importing and sortingd data

# Question A #

Linear effects of Equation

Xij=mu+Ti+ Eij

Here mu=Grand mean

Ti=Effect of population

Eij=RandomError

#Testing Of hypothesis:

# Nullhypothesis(Ho): u1=u2=u3=u4

#Alternative Hypothesis (Ha): Atleast one of u differs.

#Question B

#install.packages("tidyr")

library(tidyr)

#install.packages("dplyr")

library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
M1<-c(.34,.12, 1.23,.70,1.75,.12)

M2<-c(.91,2.94,2.14,2.36,2.86,4.55)

M3<-c(6.31,8.37,9.75,6.09,9.82,7.24)

M4<-c(17.15,11.82,10.97,17.20,14.35,16.82)

dat<-data.frame(M1,M2,M3,M4)
d<-c(M1,M2,M3,M4)

R<-c(rep(1,6),rep(2,6),rep(3,6),rep(4,6))

data2 <- pivot_longer(data=dat,c(M1,M2,M3,M4))

qqnorm(data2$value)

#conclusion:

#Therefore, we can observe the data appears to be quiet normal based on q-q norm

boxplot(d~R,main="Boxplot",xlab="Method",ylab="observations")

#conclusion:

By the boxplot we can observe that variance are not equal and fourth variance is high.

#Question C

#kruskal wallis test

kruskal.test(value~name,data = data2)
## 
##  Kruskal-Wallis rank sum test
## 
## data:  value by name
## Kruskal-Wallis chi-squared = 21.156, df = 3, p-value = 9.771e-05

#conclusion

Therefore by kruskals test P value is 9.771e-05 which is very low, so we can reject the null hypothesis.

#Question D

library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:dplyr':
## 
##     select
boxcox(value~name,data=data2)

lamda=0.5

data3<-(d^lamda)


boxplot(data3~R,main="boxplot",xlab="Method",ylab="observations")

#conclusion:

Based on boxplot we can say that variances are equal and sample size is small so that we are going for kruskals test again

anova_model<-aov(data3~R)

summary(anova_model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## R            1  32.53   32.53   251.2 1.62e-13 ***
## Residuals   22   2.85    0.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

#conclusion:

Based on this P value is low, so we can reject null hypothesis