> #4.2.1 Dried Egg Example
> x<-rbind(
+ c(9.7,8.7,NA,5.4,5.0,NA,NA,NA,NA,NA),
+ c(NA,9.6,8.8,NA,NA,5.6,NA,NA,NA,3.6),
+ c(NA,9.0,NA,7.3,NA,3.8,4.3,NA,NA,NA),
+ c(9.3,NA,8.7,NA,6.8,NA,3.8,NA,NA,NA),
+ c(10.0,NA,NA,7.5,NA,NA,NA,4.2,NA,2.8),
+ c(NA,9.6,NA,NA,NA,NA,5.1,4.6,3.6,NA),
+ c(NA,9.8,NA,NA,7.4,NA,NA,4.4,NA,3.8),
+ c(NA,NA,NA,NA,9.4,NA,6.3,NA,5.1,2.0),
+ c(9.4,9.3,8.2,NA,NA,NA,NA,NA,3.3,NA),
+ c(NA,NA,NA,8.7,9.0,6.0,NA,NA,3.3,NA),
+ c(9.7,NA,NA,NA,NA,6.7,6.6,NA,NA,2.8),
+ c(NA,NA,9.3,8.1,NA,NA,NA,NA,3.7,2.6),
+ c(9.8,NA,NA,NA,NA,7.3,NA,5.4,4.0,NA),
+ c(NA,NA,9.0,8.3,NA,NA,4.8,3.8,NA,NA),
+ c(NA,NA,9.3,NA,8.3,6.3,NA,3.8,NA,NA))
> colnames(x)<-c("a","b","c","d","e","f","g","h","i","j")
> 
> #Have to get ranks from 10*(20-x) as xtotab assumes integer and ranks inverted 
> #(see text - rank 1 given to highest score and rank 4 to lowest score)
> xtotab(10*(20-x))
  1 2 3 4
a 6 0 0 0
b 4 2 0 0
c 3 2 1 0
d 0 5 1 0
e 2 2 1 1
f 0 2 3 1
g 0 2 3 1
h 0 0 4 2
i 0 0 2 4
j 0 0 0 6
> U(10*(20-x))
     [,1] [,2] [,3] [,4]
[1,]   15    0    0    0
[2,]    0   15    0    0
[3,]    0    0   15    0
[4,]    0    0    0   15
> BlockAnalysis(10*(20-x))
$Cri
    Location    Dispersion
a -3.1176915  2.323790e+00
b -2.4248711  7.745967e-01
c -1.7320508  1.874559e-15
d -0.6928203 -2.323790e+00
e -0.6928203  8.706156e-16
f  0.6928203 -1.549193e+00
g  0.6928203 -1.549193e+00
h  1.7320508 -7.745967e-01
i  2.4248711  7.745967e-01
j  3.1176915  2.323790e+00

$partition
           df    SS       pvalue
Location    9 39.12 1.096050e-05
Dispersion  9 22.80 6.661459e-03
Residual    9 10.08 3.440477e-01
Total      27 72.00 5.788778e-06

>