Solution file for additional exercise 10.6
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Data on ratings of machines operated by different persons.
The data are unbalanced because the machines are operated
different number of times by the different persons.

- notation:
y_ijk = rating for replication k of machine i by person j,
   i = 1,2,3 (machines),
   j = 1,...,6 (person),
   k = 1,...,n_ij (replication; n_ij's are either 1, 2 or 3),
- the effect of machines is fixed (the 3 machines are of specific interest),
- the effect of persons should probably be taken as random because we have no interest in
the particular persons but in the general variability between persons, persons 
correspond to blocks, (one may also take persons as fixed)
- the effect of machines*persons should be taken as random (our interest is in
general differences between machines, not differences for specific
persons), no matter whether persons are taken as fixed or random,
- model: 
y_ijk = mu + alpha_i + B_j + (AB)_ij + eps_ijk, 
      where B_j's are assumed i.i.d. N(0,sigma_A^2),
      where AB_ij's are assumed i.i.d. N(0,sigma_AB^2), and
      where eps_ijk's are assumed i.i.d. N(0,sigma^2).
- type of analysis: two-way ANOVA with mixed effects (both fixed and random),

MTB > WOpen "h:\vhm\vhm802\data_csv\hs10_6.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\hs10_6.csv'
Worksheet was saved on 03/03/2011

MTB > Name c4 "RESI1" c5 "SRES1" c6 "TRES1"
MTB > GLM 'rating' = machine| person;
SUBC>   Random 'person';
SUBC>   Brief 2 ;
SUBC>   EMS;
SUBC>   Means machine;
SUBC>   Residuals 'RESI1';
SUBC>   SResiduals 'SRES1';
SUBC>   TResiduals 'TRES1';
SUBC>   GFourpack;
SUBC>   RType 2 .
General Linear Model: rating versus machine, person 

Factor   Type    Levels  Values
machine  fixed        3  1, 2, 3
person   random       6  1, 2, 3, 4, 5, 6

Analysis of Variance for rating, using Adjusted SS for Tests

Source          DF   Seq SS   Adj SS  Adj MS      F      P
machine          2  1648.66  1238.20  619.10  16.57  0.001 x
person           5  1008.76  1011.05  202.21   5.17  0.013 x
machine*person  10   404.32   404.32   40.43  46.34  0.000
Error           26    22.69    22.69    0.87
Total           43  3084.43

x Not an exact F-test.

S = 0.934111   R-Sq = 99.26%   R-Sq(adj) = 98.78%

Unusual Observations for rating

Obs   rating      Fit  SE Fit  Residual  St Resid
  1  52.0000  52.0000  0.9341   -0.0000         * X
  3  60.0000  60.0000  0.9341    0.0000         * X
  7  64.0000  64.0000  0.9341   -0.0000         * X
  9  68.6000  67.2000  0.6605    1.4000      2.12 R
 22  44.8000  46.8000  0.5393   -2.0000     -2.62 R
 24  65.8000  67.2000  0.6605   -1.4000     -2.12 R
 35  49.2000  46.8000  0.5393    2.4000      3.15 R

R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.

Expected Mean Squares, using Adjusted SS

   Source          Expected Mean Square for Each Term
1  machine         (4) + 2.1370 (3) + Q[1]
2  person          (4) + 2.2408 (3) + 6.7224 (2)
3  machine*person  (4) + 2.3162 (3)
4  Error           (4)

...

Variance Components, using Adjusted SS

                Estimated
Source              Value
person            24.2571
machine*person    17.0791
Error              0.8726

Least Squares Means for rating

machine   Mean
1        52.36
2        60.34
3        66.27
 
Residual Plots for rating 

MTB > GLM 'rating' = machine| person;
SUBC>   Random 'person'; 
SUBC>   SMeans C4000; 
SUBC>   Brief 0; 
SUBC>   Interact 'machine' 'person'.
MTB > GFInt 'machine' 'person'; 
SUBC>   Responses 'rating'; 
SUBC>   FMeans C4000.
Interaction Plot (fitted means) for rating 

MTB > Erase C4000.
MTB > NormTest 'SRES1'.
Probability Plot of SRES1 
The P-value for the Anderson-Darling test of normality is 0.054.

