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CMA Foundation · Fundamentals of Business Mathematics and Statistics · Correlation and Regression

For five firms, the pairs of (X, Y) values are (8, 5), (12, 9), (12, 7), (15, 7), (20, 11). Using rank 1 for the highest value and the average rank for ties, with the standard tie correction added to Σd² for each tied group, what is Spearman's coefficient?

The coefficient is 0.725. The rank differences give Σd² = 4.5, and the two tied pairs add 0.5 each, giving 5.5. Then r = 1 − 33/120 = 0.725. Ignoring the tie correction would wrongly give 0.775.

  1. A0.725Correct
  2. B0.775
  3. C0.750
  4. D0.275

Explanation

X ranks: 5, 3.5, 3.5, 2, 1. Y ranks: 5, 2, 3.5, 3.5, 1. Σd² = 0 + 2.25 + 0 + 2.25 + 0 = 4.5. Each tied pair adds (m³−m)/12 = 0.5, so two groups add 1, giving 5.5. r = 1 − 6×5.5/120 = 1 − 0.275 = 0.725. Omitting the correction gives 0.775.

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