4.42. Seven different hardwood concentrations are being studied to determine the

Question

4.42. Seven different hardwood concentrations are being studied to determine their effect on the strength of the paper produced. However, the pilot plant can only produce three runs each day. As days may differ, the analyst uses the bal- anced incomplete block design that follows. Analyze the data from this experiment (use = 0.05) and draw conclusions. Days Hardwood Concentration ( % ) 114 120 111 149 4 6 8 10 12 14 137 117 141 145 120 136 129 149 150 Days Hardwood Concentration (% ) 120 2 4 6 8 10 12 14 119 134 143 118 123 130 127

Explanation / Answer

N=matrix(c(1,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,1,1,0,1,1,0,0,0,1,1,0,
0,1,0,0,0,1,1,1,0,1,0,0,0,1),nrow=7,byrow=T)
Y=matrix(c(114,126,0,141,0,0,0,0,120,137,0,145,0,0,0,0,117,129,0,120,0,
0,0,0,149,150,0,136,120,0,0,0,143,118,0,0,119,0,0,0,123,130,117,0,134,0,0,0,127),nrow=7,,byrow=T)
Bi=colSums(Y)
Vj=rowSums(Y)
G=sum(Y)
r=3
k=3;v=7;b=7;lambda=1
Q=Vj-(1/k)*N%*%t(t(Bi))
C=r*diag(b)-t(N)%*%N/k
SStreatment=(k/(lambda*v))*(sum(Q^2))
SSblock=sum(Bi^2)/k-G^2/(b*k)
SStotal=sum(Y*Y)-G^2/(b*k)
SSerror=SStotal-SSblock-SStreatment
#BIBD ANOVA
SS=c(SStreatment,SSblock,SSerror,SStotal)
DF=c(v-1,b-1,b*k-b-v+1,b*k-1)
MST=SStreatment/(v-1)
MSE=SSerror/(b*k-b-v+1)
F=MST/MSE
Fcritical_value=qf(0.95,(v-1),(b*k-b-v+1))
if(F<Fcritical_value)
{ print("Accept Null Hypothesis")}else
{ print("Reject Null Hypothesis")}

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