[o] GLIM 4, update 8 for Sun SPARCstation / Solaris on 17 Nov 2009 at 14:11:06
 [o]                  (copyright) 1992 Royal Statistical Society, London
 [o] 

In all 3 replications (occasions), the 8 treatment combinations
are assigned to 2 blocks as follows:

Block 1: (1)	ab 	ac	bc	(ABC is confounded) 
Block 2:  a	b	c	abc


 [i] ? $uni 24$
 [i] ? $dat anesth$
 [i] ? $rea
 [i] $REA? 9 8.5 7.5 12 9.5 10.5 8.5 18.5
 [i] $REA? 9.5 7.5 7.7 11.2 10.5 12.3 8.6 20.1
 [i] $REA? 10 8.2 6.9 12.3 11.6 11.1 8.6 18.7
 [i] ? $ca a=%gl(2,1):b=%gl(2,2):c=%gl(2,4):repl=%gl(3,8)$
 [i] ? $fac a 2 b 2 c 2 repl 3$
 [i] ? $data block$
 [i] ? $read
 [i] $REA? 1 2 2 1 2 1 1 2
 [i] $REA? 1 2 2 1 2 1 1 2
 [i] $REA? 1 2 2 1 2 1 1 2
 [i] ? $fac block 2$
 [i] ? $yvar anesth$
 [i] ? $fit$
 [o]    deviance =  296.07
 [o] residual df =   23
 [o] 
 [i] ? $fit+repl$
 [o]    deviance =  295.11 (change =  -0.9633)
 [o] residual df =   21    (change =  -2     )
 [o] 
 [i] ? $fit+block$
 [o]    deviance =  289.50 (change =   -5.607)
 [o] residual df =   20    (change =   -1    )
 [o] 
 [i] ? $fit+block.repl$
 [o]    deviance =  289.46 (change =  -0.04333)
 [o] residual df =   18    (change =  -2      )
 [o] 
 [i] ? $fit+a$
 [o]    deviance =  212.42 (change =   -77.04)
 [o] residual df =   17    (change =    -1   )
 [o] 
 [i] ? $fit+b$
 [o]    deviance =  191.51 (change =   -20.91)
 [o] residual df =   16    (change =    -1   )
 [o] 
 [i] ? $fit+c$
 [o]    deviance =  130.71 (change =   -60.80)
 [o] residual df =   15    (change =    -1   )
 [o] 
 [i] ? $fit+a.b$
 [o]    deviance =  38.668 (change =   -92.04)
 [o] residual df =  14     (change =    -1   )
 [o] 
 [i] ? $fit+a.c$
 [o]    deviance =  13.042 (change =   -25.63)
 [o] residual df =  13     (change =    -1   )
 [o] 
 [i] ? $fit+b.c$
 [o]    deviance =   6.4267 (change =   -6.615)
 [o] residual df =  12      (change =   -1    )
 [o] 
 [i] ? $fit+a.b.c$
 [o]    deviance =   6.4267 (change =   0.)
 [o] residual df =  12      (change =   0 )
 [o] 
 [i] ? $disp e m$
 [o]           estimate        s.e.     parameter
 [o]      1       9.200      0.5175      1
 [o]      2      0.4000      0.5175      REPL(2)
 [o]      3      0.5000      0.5175      REPL(3)
 [o]      4       1.000      0.5175      BLOCK(2)
 [o]      5      -2.400      0.5175      A(2)
 [o]      6      -3.100      0.5175      B(2)
 [o]      7     0.06667      0.5175      C(2)
 [o]      8     0.05000      0.7318      REPL(2).BLOCK(2)
 [o]      9     -0.1500      0.7318      REPL(3).BLOCK(2)
 [o]     10       7.833      0.5975      A(2).B(2)
 [o]     11       4.133      0.5975      A(2).C(2)
 [o]     12       2.100      0.5975      B(2).C(2)
 [o]     13       0.000     aliased      A(2).B(2).C(2)
 [o] scale parameter 0.5356
 [o] 
 [o] Current model:
 [o] 
 [o]   number of observations in model is 24
 [o] 
 [o]   y-variate  ANESTH
 [o]   weight     *
 [o]   offset     *
 [o] 
 [o] 
 [o]   probability distribution is NORMAL
 [o]              link function is IDENTITY
 [o]            scale parameter is to be estimated by the mean deviance
 [o] 
 [o] linear model:
 [o]     terms: 1+REPL+BLOCK+A+B+C+REPL.BLOCK+A.B+A.C+B.C+A.B.C


ANOVA

Source                 df      SS          MS       F            
Replication            2       0.9633       
Blocks(ABC)            1       5.607
ReplxBlock             2       0.04333
                             
A                      1       77.04      77.04    143.8     
B                      1       20.91      20.91    39.040      
C                      1       60.80      60.80    113.516     
AB                     1       92.04      92.04    171.844             
AC                     1       25.63      25.63    47.853    
BC                     1       6.615      6.615    12.351       
Error                  12      6.4267     0.5356                   

Total                  23      296.07    


ABC is untestable, and the f-test value are all significant.
To overcome this problem use different defining contracts for the 
diferent replications. for example use AC for the 2nd occasion,
and BC for the 3rd.  

The 8 treatment combinations are assigned to 2 blocks as follows:

Repl. 1:
Block 1: (1)	ab 	ac	bc	(ABC is confounded) 
Block 2:  a	b	c	abc

Repl. 2:
Block 1: (1)	b 	ac	abc	(AC is confounded) 
Block 2:  a	ab	c	bc

Repl. 3:
Block 1: (1)	a 	c	ac	(BC is confounded) 
Block 2:  b	ab	bc	abc

The revised ANOVA table is:

Source                 df      SS          MS       F            
Replication            2             
Blocks(ABC)            1       
ReplxBlock             2       
                             
A                      1       
B                      1       
C                      1       
AB                     1                   
AC*                    1       
BC*                     1 
ABC*          	       1
Error                  11

Total			23     

(* From replications which are not confounded)