Spearman Rank and Pearson Product Correlation Technique
Highlights:
- Spearman Rank Correlation Technique
- Pearson Product Moment Correlation Technique
Earlier we discussed correlation technique and some of the types like the Phi coefficient and the Point Biserial. Today we would look or discuss another two types of correlation techniques. We shall look at the Spearman Rank correlation coefficient and the Pearson Product Moment correlation techniques.
Spearman Rank Correlation Technique: this is used when the two variables to be correlated are measured on an ordinal scale which is one of the levels of measurement. The data must be ranked and order.
Example: Assuming a researcher is interested in finding the relationship between ranks of students in a particular task by two judges. Use the information in the table below to determine this.
X | Y |
10 | 11 |
15 | 14 |
12 | 12 |
16 | 19 |
13 | 20 |
18 | 10 |
10 | 12 |
Spearman Rank and Pearson Product Correlation Technique
Solution
X | Y | Rx | Ry | D=Rx-Ry | D-sqr |
10 | 11 | 6.5 | 6 | 0.5 | 0.25 |
15 | 14 | 3 | 3 | 0 | 0 |
12 | 12 | 5 | 4.5 | 0.5 | 0.25 |
16 | 19 | 2 | 2 | 0 | 0 |
13 | 20 | 4 | 1 | 3 | 9 |
18 | 10 | 1 | 7 | -6 | 36 |
10 | 12 | 6.5 | 4.5 | 2 | 4 |
Total | 49.5 |
Explanation of the table.
The Rx column is gotten by ranking or assigning positions to the values of ‘x’ column starting with the highest value of x. from the table above the highest value of the ‘x’ column is 18 and it is assign a position of 1.
Second highest is 16 and its assigned a position or a rank of 2. Third highest is 15 assigned with a rank of 3, fourth highest is 13 assigned with a rank of 4. Fifth highest is 12 assigned with a rank of 5.
Now notice that we have two scores(10 and 10) in the ‘x’ column. Their rank is determined by taking the average of their relative positions. This simply mean adding the positions there are to occupy normally and dividing it by 2.
This can be gotten by (6+7)/2 = 6.5
Spearman Rank Correlation Techniques
The Ry column is also gotten by ranking or assigning positions to the values of ‘y’ column starting with the highest value of y. from the table above the highest value of the ‘y’ column is 20 and it is assign a position of 1.
Second highest is 19 assigned a position or a rank of 2. Third highest is 14 assigned with a rank of 3. Notice that 12 appeared twice, so like we did earlier, we find the average of their relative positions which is (4+5)/2 = 4.5.
Now, the next position to fill now is 6^{th}. Remember if happens that two persons took first position in a class, there will be no second position again, what would be left to fill is third position, so the same applied here. So since two persons are occupying the 4^{th} position, then there will be no fifth position again.
So we start filling from the sixth position. The sixth highest value is 11 assigned with a rank of 6 and lastly, 10 assigned with a rank of 7.
‘D = Rx-Ry’ column is gotten by subtracting the rank of y from the rank of x. like 6.5-6 =0.5, 3-3 = 0, 5-4.5 = 0.5, 2-2 =0 and so on.
‘D-Square’ column is gotten by squaring the respective values of the ‘D’ column.
Note: Always remember that when you have a tie, you simply add their position(rank) and divide it by 2 then assign the rank to both.
Now finding the Spearman rank correlation coefficient, we use the below formula,
P = 1 – (6D-sqr)/N(N-sqr -1)
= 1 – (649.5)/7(49 – 1)
= 1 – 0.88 = 0.12
This shows a weak positive relationship between the rank of the judges. In essence, the judges rarely agree in their ranking.
Pearson Product Moment Correlation Technique
This technique is used when the two variables under consideration are both continuous, or measured on an interval/ratio level of measurement.
The technique assumes the following;
- It assumes that the scores are normally distributed.
- Pearson product moment correlation technique assumes a linear relationship
- The two variables are independent of each other
Example: A researcher is interested in finding the relationship between students’ achievement in physics and mathematics. Use the table below to determine the extent of the relationship.
S/N | Physics(x) | Math(y) |
1 | 5 | 10 |
2 | 6 | 9 |
3 | 3 | 7 |
4 | 10 | 4 |
5 | 4 | 5 |
6 | 2 | 3 |
7 | 3 | 6 |
8 | 8 | 7 |
Solution
S/N | Physics(x) | Math(y) | X-sqr | Y-sqr | XY |
1 | 5 | 10 | 25 | 100 | 50 |
2 | 6 | 9 | 36 | 81 | 54 |
3 | 3 | 7 | 9 | 49 | 21 |
4 | 10 | 4 | 100 | 16 | 40 |
5 | 4 | 5 | 16 | 25 | 20 |
6 | 2 | 3 | 4 | 9 | 6 |
7 | 3 | 6 | 9 | 36 | 18 |
8 | 8 | 7 | 16 | 49 | 56 |
Total | 41 | 51 | 263 | 365 | 265 |
where, N = 8, summation ‘x’ squared = 263, summation ‘y’ squared = 365 and summation ‘xy’ = 265.
= (8265 – 4151)/Sqr[(8263 – 4141)(8365-5151)]
= 0.1
This implies a weak relationship between students’ achievement in physics and mathematics. It implies that if a student passes physics it does not mean the student will pass mathematics and if the student fails physics, it does not mean the student will fail mathematics.