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 |
X2 |
Y2 |
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=n×(∑(XY)-(∑X×∑Y))√(n×∑X2-∑(X)2)×(n×∑Y2-∑(Y)2)
r=10×(5048-(146×300))√(10×3064-∑(146)2)×(10×9250-∑(300)2)
r=10×(5048-43800)√(30640-21316)×(92500-90000)
r=10×
r = \frac{ - 387520}{\sqrt {23310000}}
r = \frac{- 387520}{4828.0431}
r = - 80.2644
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