This presentation goes through how to use Spearman's Rank for a piece of Geography coursework. It does this through a worked example, stage by stage.

1. How and why to use Spearman’s Rank… If you have done scattergraphs, Spearman’s Rank offers you the opportunity to use a statistical test to get a value which can determine the strength of the relationship between two sets of data…

2. So how do we do it? This is the equation, and looks complicated, so let’s think carefully about how we can do this… The best way to do this would be through an example. If we were looking at Settlement patterns for a town’s CBD in Geography, we may wish to compare aspects of the town, such as whether the number of people in a zone affect the type of shops that locate there (i.e. – convenience shops) To do this, we would construct a table as shown overleaf… In the above, r s refers to the overall value or rank The equation has to be done before the value is taken away from 1 In the above equation, the sign means ‘the total of’ d 2 is the first thing we will try to establish in our ranked tables (see next slides) ‘ n’ refers to the number of sites or values you will process – so if there were there 15 river sites, ‘n’ would be 15 . If there were 20 pedestrian count zones, ‘n’ would be 20 , and so on…

3. 1. Here we have laid out a table of each of the twelve zones in a town 2. Pedestrian counts for each zone here 3. Number of Convenience shops for each zone here 4. We now need to rank the data (two highlighted columns)– this is shown overleaf 22 70 12 19 64 11 6 21 10 7 24 9 8 27 8 4 19 7 3 18 6 7 12 5 15 60 4 5 25 3 2 8 2 8 40 1 D2 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone

4. You will see here that on this example, the pedestrian counts have been ranked from highest to Lowest, with the Highest value (70) Being ranked as Number 1 , the Lowest value (8) Being ranked as Number 12. 22 1 70 12 19 2 64 11 6 8 21 10 7 7 24 9 8 5 27 8 4 9 19 7 3 10 18 6 7 11 12 5 15 3 60 4 5 6 25 3 2 12 8 2 8 4 40 1 D2 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone

5. So that was fairly easy… We need to now do the next column for Convenience shops too. But hang on! Now we have a problem… We have two values that are 8, so what do we do? The next two ranks would be 4 and 5 ; we add the two ranks together and divide it by two . So these two ranks would both be called 4.5 1 22 1 70 12 2 19 2 64 11 6 8 21 10 7 7 24 9 8 5 27 8 4 9 19 7 3 10 18 6 7 11 12 5 3 15 3 60 4 5 6 25 3 2 12 8 2 8 4 40 1 D2 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone

6. This is normally the point where one of the biggest mistakes is made. Having gone from 4.5 , students will often then rank the next value as 5 . But they can’t! Why not? Because we have already used rank number 5 ! So we would need to go to rank 6 This situation is complicated further by the fact that the next two ranks are also tied. So we do the same again – add ranks 6 and 7 and divide it by 2 to get 6.5 1 22 1 70 12 2 19 2 64 11 6 8 21 10 6.5 7 7 24 9 4.5 8 5 27 8 4 9 19 7 3 10 18 6 6.5 7 11 12 5 3 15 3 60 4 5 6 25 3 2 12 8 2 4.5 8 4 40 1 D2 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone

7. Having ranked both sets of data we now need to work out the difference (d) between the two ranks. To do this we would take the second rank away from the first . This is demonstrated on the next slide 1 1 2 2 8 8 6.5 7 4.5 5 10 9 11 10 6.5 11 3 3 9 6 12 12 4.5 4 Rank (r) Rank

8. The difference between the two ranks has now been established So what next? We need to square each of these d values… Don’t worry if you have any negative values here – when we square them ( multiply them by themselves ) they will become positives 0 1 22 1 70 12 0 2 19 2 64 11 0 8 6 8 21 10 0.5 6.5 7 7 24 9 0.5 4.5 8 5 27 8 -1 10 4 9 19 7 -1 11 3 10 18 6 4.5 6.5 7 11 12 5 0 3 15 3 60 4 -3 9 5 6 25 3 0 12 2 12 8 2 -0.5 4.5 8 4 40 1 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone

9. So, the first value squared would be 0.25 (-0.5 x -0.5) 0 1 22 1 70 12 0 2 19 2 64 11 0 8 6 8 21 10 0.5 6.5 7 7 24 9 0.5 4.5 8 5 27 8 -1 10 4 9 19 7 -1 11 3 10 18 6 4.5 6.5 7 11 12 5 0 3 15 3 60 4 -3 9 5 6 25 3 0 12 2 12 8 2 -0.5 4.5 8 4 40 1 Difference (d) Rank (r) Convenience shops Rank Pedestrians Zone 0 0 0 0.25 0.25 1 1 20.25 0 9 0 0.25 D 2

11. Firstly, let’s remind ourselves of the equation... In this equation, we know the total of d 2 , which is 32 So the top part of our equation is… 6 x 32 We also know what ‘n’ is (the number of sites or zones - 12 in this case), so the bottom part of the equation is… (12x12x12) - 12

12. We can now do the equation… 6 x 32 12 3 - 12 192 1716 OK – so this gives us a figure of 0.111888111888 Is that us finished? Sadly not!

13. This is the equation, which we will by now be sick of! I have circled the part of the equation that we have done… Remember that we need to take this value that we have calculated away from 1. Forgetting to do this is probably the second biggest mistake that people make! So… 1 – 0. 111888111888 = 0.888

14. So we have our Spearman’s Rank figure….But what does it mean? -1 0 +1 0.888 Your value will always be between -1 and +1 in value. As a rough guide, our figure of 0.888 demonstrates there is a fairly positive relationship. It suggests that where pedestrian counts are high, there are a high number of convenience shops Should the figure be close to -1, it would suggest that there is a negative relationship, and that as one thing increases, the other decreases.

16. 1. This is a critical values table and the ‘n’ column shows the numbers of sites or zones you have studied. In our case, we looked at 12 zones. 2. If look across we can see there are two further columns – one labelled 0.05 , the other 0.01. The first, 0.05 means that if our figure exceeds the value , we can be sure that 95 times in 100 the figures occurred because a relationship exists, and not because of pure chance The second, 0.01 , means that if our figure exceeds this value, we can be sure that 99 times in 100 the figures occcurred because a relationship exists, and did not occur by chance. We can see that in our example our figure of 0.888 exceeds the value of 0.591 at the 0.05 level and also comfortably exceeds value at the 0.01 level too. 0.478 0.364 30 0.496 0.377 28 0.515 0.392 26 0.537 0.409 24 0.562 0.428 22 0.591 0.45 20 0.625 0.475 18 0.665 0.506 16 0.715 0.544 14 0.777 0.591 12 0.01 level 0.05 level N

17. In our example above, we can see that our figure of 0.888 exceeds the values at both the 95% and 99% levels. The figure is therefore highly significant

19. We can therefore conclude that this figure is strongly significant and that pedestrian counts and the number of convenience shops are clearly rated