## Regression Contd., Important problems Correlation and Regression

Subject: Mathematics

#### Overview

This note lets you know the product moment formula to find the Karl Pearson's coefficient and solves some important questions related to correlation and regresseion.
##### Regression Contd., Important problems Correlation and Regression

Regression Equations and regression coefficients:

Regression lines expressed in term of algebraic relations are known as the regression equations. Since there are two regression lines, So there are two regression equations.

• The regression equation of y on x expresses the variation of y for a change in the value of x.
• The regression equation of x on y expresses the variation of x for a change in the value of y.

Regression equation of y on x

Let the regression equation of y on x be

$$y=a+bx$$

$$\rightarrow\;\Sigma\;y=na+b\Sigma\;x$$

$$\rightarrow\;\frac{\Sigma}{n}\;y=a+\frac{b\Sigma\;x}{n}$$

$$\rightarrow\;\overline{y}=a+b\overline{x}$$

Subtracting two equations:

$$y-\overline{y}=b-(x-\overline{x}$$

This equation is the equation of line of regression of y on x. This equation shows that the line of regression of y on x passes through $\;(\overline{x},\overline{y})\;\;\overline{x}\;and\overline{y}$ being the arithmetic averages of x and y series, b is known as the regression coefficient of y on x. In order to differentiate it from regression coefficient of x on y, b is written as b­yx­ where

$$b_{yx}=\frac{n\Sigma\;xy-\Sigma\;x\Sigma\;y}{n\Sigma\;x^2-(\Sigma\;x)^2}=r\frac{\sigma_y}{\sigma_x}$$

Similarly the regression equation of x on y is,

$$y-\overline{x}=b_{xy}(y-\overline{y}$$

Where the line passes through $\;(\overline{x},\overline{y})$ the regression coefficient of x on y denoted by bxy ­is given by

$$b_{xy}=\frac{n\Sigma\;xy-\Sigma\;x\Sigma\;y}{n\Sigma\;y^2-(\Sigma\;y)^2}=r\frac{\sigma_x}{\sigma_y}$$

­The regression equations can easily be obtained when the deviations of the items are taken from the assumed mean.

If u=x-a and v=y-b i.e. if the deviations of the items of x series and y-series be taken from the assumed means a and b respectively then,

$$\overline{x}=a+\frac{\Sigma\;u}{n},\;\;\;\;\overline{y}=b+\frac{\Sigma\;v}{n}$$

$$b_{yx}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;u^2-(\Sigma\;u)^2}$$

$$b_{xy}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;v^2-(\Sigma\;v)^2}$$

Relation between r and b:

$$r=\sqrt{b_{yx}b_{xy}}$$

Examples

Example 1: (Correlation)

Calculate the Karl Pearson’s coefficient of correlation from the following data using product moment formula.

 X 12 9 8 10 11 13 7 Y 14 8 6 9 11 12 3

Solution:

Computation of correlation coefficient

 X Y $$x=X-\overline{x}$$ $$y=Y-\overline{Y}$$ x2 y2 xy 12 9 8 10 11 13 7 14 8 6 9 11 12 3 2 -1 -2 = 1 3 -3 5 1 -3 0 2 3 6 4 1 4 0 1 9 9 25 1 9 0 4 9 36 10 1 6 0 1 9 18 $$\Sigma\;X=70$$ $$\Sigma\;Y=63$$ $$\Sigma\;x^2=28$$ $$\Sigma\;y^2=84$$ $$\Sigma\;xy=46$$

$$\overline{X}=\frac{\Sigma\;X}{n}=\frac{70}{7}=10\;\;\;\overline{Y}=\frac{\Sigma\;Y}{n}=\frac{63}{7}=9$$

$$r=\frac{\Sigma\;xy}{\sqrt{\Sigma\;x^2}\sqrt{\Sigma\;y^2}}=\frac{46}{\sqrt{28}\sqrt{84}}$$

$$\frac{46}{\sqrt{2352}}\;=\frac{46}{48.497}$$

$$=0.95$$

Example 2: (Regression)

Obtain the regression equations of X on Y and Y on X taking origin as 2 and 200 for X and Y respectively.

