Three judges evaluated the performance of 11 students in a cultural program and assigned ranks to each student. Spearman's rank order correlation was used to determine the level of agreement between the ranks assigned by the two judges. The analysis found a significant positive correlation (r=0.89) between the ranks, indicating a high level of agreement between the judges. However, when a second analysis was done on ranks assigned by different judges the following year, it found a significant negative correlation (r=-0.88), suggesting the two judges that year assessed students' performances in contradictory ways. While both analyses found statistically significant correlations, only the first showed a practical agreement between the judges.

1. Spearman’s Rank order Correlation: A method
of correlation between two ordinal variables
Dr. Rajeev Kumar
M.S.W., (TISS, Mumbai), M.Phil., (CIP, Ranchi), UGC-JRF, Ph.D. (IIT Kharagpur)
Lecture-10 (Research Methodology)

2. A Case: In a college, 11 students participated in a cultural programme. The performance of all
eleven students were evaluated by two judges by assigning each students a rank according to
his/her performance
Name of contestants The first Judge
Assigned rank
The second Judge
Assigned rank
Ramesh 8 5
Pooja 10 11
Joseph 9 10
Amir 3 2
Sonu 11 9
Rahul 4 3
Manisha 7 8
Santosh 2 4
Pratibha 1 1
Manoj 6 7
Sony 5 6
2

3. Now, what to do?
 Can you see any relation between ranks to all students given by two
judges?
 Do you see any agreement between two judges?
To find this answer, we formulate two hypotheses
Alternative hypothesis
Ha: There will be a significant correlation between ranks given by both the
judges.
H0: There will not be any significant correlation between the ranks given by
both the judges.
3

4. Spearman’s rank order correlation
To find the agreement between both the judges, we need to apply the
Spearman’s rank order correlation.
The Spearman Rank correlation is applied between two ranks, which are
ordinal variables. It can be applied to continuous data also, but after
converting them into ranks.
Whereas, Pearson’s Correlation is applied between to continuous variables
(Either Interval or Ratio).
The value of Spearman’s Correlation ranges from -1 to +1 likewise Pearson
correlation. Rules of interpretations are similar to Pearson’s correlation.
4

5. Name of
contestants
The
first
Judge
Assign
ed rank
(d1)
The
second
Judge
Assigned
rank (d2)
D= (D1-D2) D² Calculation of spearman’s
rank correlation
Ramesh 8 5 3 9 ΣD²= 24
6ΣD²= 144
Rho= 1- 6ΣD²/ n (n²-1
Pooja 10 11 -1 1 n=11, n²= 121 Rho= 1-0.1090=0.89
Joseph 9 10 -1 1 n (n²-1)= 11(121-1)= 1320 Rho= 0.89
Amir 3 2 1 1 6ΣD²/ n (n²-1)= 144/1320
Sonu 11 9 2 4 6ΣD²/ n (n²-1)= .1090
Rahul 4 3 1 1 1-0.1090=0.89
Manisha 7 8 -1 1
Santosh 2 4 -2 4
Pratibha 1 1 0 0
Manoj 6 7 -1 1
Sony 5 6 -1 1
ΣD²= 24
5
Step-1: Arrange the data in a table

6. Step-2: subtract the ranks and obtain
the value ‘D’. D= D1-D2
6

7. Step-3: Square the value of ‘D’ and
apply the formula.
7

8. Step-4: see the critical table 8

9. Step-5: Interpret the result
 The critical value (two-tailed) of Spearman rank test at (p ≤0.05) for n=11 is .618
 The critical value (two-tailed) of Spearman rank test at (p ≤0.01) for n=11 is .755
 Our obtained test value r= 0.89. This is a positive correlation and it is higher than .618
and .755. Therefore, the correlation value .89 is highly significant at (p ≤0.01).
 It means, there is an agreement of ranking assigned by both the judges. Though, due
to subjective assessment, their ranks look different but both of them evaluate the
students equally. There is no difference is the assessment of both the judges.
9
12-11-2021

10. The nature of the result
 There is consistency in the result given
by judges, we can accept them and
award the wining candidates.
 As there is a significant positive
correlation.
Ha: There will be a significant correlation
between ranks given by both the judges.
H0: There will not be any significant
correlation between the ranks given by
both the judges.
 Therefore, the alternative hypothesis is
accepted and null hypothesis is rejected.
10

11. Case-2: what if the story would have been different?
Next year, once again cultural programme took place. Two different
judges were invited. Now look at their ranks. What do you think?
Name of contestants The first Judge
Assigned ranks
The second judge
assigned ranks
Ramesh 8 4
Pooja 10 1
Joseph 9 3
Amir 3 9
Sonu 11 2
Rahul 4 7
Manisha 7 6
Santosh 2 10
Pratibha 1 11
Manoj 6 5
Sony 5 8
11

12. Same hypotheses
There are same students, and similar performance, but different judges.
We formulate the same hypotheses
Ha: There will be a significant correlation between ranks given by both
the judges.
H0: There will not be any significant correlation between the ranks given
by both the judges.
12

13. Lets calculate and find out the result. 13

14. Now there is new interpretation
 The critical value (two-tailed) of Spearman rank test at (p ≤0.05) for n=11 is
.618
 The critical value (two-tailed) of Spearman rank test at (p ≤0.01) for n=11 is
.755
 Our obtained test value r= - 0.88. This is a negative correlation and it is
higher than .618 and .755. Therefore, the correlation value -.88 is highly
significant at (p ≤0.01) but in negative direction.
 But this time, there is no agreement between both the judges. Both are
assessing the students in contradictory way. Their results are contrast to
each other. Their results can not be considered for the award ceremony.
14

