Spearman's Rank CorrelationDate: 02/17/99 at 17:45:10 From: Bobby Subject: Spearman's Rank Correlation I would like to know where the 6 in the formula of Spearman's rank correlation originated. How did it turn out to be 6? I have tried consulting people and encyclopedias but without answers. Thank you very much. Date: 02/17/99 at 18:19:49 From: Doctor Pat Subject: Re: Spearman's Rank Correlation I am not sure, but I believe that the 6 comes from the sum of squares of integers in the denominator. Remember that the Spearman rho is (if there are no ties) equivalent to the Pearson's r when the ranks are treated as x,y coordinates. The denominator has (n-1)sx*sy. The sums of the squares of the x variables, since they are ranks, must be consecutive integers. Similar method and reasoning works for the y values. When we find the variances by summing these squares we get a six in each denominator, and taking the square root of 6*6 gives us a 6 outside the radical. To simplify the mathematics, multiply numerator and denominator by 6 and it shows up at the top... Good luck! - Doctor Pat, The Math Forum http://mathforum.org/dr.math/ Date: 02/18/99 at 16:17:28 From: Doctor Pat Subject: Re: Spearman's Rank Correlation After I sent you the first response, I posted a note on a statistics list. Here is a MUCH better response to your question that I received from a correspondant of mine who is an excellent statistician. -------------------------------------------------------------------- The Spearman rank correlation coefficient is based upon the sum (differences^2) in the two rankings. Consider the extreme situations of the rankings. If there are N ranks and the two rankings are identical in every place, then sum(differences^2) = 0. If they are the reverse of each other, for example, if one is 1, 2, 3, 4, 5 and the other is 5, 4, 3, 2, 1, then the sum(differences^2) = N(N^2 - 1)/3. The closer to 0 the sum(differences^2) is, the more alike the rankings. The closer to N(N^2 - 1)/3 the sum(differences^2), the more disparate the rankings. But we want to have a common scale to judge by, not one that depends upon N. Moreover, we would like the scale to go from -1 to 1, so we want to find a transformation that takes 0 to 1 and N(N^2 - 1)/3 to -1. Write the equation of the line passing through the points (0, 1) and (N(N^2 - 1)/3, -1). This line has a slope of -6/(N(N^2 - 1)) and an intercept of 1. - Doctor Pat, The Math Forum http://mathforum.org/dr.math/ |
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