Mann Whitney U

Mann Whitney U AIM: TO BE ABLE TO APPLY THE MANN WHITNEY U TEST DATA AND EVALUATE ITS EFFECTIVENESS.

What is it? - a technique that tests if there is a significant difference between the medians of two sets of data - it tells us if the differences between data sets are significant or whether they are due to chance - it is non parametric (it assumes data is not normally distributed) - however, it does assume that there is a similar dispersion of both sets of data (you can check this on a dispersion graph) Examples: ◦ Is there a significant difference between pebble sizes at an upstream and a downstream site? ◦ Is there a significant difference in traffic flows before and after the new Tesco store was built? ◦ Comparing differences in biodiversity in 2 ecosystems.

When can the test be used? The test can be used if you want to investigate the differences between two sets of similar data. Some conditions apply:- you can only compare 2 sets of data, no more than that - the data must be ordinal (it can be ranked in order) - you need a minimum of 5 values in each data set - it is not advisable to use more than 20 values in each data set as the exercise becomes unwieldy (so only really suitable for small sets of data)

The Mann Whitney U test starts with a null hypothesis : There is no significant difference in the medians of the two sets of data. Once you have calculated the value of U, you then have to compare it to the critical values. If the value of U is less than or equal to the critical value then you can reject the null hypothesis and accept that there is a difference in the 2 sets of data. Strengths

Weaknesses

• You can use two data sets that have different sizes, e.g. one data set could have 10 values and the other only 8. • You can state whether the relationship is significant or if your results occurred by chance. • You can see clearly whether there is a difference in the median of 2 sets of data • It is fairly easy to convert original data to ranked form.

It is a lengthy calculation and prone to human error It does not explain why the difference in the two sets of data occurs. In ranking the data the range of raw data is lost.

Mann Whitney U – Worked Example You conduct a questionnaire survey of homes in the Heathrow flight path, and also a control population of homes in South west London. You ask the question “How intrusive is plane noise in your daily life?” to a group of people in each area. Step 1: Create a null hypothesis (H0)

“There is no difference in attitudes to plane noise between the two areas”

Mann Whitney U – Worked Example Noise complaints 1 = no complaint 5 = very unhappy

Step 2: Tabulate the results Homes near airport

Rank

Homes in S W London

5

3

4

2

4

4

3

1

5

2

4

1

5

Rank

Mann Whitney U – Worked Example Step 3: Rank the results treating the data as one sample. Smallest = 1 and so on. If 2 values are the same take an average. Homes near airport Rank Homes in S W London Rank 5

3

4

2

4

4

3

1

5

2

4

1

5

Mann Whitney U – Worked Example Step 4: Calculate the sum of the ranks for each column. Lets call them Rx and Ry Homes near airport

Rank (rx)

Homes in S W London

Rank (ry)

5

12

3

5.5

4

8.5

2

3.5

4

8.5

4

8.5

3

5.5

1

1.5

5

12

2

3.5

4

8.5

1

1.5

5

12

Total

∑rx = 67

∑ry =24

Mann Whitney U – Worked Example Step 5: The calculations! ∑rx = 67 ∑ry = 24 Ux = NxNy +

Uy = NxNy +

The Mann Whitney U calculation for sample y

Nx(Nx + 1) - ∑rx 2 Ny(Ny + 1) 2

The number in the sample

- ∑ry

The sum of the ranks for sample x

Mann Whitney U – Worked Example Ux = NxNy +

Step 5: The calculations! ∑rx = 67 ∑ry = 24 Ux =

Nx(Nx + 1) 2

- ∑rx

Uy = NxNy + Ny(Ny + 1) - ∑r y 2

Uy =

Mann Whitney U – Worked Example Step 6: Look up the smaller of the 2 values in a 5% significance table Ux = 3 Uy = 39

If it is equal to or smaller than the critical value you can reject H0

Size of smallest sample (n1)

and accept H1

NB. The difference between Spearman’s rank and Mann Whitney U Spearman’s rank – your figure has to exceed the critical value at the 95% level of significance to reject the null hypothesis. Mann Whitney U – your figure has to be equal to or less than the critical value at the 95% significance level to reject the null hypothesis.

Mann Whitney U – Worked Example Step 7: Interpreting the result 

Smallest U value = 3



Critical Value at 0.05 level = 6



The U value is ____________ than the critical value. Therefore we should __________ the null hypothesis



We can be _____ % certain that our result did/did not occur by chance and that people living in the Heathrow flight path are/are not negatively impacted by the airplanes.



NB – To check for 99% significance we would have to consult a different table.

5(a) Interpret the result of the Mann Whitney U test in Figure 1 - what are the U1 and U2 values? Is the lowest of the 2 higher or lower than the critical value of 27? - What is the probability that this has occurred by chance? - Which hypothesis can be accepted and which rejected? - If we wanted 99% significance what would we need?

5(b) Suggest why the Mann Whitney U test is suitable to interpret this set of data - what type of test is it? What does it assume about the data? - Can it be used for ordinal data? (data that can be ranked) - what size data set is it suitable for for? - what does the test tell us about the relationship between the 2 sets of data? - What else?

techniques may help in the analysis of data and increase geographical understanding - you need to identify at least 2 statistical tests. (Mann Whitney U, Spearman’s Rank, Chi Squared, measures of central tendancy/dispersion) - a bit about how they work and measure statistical significance - how the results of the tests can be tested for statistical significance and when to accept/reject null hypothesis - what happens next if result is found not to be statistically significant? -how have you used statistical tests in your own fieldwork? What did they show? How did they increase your geographical understanding?