## 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 (°)
2015
3521
3236
1612
4110
2318

### 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.

(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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