Statistics Definitions > Spearman Rank Correlation / Spearman’s Rho

## What is Spearman Rank Correlation / Spearman’s Rho?

The Spearman rank correlation coefficient, r_{s}, is the nonparametric version of the Pearson correlation coefficient. Your data must be ordinal, interval or ratio. Spearman’s returns a value from -1 to 1, where:

+1 = a perfect positive correlation between ranks

-1 = a perfect negative correlation between ranks

0 = no correlation between ranks.

**Contents**:

## Spearman Rank Correlation: Worked Example (No Tied Ranks)

The formula for the Spearman rank correlation coefficient when there are no tied ranks is:

**Sample Question: **

The scores for nine students in physics and math are as follows:

Physics: 35, 23, 47, 17, 10, 43, 9, 6, 28

Mathematics: 30, 33, 45, 23, 8, 49, 12, 4, 31

Compute the student’s ranks in the two subjects and compute the Spearman rank correlation.

Step 1: Find the ranks for each individual subject. I used the Excel rank function to find the ranks. If you want to rank by hand, order the scores from greatest to smallest; assign the rank 1 to the highest score, 2 to the next highest and so on:

Step 2: Add a third column, d, to your data. The d is the difference between ranks. For example, the first student’s physics rank is 3 and math rank is 5, so the difference is 3 points. In a fourth column, square your d values.

Step 4: Sum (add up) all of your d-squared values.

4 + 4 + 1 + 0 + 1 + 1 + 1 + 0 + 0 = 12. You’ll need this for the formula (the Σ d^{2 is just “the sum of d-squared values”).}

Step 5: Insert the values into the formula. These ranks are not tied, so use the first formula:

= 1 – (6*12)/(9(81-1))

= 1 – 72/720

= 1-0.1

= 0.9

**The Spearman Rank Correlation for this set of data is 0.9.**

## Spearman Rank Correlation: Worked Example (Tied Ranks)

Tied ranks are where two items in a column have the same rank. A couple of different formulas exist for dealing with tied ranks. Perhaps the easiest way is to use the mean of the tied ranks.

Let’s say two items in the above example tied for ranks 5 and 6. You would assign each data point a mean rank of 5.5:

Use the same formula, this time inserting the d-squared value for the tied ranks (14.5 from the new data):

= 1 – (6*14.5)/(9(81-1))

= 1 – 87/720

= 1 – 0.120833333

= 0.879

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Thank you for the step by step guidelines. I’m beginning to enjoy my subject because of your assistance. VerY stunning, correct an prompt.

I’m beginning to enjoy my subject because of your assistance. VerY stunning, correct an prompt.

thanks for the help If three judges are judging contestants, using spearmans rank correlaton how do u find the best Or the top three winners in the contest of 7 people

Hello, Joy,

Can you post this on our homework help forum? One of our mods will be happy to help. Thanks!

This website has helped me so so much with my work. Amazingly straight forward

Also advice how to calculate number of observations (n) when we have the value of Co-relation (R)

If you have just r, n could be anything. For example, a set of 100, 1000, or 10000 numbers could all have the same r value. I think you’d need a little more info to come up with n.

I successfully worked through the above example, using your step by step. However, I have tied ranks in my dataset, and am unable to decipher the formula for that. Do you have a step-by-step for data with tied ranks?

Penny,

Not yet — it’s in the plan for the future though.

The formula is very similar to how you work the Pearson Correlation (with summations of differences). It’s very cumbersome — good luck :)

Thanks but how I wish there was a worked out example with tied scores!

hi i have one question regarding to spearman rank correlation .how can i compute for the given three rank using spearman rank correlation coefficient in order to find the nearest approach

Hello, Semir,

Could you rephrase your question, please? I don’t understand quite what you mean when you say “find the nearest approach”.

Zippie: I added an example. Hope it helps :)

Thnk u soooo much it’s really very helpful