 X 1 2 3 4 5 Y 166 184 142 180 338

Solution:

 X Y u=X-A=X-2 v=Y-B=Y-200 uv u2 v2 1 2 3 4 5 166 184 142 180 338 -1 0 1 2 3 -34 -16 -58 -20 138 34 0 -58 -40 414 1 0 1 4 9 1156 256 3364 400 19044 Total 5 10 350 15 24220

$$The\;regression\;equation\;of\;X\;on\;Y\;is\;X-\overline{X}\;=b_{xy}(Y-\overline{Y}$$

$$\overline{x}=a+\frac{\Sigma\;u}{n}=2+\frac{5}{5}=3$$

$$\overline{y}=b+\frac{\Sigma\;v}{n}=200=\frac{10}{5}=202$$

$$b_{xy}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;v^2-(\Sigma\;v)^2}$$

$$=\;\frac{5(350)-5(10)}{5(24220)-(10)^2}=\frac{1700}{121000}=0.014$$

The required equation is,

$$X-3=0.014(Y-202)$$

$$\rightarrow\;X=0.014Y+0.172$$

Again,

$$The\;regression\;equation\;of\;Y\;on\;X\;is\;Y-\overline{Y}\;=b_{yx}(X-\overline{X}$$

$$\overline{x}=3$$

$$\overline{y}=202$$

$$b_{yx}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;u^2-(\Sigma\;u)^2}$$

$$=\;\frac{5(350)-5(10)}{5(15)-(5)^2}=\frac{1700}{50}=34$$

The required equation is,

$$Y-202=34(X-3)$$

$$\rightarrow\;Y=34Y+100$$

Example 3: (Regression)

For a certain bivariate data

 X Y Mean 10 18 Std 2.5 2.0

And the coefficient of correlation between X and Y is 0.8. Determine the following:

• The regression of Y on X and the regression of X on Y
• Estimated value of Y for X = 15

Solution:

We know,

$$b_{xy}=r\frac{\sigma_x}{\sigma_y}=0.8\frac{2.5}{2.0}=1$$

$$b_{yx}=r\frac{\sigma_y}{\sigma_x}=0.8\frac{2}{2.5}=0.64$$

the regression equation of Y on X is,

$$y-\overline{y}=b_{yx}(x-\overline{x}$$

$$Y-10=1(X-18)$$

the regression equation of X on Y is,

$$X-\overline{X}=b_{xy}(Y-\overline{Y}$$

$$X-18=0.64(Y-10)$$

Thus for X=15, Y=(15-18)+10=7

Example 4 : (Correlation)

Calculate the coefficient of correlation from the following data of price and demand:

 Price (Rs) 14 16 19 22 24 30 Demand(kg) 24 22 20 24 23 26

Solution:

Computation of correlation coefficients:

 Price(x) u=x-19 u2 Demand(y) v=y-23 v2 uv 14 16 19 22 24 30 -5 -3 0 3 5 11 25 9 0 9 25 121 24 22 20 24 23 26 1 -1 -3 1 0 3 1 1 9 1 0 9 -5 -3 0 3 0 33 $$\Sigma\;u=11$$ $$\Sigma\;u^2=189$$ $$\Sigma\;v=1$$ $$\Sigma\;v^2=21$$ $$\Sigma\;uv=34$$

$$r=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{\sqrt{n\Sigma\;u^2-(\Sigma\;u)^2}\sqrt{n\Sigma\;v^2-(\Sigma\;v)^2}}$$

$$=\frac{6\times\;34-11\times\;1}{\sqrt{6\times\;189-11^2}\sqrt{6\times\;21-1^2}}$$

$$=\frac{204-11}{\sqrt{1134-221}\sqrt{126-1}}$$

$$=\frac{193}{\sqrt{1013\times\;125}}$$

$$=\frac{193}{\sqrt{126625}}$$

$$\frac{13}{355.84}$$

$$=0.542$$

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )

##### Things to remember
• the regression equation of x on y is,

$$y-\overline{x}=b_{xy}(y-\overline{y}$$

• If u=x-a and v=y-b i.e. if the deviations of the items of x series and y-series be taken from the assumed means a and b respectively then,

$$\overline{x}=a+\frac{\Sigma\;u}{n},\;\;\;\;\overline{y}=b+\frac{\Sigma\;v}{n}$$

$$b_{yx}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;u^2-(\Sigma\;u)^2}$$

$$b_{xy}=\frac{n\Sigma\;uv-\Sigma\;u\Sigma\;v}{n\Sigma\;v^2-(\Sigma\;v)^2}$$

• $$r=\sqrt{b_{yx}b_{xy}}$$
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