o working with our size-finger length-htime
SF	htime
11	-.05	.15
12	.39	.23
13	-.12	-.40
21	-.21	-.82
22	-.94	.30
23	.00	-.20
31	-1.19	-1.00
32	-1.46	-1.95
33	-1.81	-.72

o interested in 2 factor experiment
  - for us :  does size matter?  does finger length
    matter?
  - model
Y(ijt) = m +a(i) + b(i) + (ab)(ij) +e(ijt)
  - is there interaction
  - what is estimable?
  - finish orthogonal (trend) contrasts & use
  - then 2 way additive
  - then 2 way complete
ALL MUST UNDERGO ASSUMPTION CHECKING AND
  DO APPROPRIATE FOLLOW UPS

o I use definitional rather than computation
 formulas.  will use equal sample sizes (see
  text for unequal). for now will ignore
  least squares derivation (revisit later)
 o trend 
  - A.2
  - what do you notice about trend and
   trend x trend contrasts?
 inner product = 0 ie.
c(1)*C(1)'  + ... c(ij)*c(ij)' = 0
Definition of orthogonal
 - on equal sample sizes and equally
  spaced factor levels
  - why special?
theorem:  in the case wehre we have a SS based
 on k df then k-1 orthogonal contrasts exist
 so that ssc(1)+...+ssc(k-1) = SS
where 
                         ------
  ssc(q) = (sum of c(ij)* Y (ij .)-squared
           ------------------------------
           (sum of (c(ij)-squared/r(ij))
  - like for SS (these derived from normal
 equations)
              -------    ----  
SSA = br (sum (Y(i..) - Y(...))-squared)
              -------    ----  
SSB = ar (sum (Y(.j.) - Y(...))-squared)

              -------    ---- 
SSAB = (sum (Y(ij.) - Y(i..) -
  -----    ------  
   Y(.j.) + Y(...))-squared)

              -------    ----  
SSTOT =  (sum (Y(ijt) - Y(...))-squared)

  - want to test H0: factor A treatment effects same
  OR H0: factor B treatment effects same
  OR H0: interaction is zero
think of 1-way ANOVA
                  ----
MSE = sum((Y(it) - Y(i.)-squared) / ( n - v)
here # gps is 9 (in cell means model) and 
  sample size is 9...oops! zero df
  - Need replicates or do 'orthogonal trick' =
  sacrifice some interaction contrasts to
  get an error term (if feasible)
  - Example  pool SSAB do SSAB/SSE against F;
 if non-significant do main effects
Note: my policy and Dean's is to check 
interaction and if significant only do
cell mean comparisons individually..using MSE
from global test.
o what if we were sure there was
 no interaction (without data snooping)
  - use
Y(ijt) = m + a(i) + b(j) + e(ijt)
and now can do usual 2-wa ANOVA,
 i.e. enough df
  - the least squares estimate
  formulas are above
  - the table :
Source	SS	df		MS	F
----------------------------------------
A		SSA	a-1		MSA	MSA/MSE
B		SSB	b-1		MSB	MSB/MSE
Error		SSE	n-a-b+1	MSE
Total		SStot	n-1	
  - Example
Note: get SSE = SStot - rest
  - followups
form: sum (c(j)*b(j)) contained in
          -------  
  sum(c(j)*Y(.j.)) +- w * 
     sqrt(MSE*sum(c(j)-squared/ar)
 sum(c(i)*Y(i..)) +- w * 
     sqrt(MSE*sum(c(i)-squared/br)
w is
Bonferonni= t(df=same as MSE, alpha/2m) A4
Scheffe=sqrt((a-1)*F(dfn=a-1,
 dfd=same as MSE, alpha) A6
Tukey=q(a,df=same as MSE,alpha) A8
Hsu & Dunnett1=t(.5)(a-1,
 df same as MSE, alpha) A9
Dunnett2 = t(.5)(a-1, 
 df same as MSE, alpha) A10
o 2-way complete model
  - the table
Source	SS	df		MS	F
--------------------------------
A		SSA	a-1		MSA	MSA/MSE
B		SSB	b-1		MSB	MSB/MSE
AB		SSAB (a-1)(b-1)	MSAB
Error		SSE	n-ab		MSE
Total		SStot	n-1	
  - Example
Note: get SSE = SStot - rest


SSAB = sum ((Y(ij.) - Y(...)-squared)
see above for other SS
  - MUST FIRST TEST INTERACTION
AND FIND NONSIGNIFICANT TO PROCEED
  - followups
see above