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The Mann-Whitney U
test

Peter Shaw
Introduction



We meet our first inferential test.
You should not get put off by the
messy-looking formulae – it’s
usually run on a PC anyway.
The important bit is to understand
the philosophy of the test.
Imagine..

That you have acquired a set of
measurements from 2 different sites.



Maybe one is alleged to be polluted, the
other clean, and you measure residues in
the soil.
Maybe these are questionnaire returns
from students identified as M or F.
You want to know whether these 2 sets
of measurements genuinely differ. The
issue here is that you need to rule out
the possibility of the results being
random noise.
The formal procedure:

Involves the creation of two competing
explanations for the data recorded.


Idea 1:These are pattern-less random
data. Any observed patterns are due to
chance. This is the null hypothesis H0
Idea 2: There is a defined pattern in the
data. This is the alternative hypothesis
H1

Without the statement of the competing
hypotheses, no meaning test can be run.
Occam’s razor



If competing explanations exist, chose
the simpler unless there is good reason
to reject it.
Here, you must assume H0 to be true
until you can reject it.
In point of fact you can never
ABSOLUTELY prove that your
observations are non-random. Any
pattern could arise in random noise, by
chance. Instead you work out how likely
H0 is to be true.
Example
You conduct a questionnaire survey of homes in the
Heathrow flight path, and also a control population of
homes in South west London. Responses to the question
“How intrusive is plane noise in your daily life” are
tabulated:









Noise complaints 1= no complaint, 5 = very unhappy
Homes near airport
Control site
5
3
4
2
4
4
3
1
5
2
4
1
5
Stage 1: Eyeball the
data!



These data are ordinal, but not normally
distributed (allowable scores are 1, 2, 3, 4 or
5).
Use Non-parametric statistics
It does look as though people are less happy
under the flightpath, but recall that we must
state our hypotheses H0, H1
H0: There is no difference in attitudes to plane
noise between the two areas – any observed
differences are due to chance.
 H1: Responses to the question differed
between the two areas.

Now we assess how
likely it is that this
pattern could occur by
chance:


This is done by performing a calculation.
Don’t worry yet about what the calculation
entails.
What matters is that the calculation gives an
answer (a test statistic) whose likelihood
can be looked up in tables. Thus by means
of this tool - the test statistic - we can work
out an estimate of the probability that the
observed pattern could occur by chance in
random data
One philosophical
hurdle to go:




The test statistic generates a probability - a
number for 0 to 1, which is the probability of H0
being true.
If p = 0, H0 is certainly false. (Actually this is
over-simple, but a good approximation)
If p is large, say p = 0.8, H0 must be accepted
as true.
But how about p = 0.1, p = 0.01?
Significance



We have to define a threshold, a boundary, and
say that if p is below this threshold H0 is rejected
otherwise H1 is accepted.
This boundary is called the significance level. By
convention it is set at p=0.05 (1:20), but you can
chose any other number - as long as you specify
it in the write-up of your analyses.
WARNING!! This means that if you analyse 100
sets of random data, the expectance (log-term
average) is that 5 will generate a significant test.
The procedure:
Set up H0, H1.








Decide significance level
p=0.05
Data
5
4
4
3
5
4
5
3
2
4
1
2
1
Test statistic
U = 15.5
Probability of
H0 being true
p = 0.03
Is p above critical level?
Y
N
Accept H0
Reject H0
This particular test:



The Mann-Whitney U test is a non-parametric
test which examines whether 2 columns of data
could have come from the same population (ie
“should” be the same)
It generates a test statistic called U (no idea why
it’s U). By hand we look U up in tables; PCs
give you an exact probability.
It requires 2 sets of data - these need not be
paired, nor need they be normally distributed,
nor need there be equal numbers in each set.
How to do it









1: rank all data into
2 Harmonize ranks where the
ascending order, then
re-code the data set
replacing raw data
with ranks.
same value occurs more than
once
Data
5
4
4
3
5
4
5
3
2
4
1
2
1








Data
5
4
4
3
5
4
5
#13
#10
#9
#6
#12
#8
#11

3
2
4
1
2
1
#5
#4
#7
#2
#3
#1







Data
5
4
4
3
5
4
5
#13 = 12
#10 = 8.5
#9 = 8.5
#6 = 5.5
#12 = 12
#8 = 8.5
#11 = 12
3
2
4
1
2
1
#5
#4
#7
#2
#3
#1
=
=
=
=
=
=
5.5
3.5
8.5
1.5
3.5
1.5
Once data are ranked:


Add up ranks for each column; call these rx and ry
(Optional but a good check:
 rx

+ ry = n2/2 + n/2, or you have an error)
Calculate


Ux = NxNy + Nx(Nx+1)/2 - Rx
Uy = NxNy + Ny(Ny+1)/2 - Ry

take the SMALLER of these 2 values and look up in tables. If U
is LESS than the critical value, reject H0

NB This test is unique in one feature: Here low values of the
test stat. Are significant - this is not true for any other test.
In this case:












Data
5
4
4
3
5
4
5
#13 = 12
#10 = 8.5
#9 = 8.5
#6 = 5.5
#12 = 12
#8 = 8.5
#11 = 12
___
rx=67
3
2
4
1
2
1
#5
#4
#7
#2
#3
#1
=
=
=
=
=
=
5.5
3.5
8.5
1.5
3.5
1.5
___
ry=24
Check: rx + ry + 91
13*13/2 + 13/2 = 91 CHECK.
Ux = 6*7 + 7*8/2 - 67 = 3
Uy = 6*7 + 6*7/2 - 24 = 39
Lowest U value is 3.
Critical value of U (7,6) = 4 at p =
0.01.
Calculated U is < tabulated U so
reject H0.
At p = 0.01 these two sets of data
differ.
Tails.. Generally use
2 tailed tests
2 tailed test: These
populations DIFFER.
1 tailed test: Population X is
Greater than Y (or Less than Y).
Lower tail of distribution
Upper tail of distribution
Kruskal-Wallis: The U test’s big cousin
When we have 2 groups to compare (M/F, site 1/site 2, etc) the U test is
correct applicable and safe.
How to handle cases with 3 or more groups?
The simple answer is to run the Kruskal-Wallis test. This is run on a PC,
but behaves very much like the M-W U. It will give one significance
value, which simply tells you whether at least one group differs from one
other.
Males Females
Do males differ
from females?
Site 1 Site 2 Site 3
Do results differ
between these
sites?
Your coursework:
I will give each of you a sheet with data collected from 3 sites. (Don’t
try copying – each one is different and I know who gets which dataset!).
I want you to show me your data processing skills as follows:
1: Produce a boxplot of these data, showing how values differ between
the categories.
2: Run 3 separate Mann-Whitny U tests on them, comparing 1-2, 1-3 and
2-3. Only call the result significant if the p value is < 0.01
3: Run a Kruskal-Wallis anova on the three groups combined, and
comment on your results.