingredient has a real effect on reaction time.

Tukey’s test confirms that Ingredients D and E produce significantly shorter reaction times than A and C. Ingredients D and E are not significantly different from each other, and Ingredient B shows no significant difference from any group.