81
11. RCBD III; Latin squares
Levenes test for equal variances in each block. The
blocks correspond to the 4 columns of the data, so
we rst compute the medians of the columns, sub-
tract ¯hen from ench column o¯d take absolute values:
= ¯ 1 ¯. The do an anova to
h i
see if is the same in each block.
data.matrix F)
blocks 3 0.022500 0.007500 0.5806 0.6389
Residuals 12 0.155000 0.012917
Using treatments rather than blocks gives = 698
(you should check this). 82
The analogue of the Kruskal-Wallis test, for a RCBD,
is Friedmans test. The observations are replaced by
their ranks within each block, and the usual ANOVA
is run. This method does not take account of the
fact that the denominator of the is not random (as
before); the function friedman.test will do so. The
di erences are generally slight.
> friedman.test(y,treatments,blocks)
Friedman rank sum test
data: y, treatments and blocks
Friedman chi-squared = 8.8462, df = 3,
p-value = 0.03141 83
So - the assumptions seem to be met, and at least
some of the di erences in the treatment means, i.e.
in the mean readings = + , are signicant -
the readings of the hardness testing device depend on
which tip is being used. This is bad news for the
engineers. Is there any one tip responsible for the
di erences? We should look at all of the di erences
= ¯ ¯ to see which are signicant.
Method 1: Fishers LSD (Least Signicant Dif-
ference). A 100(1 )% condence interval on
one di erence is
s µ ¶
1 1
¯ ¯ ± 2 ( ) +
s µ ¶
2
= ¯ ¯ ± 2 9 00889
4
With = 05, the 95% interval is ¯ ¯ ± 151.
Converting this to a hypothesis test, we see that
the hypothesis of equality is rejected if
¯ ¯
¯¯ ¯ ¯ = 151 84
Since
|¯ ¯ | = 025 151
1 2
|¯1 ¯3|

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