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| Land Use Category | On Flood plain | Off Flood Plain | Total |
|---|---|---|---|
| Pre War Low Cost Housing | 15 | 24 | 39 |
| Pre War High Cost Housing | 2 | 28 | 30 |
| Post War Low Cost Housing | 6 | 12 | 18 |
| Post War High Cost Housing | 11 | 32 | 43 |
| Retail / Admin - High Order | 5 | 7 | 12 |
| Retail / Admin - Low Order | 24 | 19 | 43 |
| Industry | 5 | 12 | 17 |
| Transport | 28 | 33 | 61 |
| Car Parks | 14 | 4 | 18 |
| Open Space (inc. agriculture) | 80 | 13 | 93 |
| Community Services | 10 | 16 | 26 |
| Total | 200 | 200 | 400 |
Construct a mirror graph of results. This may look similar to the example below:
If you have collected data on flood likelihood and flood severity, you can calculate the risk score for each survey square using the following equation.
Risk = likelihood x severity
You can then map the risk of flooding for the study area. The example below shows how this could be done using a spreadsheet program.

One way of displaying the results of the bipolar analysis is to construct a modified bar chart with the bipolar score on the x-axis and the different flood management techniques shown on the y-axis. The example below shows what this might look like.
Association tests (like the chi squared test) can be useful. For example, if you have collected data on the likelihood of flooding and on land use (as described on the Data collection page) you can test the significance of the association between them. Need more information about this test?
The null hypothesis is that there is no significant association between flood likelihood and land use.
The observed data was collected from 200 squares in a mixed residential / town centre area, both on and off the flood plain of the River Avon.
Each square was categorised as being of either at high likelihood of flooding (more than or equal to 2.5) or low likelihood of flooding (less than 2.5), and was categorised in one of 5 pre-determined land-use categories.
| High likelihood | Low likelihood | |
|---|---|---|
| Land use | OBSERVED DATA | OBSERVED DATA |
| High-value retail | 0 | 12 |
| High-value housing / low-value retail | 4 | 20 |
| Medium-value housing | 24 | 18 |
| Low-value housing | 37 | 12 |
| Open space | 62 | 11 |
First calculate expected values.
Expected values = (row total x column total) ÷ (grand total)
| High likelihood | Low likelihood | TOTAL | |||
|---|---|---|---|---|---|
| Land use | OBSERVED | EXPECTED | OBSERVED | EXPECTED | |
| High-value retail | 0 | 7.6 | 12 | 4.4 | 12 |
| High-value housing / low-value retail | 4 | 15.2 | 20 | 8.8 | 24 |
| Medium-value housing | 24 | 26.7 | 18 | 15.3 | 42 |
| Low-value housing | 37 | 31.1 | 12 | 17.9 | 49 |
| Open space | 62 | 46.4 | 11 | 26.7 | 73 |
TOTAL |
127 | 73 | 200 | ||

Now calculate the sum of (observed - expected)2 ÷ (expected)
In our example, chi squared = 70.4
Now compare the calculated value of chi squared against the critical values for (columns of observed values - 1) x (rows of observed values-1) degrees of freedom.
In our example, at the p=0.05 probability level where degrees of freedom = 3, the critical value of chi squared = 9.49.
Since 70.4 > 9.49, the null hypothesis is rejected at the p=0.05 level.
Therefore, we are 95% certain that there is an association between flood likelihood and land use.
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