Page 188 - DMGT404 RESEARCH_METHODOLOGY

Research Methodology

Notes Since the Spearman’s formula is based upon the assumption of different ranks to different
individuals, therefore, its correction becomes necessary in case of tied ranks. It should be noted
that the means of the ranks will remain unaffected. Further, the changes in the variances are
usually small and are neglected. However, it is necessary to correct the term d i 2 and accordingly
2
m m   1
the correction factor , where m denotes the number of observations tied to a particular
12
 2 4 1 
rank, is added to it for every tie. We note that there will be two correction factors, i.e.,
12
 3 9 1 
and in the above example.
12
9.1.9 Limits of Rank Correlation

A positive rank correlation implies that a high (low) rank of an individual according to one
characteristic is accompanied by its high (low) rank according to the other. Similarly, a negative
rank correlation implies that a high (low) rank of an individual according to one characteristic
is accompanied by its low (high) rank according to the other. When r = +1, there is said to be
perfect consistency in the assignment of ranks, i.e., every individual is assigned the same rank
with regard to both the characteristics. Thus, we have d 2 i = 0 and hence, r = 1.

Similarly, when r = – 1, an individual that has been assigned 1st rank according to one characteristic
must be assigned nth rank according to the other and an individual that has been assigned 2nd
rank according to one characteristic must be assigned (n – 1)th rank according to the other, etc.
Thus, the sum of ranks, assigned to every individual, is equal to (n + 1), i.e., X + Y = n + 1 or
i i
Y = (n + 1) – X , for all i = 1, 2, ...... n.
i i
Further, d = X – Y = X – (n + 1) + X = 2X – (n + 1)
i i i i i i
Squaring both sides, we have

2
2
d  2X  n  1   4X  n   1  4 n  1 X i
2
2
i
i
i
Taking sum over all the observations, we have
2
4 X 
2
 d   i 2  n n   1   4 n   1  X  4n n  1 6 2n   1   n n  1  4n  n  1  2
2
i
i
2
2
2    n n  1 n   1  n n   1
  n n   1   3 2n  1  n   1   2 n   1  3  3


Substituting this value in the formula for rank correlation we have
2
6n n   1 1
  1     1
2
3  n n   1
Hence, the Spearman’s coefficient of correlation lies between – 1 and + 1.
Example: The following table gives the marks obtained by 10 students in commerce and
statistics. Calculate the rank correlation.
Marks in Statistics 35 90 70 40 95 45 60 85 80 50
Marks in Commerce 45 70 65 30 90 40 50 75 85 60

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