Assignment_12
Rezwan
2026-10-11
Answer to Question 1
Yes, this is a valid 5 × 5 Latin Square Design. Each ingredient (A, B, C, D, and E) appears exactly once in every row (Batch) and exactly once in every column (Day). So, the design satisfies the requirements of a Latin Square Design.
Answer to Question 2
The statistical model for the Latin Square Design is:
\[X_{ijk} = \mu + \tau_i + \beta_j + \alpha_k + \epsilon_{ijk}\]
Where:
- \(X_{ijk}\) = observed reaction time for treatment \(i\), block 1 level \(j\), and block 2 level \(k\)
- \(\mu\) = overall mean reaction time
- \(\tau_i\) = effect of treatment \(i\) (Ingredient)
- \(\beta_j\) = effect of Block 1 (Batch)
- \(\alpha_k\) = effect of Block 2 (Day)
- \(\epsilon_{ijk}\) = random error
Answer to Question 3
The reaction time is the response variable. Ingredient is the treatment factor, while Batch and Day are blocking factors.
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)
data <- data.frame(Batch, Day, Ingredient, Time)
data## Batch Day 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 8We want to determine whether the five ingredients have the same mean reaction time.
Null hypothesis:
\[H_0: \mu_A = \mu_B = \mu_C = \mu_D = \mu_E\]
Alternative hypothesis:
\[H_a: \text{At least one ingredient mean differs}\]
The significance level is:
\[\alpha = 0.05\]
A Latin Square Design controls variation from two blocking factors, Batch and Day.
model <- aov(Time ~ Ingredient + Batch + Day, data = data)
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 ' ' 1The ANOVA results for Ingredient are:
\[F = 11.309\]
\[p = 0.000488\]
Since:
\[0.000488 < 0.05\]
We reject the null hypothesis. There is sufficient statistical evidence to conclude that the mean reaction times are not the same for all five ingredients.
The average reaction time for each ingredient is calculated to understand the treatment differences.
aggregate(Time ~ Ingredient, data = data, FUN = mean)## Ingredient Time
## 1 A 8.4
## 2 B 5.6
## 3 C 8.8
## 4 D 3.4
## 5 E 3.2These are descriptive differences. Additional pairwise comparisons would be required to establish which specific ingredient means differ significantly.
The ANOVA assumptions can be examined using residual diagnostic plots.
par(mfrow = c(2, 2))
plot(model)
par(mfrow = c(1, 1))The residuals are scattered around zero without a strong pattern. The red line shows slight curvature, but there is no major indication of nonlinearity or unequal variance. Most residuals follow the reference line reasonably well, with some deviations at the tails. Therefore, the normality assumption appears approximately satisfied.
The diagnostic plots do not show serious violations of the ANOVA assumptions. Therefore, the ANOVA results appear reasonably reliable.
At the 0.05 significance level, the ANOVA indicates a statistically significant difference in mean reaction time among the five ingredients. Therefore, the null hypothesis is rejected. Although the Batch and Day effects are not statistically significant, they were included as blocking factors to account for potential nuisance variation. The results suggest that ingredient selection affects the reaction time of the chemical process.
Complete R Code
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)
data <- data.frame(Batch, Day, Ingredient, Time)
data
model <- aov(Time ~ Ingredient + Batch + Day, data = data)
summary(model)
aggregate(Time ~ Ingredient, data = data, FUN = mean)
par(mfrow = c(2, 2))
plot(model)
par(mfrow = c(1, 1))