Flipped assignment 10

Ouestion number (a)

The linear effect equation is:

\(Y_{ij} =\mu+\tau_{ij}+\epsilon_{ij}\)

where,

\(\mu\) = means, \(\tau\) = effect, \(\epsilon\) = error

The required hypothesis is :

\(H\_{0}:\tau\_{i}=0\) \(H_{0}:\tau_{i} \neq 0\)

Ouestion number (b)

model<- c(0.34, 0.12, 1.23, 0.70, 1.75,.12,0.91, 2.94, 2.14, 2.36, 2.86, 4.55, 6.31, 8.37, 9.75, 6.09, 9.82, 7.24, 17.15, 11.82, 10.97, 17.20, 14.35, 16.82)
factor<-c(rep(1,6), rep(2,6), rep(3,6), rep(4,6))
df<-data.frame(model,factor)
df

##    model factor
## 1   0.34      1
## 2   0.12      1
## 3   1.23      1
## 4   0.70      1
## 5   1.75      1
## 6   0.12      1
## 7   0.91      2
## 8   2.94      2
## 9   2.14      2
## 10  2.36      2
## 11  2.86      2
## 12  4.55      2
## 13  6.31      3
## 14  8.37      3
## 15  9.75      3
## 16  6.09      3
## 17  9.82      3
## 18  7.24      3
## 19 17.15      4
## 20 11.82      4
## 21 10.97      4
## 22 17.20      4
## 23 14.35      4
## 24 16.82      4

boxplot(model~factor)

• The sample size is too small to conduct a reasonable test for normality.

• From the boxplot we obtained as a result, we can see that the variance is not constant.

  Question number (c)

aov.model<-aov(model~factor, data=df)
summary(aov.model)

##             Df Sum Sq Mean Sq F value  Pr(>F)    
## factor       1  672.0   672.0   149.9 2.7e-11 ***
## Residuals   22   98.6     4.5                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(aov.model)

From the plots, we can conclude that the residuals are not normally distributed and the data does not have constant variance.

Question number (d)

library(MASS)
boxcox(model~factor, data=df)

model2<-(sqrt(model)-1)/0.5
df2<-data.frame(model2,factor)
aov.model<-aov(model2~factor, data=df2)
summary(aov.model)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## factor       1  130.1  130.12   251.2 1.62e-13 ***
## Residuals   22   11.4    0.52                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

plot(aov.model)

After transformation variance among different estimation methods has become much more stable.

Question number (e)

kruskal.test(model~factor, data=df)

## 
##  Kruskal-Wallis rank sum test
## 
## data:  model by factor
## Kruskal-Wallis chi-squared = 21.156, df = 3, p-value = 9.771e-05

Since the p-value is less than that of the alpha(0.05), we reject Ho.

Complete R- Code:

model<- c(0.34, 0.12, 1.23, 0.70, 1.75,.12,0.91, 2.94, 2.14, 2.36, 2.86, 4.55, 6.31, 8.37, 9.75, 6.09, 9.82, 7.24, 17.15, 11.82, 10.97, 17.20, 14.35, 16.82)
factor<-c(rep(1,6), rep(2,6), rep(3,6), rep(4,6))
df<-data.frame(model,factor)
df

boxplot(model~factor)


aov.model<-aov(model~factor, data=df)
summary(aov.model)
plot(aov.model)

library(MASS)
boxcox(model~factor, data=df)

model2<-(sqrt(model)-1)/0.5
df2<-data.frame(model2,factor)
aov.model<-aov(model2~factor, data=df2)
summary(aov.model)
plot(aov.model)

kruskal.test(model~factor, data=df)