Statistical Methods in Medical Research 4ed

292 Analysing non-normal data
It is useful to apply a continuity correction of 1 unit, since the possible values of S turn
out to be separated by an interval of 2 units. The standardized normal deviate is
22=11 18 ˆ 1 97, for which P ˆ 0 049, rather close to the exact value.
For this set of data Spearman's rank correlation coefficient is 0 68. The approximate t
test from (7.19) gives t ˆ 2 66 on 8 DF …P ˆ 0 029†. A more exact significance level,
allowing for the discreteness of the distribution, is 0 031. Both these values are smaller,
indicating a more significant result, than for Kendall's test. However, different measures
of association which are not logically equivalent will necessarily achieve different levels of
significance, and the reader will hardly need to be reminded that the method chosen
for final presentation should not be selected merely as the one showing the most significance.
10.6 Permutation and Monte Carlo tests
In the preceding sections the significance levels of various distribution-free tests
have been given by reference to tables specially constructed for the purpose. In
most cases these tables only apply to small sample sizesÐfor larger samples
approximations to the appropriate null distribution are generally used (as given
in Table 10.1). However, it is instructive to consider in more detail how the
special tables have been constructed. This is because an explanation of the tables
will illustrate that many of the distribution-free tests can be justified as instances
of a more general class of tests known as permutation tests.
As an example consider the comparison of two independent samples of sizes
4 (sample 1) and 5 (sample 2) using the Wilcoxon rank sum test. As above, the
sum of the ranks from sample 1 is denoted by T1. Reference to Table A7
shows that the difference between the groups is significant at the 5% level if
T 0 11. As explained in the legend to Table A7, T 0 ˆ T1 if T1 E…T1† and
T 0 ˆ 2E…T1† T1…ˆ 40 T1 in this example) if T1 > E…T1†: T 0 is considered
because the test is two-sided. If, for simplicity, it is assumed that there are no
ties, then the ranks of the four values from sample 1 will be four values chosen
from f1, 2, 3, 4, 5, 6, 7, 8, 9g. Under the null hypothesis that both samples were
drawn from the same population, any set of four of these values is equally likely.
9
There are, therefore, 4 ˆ 126 different sets of four ranks that could have come
n
from sample 1 (see §3.6 for a definition of the binomial coefficient r ). From
Table 10.1 it can be seen that the only possible values for T 0 are from 10 to
E…T1† ˆ20, that is, 11 distinct values. Consequently some values of T 0 will arise
from several sets of ranks. Indeed, because each set of ranks is equally likely, the
value of P…T 0 ˆ t† under the null distribution is simply the number of sets of
ranks giving T 0 ˆ t, divided by 126.
The smallest value of T 0 is 10, which occurs only when sample 1 has the four
smallest ranks, or the four largest ranks. The second smallest value of T 0 , 11, also
can be obtained in only two ways, namely when the ranks for sample 1 are either
f1, 2, 3, 5g or f5, 7, 8, 9g. The next smallest value, 12, can occur in four ways,