Karl Pearson’s Coefficient Method (Correlation)

Karl Pearson’s Coefficient Method

Karl Pearson’s method of Coefficient of correlation is also known as Pearsonian coefficient or correlation or product correlation method.

Let X and Y be two random variables, then the correlation of coefficient between the variables X and Y is denoted by r(X, Y) or simply by r_XY, and is defined as:\[{{r}_{XY}}=r\left( X,Y \right)=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}\]\[Where,Cov\left( X,Y \right)=E\left[ \left\{ X-E\left( X \right) \right\}\left\{ Y-E\left( Y \right) \right\} \right]\]\[\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)={{\mu }_{11}}\]\[{{\sigma }_{X}}^{2}=E{{\left\{ X-E\left( X \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}}\]\[{{\sigma }_{Y}}^{2}=E{{\left\{ Y-E\left( Y \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}}\]\[\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)}{{{\left[ \left\{ \frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}} \right\}\left\{ \frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}} \right\} \right]}^{\frac{1}{2}}}}\]Other equivalent forms of Coefficient of correlation formulas are:\[Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)\]\[\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}-\bar{y}\frac{1}{n}\sum{{{x}_{i}}}}-\bar{x}\frac{1}{n}\sum{{{y}_{i}}}+\bar{x}\bar{y}\]\[\therefore Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}\]\[{{\sigma }_{X}}^{2}=\frac{1}{n}\sum{{{x}_{i}}^{2}}-{{\bar{x}}^{2}}\]\[{{\sigma }_{Y}}^{2}=\frac{1}{n}\sum{{{y}_{i}}^{2}}-{{\bar{y}}^{2}}\]\[\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{x}_{i}}^{2}}-{{{\bar{x}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{y}_{i}}^{2}}-{{{\bar{y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}\]

Note:

The values of correlation coefficient cannot exceed unity numerically. It always lies in between -1 and +1. That is -1 ≤ r(X, Y) ≤ 1. If r =+1, the correlation is perfect and positive and if r = -1, correlation is perfect and negative.

1. It is not affected by change of origin or change of scale.
2. It is a relative measure. It does not have any unit attached to it.

Example 01

Calculate the coefficient of correlation by Karl Pearson’s method based on following values

T1	75	60	45	30	15
T2	150	175	200	225	250

Solution:

T1	X = T1/15	X2	T2	Y = T2/25	Y2	XY
75	5	25	150	6	36	30
60	4	16	175	7	49	28
45	3	9	200	8	64	24
30	2	4	225	9	81	18
15	1	1	250	10	100	10
Total	15	55		40	330	110

\[\bar{X}=\frac{\sum{X}}{5}=\frac{15}{5}=3\]\[\bar{Y}=\frac{\sum{Y}}{5}=\frac{40}{5}=8\]\[\frac{1}{n}\sum{{{X}^{2}}=\frac{1}{5}\times 55=11}\]\[\frac{1}{n}\sum{{{Y}^{2}}=\frac{1}{5}\times 330=66}\]\[\frac{1}{n}\sum{XY=\frac{1}{5}\times 110=22}\]\[\therefore {{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}\]\[\therefore {{r}_{XY}}=\frac{22-3\times 8}{{{\left[ \left\{ 11-{{3}^{2}} \right\}\left\{ 66-{{8}^{2}} \right\} \right]}^{\frac{1}{2}}}}=\frac{-2}{2}=-1\]

Example 02

A computer while calculating correlation coefficient between two variables X and Y from 25 pairs of observation obtained the following results: n = 25, ΣX = 125, ΣX² = 650, ΣY = 100, ΣY² = 460, ΣXY = 508. It was later noticed that there was some mistake while copied down two pairs as

X	6	8
Y	14	6

While the correct values were

X	8	6
Y	12	8

Obtain the correct value of the correlation coefficient.

Solution:

Corrected ΣX = 125 – 6 – 8 + 8 + 6 = 125
Corrected ΣY = 100 – 14 – 6 + 12 + 8 = 100
Corrected ΣX² = 650 – 6² – 8² + 8² + 6² = 650
Corrected ΣY² = 460 – 14² – 6² + 12² + 8² = 436
Corrected ΣXY = 508 – 6(14) – 8(6) + 8(12) + 6(8) = 520\[\bar{X}=\frac{\sum{X}}{25}=\frac{125}{25}=5\]\[\bar{Y}=\frac{\sum{Y}}{25}=\frac{100}{25}=4\]\[\therefore \text{Corrected}~~{{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}\]\[\therefore {{r}_{XY}}=\frac{\frac{1}{25}\times 520-5\times 4}{{{\left[ \left\{ \frac{1}{25}\times 650-{{5}^{2}} \right\}\left\{ \frac{1}{25}\times 436-{{4}^{2}} \right\} \right]}^{\frac{1}{2}}}}\]\[\therefore {{r}_{XY}}=\frac{\frac{4}{5}}{1\times \frac{6}{5}}=\frac{2}{3}=0.67\]

Example 03

Calculate the correlation coefficient for the following

Firm	1	2	3	4	5	6	7	8	9	10
Sales	50	50	55	60	65	65	65	60	60	60
Expenses	11	13	14	16	16	15	15	14	13	13

Solution:

Here,
\[\bar{X}=\frac{\sum{X}}{N}=\frac{580}{10}=58\]\[\bar{Y}=\frac{\sum{Y}}{N}=\frac{140}{10}=14\]

Firm	Sale X	\[x=X-\bar{X}\]	\[{{x}^{2}}\]	Expenses Y	\[y=Y-\bar{Y}\]	\[{{y}^{2}}\]	\[xy\]
1	50	-8	64	11	-3	9	24
2	50	-8	64	13	-1	1	8
3	55	-3	9	14	0	0	0
4	60	2	4	16	2	4	4
5	65	7	49	16	2	4	14
6	65	7	49	15	1	1	7
7	65	7	49	15	1	1	7
8	60	2	4	14	0	0	0
9	60	2	4	13	-1	1	-2
10	60	-8	64	13	-1	1	8
N = 10	\[\sum{X=580}\]	\[\sum{x=0}\]	\[\sum{{{x}^{2}}=360}\]	\[\sum{Y=140}\]	\[\sum{y=0}\]	\[\sum{{{y}^{2}}=22}\]	\[\sum{xy=70}\]

\[\therefore r(x,y)=\frac{\sum{xy}}{\sqrt{\sum{{{x}^{2}}.\sum{{{y}^{2}}}}}}=\frac{70}{\sqrt{360\times 22}}=0.787\]

<< Previous Topic Scatter Diagram Method

Next Topic >> Interquartile and Semi-Interquartile Range

;