Answer to question no-1

This is a valid latin square design. Since all the letters (A,B,C,D,E) appears only once in each row and each column.

Answer to question no-2

The model equation for the latin sqaure design is as follows:

Yij = µ + τi + βj + αk + εijk

Where,

µ = Grand Mean

τi = Treatment effect

βj = Block-1 effect

αk = Block-2 effect

εijk = Random error

i = number of treatments

j = number of blocks

k = replications

Answer to question no-3

Entering Data:

batch <- c(rep(1,5), rep(2,5), rep(3,5), rep(4,5),rep(5,5))
day <- c(rep(seq(1,5),5))
letter <- 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)

batch <- as.factor(batch)
letter <- as.factor(letter)
day <- as.factor(day)

For our ANOVA analysis, we have the following hypothesis:

The Null hypothesis: Ho: τ1 = τ2 = τ3 = τ4 = τ5 = 0 ( \(\forall\) i)

Alternative hypothesis: Ha: One of the τi ≠ 0 ( \(\exists\) i)

Now we perform our ANOVA analysis:

anova <- aov(obs~batch+letter+day)
summary(anova)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## batch        4  15.44    3.86   1.235 0.347618    
## letter       4 141.44   35.36  11.309 0.000488 ***
## 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

From our ANOVA analysis we see that only the letter group has P-value significantly smaller than our threshold alpha = 0.05. This means that we reject our null hypothesis that all the effects of the treatment are equal to each other and equal to zero.

So we conclude that the ingredients (A,B,C,D,E) have significant effect on the reaction time of the chemical process.

Source Codes:

batch <- c(rep(1,5), rep(2,5), rep(3,5), rep(4,5),rep(5,5))
day <- c(rep(seq(1,5),5))
letter <- 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)

batch <- as.factor(batch)
letter <- as.factor(letter)
day <- as.factor(day)

anova <- aov(obs~batch+letter+day)
summary(anova)