Calculate the coefficient of correlation
The
data below gives the marks obtained by 10 pupils taking Maths and Physics tests
Pupil |
A |
B |
C |
D |
E |
F |
G |
H |
I |
J |
Maths Marks |
20 |
23 |
8 |
29 |
14 |
11 |
11 |
20 |
14 |
14 |
Physics Marks
|
30 |
35 |
21 |
33 |
33 |
26 |
22 |
31 |
33 |
36 |
Calculate the coefficient of correlation.
Solution:
Pupil |
Maths Marks |
Physics Marks |
XY |
`X^2` |
`Y^2` |
A |
20 |
30 |
600 |
400 |
900 |
B |
23 |
35 |
805 |
529 |
1225 |
C |
8 |
21 |
168 |
64 |
441 |
D |
29 |
33 |
899 |
841 |
1089 |
E |
14 |
33 |
462 |
196 |
1089 |
F |
11 |
26 |
286 |
121 |
676 |
G |
11 |
22 |
242 |
121 |
484 |
H |
20 |
31 |
620 |
400 |
961 |
I |
14 |
33 |
462 |
196 |
1089 |
J |
14 |
36 |
504 |
196 |
1296 |
∑ |
146 |
300 |
5048 |
3064 |
9250 |
Here n = 10
`r = \frac{n \times (\sum (XY ) - (\sum X \times \sum Y))}{\sqrt {(n \times \sum X^2 - \sum ( X )^2) \times (n \times \sum Y^2 - \sum ( Y )^2)}}`
`r = \frac{10 \times (5048 - (146 \times 300))}{\sqrt {(10 \times 3064 - \sum (146)^2) \times (10 \times 9250 - \sum ( 300)^2 ) }}`
`r = \frac{10 \times (5048 - 43800)}{\sqrt {(30640 - 21316) \times (92500 - 90000)}}`
`r = \frac{10 \times - 38752}{\sqrt {9324 \times 2500}}`
`r = \frac{ - 387520}{\sqrt {23310000}}`
`r = \frac{- 387520}{4828.0431}`
`r = - 80.2644`
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