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Worked Examples · Example 3
Q.

Example 3. Five persons are assessed by three judges in a beauty contest. We have to find out which pair of judges has the nearest approach to common perception of beauty.

JudgeCompetitor 1Competitor 2Competitor 3Competitor 4Competitor 5
A12345
B24153
C13524

Calculate the rank correlation coefficient for each pair of judges.

Punjab PsebTextbookSubjective· 5mImportance★★★★★est
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Ranks are already given. Computing ∑d2\sum d^{2} for each pair and applying Spearman's formula gives rs(A,B)=+0.3r_s(A,B)=+0.3, rs(A,C)=+0.5r_s(A,C)=+0.5, rs(B,C)=−0.4r_s(B,C)=-0.4. The A–C pair shares the closest common perception of beauty.

Concept first

Beauty cannot be measured on a numerical scale, so we use Spearman's rank correlation coefficient, which correlates the ranks the judges award:

rs=1−6∑d2N(N2−1),d=difference in ranks,  N=5r_s=1-\frac{6\sum d^{2}}{N(N^{2}-1)},\qquad d=\text{difference in ranks},\; N=5

The pair whose rsr_s is the highest positive value agrees most closely.

Pair A & B

Comp.ABddd2d^{2}
112-11
224-24
33124
445-11
55324
Σ14

rs=1−6(14)5(24)=1−84120=+0.3r_s=1-\dfrac{6(14)}{5(24)}=1-\dfrac{84}{120}=+0.3

Pair A & C

Comp.ACddd2d^{2}
11100
223-11
335-24
44224
55411
Σ10

rs=1−6(10)120=1−60120=+0.5r_s=1-\dfrac{6(10)}{120}=1-\dfrac{60}{120}=+0.5

Pair B & C

Comp.BCddd2d^{2}
12111
24311
315-416
45239
534-11
Σ28

rs=1−6(28)120=1−168120=−0.4r_s=1-\dfrac{6(28)}{120}=1-\dfrac{168}{120}=-0.4

Interpretation

Only A–C (+0.5+0.5) and A–B (+0.3+0.3) show agreement; B–C is negatively correlated (−0.4-0.4). The largest positive coefficient belongs to A and C.

A note on the answer key …

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