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Worked Examples · Example 5
Q.

Example 5. The values of X and Y are given as follows. Work out the ranks (giving common ranks to repeated items) and calculate the Spearman's rank correlation coefficient.

XY
120075
115065
100050
990100
80090
78085
76090
75040
73050
70060
62050
60075
Telangana TsbieTextbookSubjective· 5mImportance★★★★★est
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X-values are all distinct; Y has ties (90 twice, 75 twice, 50 thrice), so those items get common (average) ranks and a tie-correction m3−m12\frac{m^{3}-m}{12} is added for each group. Here ∑d2=198\sum d^{2}=198 and the total correction =3=3, giving rs≈+0.297r_s\approx +0.297.

Concept first

When items repeat, each is given the average of the ranks it would occupy, and Spearman's formula carries a correction term for every tie group of size mm:

rs=1−6[∑d2+∑m3−m12]N3−N,N=12r_s=1-\frac{6\left[\sum d^{2}+\sum\dfrac{m^{3}-m}{12}\right]}{N^{3}-N},\qquad N=12

Ranking Y (highest = 1): 100(1)100(1); 90,90→90,90\to ranks 2,3 ⇒2.5\Rightarrow 2.5 each; 85(4)85(4); 75,75→75,75\to ranks 5,6 ⇒5.5\Rightarrow 5.5 each; 65(7)65(7); 60(8)60(8); 50,50,50→50,50,50\to ranks 9,10,11 ⇒10\Rightarrow 10 each; 40(12)40(12).

The working table

XXRXR_XYYRYR_Yddd2d^{2}
12001755.5-4.520.25
11502657-5.025.00
100035010-7.049.00
990410013.09.00
8005902.52.56.25
78068542.04.00
7607902.54.520.25
75084012-4.016.00
73095010-1.01.00
700106082.04.00
6201150101.01.00
60012755.56.542.25

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