Question

The two lines of regressions are $x + 2y - 5 = 0$ and $2x + 3y - 8 = 0$ and the variance of $x$ is $12$. Find the variance of $y$ and the coefficient of correlation.

Answer 1

Hint: Here, it is given the regressions of two lines and the variance of $x$. We have to find the variance of $y$ and coefficient of correlation. We are going to find the coefficient correlation by using the given regressions of lines and to find the variance of $y$, we have to use the variance of $x$.

Formula used: ${b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}$
${b_{xy}} = r\dfrac{{{\sigma _x}}}{{{\sigma _y}}}$
Correlation coefficient,
${r^2} = {b_{xy}} \times {b_{yx}} \Rightarrow r = \sqrt {{b_{xy}} \times {b_{yx}}} $

Complete step-by-step solution:
It is given in the question, the regressions of two lines are $x + 2y - 5 = 0$ and $2x + 3y - 8 = 0$
The variance of $x$ is $12$.
Let us consider the equations,
$x + 2y - 5 = 0 - - - - - (1)$
$2x + 3y - 8 = 0 - - - - - (2)$
From equation (1), divide the equation by $2$ to get the regression line of $y$ on $x$,
Hence, $x + 2y - 5 = 0$
$ \Rightarrow \dfrac{x}{2} + y - \dfrac{5}{2} = 0$
Rewriting the equation we get,
$ \Rightarrow y = - \dfrac{x}{2} + \dfrac{5}{2}$
The is equation is the regression line of $y$ on $x$.
From that we can get,
${b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}} = - \dfrac{1}{2} - - - - ( * )$
From equation (2), divide the equation by $2$ to get the regression line of $x$ on $y$,
Hence, $2x + 3y - 8 = 0$
$ \Rightarrow x + \dfrac{3}{2}y - \dfrac{8}{2} = 0$
Rewriting the equation we get,
$ \Rightarrow x = - \dfrac{3}{2}y + 4$
The is equation is the regression line of $x$ on $y$.
From that we can get,
${b_{xy}} = r\dfrac{{{\sigma _x}}}{{{\sigma _y}}} = - \dfrac{3}{2}$
Now we are going find correlation of coefficient,
Since, ${r^2} = {b_{xy}} \times {b_{yx}} \Rightarrow r = \sqrt {{b_{xy}} \times {b_{yx}}} $
Subtituting the values,
$ \Rightarrow r = \sqrt { - \dfrac{3}{2} \times - \dfrac{1}{2}} $
Multiplying the terms
$ \Rightarrow \sqrt {\dfrac{3}{4}} $
Taking square root we get,
$ \Rightarrow \pm \dfrac{{\sqrt 3 }}{2}$
${b_{xy}}$ and ${b_{yx}}$ being both are negative,
Hence, $r$ is also negative.
$r = - \dfrac{{\sqrt 3 }}{2}$
$\therefore $ The correlation coefficient $r = - \dfrac{{\sqrt 3 }}{2}$
Consider the given variance of $x$ is $12$,
That is, $\sigma _x^2 = 12$,
$ \Rightarrow {\sigma _x} = \sqrt {12} $
Since already we known that,
${b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}} = - \dfrac{1}{2} - - - - ( * )$
Hence we have the values of ${\sigma _x}$ and $r$, so substituting those in (*), we get,
$ \Rightarrow - \dfrac{{\sqrt 3 }}{2} \times \dfrac{{{\sigma _y}}}{{\sqrt {12} }} = - \dfrac{1}{2}$
Rearranging the terms to solve for ${\sigma _y}$,
$ \Rightarrow {\sigma _y} = - \dfrac{1}{2} \times \left( { - \dfrac{2}{{\sqrt 3 }}} \right) \times \sqrt {12} $
Simplifying their terms we get,
$ \Rightarrow {\sigma _y} = - \dfrac{1}{2} \times \left( { - \dfrac{2}{{\sqrt 3 }}} \right) \times 2\sqrt 3 $
Hence,
$ \Rightarrow {\sigma _y} = 2$
Squaring on both sides,
$ \Rightarrow \sigma _y^2 = 4$
Hence we got the variance of $y$,
$\therefore \sigma _y^2 = 4$
The Variance of y is 4.

Note: We have to remember that the composition of regression lines is based on the least square assumptions. Regression analysis is generally based on the summation of squares of deviations of observed values from the lines of the best fit. A line of regression gives the best average value of one variable from any given value of the other variable.