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##      speed           dist       
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#we reject Ho if  T2>c2
#Ho: The are appears to be parallel
n1=32
n2=32
n=(1/n1)+(1/n2)
y1_bar=matrix(c(15.97,15.91,27.19,22.75),nrow=4,byrow=TRUE)
y2_bar=matrix(c(12.34,13.91,16.66,21.94),nrow=4,byrow=TRUE)
sp=matrix(c(7.164,6.047,5.693,4.701,6.047,15.89,8.492,5.856,5.693,8.492,29.36,13.98,4.701,5.856,13.98,22.32),nrow=4,byrow=TRUE)
y1_bar
##       [,1]
## [1,] 15.97
## [2,] 15.91
## [3,] 27.19
## [4,] 22.75
y2_bar
##       [,1]
## [1,] 12.34
## [2,] 13.91
## [3,] 16.66
## [4,] 21.94
c=matrix(c(-1,1,0,0,0,-1,1,0,0,0,-1,1),nrow = 3,byrow=TRUE)
t2=t(y1_bar-y2_bar)%*%t(c)%*%solve((n)*c%*%sp%*%t(c))%*%c%*%(y1_bar-y2_bar)
t2
##          [,1]
## [1,] 74.25857
c_sq=((n1+n2-2)*3*qf(0.99,3,n1+n2-4))/(n1+n2-4)#p=4
c_sq
## [1] 12.79026
#Here T2>c2 ,se we reject Ho (T2=74.25857 and c2=12.79026)
#So, the profiles are not parallel
#if the profiles are appears to be parallel,test it for the level of profiles
n1=32
n2=32
y_bar=n1/(n1+n2)*y1_bar+n2/(n1+n2)*y2_bar
y_bar
##        [,1]
## [1,] 14.155
## [2,] 14.910
## [3,] 21.925
## [4,] 22.345
T2=(n1+n2)*t(y_bar)%*%t(c)%*%solve(c%*%sp%*%t(c))%*%c%*%y_bar#statistic
T2
##          [,1]
## [1,] 254.1824
c2=((n1+n2-1)*3*qf(0.99,3,n1+n2-4+1))/(n1+n2-4+1)
c2
## [1] 12.76507
#here T2=254.1824 greater than c2=12.76507
#hence we reject H0:c*mu=0
#c)
one_dash=matrix(1,ncol=4,byrow=TRUE)
one_dash
##      [,1] [,2] [,3] [,4]
## [1,]    1    1    1    1
T2=((one_dash%*%(y1_bar-y2_bar))/(sqrt(n*one_dash%*%sp%*%t(one_dash))))**2
T2
##          [,1]
## [1,] 28.04918
t2=qt(0.99/2,n1+n2-2)
t2
## [1] -0.01258412
#here t2>t2
#so we reject the null hypothesis