Question 1

Null Hypothesis:

Ho:μ1=μ2=μ3=μ3=μ4

Alternative Hypothesis:

Ha : atleast one μ’s differs

The null hypothesis for fixed effect can be written as

Ho:τi=0 for all i

alternative hypothesis:

Ha : τi≠0 for some i

Linear Effects model for RCBD can be represented as :

yij=μ+τi+βj+ϵij

μ is the grand mean value

τi is the fixed effects for treatment i

Where βj is the block effect for j

Where ϵij is the random error

Getting our our data we have

chemical<- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolt<- c(rep(seq(1,5),4))
obs<- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
library(GAD)
chemical <- as.fixed(chemical)
bolt<- as.fixed(bolt)
model <- lm(obs~chemical+bolt)
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    
## bolt      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

since the p-value 0.1211<0.15 we reject the null hypothesis, stating that the chemical treatment has no significant effect on the strength of the type of cloth

Question 2

Null Hypothesis is the Ho:μ1=μ2=μ3=μ4

Alternative Hypothesis: Ha : Atleast one μ’s differs The null hypothesis for fixed effect can be written as

linear effect for Completely random design is yij=μ+τi+ϵij μ is the grand mean τi is the fixed effects for treatment i ϵij is the random error

chemical1<- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
rep<- c(rep(seq(1,5),4))
obs <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)
chemical1<- as.fixed(chemical1)
model2 <- lm(obs~chemical1)
gad(model2)
## Analysis of Variance Table
## 
## Response: obs
##           Df Sum Sq Mean Sq F value Pr(>F)
## chemical1  3  12.95  4.3167  0.3863 0.7644
## Residual  16 178.80 11.1750

Since p-value 0.7644>0.15 , we fail to reject null hypothesis, stating that the chemical treatment has no effect on the strength of the type of cloth

Question 3

In the Question 01: we blocked bolts, and we noticed we had a p-value of (0.1211) which was less than (reference value of alpha=0.15), we then rejected the null hypothesis.

In the Question 2: however, we did not account for blocking on bolts (replication), which resulted in a p value of 0.7644 (Insignificant at alpha=0.15) and hence we cannot reject the null hypothesis.

This indicates that when we block, we included nuisance variability, forcing us to reject the null hypothesis, which is what we must do when we are aware of nuisance variability.

We can assert that if we do not block, our decisions will be greatly impacted. Therefore, regarding the question of whether to block or not to block,

we must block whenever it is required to avoid making a mistake decision.

Yes, I believe it represented a considerable amount of nuisance variability as it altered our judgment to reject the null hypothesis with blocking and fail to reject it without blocking