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How and why to use Spearman’s Rank…,[object Object],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…,[object Object]
So how do we do it?,[object Object],This is the equation, and looks complicated, so let’s think carefully about how we can do this… ,[object Object],In the above, rsrefers to the overall value or rank,[object Object],The equation has to be done before the value is taken away from 1,[object Object],In the above equation, the            sign means ‘the total of’,[object Object],d2 is the first thing we will try to establish in our ranked tables (see next slides),[object Object],‘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…,[object Object],The best way to do this would be through an example. ,[object Object],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),[object Object],To do this, we would construct a table as shown overleaf…,[object Object]
1. Here we have laid out a table of each of the twelve zones in a town,[object Object],2. Pedestrian counts for each zone here,[object Object],3. Number of Convenience shops for each zone here,[object Object],4. We now need to rank the data (two highlighted columns)– this is shown overleaf,[object Object]
You will see here that ,[object Object],on this example, the,[object Object],pedestrian counts,[object Object],have been ranked ,[object Object],from highest to ,[object Object],Lowest, with the ,[object Object],Highest value (70),[object Object],Being ranked as ,[object Object],Number 1, the ,[object Object],Lowest value (8) ,[object Object],Being ranked as ,[object Object],Number 12.,[object Object]
So that was fairly easy…,[object Object],We need to now do the next column for Convenience shops too.,[object Object],But hang on!,[object Object],Now we have a problem…,[object Object],We have two values that are 8, so what do we do?,[object Object],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,[object Object]
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. ,[object Object],But they can’t! Why not?,[object Object],Because we have already used rank number 5! So we would need to go to rank 6,[object Object],This situation is complicated further by the fact that the next two ranks are also tied.,[object Object],So we do the same again – add ranks 6 and 7 and divide it by 2 to get 6.5,[object Object]
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.,[object Object],This is demonstrated on the next slide,[object Object]
The difference between the two ranks has now been established ,[object Object],So what next? We need to square each of these d values…,[object Object],Don’t worry if you have any negative values here – when we square them (multiply them by themselves) they will become positives,[object Object]
So, the first value squared would be 0.25 (-0.5 x -0.5),[object Object]
So what do we with these ‘d2’ figures?,[object Object],	First we need to add all of the figures in this d2 column together,[object Object],This gives us…. 32,[object Object],Now we can think about doing the actual equation!,[object Object]
Firstly, let’s remind ourselves of the equation...,[object Object],In this equation, we know the total of d2, which is 32,[object Object],So the top part of our equation is…,[object Object],6 x 32,[object Object],We also know what ‘n’ is (the number of sites or zones - 12 in this case), so the bottom part of the equation is…,[object Object],(12x12x12) - 12,[object Object]
We can now do the equation…,[object Object],192,[object Object],1716,[object Object],6 x 32,[object Object],123 - 12,[object Object],OK – so this gives us a figure of 0.111888111888,[object Object],Is that us finished?,[object Object],Sadly not!,[object Object]
This is the equation, which we will by now be sick of!,[object Object],I have circled the part of the equation that we have done…,[object Object],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!,[object Object],So…,[object Object],1 – 0. 111888111888 = 0.888,[object Object]
So we have our Spearman’s Rank figure….But what does it mean?,[object Object],-1,[object Object],0,[object Object],+1,[object Object],0.888,[object Object],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,[object Object],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.,[object Object]
However…,[object Object],	Just looking at a line and making an estimation isn’t particularly scientific. It might be that just by chance you have found a relationship in your sample of data.  ,[object Object]
To be sure that chance is not the reason for the relationship, we need to look in critical values tables to see the level of significance and strength of the relationship. This is shown overleaf…,[object Object]
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.,[object Object],2. If look across we can see there are two further columns – one labelled 0.05, the other 0.01.,[object Object],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,[object Object],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.,[object Object],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.,[object Object]
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,[object Object]
Finally…,[object Object],You need to think how you can use this yourself…I would advise that you do scattergraphs for the same sets of data so that you have a direct comparison,[object Object]
We can therefore conclude,[object Object],that the relationship observed in the sample has been caused ,[object Object],by something other than,[object Object],chance and may reflect a real causality in the total ,[object Object],population of all shops that could be studied.,[object Object]

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Spearman after priory man

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