1
-Sum Test, also known as the Mann-Whitney test
• Rank the data. That is, replace the data values by their ranks, from smallest
to largest. For example, the pH samples are:
Loc
1 8.53 8.52 8.01 7.99 7.93 7.89 7.85 7.82 7.80
2 7.85 7.73 7.58 7.40 7.35 7.30 7.27 7.27 7.23
are replaced by the ranks
Loc
1 18 17 16 15 14 13 11.5 10 9
2 11.5 8 7 6 5 4 3 2 1
The tied values (7.85, 7.85) would have had ranks 11 and 12 were they slightly
different. In the case of ties, assign the “mid-rank” [here (11+12)/2] to both
values.
• Calculate W , which is the sum of the ranks in the first group. In this case,
W = 123.5.
• In small samples we compare W to the distribution of values of W under all
possible allocations of the ranks to the two samples.
• In larger samples we use a normal approximation to this distribution.
• Let n1 be the number of observations in the first group, n2 the number in the
second group, and N = n1 + n2. Here n1 = n2 = 9 and N = 18.
• Under the null hypothesis that the two distributions are the same, the mean
and variance of W are
µW = n1(N + 1)/2
and
σ2W = n1n2(N + 1)/12
• the observed value of the test statistic is
Zobs =
W − µW
σW
=
(123.5− 9× 19/2)√
9× 9× 19/12
= 3.355
• As with the permutation test, the null hypothesis is that the population dis-
tributions are the same, and the two sided alternative is that the distributions
are different.
2
• The hypotheses can also be written in terms of means. Where µ1 and µ2 are
the means of the two populations, the null hypothesis is H0 : µ1 = µ2. The
possible alternatives and p-values are:
HA p-value
µ1 6= µ2 2P (Z > |Zobs|)
µ1 > µ2 P (Z > Zobs)
µ1 < µ2 P (Z < Zobs)
• For example, with the two sided alternative, the p-value is
2P (Z > |3.355|) = 2(.0004) = .0008
3
Comparison of Wilcoxon, Permutation
and t-tests
• What p-value did the permutation test give in this case?
The test statistic for the permutation test was |X̄1 − X̄2| = 0.595556.
With a two sided alternative, there were
18
9
= 48620 possible arrangements
of the data into two groups of size 9, of which 8 gave at least as much evidence
against H0 as did the value 3.335. Hence the p-value for the permutation test
was
P = 8/48620 = .0002.
• The p-value for the pooled t-test was 0.0002.
• In this case, the Wilcoxon, permutation and t-tests all show very strong evidence
against the null hypothesis of equal distributions.
• Typically, the p-values for the permutation and Wilcoxon test will be fairly
close if the sample size is moderately large.
• If the data are approximately normally distributed and the assumption of equal
variances holds, the p-value for the pooled t-test will be fairly close to those for
the Wilcoxon and permutation tests.
4
another example: n1 = n2 = 3 with data
Treatment 175 250 260
Control 255 275 300
• The treatment ranks are 1,2,4.
• The control ranks are 3,5,6.
• The sum of the treatment ranks is W = 7.
• In this case is it possible to consider all possible permutations of the ranks
between the two samples
1 2 3 4 5 6 W
1 1 1 2 2 2 6
1 1 2 1 2 2 7 = Wobs
1 1 2 2 1 2 8
1 1 2 2 2 1 9
1 2 1 2 2 1 10
1 2 1 2 1 2 9
1 2 1 1 2 2 8
1 2 2 1 1 2 10
1 2 2 2 1 1 12
1 2 2 1 2 1 11
2 1 1 1 2 2 9
2 1 1 2 1 2 10
2 1 1 2 2 1 11
2 1 2 1 1 2 11
2 1 2 1 2 1 12
2 1 2 2 1 1 13
2 2 1 1 1 2 12
2 2 1 1 2 1 13
2 2 1 2 1 1 14
2 2 2 1 1 1 15
5
• The distribution of W under Ho is summarized in the table below
W 6 7 8 9 10 11 12 13 14 15
Prob 1
20
1
20
2
20
3
20
3
20
3
20
3
20
2
20
1
20
1
20
• The probability of getting a value for W as small or smaller than Wobs = 7
is 2/20, and the probability of getting a value as extreme or more extreme is
P = 4/20 = .2
• Under the null hypothesis, the mean of W is 3 × 7/2 = 10.5 and the variance
of W is 3× 3× 7/12 = 5.25.
• Using the normal approximation to the null distribution of W , the observed
test statistic is
Zobs = (7− 10.5)/
√
5.25 = −1.53.
• With the two sided alternative, the p-value is
2P (Z > | − 1.53|) = 2P (Z > 1.53) = 2(.063) = .126
• Even with such small samples, the two approaches give quite similar answers.