The regression equations are

$y=16+1.6x$

and

$x=-1+0.4y$

Example 6.10: For the following data, find the regression line of y on x:

Solution: We have

Total:

$\sum x_i=33,\sum y_i=84,\sum x_i^2=219,\sum x_i{y}_i=451$

$\overline{x}=\frac{\sum x_i}{n}=\frac{33}{7}=4.714$

$\overline{y}=\frac{\sum y_i}{n}=\frac{84}{7}=12$

and

$b=\frac{n\sum x_i{y}_i-\sum x_i\sum y_i}{\left[n\sum x_i^2-{\left(\sum x_i\right)}^2\right]}=\frac{7(451)-(33)(84)}{7(2119)-{(33)}^2}=0.867$

The regression equation of y on x is

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

i.e.,

$y-12=0.867(x-4.714)$

or

$y=0.867x+7.9129$

Example 6.11: From the following data, fit two regression equations by finding actual means (of x and y), i.e., by actual means method.

Solution: We change the origin and find the regression equations as follows:

We have

$\overline{x}=\frac{\sum x_i}{n}=\frac{1+2+3+4+5+6+7}{7}=\frac{28}{7}=4$

$\overline{y}=\frac{\sum y_i}{n}=\frac{2+4+7+6+5+6+5}{7}=\frac{35}{7}=5$

x	y	$X=x–\overline{x}$	$Y=y–\overline{y}$	$X^2$	$Y^2$	XY
1	2	–3	–3	9	9	9
2	4	–2	–1	4	4	2
3	7	–1	2	1	1	–2
4	6	0	1	0	0	0
5	5	1	0	1	1	0
6	6	2	1	4	4	2
7	5	3	0	9	9	0
28	35	0	0	28	16	11

We have

$\begin{array}{l}\sum x_i=28,\sum y_i=35,\sum =0,\sum Y_i=0,\sum X_i^2=28,\\ \sum Y_i^2=16,\sum X_i{Y}_i=11\end{array}$

$b_{yx}=\frac{\sum X_i{Y}_i}{\sum X_i^2}=\frac{11}{28}=0.3928=0.393(\mathrm{Approximately})$

$b_{xy}=\frac{\sum X_i{Y}_i}{\sum Y_i^2}=\frac{11}{16}=0.6875=0.688(\mathrm{Approximately})$

The regression equation of y on x is

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

$(y-5)=0.393(x-4)$

or

$y=0.393x+3.428$

And the regression equation of x on y is

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

$(x_i-4)=0.688(y_i-5)$

or

$x=0.688y+0.56$

The required equations are

$y=0.393x+3.428$

$x=0.688y+0.56$

Example 6.12: From the following results obtain the two regression equations and estimate the yield of crops when the rainfall is 29 cm and the rainfall when the yield is 600 kg.

Coefficient of correlation between yield and rainfall is 0.52.

Solution: We have

$\overline{x}=26.7,\overline{y}=508.4$

${\sigma }_x=4.6,{\sigma }_y=36.8$

and r=0.52

$b_{yx}=r\frac{{\sigma }_y}{{\sigma }_x}=(0.52)\frac{36.8}{4.6}=4.16$

$b_{xy}=r\frac{{\sigma }_x}{{\sigma }_y}=(0.52)\frac{4.6}{36.8}=0.065$

Regression equation of y on x

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

i.e.,

$y=508.4=4.16(x-26.7)$

or

$y=397.328+4.16x$

when

$x=29,$

we have

$y=397.328+4.16(29)=517.968\mathrm{kg}$

Regression equation of x on y

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

or

$x-26.7=-0.065(y-508.4)$

or

$x=-6.346+0.065x$

When y=600 kg.

$\begin{array}{cc}x& =-6.346+0.065(-600)\\ & =32.654\mathrm{cm}\end{array}$

The regression equations are

$y=397.328+4.16x$

$x=-6.346+0.065y$

When the rainfall is 29 cm the yield of the crop is 517.968 kg and when the yield is 600 kg the rainfall is 32.654 cm.

Example 6.13: Find the most likely price of a commodity in Bombay corresponding to the price of Rs. 70 at Calcutta from the following:

Correlation coefficient between the prices of commodity in the two cities is 0.8.

