• Revision:Spearmans rank correlation

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Geography > Spearman's Rank Correlation

Spearman's Rank Correlation is a technique used to test the direction and strength of the relationship between two variables. In other words, its a device to show whether any one set of numbers has an effect on another set of numbers.

It uses the statistic R_s which falls between -1 and +1.

Procedure for using Spearman's Rank Correlation

  1. State the null hypothesis i.e. "There is no relationship between the two sets of data."
  2. Rank both sets of data from the highest to the lowest. Make sure to check for tied ranks.
  3. Subtract the two sets of ranks to get the difference d.
  4. Square the values of d.
  5. Add the squared values of d to get \sigma d^2
  6. Use the formula R_s = 1 - (6\sigma d^2/n^3-n) where n is the number of ranks you have.
  7. If the R_s value...
... is -1, there is a perfect negative correlation.
...falls between -1 and -0.5, there is a strong negative correlation.
...falls between -0.5 and 0, there is a weak negative correlation.
... is 0, there is no correlation
...falls between 0 and 0.5, there is a weak positive correlation.
...falls between 0.5 and 1, there is a strong positive correlation
...is 1, there is a perfect positive correlation
between the 2 sets of data.
  1. If the R_s value is 0, state that null hypothesis is accepted. Otherwise, say it is rejected.

Practical Example of Spearman's Rank Correlation

Question: Use the Spearman's Rank Correlation to establish whether there is any relationship between the distance away from school students live and the IB Geography grades they attain.

  • Red type indicates what you have been given. Black type indicates the working done.
  1. Null Hypothesis: There is no relationship between the two sets of data.
Distance From School (in miles) r IB Geography Grades Attained r d d^2
3 2 4 4 2 4
7 1 4 4 3 9
2 3 7 1 2 4
2 3 6 2 1 1
1 5 5 3 2 4
 \sigma d^2 = 22
  1. R_s = 1-(6 \sigma d^2 / n^3-n)
\sigma d^2 = 22 therefore 6\sigma d^2 = 132
n = 5 therefore n^3 - n = 120
  1. R_s = 1 - (132/120)
  2. R_s = 1 - 0.91
  3. R_s = 0.09
  4. There is a weak positive correlation between the two sets of data. The null hypothesis is rejected.


This is a very poor example. There are three mistakes. 1) Firstly, the sign for standard deviation (σ) is used, it should be the sum of sign: Σ. 2) Some figures in column d should be negative (the reason for squaring d is to remove the negatives). 3) The final figures do not add up.

Not exactly great for a revision guide.

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