Properties of coefficient of correlation

4,490 views 8 slides Feb 29, 2020
Slide 1
Slide 1 of 8
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8

About This Presentation

Properties of coefficient of correlation


Slide Content

Properties of coefficient of correlation:-

1) The coefficient of correlation always lies between -1 and
+1 i.e, −1≤??????≤+1

2) The correlation coefficient is symmetrical with respect to
X and Y i.e ??????
��= ??????
��


3) The coefficient of correlation is the geomatric mean of the
two regression coefficient.
r = √??????×?????? Or r = √??????
��×??????
��

4) It does not depend upon the units employed

5) It is independent of orgin and unit of measurement


6) The coefficient of cerrelation is unaffected by change of
origin and scale i.e ??????
��=??????
��

7) The coefficient of cerrelation is a pure number.

Example-7:
i) Calculate regression co-efficient by� and ??????�� and
calculate correlation with the help of regression
coefficients for the following pairs of observations.
ii) Calculate Karl Pearson’s coefficient of correlation and
then verify that.

X 1 2 3 4 5 6 7 8
Y 12 14 16 18 20 22 24 26
Solution:

We know that the correlation coefficient is the geometric mean of the
two regression coefficients.
??????=√??????��×??????��

??????��=
??????∑��−∑�∑�
??????∑�
2
−(∑�)
2



??????��=
??????∑��−∑�∑�
??????∑�
2
−(∑�)
2


The necessary calculations for regression coefficients are given
below.

� � �� �
�
�
�

1 12 12 1 144
2 14 28 4 196
3 16 48 9 256
4 18 72 16 324
5 20 100 25 400
6 22 132 36 484
7 24 168 49 576
8 26 208 64 676
∑x=36 ∑y=152 ∑xy=768 ∑x
2
=204 ∑y
2
=3056

??????��=
??????∑��−∑�∑�
??????∑�
2
−(∑�)
2


??????��=
8(768)−(36)(152)
8(204)−(36)
2


??????��=
6144−5472
1632−1296


??????��=
672
336


??????��=2

??????��=
??????∑��−∑�∑�
??????∑�
2
−(∑�)
2


??????��=
8(768)−(36)(152)
8(3056)−(152)
2


??????��=
6144−5472
24448−23104


??????��=
672
1344


??????��=0.5

We know that correlation coefficient is the geometric mean of the two
regression coefficients i.e.

??????=√??????��×??????��

??????=√2×0.5

??????=�

(ii) Karl Pearson’s co-efficient of correlation.
??????=
??????∑��−∑�∑�
√(??????∑�
2
−(∑�)
2
)(??????∑�
2
−(∑�)
2
)


??????=
8(768)−(36)(152)
√(8(204)−(36)
2
)(8(3056)−(15)
2
)


??????=
6144−5472
√(1632−1296)(24448−23104)


??????=
672
√(336)(1344)


??????=
672
√451584


??????=
672
672


??????=�

Hence Proved
??????=√??????��×??????��

Example-8:
If ??????
��=51.9 and ??????
��=0.019
Find coefficient of determination

Solution:
??????
��
2
=??????
��×??????
��×100

??????
��
2
=(51.9)(0.019)×100

??????
��
�
=��.��%

It means that 98.61% of the variation in the �-variable is explained or
accounted for ??????� variation in the �-variable.


Example-9:
For the following two sets, the regression lines for each set are
respectively.
i) �=1.94�+10.83 (� �� �) and
�=0.15�+6.18 (� �� �)

ii) �=−1.96�+15 (� �� �) and
�=−0.45�+7.16 (� �� �)

Find coefficient of correlation in each case.

Solution:
i) Regression coefficient � on � (??????��)=1.94

Regression coefficient � on � (??????��)=0.15

??????=√??????��×??????��

??????=√(1.94)(0.15)

??????=�.��

ii) Regression coefficient � on � (??????��)=−1.96

Regression coefficient � on � (??????��)=−0.45

??????=√??????��×??????��

??????=√(1.96)(0.45)

??????=−�.��

It is to be noted when both regression coefficients are negative then
“??????” is also negative.


Example-10:
If �=18, ∑�=638, ∑�=41, ∑��=1569.5, ∑�
2
=25814,
∑�
2
=101.45

i) Find simple coefficient of correlation.

ii) If �=
�−3
5
and �=
�
20
then what would be the coefficient of
correlation between � and �.


Solution:
??????=
�∑��−(∑�)(∑�)
√{�(∑�
2
)−(∑�)
2
}{�(∑�
2
)−(∑�)
2
}


??????=
18(1569.5)−(638)(41)
√{18(25814)−(638)
2
}{18(101.45)−(41)
2
}

??????=
28251−26158
√(464652−407044)(18261−1681)

??????=
2093
√(57608)(145.1)


??????=
2093
√8358920.8


??????=
2093
2891.18


??????=�.��

(ii)Correlation is unaffected by the change of origin and scale.
i.e.
??????
��=??????
��

??????
��=�.��