Spearman's Rank Correlation Coefficient
Practise Spearman's Rank Correlation Coefficient for GCSE Statistics with this free worksheet and full mark scheme — Higher tier exam-style questions with worked answers for AQA and Edexcel. Spearman's rank correlation coefficient measures the strength of a relationship using ranked data, from -1 to +1.
Free downloads
These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA and Edexcel specifications. Every worksheet comes with a full mark scheme.
This is a Higher tier only topic, so there's no Foundation paper — only the Higher worksheet and mark scheme below.
Topic overview
Spearman's rank correlation coefficient measures the strength and direction of correlation using ranks rather than the original values. This is a Higher-only topic.
The formula is \(r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)}\), where \(d\) is the difference between the two ranks for each item and \(n\) is the number of items.
The result always lies between \(-1\) and \(1\). A value near \(1\) indicates strong positive rank correlation, near \(-1\) strong negative, and near \(0\) little or no correlation. Because it uses ranks, it can be applied to ordinal data and is not distorted by outliers in the way that a calculation using actual values would be.
Revision notes
The formula
\(r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)}\).
Rank each variable separately, find the difference \(d\) between the two ranks for each item, square each difference and add them to get \(\sum d^2\). Then substitute.
Interpreting the result
The value always lies between \(-1\) and \(1\).
Near \(1\): strong positive rank correlation. Near \(-1\): strong negative. Near \(0\): little or no correlation. A result outside this range means an arithmetic error has been made.
Why use ranks
Ranking allows the method to be used with ordinal data, where actual values may not exist.
It is also not distorted by outliers, because an extreme value simply becomes the highest or lowest rank rather than pulling the calculation.
Key points
- Spearman's coefficient uses ranks, not values.
- \(r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)}\).
- \(d\) is the difference between the two ranks.
- The result lies between \(-1\) and \(1\).
- Near 1 means strong positive correlation.
- It can be used with ordinal data.
Worked examples
Example 1
For 5 items, \(\sum d^2 = 4\). Work out Spearman's rank correlation coefficient. [3 marks]
Working
Example 2
A calculation gives \(r_s = 1.4\). Explain what this shows. [2 marks]
Working
Example 3
Explain one advantage of using ranks rather than the actual values. [2 marks]
Working
Common mistakes
Forgetting to square the differences.
The formula uses \(\sum d^2\), not \(\sum d\).
Accepting a result outside \(-1\) to \(1\).
That indicates an arithmetic error.
Ranking both variables in different directions.
Rank both the same way, either highest first or lowest first, consistently.
Forgetting to subtract from 1.
The formula is 1 minus the fraction.
Exam tips
- Rank both variables in the same direction.
- Square each difference before summing.
- Check the result lies between −1 and 1.
- Remember this is a Higher-only topic.
Key terms
- Spearman's coefficient
- A measure of rank correlation between two variables.
- Rank
- The position of a value when placed in order.
- \(\sum d^2\)
- The total of the squared rank differences.
- Ordinal data
- Data with a natural order, suitable for ranking.
Related topics
Written and reviewed against the current AQA and Edexcel specifications. Spotted an error? Let us know.