1 Assignment on Randomised Complete Block Design (RCBD)

1.1 Question 1

  • Linear Effects Equation for RCBD:

    • \(y_{i,j,k} = \mu + \tau_{i} + \beta_{j} + \epsilon _{i,j,k}\)

      Where \(\mu\): Population Mean; \(\tau_{i}\): Treatment effect for population i;
      \(\beta_{j}\) : Block effect for jth block; \(\epsilon_{i,j,k}\): Error corresponding to ith population, jth block,and kth number of replication.

  • Hypothesis:
    • \(H_{0}: \tau_{i} = 0\) \(\forall {i}\)
    • \(H_{a}: \tau_{i} \neq 0\) \(\forall i\)
library(GAD)
Obs<- c(73, 68, 74, 71, 67,
        73, 67, 75, 72, 70,
        75, 68, 78, 73, 68,
        73, 71, 75, 75, 69)
Chemical<- c(rep(1:4, each = 5))
Chemical
##  [1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4
Bolts<- c(rep(seq(1,5),4))
Bolts
##  [1] 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
#Q1
Chemical<- as.fixed(Chemical)
Bolts<- as.fixed(Bolts)

model<- lm(Obs~Chemical+Bolts)
model
## 
## Call:
## lm(formula = Obs ~ Chemical + Bolts)
## 
## Coefficients:
## (Intercept)    Chemical2    Chemical3    Chemical4       Bolts2       Bolts3  
##       72.35         0.80         1.80         2.00        -5.00         2.00  
##      Bolts4       Bolts5  
##       -0.75        -5.00
gad(model)
## Analysis of Variance Table
## 
## Response: Obs
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## Chemical  3  12.95   4.317  2.3761    0.1211    
## Bolts     4 157.00  39.250 21.6055 2.059e-05 ***
## Residual 12  21.80   1.817                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  • Comment: The P-value (0.1211) is less than the \(\alpha\) = 0.15. Therefore, we reject the null hypothesis, considering significant block effect.

1.2 Question 2.

  • Linear Effects Equation for CRD:

    • \(y_{i,j} = \mu + \tau_{i} + \epsilon _{i,j}\)

      Where \(\mu\): Population Mean; \(\tau_{i}\): Treatment effect for population i;
      \(\epsilon_{i,j}\): Error corresponding to ith population and jth block.

  • Hypothesis:
    • \(H_{0}: \tau_{i} = 0\) \(\forall {i}\)
    • \(H_{a}: \tau_{i} \neq 0\) \(\forall i\)

Considering the block effect as insignificant.

model2<- lm(Obs~Chemical)
gad(model2)
## Analysis of Variance Table
## 
## Response: Obs
##          Df Sum Sq Mean Sq F value Pr(>F)
## Chemical  3  12.95  4.3167  0.3863 0.7644
## Residual 16 178.80 11.1750
  • comment: As the p value(0.7644) obtained is higher than the \(\alpha\) = 0.15, which results in failing to reject the null hypothesis.

1.3 Question 3.

  • In question 1, we considered the block effect significant while performing ANOVA, we got the p-value less than 0.15. Therefore, we rejected null hypothesis, so there is significant difference in chemicals.

  • Whereas in question 2, we ignored the block effect while performing ANOVA, we got the p-value higher than 0.15. Therefore, we failed to reject the null hypothesis, so there is no significant chemical difference.

  • We believe that the Bolt of cloth represents significant nuisance variability, as it has forced us to change the judgment to reject the null hypothesis with blocking and failed to reject it without blocking.

2 Complete R Code

library(GAD)

Obs<- c(73, 68, 74, 71, 67,
        73, 67, 75, 72, 70,
        75, 68, 78, 73, 68,
        73, 71, 75, 75, 69)
?rep
Chemical<- c(rep(1:4, each = 5))
Chemical
Bolts<- c(rep(seq(1,5),4))
Bolts

#Q1
Chemical<- as.fixed(Chemical)
Bolts<- as.fixed(Bolts)

model<- lm(Obs~Chemical+Bolts)
model
gad(model)

#Q2
model2<- lm(Obs~Chemical)
gad(model2)