Question 1:
Null Hypothesis Testing
H0:μ1=μ2=μ3=μ4
Linear Equation
yi,j=μ+τi+βj+ϵi,j
library(GAD)
chemical is a factor (4 levels), bolts is observed (n=4).
chemical<-c(rep(1,4),rep(2,4),rep(3,4),rep(4,4))
chemical
## [1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
obs<-c(73,68,74,71,
73,67,75,72,
75,68,78,73,
73,71,75,75)
obs
## [1] 73 68 74 71 73 67 75 72 75 68 78 73 73 71 75 75
chemical<-as.fixed(chemical)
Running the Model with Blocked Observations:
model<-lm(obs~chemical)
gad(model)
## $anova
## Analysis of Variance Table
##
## Response: obs
## Df Sum Sq Mean Sq F value Pr(>F)
## chemical 3 14.187 4.7292 0.4739 0.7062
## Residuals 12 119.750 9.9792
Question 2
Null Hypothesis Testing
H0:μ1=μ2=μ3=μ4
Linear Equation
yi,j=μ+τi+ϵi,j
Solution (4 levels), bolts (n=4), sequencing in 4s.
solution<-c(rep(1,4),rep(2,4),rep(3,4),rep(4,4))
solution
## [1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
bolts<-c(seq(1,4),seq(1,4),seq(1,4),seq(1,4))
bolts
## [1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
obs<-c(73,68,74,71,
73,67,75,72,
75,68,78,73,
73,71,75,75)
obs
## [1] 73 68 74 71 73 67 75 72 75 68 78 73 73 71 75 75
solution<-as.random(solution)
bolts<-as.random(bolts)
model<-lm(obs~solution+bolts)
gad(model)
## $anova
## Analysis of Variance Table
##
## Response: obs
## Df Sum Sq Mean Sq F value Pr(>F)
## solution 3 14.187 4.729 2.7349 0.1057217
## bolts 3 104.188 34.729 20.0843 0.0002515 ***
## Residuals 9 15.563 1.729
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
as.random(bolts)
used for unblocked (no blocking)
design
Question 3
In the first case (with blocking), P = 0.7062 > 0.15, so we failed to reject the null hypothesis—no significant treatment differences. In the second case (without blocking), P = 0.105 < 0.15, so we rejected the null hypothesis—indicating treatment differences. This shows that blocking helped control nuisance variability, reducing the residual mean square from 9.979 (CRD) to 1.729 (RCBD), which increased the F-value and improved test sensitivity.