Assignment 11 - Randomized Complete Block Design

Question 1 - RCBD

A chemist wishes to test the effect of four chemical agents on tensile strength. Because there may be variability from one bolt of cloth to another, the five bolts are treated as blocks.

Data

library(GAD)

chemical <- factor(rep(1:4, each = 5))
bolt <- factor(rep(1:5, times = 4))

strength <- c(
  73, 68, 74, 71, 67,
  73, 67, 75, 72, 70,
  75, 68, 78, 73, 68,
  73, 71, 75, 75, 69
)

dat1 <- data.frame(chemical, bolt, strength)

dat1
##    chemical bolt strength
## 1         1    1       73
## 2         1    2       68
## 3         1    3       74
## 4         1    4       71
## 5         1    5       67
## 6         2    1       73
## 7         2    2       67
## 8         2    3       75
## 9         2    4       72
## 10        2    5       70
## 11        3    1       75
## 12        3    2       68
## 13        3    3       78
## 14        3    4       73
## 15        3    5       68
## 16        4    1       73
## 17        4    2       71
## 18        4    3       75
## 19        4    4       75
## 20        4    5       69

Linear Effects Model

Because chemical is the treatment of interest, it is a fixed effect. Bolt is treated as a random block effect.

\[ Y_{ij}=\mu+\alpha_i+\beta_j+\epsilon_{ij} \]

where \(\alpha_i\) is the fixed chemical effect, \(\beta_j\) is the random bolt effect, and \(\epsilon_{ij}\) is random error.

Hypotheses for Chemical Effect

\[ H_0:\alpha_1=\alpha_2=\alpha_3=\alpha_4=0 \]

equivalently,

\[ H_0:\mu_1=\mu_2=\mu_3=\mu_4 \]

\[ H_a:\text{At least one chemical mean differs} \]

The significance level is:

\[ \alpha=0.15 \]

GAD Analysis

chemical.fixed <- as.fixed(dat1$chemical)
bolt.random <- as.random(dat1$bolt)

model1 <- lm(strength ~ chemical.fixed + bolt.random)

gad(model1)
## $anova
## Analysis of Variance Table
## 
## Response: strength
##                Df Sum Sq Mean Sq F value    Pr(>F)    
## chemical.fixed  3  12.95   4.317  2.3761    0.1211    
## bolt.random     4 157.00  39.250 21.6055 2.059e-05 ***
## Residuals      12  21.80   1.817                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Results and Conclusion

The treatment test gives approximately

\[ F=2.3761 \]

with

\[ p=0.1211 \]

Since

\[ 0.1211<0.15 \]

we reject \(H_0\).

Conclusion: At the 0.15 significance level, there is sufficient evidence that the chemical agent affects tensile strength.

The block effect is also significant:

\[ F=21.6055,\qquad p=0.0000206 \]

Thus, bolt-to-bolt variability is a significant source of nuisance variability.


Question 2 - Completely Randomized Design

Now suppose the experiment is performed without blocking on the bolts. The observations are treated as a completely randomized design.

Data

chemical2 <- factor(rep(1:4, each = 5))

strength2 <- c(
  73, 68, 74, 71, 67,
  73, 67, 75, 72, 70,
  75, 68, 78, 73, 68,
  73, 71, 75, 75, 69
)

dat2 <- data.frame(
  chemical = chemical2,
  strength = strength2
)

dat2
##    chemical strength
## 1         1       73
## 2         1       68
## 3         1       74
## 4         1       71
## 5         1       67
## 6         2       73
## 7         2       67
## 8         2       75
## 9         2       72
## 10        2       70
## 11        3       75
## 12        3       68
## 13        3       78
## 14        3       73
## 15        3       68
## 16        4       73
## 17        4       71
## 18        4       75
## 19        4       75
## 20        4       69

Linear Effects Model

The completely randomized design model is

\[ Y_{ij}=\mu+\alpha_i+\epsilon_{ij} \]

where \(\alpha_i\) is the fixed effect of chemical.

Hypotheses

\[ H_0:\mu_1=\mu_2=\mu_3=\mu_4 \]

\[ H_a:\text{At least one population mean differs} \]

The significance level is:

\[ \alpha=0.15 \]

GAD Analysis

chemical2.fixed <- as.fixed(dat2$chemical)

model2 <- lm(strength ~ chemical2.fixed,
             data = dat2)

gad(model2)
## $anova
## Analysis of Variance Table
## 
## Response: strength
##                 Df Sum Sq Mean Sq F value Pr(>F)
## chemical2.fixed  3  12.95  4.3167  0.3863 0.7644
## Residuals       16 178.80 11.1750

Results and Conclusion

The treatment test gives approximately

\[ F=0.3863 \]

with

\[ p=0.7644 \]

Since

\[ 0.7644>0.15 \]

we fail to reject \(H_0\).

Conclusion: At the 0.15 significance level, there is insufficient evidence that the four chemical agents have different mean tensile strengths when the experiment is analyzed as a completely randomized design.


Question 3 - Comparison of the Two Designs

The RCBD analysis detected a significant chemical effect at the 0.15 significance level, while the completely randomized analysis did not.

For the RCBD:

\[ p_{\text{chemical}}=0.1211 \]

For the CRD:

\[ p_{\text{chemical}}=0.7644 \]

The difference occurs because blocking accounts for the substantial variability among the five bolts.

The bolt effect was highly significant:

\[ F=21.6055,\qquad p=0.0000206 \]

Therefore, the bolt of cloth is a significant source of nuisance variability.

Conclusion: Blocking on bolt is beneficial for this experiment because it removes substantial nuisance variation from the error term and provides a more sensitive test of the chemical treatment effect.

Complete Code

library(GAD)

chemical <- factor(rep(1:4, each = 5))
bolt <- factor(rep(1:5, times = 4))

strength <- c(
  73, 68, 74, 71, 67,
  73, 67, 75, 72, 70,
  75, 68, 78, 73, 68,
  73, 71, 75, 75, 69
)

dat1 <- data.frame(chemical, bolt, strength)

chemical.fixed <- as.fixed(dat1$chemical)
bolt.random <- as.random(dat1$bolt)

model1 <- lm(strength ~ chemical.fixed + bolt.random)

gad(model1)

chemical2 <- factor(rep(1:4, each = 5))

strength2 <- c(
  73, 68, 74, 71, 67,
  73, 67, 75, 72, 70,
  75, 68, 78, 73, 68,
  73, 71, 75, 75, 69
)

dat2 <- data.frame(
  chemical = chemical2,
  strength = strength2
)

chemical2.fixed <- as.fixed(dat2$chemical)

model2 <- lm(strength ~ chemical2.fixed,
             data = dat2)

gad(model2)