MTB > Name c7 "ByVar1" c8 "ByVar2" c9 "Mean1"
MTB > Statistics 'rating';
SUBC>   By 'machine' 'person';
SUBC>   GValues 'ByVar1'-'ByVar2';
SUBC>   Mean 'Mean1'.
MTB > Name c10 "RESI2"
MTB > Twoway 'Mean1' 'ByVar1' 'ByVar2';
SUBC>   Residuals 'RESI2';
SUBC>   Means 'ByVar1' 'ByVar2'.
Two-way ANOVA: Mean1 versus ByVar1, ByVar2 

Source  DF       SS       MS      F      P
ByVar1   2   584.74  292.368  20.16  0.000
ByVar2   5   410.41   82.082   5.66  0.010
Error   10   144.99   14.499
Total   17  1140.14

S = 3.808   R-Sq = 87.28%   R-Sq(adj) = 78.38%

                 Individual 95% CIs For Mean Based on
                 Pooled StDev
ByVar1     Mean  ---------+---------+---------+---------+
1       52.3611  (-----*-----)
2       60.3389                (-----*----)
3       66.2722                          (----*-----)
                 ---------+---------+---------+---------+
                       54.0      60.0      66.0      72.0

                 Individual 95% CIs For Mean Based on
                 Pooled StDev
ByVar2     Mean  -----+---------+---------+---------+----
1       61.0667                 (------*------)
2       57.9000             (------*------)
3       66.0000                        (------*------)
4       59.7333               (------*------)
5       62.6667                    (------*------)
6       50.5778  (------*------)
                 -----+---------+---------+---------+----
                   49.0      56.0      63.0      70.0

MTB > NormTest 'RESI2'.
Probability Plot of RESI2 
The P-value for the Anderson-Darling test of normality is 0.353.

MTB > Name c11 "ByVar3" c12 "Mean3"
MTB > Statistics 'rating';
SUBC>   By 'person';
SUBC>   GValues 'ByVar3';
SUBC>   Mean 'Mean3'.
MTB > NormTest 'Mean3'.
Probability Plot of Mean3 
The P-value for the Anderson-Darling test of normality is 0.531.

Comments and answers to questions:
----------------------------------
The residuals look a bit strange because of 5 scattered residuals
with values that are somewhat more extreme than all the other residuals.
The two values for machine 2 and employee 3 don't agree well, and there
is also substantial disagreement between the three values of employee 6
at machine 1. However, the residuals are not very large, and there would
not seem to be any reason to exclude them. One would be advised to look
at the experimental protocols to see if some errors had occurred. Some
observations do not have standardised residuals, because there was no
replication at the corresponding person*machine level; it's not an error
or a serious problem. The analysis of machine*person means did not show
any problems with residuals. Note that a likelihood-based analysis identifies
an extreme (small) residual for person=6 and machine=2. The residual is also
low in the analysis based on the means, but not as extreme as in the likelihood-
based analysis. It could be suggested to inspect the circumstances of this
particular set of values (which are pretty consistent). Also generally, person 6 
is somewhat lower than the others, and one might consider omitting him
from the sample due to the extreme variations across machines. The
overall value for person 6 is not beyond what could be expected from a 
normal distribution sample.

The estimated variance components are given above. There is a strong
effect of machines (P=0.001 with the approximate test of the ANOVA
table), and the least squares means show machine 3 to perform best and
machine 1 to perform worst. The interaction plot shows roughly parallel
curves, except that machine 2 performs very poorly with employee 6.
No standard errors are given for the least squares means, and Minitab
will not perform any pairwise comparisons. As judged from the means and the
strong significance of machines, one would guess all machines to be 
statistically different. A likelihood-based analysis (xtmixed in Stata
or proc mixed in SAS) gives the correct standard errors and provides
pairwise comparisons.