Statistical Methods in Medical Research 4ed

156 Analysing variances
Comparison of two counts
Suppose that x1 is a count which can be assumed to follow a Poisson distribution
with mean m 1. Similarly let x2 be a count independently following a
Poisson distribution with mean m 2. How might we test the null hypothesis that
m 1 ˆ m 2?
One approach would be to use the fact that the variance of x1 x2 is m 1 ‡ m 2
(by virtue of (3.19) and (4.9)). The best estimate of m 1 ‡ m 2 on the basis of
the available information is x1 ‡ x2. On the null hypothesis E…x1 x2† ˆ
m 1 m 2 ˆ 0, and x1 x2 can be taken to be approximately normally distributed
unless m 1 and m 2 are very small. Hence,
z ˆ x1 x2
p …5:7†
…x1 ‡ x2†
can be taken as approximately a standardized normal deviate.
Asecond approach has already been indicated in the test for the comparison
of proportions in paired samples (§4.5). Of the total frequency x1 ‡ x2, a portion
x1 is observed in the first sample. Writing r ˆ x1 and n ˆ x1 ‡ x2 in (4.17) we have
1
z ˆ x1 2 …x1 ‡ x2†
p 1
ˆ
2 …x1 ‡ x2†
x1 x2
p
…x1 ‡ x2†
as in (5.7). The two approaches thus lead to exactly the same test procedure.
Athird approach uses a rather different application of the x2 test from that
described for the 2 2 table in §4.5, the total frequency of x1 ‡ x2 now being
divided into two components rather than four. Corresponding to each observed
frequency we can consider the expected frequency, on the null hypothesis, to be
1
2 …x1 ‡ x2†:
Observed x1 x2
Expected
1
2 …x1 ‡ x2†
1
2 …x1 ‡ x2†
Applying the usual formula (4.30) for a x 2 statistic, we have
X 2 ˆ ‰x1
1
2 …x1 ‡ x2†Š 2
1
2 …x1 ‡ x2†
ˆ …x1 x2† 2
:
x1 ‡ x2
As for (4.30) X 2 follows the x 2 …1†
‡ ‰x2
1
2 …x1 ‡ x2†Š 2
1
2 …x1 ‡ x2†
…5:8†
distribution, which we already know to be
the distribution of the square of a standardized normal deviate. It is therefore
not surprising that X 2 given by (5.8) is precisely the square of z given by
(5.7). The third approach is thus equivalent to the other two, and forms a
particularly useful method of computation since no square root is involved
in (5.8).