Statistical Methods in Medical Research 4ed

234 Comparison of several groups
the standard deviation, then if the variances differ between groups so also will
the mean absolute deviations. Thus the variance ratio test for the equality of
group means is a test of the homogeneity of variances. Carroll & Schneider (1985)
showed that it is preferable to measure the deviations from the group medians to
cope with asymmetric distributions.
8.8 Comparison of several counts: the Poisson
heterogeneity test
Suppose that k counts, denoted by x1, x2, ..., xi, ..., xk are available. It may be
interesting to test whether they could reasonably have been drawn at random
from Poisson distributions with the same (unknown) mean m. In many microbiological
experiments, as we saw in §3.7, successive counts may be expected to
follow a Poisson distribution if the experimental technique is perfect. With
imperfect technical methods the counts will follow Poisson distributions with
different means. In bacteriological counting, for example, the suspension may be
inadequately mixed, so that clustering of the organisms occurs; the volumes of
the suspension inoculated for the different counts may not be equal; the culture
media may not invariably be able to sustain growth. In each of these circumstances
heterogeneity of the expected counts is present and is likely to manifest
itself by excessive variability of the observed counts. It seems reasonable, therefore,
to base a test on the sum of squares about the mean of the xi. An
appropriate test statistic is given by
X 2 P 2
…x x†
ˆ , …8:31†
x
which, on the null hypothesis of constant m, is approximately distributed as
x2 …k 1† . The method is variously called the Poisson heterogeneity or dispersion
test.
The formula (8.31) may be justified from two different points of view. First, it
is closely related to the test statistic (5.4) used for testing the variance of a normal
distribution. On the present null hypothesis the distribution is Poisson, which we
know is similar to a normal distribution if m is not too small; furthermore,
s2 ˆ m, which can best be estimated from the data by the sample mean x.
Replacing s2 0 by x in (5.4) gives (8.31). Secondly, we could argue that, given
the total count P x, the frequency `expected' at the ith count on the null
hypothesis is P x=k ˆ x. Applying the usual formula for a x2 index,
P 2
‰…O E† =EŠ, immediately gives (8.31). In fact, just as the Poisson distribution
can be regarded as a limiting form of the binomial for large n and small p, so the
present test can be regarded as a limiting form of the x2 test for the 2 k table
(§8.5) when R=N is very small and all the ni are equal; under these circumstances
it is not difficult to see that (8.29) becomes equivalent to (8.31).