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CA Foundation · Quantitative Aptitude · Correlation and Regression

In a ranking of 10 firms by two analysts, some ranks are tied. In series X there are two separate ties of 2 items each, and in series Y there is one tie of 3 items. The sum of squared rank differences (using average ranks) is Σd² = 30. What is the rank correlation coefficient after applying the tie correction?

The corrected coefficient is 0.80. The tie corrections total 0.5 + 0.5 + 2 = 3, which is added to Σd² = 30 to give 33. Then R = 1 - 6×33/(10×99) = 0.8. Leaving out the correction would wrongly give 0.82.

  1. A0.80Correct
  2. B0.82
  3. C0.78
  4. D0.20

Explanation

Correction factors are (m³-m)/12: for each tie of 2 it is 0.5, for the tie of 3 it is 2. Total = 0.5 + 0.5 + 2 = 3. Adjusted numerator = 30 + 3 = 33. R = 1 - 6(33)/(10 × 99) = 1 - 198/990 = 0.80. Ignoring the correction gives 0.82, which is incorrect.

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