Solution file for additional exercise 2.5:
------------------------------------------

First part of the solution examines the 1-way ANOVA assumptions by
descriptive statistics for each group of origins:

MTB > WOpen "R:\data_csv\hs02_5.csv";
SUBC>   FType;
SUBC>     CSV;
SUBC>   DecSep;
SUBC>     Period;
SUBC>   Field;
SUBC>     Comma;
SUBC>   TDelimiter;
SUBC>     DoubleQuote.
Retrieving worksheet from file: ‘R:\data_csv\hs02_5.csv’
Worksheet was saved on 12/01/2011

MTB > Describe 'calv1serv';
SUBC>   By 'origin';
SUBC>   Mean;
SUBC>   SEMean;
SUBC>   StDeviation;
SUBC>   QOne;
SUBC>   Median;
SUBC>   QThree;
SUBC>   Minimum;
SUBC>   Maximum;
SUBC>   Skewness;
SUBC>   Kurtosis;
SUBC>   N;
SUBC>   NMissing.
Descriptive Statistics: calv1serv 

Variable   origin   N  N*   Mean  SE Mean  StDev  Minimum     Q1  Median
calv1serv  1       13   0   79.8     10.2   37.0     43.0   53.5    69.0
           2       38   0  91.76     7.58  46.71    41.00  63.00   77.00
           3       73   0  76.93     3.83  32.76    33.00  53.00   70.00

Variable   origin      Q3  Maximum  Skewness  Kurtosis
calv1serv  1        100.0    162.0      1.28      0.75
           2       102.00   221.00      1.31      0.95
           3        91.50   205.00      1.85      4.48

MTB > PPlot 'calv1serv';
SUBC>   Normal;
SUBC>   Symbol;
SUBC>   FitD;
SUBC>   Grid 2;
SUBC>   Grid 1;
SUBC>   MGrid 1;
SUBC>   Panel 'origin'.
Probability Plot of calv1serv 

Comments on descriptive statistics:
-----------------------------------
Quite clearly a normal distribution does not describe these data well at
all, because they are markedly right-skewed (the computed skewness in the 3
groups range between 1.3 and 1.8). The normal plots are clearly curved, with a
similar shape in all three groups. The tests for normal distribution have P-values
below 0.015 (and much lower for the two groups of substantial size). Therefore,
the need of transformation is obvious. Note finally, that the standard
deviation is largest in the group with the largest mean, something that
we can hope to correct by transformation (although the major concern is
not with the homoscedasticity).

The easiest way to carry out a Box-Cox transformation analysis in Minitab is to use the
Regression-General Regression menu, in which we must specify the predictor (origin) as
categorical.

MTB > GReg 'calv1serv' = origin;
SUBC>   Categorical 'origin';
SUBC>   Constant;
SUBC>   Confidence 95.0;
SUBC>   Coding -1;
SUBC>   Boxcox;
SUBC>   TEquation;
SUBC>   TCoef;
SUBC>   TSummary;
SUBC>   TANOVA;
SUBC>   TDiag 0.
 
General Regression Analysis: calv1serv versus origin 

Box-Cox transformation of the response with rounded lambda = -0.5
The 95% CI for lambda is (-1.095, -0.265)
...

Comments:
---------
The analysis does not give us the estimated optimal lambda-value, only
the rounded value at -0.5 and the CI. We could base our evaluation of
the model on inverse square-root transformed scale on the residuals
obtained with the General Regression command, but we will instead use
descriptive statistics in the same way as above. Note that we can do
this only because of the simplicity of the 1-way ANOVA model; in more
complex cases we would rely on the residuals.

MTB > Let c3 = -1/sqrt('calv1serv')
MTB > name c3 'invroot'
MTB > Describe  'invroot';
SUBC>   By 'origin';
SUBC>   Mean;
SUBC>   SEMean;
SUBC>   StDeviation;
SUBC>   QOne;
SUBC>   Median;
SUBC>   QThree;
SUBC>   Minimum;
SUBC>   Maximum;
SUBC>   Skewness;
SUBC>   Kurtosis;
SUBC>   N;
SUBC>   NMissing.
Descriptive Statistics: invroot 

Variable  origin   N  N*      Mean  SE Mean    StDev   Minimum        Q1
invroot   1       13   0  -0.11904  0.00647  0.02333  -0.15250  -0.13726
          2       38   0  -0.11285  0.00394  0.02428  -0.15617  -0.12602
          3       73   0  -0.12010  0.00246  0.02098  -0.17408  -0.13736

Variable  origin    Median        Q3   Maximum  Skewness  Kurtosis
invroot   1       -0.12039  -0.10139  -0.07857      0.33     -0.66
          2       -0.11399  -0.09906  -0.06727      0.13     -0.74
          3       -0.11952  -0.10454  -0.06984      0.17     -0.26

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

Comments:
---------
The transformation worked well. The three groups have almost the same
standard deviations and distributions with no substantial departures
from normality (and non-significant P-values of the normality tests).