15. But what about the hypotheses?
Ha: There will be a significant correlation
between ranks given by both the judges.
H0: There will not be any significant
correlation between the ranks given by both
the judges.
There is significant correlation, but in the
negative direction. Therefore, the alternative
hypothesis (two tailed) is accepted and null
hypothesis is rejected.
Two tailed: it could be in both the directions –
negative or positive
There is a statistical significance but not the
practical significance. Because judges did not
assessed the students equally.
The nature of the result
15

16. What if variables are continuous?
 There are 11 students, who secured marks in study (x) ( out of100) and
extra-curricular activities (y) (out of 50).
16
Students X= Marks in studies Y= marks in extra-
curricular activities (50)
Ramesh 89 45
Pooja 91 47
Joseph 65 32
Amir 43 22
Sonu 72 34
Rahul 56 26
Manisha 59 28
Santosh 49 24
Pratibha 95 46
Manoj 78 38
Sony 81 43

17. Step-1: arrange the data in a table
Students X (study
marks)
D1 Y (extra-
curricular
marks)
D2 D=D1-D2 D²
Ramesh 89 45
Pooja 91 47
Joseph 65 32
Amir 43 22
Sonu 72 34
Rahul 56 26
Manisha 59 28
Santosh 49 24
Pratibha 95 46
Manoj 78 38
Sony 81 43
17

18. Step-2: calculate the ranks of (x) and (Y)
Students X (study
marks)
D1 Y (extra-
curricular
marks)
D2 D=D1-D2 D²
Ramesh 89 3 45 3
Pooja 91 2 47 1
Joseph 65 7 32 7
Amir 43 11 22 11
Sonu 72 6 34 6
Rahul 56 9 26 9
Manisha 59 8 28 8
Santosh 49 10 24 10
Pratibha 95 1 46 2
Manoj 78 5 38 5
Sony 81 4 43 4
18

19. Step-3: Now square the D
Students X (study
marks)
D1 Y (extra-
curricular
marks)
D2 D=D1-D2 D²
Ramesh 89 3 45 3 0
Pooja 91 2 47 1 1
Joseph 65 7 32 7 0
Amir 43 11 22 11 0
Sonu 72 6 34 6 0
Rahul 56 9 26 9 0
Manisha 59 8 28 8 0
Santosh 49 10 24 10 0
Pratibha 95 1 46 2 -1
Manoj 78 5 38 5 0
Sony 81 4 43 4 0
19

20. Step-4: calculate the differences of ranks. D=D1-D2
Students X (study
marks)
D1 Y (extra-
curricular
marks)
D2 D=D1-D2 D²
Ramesh 89 3 45 3 0 0
Pooja 91 2 47 1 1 1
Joseph 65 7 32 7 0 0
Amir 43 11 22 11 0 0
Sonu 72 6 34 6 0 0
Rahul 56 9 26 9 0 0
Manisha 59 8 28 8 0 0
Santosh 49 10 24 10 0 0
Pratibha 95 1 46 2 -1 1
Manoj 78 5 38 5 0 0
Sony 81 4 43 4 0 0
ΣD² =2
20

21. Step-5: calculate the Spearman rho
r= 0.99 is highly positive significant (p≤0.000001), practically, no need to see
the critical table. All the students are performing equally good in studies and
extra-curricular activities. Both studies and extra activities are promoting each
other.
21

22. Step-3: Lets try the same correlation with Pearson method.
Students X
(study
marks)
(x-xˉ) Y
(extra-
curricula
marks)
(Y-Yˉ) (x-xˉ) (Y-Yˉ) (x-xˉ)² (Y-Yˉ)²
Ramesh 89 18.27 45 10 182.72 333.89 100
Pooja 91 20.27 47 12 243.27 410.98 144
Joseph 65 -5.72 32 -3 17.18 32.80 9
Amir 43 -27.72 22 -13 360.45 768.80 169
Sonu 72 1.27 34 -1 -1.27 1.60 1
Rahul 56 -14.72 26 -9 132.54 216.89 81
Manisha 59 -11.72 28 -7 82.09 137.52 49
Santosh 49 -21.72 24 -11 239 472.07 121
Pratibha 95 24.27 46 11 267 589.18 121
Manoj 78 7.27 38 3 21.81 52.89 9
Sony 81 10.27 43 8 82.18 105.52 64
Xˉ=
70.72
Yˉ=35 = 1627
Σ(x-xˉ) (Y-Yˉ)
=3122.18
Σ(x-xˉ)²
=868
Σ(Y-Yˉ)²
22

23. Calculation of correlation with Pearson
method
23
 We can see, through Pearson method also, r=0.9883, which is approximately 0.99.
Therefore, the value of ‘r’ from Pearson and Spearman both the methods are very
close or similar. In research practice, we can apply both the methods alternatively.

24. Now look at this formula
1. What will happen, if D² increase and decrease?
2. What will happen, if the sample size decrease?
3. What will happen, if sample size increase?
4. What will happen, if whole 6ΣD²/ N(N²-1)
increase and decrease?
5. What will happen, if the value of 6ΣD²/ N(N²-1)
become1 or close to 1?
6. What will happen, if the value of 6ΣD²/ N(N²-1) is
more than 1?
7. In what conditions the value of D² will increase?
8. In what conditions the value of D² will decrease?
24 12-11-2021

25. Congratulation!
Now your concepts are very clear!
Thanks for your interest and attention
IN THE NEXT SESSION, WE WILL LEARN HOW TO SOLVE THE SPEARMAN’S
CORRELATION WITH TIED RANKS AND PARTIAL CORRELATION. AND IT
WILL BE THE LAST SESSION ON CORRELATIONS.