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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.

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introduction
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
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
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
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
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
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
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
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
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
The procedure:

Decide significance level p=0.05

Set up H0, H1.

  • Data
  • 5 3
  • 4 2
  • 4 4
  • 3 1
  • 5 2
  • 4 1
  • 5

Test statistic

U = 15.5

Probability of H0 being true

p = 0.03

Is p above critical level?

Y N

Reject H0

Accept H0

this particular test
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
How to do it

2 Harmonize ranks where the same value occurs more than once

  • 1: rank all data into ascending order, then re-code the data set replacing raw data with ranks.
  • Data
  • 5 3
  • 4 2
  • 4 4
  • 3 1
  • 5 2
  • 4 1
  • 5
  • Data
  • 5 #13 3 #5
  • 4 #10 2 #4
  • 4 #9 4 #7
  • 3 #6 1 #2
  • 5 #12 2 #3
  • 4 #8 1 #1
  • 5 #11
  • Data
  • 5 #13 = 12 3 #5 = 5.5
  • 4 #10 = 8.5 2 #4 = 3.5
  • 4 #9 = 8.5 4 #7 = 8.5
  • 3 #6 = 5.5 1 #2 = 1.5
  • 5 #12 = 12 2 #3 = 3.5
  • 4 #8 = 8.5 1 #1 = 1.5
  • 5 #11 = 12
once data are ranked
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
In this case:
  • Data
  • 5 #13 = 12 3 #5 = 5.5
  • 4 #10 = 8.5 2 #4 = 3.5
  • 4 #9 = 8.5 4 #7 = 8.5
  • 3 #6 = 5.5 1 #2 = 1.5
  • 5 #12 = 12 2 #3 = 3.5
  • 4 #8 = 8.5 1 #1 = 1.5
  • 5 #11 = 12
  • ___ ___
  • rx=67 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
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).

Upper tail of distribution

Lower tail of distribution

slide17

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.

Females

Males

Site 2

Site 3

Site 1

Do males differ from females?

Do results differ between these sites?

slide18

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.