Supplementary Exercise 13.19 of IPS7e
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A 2*4 ANOVA with 5 observations per cell; N=2*4*5=40 observations.

(a) The F-test for interaction has the following degrees of freedom
  DFAB = (2-1)*(4-1) = 3
  DFE = DFT - DFA - DFB - DFAB = 39-1-3-3 = 32 (= N-IJ = 40-2*4)
Table F of PSLS does not have the denominator DF=32, only 25 and 50; same thing
for Table E of IPS, with entries for 30 and 40. We use the largest DF in the table
below the actual DF, i.e. DF=25 for the PSLS table and DF=30 for the IPS table.
The PSLS table values (for F(3,25)) are: 90%-perc=2.32, 95%-perc=2.99, ..., 
99.9%-perc=7.45. For assessing significance it is conservative to use a smaller 
denominator degrees of freedom.

(b) Sketch a right-skewed distribution on the positive axis with these
values, or use the Probability Distribution menu in Minitab.

(c) The observed value Fobs=2.59 falls between the 90%- and 95%-percentiles. 
Therefore 0.05<P<0.10. The exact P-value in F(3,32) is 0.07 (computed using 
Minitab/Stata/R).

(d) There is no evidence at the 5% level for an interaction, but the
P-value is close. Therefore I would expect some non-parallelism in the
graph. A strong non-parallelism would intuitively correspond to a
significant P-value of the test. However, generally speaking statistical 
significance cannot be determined from the points of the graph alone, 
because one also needs to consider the number of replications (here 5) 
and the error variation; these two quantities determine the standard error 
of group means).


Minitab command to compute the P-value for (c):

MTB > CDF 2.59;
SUBC>   F 3 32.
Cumulative Distribution Function 

F distribution with 3 DF in numerator and 32 DF in denominator

   x  P( X <= x )
2.59     0.930033