Data analysis
Well-constructed data analysis will help you spot patterns, trends and relationships in your results. Data analysis can be qualitative and/or quantitative, and may include statistical tests. An example of a statistical test is outlined below.
Mann Whitney U test
Mann Whitney U is a statistical test that is used either to test whether there is a significant difference between the medians of two sets of data.
The Mann Whitney U test can only be used if there are at least 6 pairs of data. It does not require a normal distribution.
There are 3 steps to take when using the Mann Whitney U test
Step 1. State the null hypothesis
There is no significant difference between _______ and _______
Step 2. Calculate the Mann Whitney U statistic
\(U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1) \;- ∑ R_2\) \(U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1) \;- ∑ R_1\)- \(n_1\) is the number of values of \(x_1\)
- \(n_2\) is the number of values of \(x_2\)
- \(R_1\) is the ranks given to \(x_1\)
- \(R_2\) is the ranks given to \(x_2\)
Step 3. Test the significance of the result
Compare the calculated value of U against the critical value for U at a confidence level of 95% (ie significance value of p = 0.05).
If the calculated value of U is equal to or smaller than the critical value (at p=0.05) then REJECT the null hypothesis. There is a SIGNIFICANT difference between the 2 data sets.
If U is greater than the critical value, then ACCEPT the null hypothesis. There is NOT a significant difference between the 2 data sets.
Worked example
A geographer was interested in whether there was a difference in cliff gradient between places with a beach and places with no beach. Here are the results.
| Cliff gradient where there is no beach (°) | Cliff gradient where there is a beach (°) |
|---|---|
| 20 | 15 |
| 35 | 21 |
| 32 | 36 |
| 16 | 12 |
| 41 | 10 |
| 23 | 18 |
Step 1. State the null hypothesis
There is no significant difference in cliff gradient between places with a beach and places with no beach.
Step 2. Calculate the Mann Whitney U statistic
(a) Give each result a rank. Calculate the sum of the ranks for the two columns.
| NO BEACH | NO BEACH | BEACH | BEACH |
|---|---|---|---|
| Cliff gradient (°) | Rank | Cliff gradient (°) | Rank |
| 20 | 6 | 15 | 3 |
| 35 | 10 | 21 | 7 |
| 32 | 9 | 36 | 11 |
| 16 | 4 | 12 | 2 |
| 41 | 12 | 10 | 1 |
| 23 | 8 | 18 | 5 |
| TOTAL | 49 | TOTAL | 29 |
(b) Calculate \(∑R_1\) and \(∑R_2\)
\(∑R_1\) is the sum of the ranks in the first column (no beach) = \(49\)
\(∑R_2\) is the sum of the ranks in the first column (beach) = \(29\)
\(n_1 = 6\) and \(n_2 = 6\)
(c) Calculate \(U_1\) and \(U_2\)
Remember that
\(U_1 = ((n_1 × n_2) + 0.5 × n_2 (n_2 + 1) \;- ∑R_2\) \(U_2 = ((n_1 × n_2) + 0.5 × n_1 (n_1 + 1) \;- ∑R_1\)So using the numbers from this example
\(U_1 = ((6 × 6) + 0.5 × 6 (6 + 1)\;- 29\;\;\;\)So\(\;U_1 = 28\)
\(U_2 = ((6 × 6) + 0.5 × 6 (6 + 1)\;- 49\;\;\;\)So\(\;U_2 = 8\)
Step 3. Test the significance of the result
In this example, \(U_1 = 28\) and \(U_2 = 8\)
Select the smaller of the 2 values, which in this example is \(U_2 = 8\)
Use a set of Mann-Whitney U critical value tables (example here) to find the critical value at the 95% confidence level (look for ‘Alpha = 0.05 (two-tailed)’ in the table).
The critical value at the 95% confidence level where \(n_1=6\) and \(n_2=6\) is \(5\). Since our calculated value of \(\)8 is more than the critical value of 5, the null hypothesis is not rejected.
In conclusion, there is no significant difference in cliff gradient between places with a beach and places with no beach.
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