Solution: we have

$\overline{x}=65,\overline{y}=67$

${\sigma }_x=2.5,{\sigma }_y=3.5$

and $r=0.8$

$(y,\overline{y})=r\frac{{\sigma }_y}{{\sigma }_x}(x_i-\overline{x})=0$

$\Rightarrow y-67=(0.8).\left(\frac{3.5}{2.5}\right)(x-65)$

$\Rightarrow y=67+1.12x-72.8$

$\Rightarrow y=-5.8+1.12x$

when $x=70,$

$\Rightarrow y=-5.8+1.12\times 70=-5.8+78.4$

$\Rightarrow y=72.60$

The price of the commodity in Bombay corresponding to Rs. 70 at Calcutta is Rs. 72.60.

Example 6.14: The regression equation calculated from a given set of observations

$x=-0.4y+6.4$

and

$y=-0.6x+4.6$

Calculate $\overline{x},\overline{y}$ and r_xy

Solution: We have

$x=-0.4y+6.4$ (6.19)

(6.19)

and

$y=-0.6x+4.6$ (6.20)

(6.20)

From Eq. (6.20), we have $y=-0.6(-0.4y+6.4)+4.6$

$\Rightarrow y=0.24y-3.84+4.6$

$\Rightarrow 0.76y=0.76$

$\Rightarrow y=1$

From Eq. (6.19), we have $x=-0.4\times 1+6.4=6.0$

But $(\overline{x},\overline{y})$ in the point of intersection of Eqs. (6.19) and (6.20)

Hence

$(\overline{x},\overline{y})=(1,6)$

$\overline{x}=1,\overline{y}=6$

Clearly, Eq. (6.19) in the regression equation x on y and Eq. (6.20) is the regression equation y on x.

We have

$b_{yx}=-0.6,b_{xy}=-0.4$

and

$r^2=(-0.4)(-0.6)=0.24$

$r=r_{xy}=\pm \sqrt{0.24}$

Since $b_{yx}$ and $b_{yx}$ are both negative $r=r_{xy}$ is negative

$r_{xy}=\pm \sqrt{0.24}$

Example 6.15: Show that the coefficient of correlation is the Geometric mean (GM) of the coefficient of regression.

Solution: The coefficients of regression are $r({\sigma }_x/{\sigma }_y)$ and $r({\sigma }_y/{\sigma }_x)$.

GM of the regression coefficients is

$\begin{array}{c}=\sqrt{r\frac{{\sigma }_x}{{\sigma }_y}\cdot r\frac{{\sigma }_y}{{\sigma }_x}}=\sqrt{r^2}=r\\ =\mathrm{correlation}\mathrm{of}\mathrm{correlation}\end{array}$

Example 6.16: In a partially destroyed laboratory record of an analysis of correlation data, the following results are legible:

Variance of x=9

Regression equation: $8x-10y+66=0,40x-18y=214$

What were

Solution: Variance of x=9

i.e., ${\sigma }_x^2=9\Rightarrow {\sigma }_x=3$

Solving the regression equations

$8x-10y+66=0$ (6.21)

(6.21)

$40x-18y=214$ (6.22)

(6.22)

We obtain

$x=13,y=17$

Since the pair of intersection of the regression lines is $(\overline{x},\overline{y}),$ we have

$(\overline{x},\overline{y})=(x,y)=(13,17)$

Therefore $\overline{x}=13,\overline{y}=17$

The regression Eqs. (6.21) and (6.22) can be written as

$y=0.8x+6.6$ (6.23)

(6.23)

$x=0.45y+5.35$ (6.24)

(6.24)

The regression coefficient of y on x is

$r\frac{{\sigma }_y}{{\sigma }_x}=0.8$ (6.25)

(6.25)

And the regression coefficient of x on y is

$r\frac{{\sigma }_x}{{\sigma }_y}=0.45$ (6.26)

(6.26)

Multiplying Eqs. (6.25) and (6.26), we get

$r^2=0.45\times 0.8$

$\Rightarrow r^2=0.36$

$\Rightarrow r=0.6$

Putting the values of r and ${\sigma }_x$ in Eq. (6.25), we get the value of ${\sigma }_y$ as follows:

$r\frac{{\sigma }_y}{{\sigma }_x}=0.8$

$(0.6)\frac{{\sigma }_y}{3}=0.8$

${\sigma }_y=\frac{0.8}{0.2}=4$

Example 6.17: If one of the regression coefficients is greater than unity. Show that the other regression coefficient is less than unity.

