Lecture 10_spearman's rank correlation
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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.
Lecture 10_spearman's rank correlation
- 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.