Solution file for additional exercise 6.1
-----------------------------------------

Data: measurements of iron in the livers of white rats.
Rats were randomly allocated to 5 diets (A-E), with 10 rats per diet.

The data constitute 5 independent samples with continuous outcome, 
and the model immediately suggested is a one-way ANOVA. 

MTB > WOpen "h:\VHM\VHM802\Data_csv\hs06_1.csv";
SUBC>   FType;
SUBC>     CSV;
SUBC>   DecSep;
SUBC>     Period;
SUBC>   Field;
SUBC>     Comma;
SUBC>   TDelimiter;
SUBC>     DoubleQuote.
Retrieving worksheet from file: 'h:\VHM\VHM802\Data_csv\hs06_1.csv'
Worksheet was saved on 12/02/2011

MTB > Oneway 'iron' 'diet';
SUBC>   GBoxplot.
One-way ANOVA: iron versus diet 

Source  DF      SS     MS      F      P
diet     4  127.82  31.96  12.44  0.000
Error   45  115.60   2.57
Total   49  243.42

S = 1.603   R-Sq = 52.51%   R-Sq(adj) = 48.29%

                         Individual 95% CIs For Mean Based on
                         Pooled StDev
Level   N   Mean  StDev    +---------+---------+---------+---------
A      10  5.693  2.406                           (----*-----)
B      10  2.906  1.983             (-----*----)
C      10  2.823  1.621             (----*----)
D      10  1.584  0.562       (----*----)
E      10  1.090  0.428    (----*-----)
                           +---------+---------+---------+---------
                         0.0       2.0       4.0       6.0
Pooled StDev = 1.603

Comments:
---------
The standard deviations are clearly not equal in the 5 groups, but seem
to increase almost linearly with the mean. We therefore try a log
transformation.

MTB > let c3=ln(c1)
MTB > name c3 'ln(iron)'
MTB > Oneway 'ln(iron)' 'diet';
SUBC>   GBoxplot;
SUBC>   GFourpack.
One-way ANOVA: ln(iron) versus diet 

Source  DF      SS     MS      F      P
diet     4  14.902  3.725  15.72  0.000
Error   45  10.665  0.237
Total   49  25.567

S = 0.4868   R-Sq = 58.29%   R-Sq(adj) = 54.58%

                           Individual 95% CIs For Mean Based on
                           Pooled StDev
Level   N    Mean   StDev  -----+---------+---------+---------+----
A      10  1.6517  0.4579                             (-----*----)
B      10  0.8741  0.6465                (-----*----)
C      10  0.8939  0.5584                 (----*----)
D      10  0.4056  0.3447         (----*----)
E      10  0.0259  0.3559  (----*-----)
                           -----+---------+---------+---------+----
                              0.00      0.60      1.20      1.80
Pooled StDev = 0.4868
 
Residual Plots for ln(iron) 

MTB > PPlot 'lniron';
SUBC>   Normal;
SUBC>   Symbol;
SUBC>   FitD;
SUBC>   Grid 2;
SUBC>   Grid 1;
SUBC>   MGrid 1;
SUBC>   Panel 'diet'.
Probability Plot of lniron 

Comments:
---------
The 1-way ANOVA model:
  ln(iron_i) = mu_diet(i) + eps_i, i=1,...,50
  where the errors eps_i are N(0,sigma^2)

seems to describe the data well. The standard deviations are roughly the
same in the 5 groups, the values in each group do not show strong
deviations from normality (difficult to assess with only 10 obs.), and
the residual plot and normal plot look fine.

A Box-Cox analysis points to an optimal power of lambda=-0.3, and we may
also consider the rounded "nice" value of -0.5 (this is Minitab's choice
in the General Regression menu, with a 95% CI for lambda of (-0.685,0.075)). 
At both scales the analysis of residuals do not detect any major problems 
with model assumptions. For simplicity, it might be
preferable to work with the log transformation (and that's what we'll do
for this solution).

The ANOVA table is given above, and the F-statistic for testing all
groups having equal means is F=15.72 and clearly significant in F(4,45).

The text defines 4 contrasts by their coefficients. These are seen to
be pairwise orthogonal, because for any pair the sum of products of
coefficients is zero. For example, for contrasts 1 and 2: 
  1*1 + 1*(-1) + 0*2 + 0*0 + 0*0 = 0
or for contrasts 2 and 3:
  1*2 + 1*2 + (-2)*2 + 0*3 + 0*3 = 0.

Contrast           Estimate   SE    SS   SS(%)   t     P(t)   F(Schef) P(Schef)
-------------------------------------------------------------------------------
beef vs pork         0.778  .2177  3.03  20.3   3.57  0.0001    3.19    0.022  
mammals vs poultry   0.738  .3771  0.91   6.1   1.96  0.057     0.96    0.44
animal vs vegetab.   5.545  .8432 10.25  68.8   6.58  0.0000   10.8     0.0000
beans vs oats        0.380  .2177  0.72   4.8   1.75  0.088     0.76    0.56 
-------------------------------------------------------------------------------

formulae:  SS=(estimate^2)/[(w_1^2+...+w_5^2)/10]
           SS (%) = SS / SSTrT (SSTrT=14.902)
           t=Est/SE=sqrt(SS/MSE) (MSE=0.237)
           P(t) ~ t(45)
           F(Scheffe)=SS/4/MSE (or t^2/4)
           P(Scheffe) ~ F(4,45)

Conclusions:
------------
Just looking at the SS-values for the contrasts, it is clear that the
last one (animal vs. vegetable) is accounting for a large proportion (69%) 
of the variation between diets.

Using ordinary t-tests, all 4 contrasts are interesting
- two of them are clearly significant and the other two are borderline
significant. This method assumes that all contrasts were preplanned, and
works with individual error levels of 5%.

It is possible to do a Bonferroni correction for carrying out 4 tests by
multiplying all P-values by 4 (not shown). This approach would make the 2 
"interesting" contrasts non-significant. Note however that the method 
still assumes that all contrasts were preplanned. We could also do a Holm 
correction, by which we would then multiply the P-values by 4,3,2,1; the
conclusions are the same.

Finally, the Scheffe method takes into account both the multiple tests and
contrasts being suggested by the data. It is the most conservative of
the 3 methods but also the most "safe". It shows the animal vs vegetable
contrast to be still highly significant, and also the beef vs pork
contrast is still significant. There is evidence that beef and pork
diets are different, and that animal and vegetable diets are clearly
different.