Solution: Let one of the regression coefficient, say $b_{yx}>1$

Then

$b_{yx}>1\Rightarrow b_{yx}<1$

Since,

$b_{yx}·b_{xy}=r^2\le 1$

We have

$b_{xy}\le \frac{1}{b_{yx}}$

$b_{xy}<1\left(\because \frac{1}{b_{yx}}<1\right)$

Example 6.18: Show that the Arithmetic mean of the regression coefficients is greater than the correlation coefficient.

Solution: We have to show that $\frac{b_{xy}+b_{yx}}{2}>r$

Consider

${({\sigma }_y-{\sigma }_x)}^2>0$

Clearly

${({\sigma }_y-{\sigma }_x)}^2>0$

(Since square of two real qualities is always >0)

${\sigma }_x^2+{\sigma }_y^2-2{\sigma }_x{\sigma }_y>0$

or

$\frac{{\sigma }_x^2}{{\sigma }_y{\sigma }_x}+\frac{{\sigma }_y^2}{{\sigma }_y{\sigma }_x}>2$

or

$\frac{{\sigma }_x}{{\sigma }_y}+\frac{{\sigma }_y}{{\sigma }_x}>2$

or

$r\frac{{\sigma }_x}{{\sigma }_y}+r\frac{{\sigma }_y}{{\sigma }_x}>2$

or

$b_{xy}+b_{yx}>2r$

or

$\frac{b_{xy}+b_{yx}}{2}>r$

Hence proved.

Example 6.19: Given that $x=4y+5$ and $y=kx+4$ are two lines of regression. Show that $0\le k\le 1/4$. If $k=1/8$, find the means of the variables and ratio of their variables.

Solution: $x=4y+5\Rightarrow b_{xy}=4$

$y=kx+4\Rightarrow b_{yx}=k$

$r^2=b_{xy}\cdot b_{yx}=4\cdot k$

but

$-1\le r\le 1$

$\Rightarrow 0\le r^2\le 1$

$\Rightarrow 4\le 4k\le 1$

$\Rightarrow 0\le k\le \frac{1}{4}$

when $k=\frac{1}{8}$, $r^2=4·\frac{1}{8}=\frac{1}{2}\Rightarrow r=0.7071$

$y=kx+4$

or

$y=\frac{1}{8}x+4$

or

$8y=x+32$

or

$8y=4y+5+32$

or

$4y=37\Rightarrow y=9.25$

Now

$x=4y+5\Rightarrow x=4(9.25)+5$

or

$x=42$

Therefore $(\overline{x},\overline{y})=(x,y)$ (the point of intersection is $(\overline{x},\overline{y})$)

$=(42,9.25)$

i.e.,

$\overline{x}=42,\overline{y}=9.25$

Hence

$\frac{b_{xy}}{b_{yx}}=\frac{r\frac{{\sigma }_x}{{\sigma }_y}}{r\frac{{\sigma }_y}{{\sigma }_x}}=\frac{{\sigma }_x·{\sigma }_x}{{\sigma }_y·{\sigma }_y}=\frac{4}{\left(\frac{1}{8}\right)}=32$

i.e.,

$\frac{{({\sigma }_x)}^2}{{({\sigma }_y)}^2}=32$

The ratio of the variances is 32:1.

Multiple correlation is a statistical technique that predicts value of one variable on the basis of two or more other variables.

Regression analysis involves identifying the relationship between a dependent variable and one or more independent variables. In simple linear regression, the model is used to describe the relationship between a single dependent variable y and a single independent variable x. Multiple regression is concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parameters multiple correlation coefficients is denoted by R. It is never negative ab=nd there are several ways to compute its value. R lies between 0 and 1, and it tells the only strength of association between the dependent and independent variables.

If R=0 then there is no linear association relationship, if R=1 there is stronger linear association, i.e., perfect linear relationship between x and y. Correlation and regression analysis are related in the sense that both deal with relationships among variables.