Question

# The regression equation $y$ on $x$ is $y = \dfrac{2}{9}x$ and the regression equation of $x$ on $y$ is $x = \dfrac{y}{2} + \dfrac{7}{6}$ .Find: $\left( a \right){\text{ Correlation coefficient between x and y}}$$\left( b \right){\text{ }}\sigma _y^2{\text{ if }}\sigma _x^2 = 4$

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Hint: So we have the equation of regression equation given for each and by comparing the equation with $y - \bar y = {b_{yx}}\left( {x - \bar x} \right)$ and for $X$ on $Y$ it will be $x - \bar x = {b_{xy}}\left( {y - \bar y} \right)$ , we will get ${b_{yx}}\& {b_{xy}}$ . And then by using the formula of correlation coefficient we will get the value. And for the second question we have the relation ${b_{yx}} = r \cdot \dfrac{{{\sigma _y}}}{{{\sigma _x}}}$ , we will get the value for ${\sigma _y}$ .

Formula used:
Correlation coefficient,
$r = \sqrt {{b_{yx}} \cdot {b_{xy}}}$
Here,
$r$ , will be the correlation coefficient
${b_{yx}}\& {b_{xy}}$ , will be the regression equation value

So we have the regression equation $y$ on $x$ is $y = \dfrac{2}{9}x$ .
So now on comparing the above equation with the formula $y - \bar y = {b_{yx}}\left( {x - \bar x} \right)$, we get
$\Rightarrow {b_{yx}} = \dfrac{2}{9}$
Similarly the regression equation of $X$ on $Y$ is $x = \dfrac{y}{2} + \dfrac{7}{6}$.
So on comparing the above equation with the formula $x - \bar x = {b_{xy}}\left( {y - \bar y} \right)$ , we get
$\Rightarrow {b_{xy}} = \dfrac{1}{2}$
As we know the correlation coefficient between $x$ and $y$ is
$r = \sqrt {{b_{yx}} \cdot {b_{xy}}}$
So on substituting the values, we get
$\Rightarrow r = \sqrt {\dfrac{2}{9} \cdot \dfrac{1}{2}}$
And on solving the above square root, we get
$\Rightarrow r = \pm \dfrac{1}{3}$
So if $r = \dfrac{1}{3}$ then ${b_{yx}}\& {b_{xy}}$ will be positive.
Therefore, the correlation between $x$ and $y$ is $\dfrac{1}{3}$
As we know we have the values given as $\sigma _x^2 = 4$
So on solving it we get
$\Rightarrow \sigma _x^{} = 2$
So, we have the relation given by ${b_{yx}} = r \cdot \dfrac{{{\sigma _y}}}{{{\sigma _x}}}$ .
On substituting the values, we have
$\Rightarrow \dfrac{2}{9} = \dfrac{1}{3} \cdot \dfrac{{{\sigma _y}}}{2}$
Now taking the constant term one side and solving it, we will get the value as
$\Rightarrow {\sigma _y} = \dfrac{{12}}{9}$
And on making the fraction into the simplest form we get
$\Rightarrow {\sigma _y} = \dfrac{4}{3}$
Hence, $\sigma _y^2 = \dfrac{{16}}{9}$ will be the value.

Note: The level of affiliation is estimated by a correlation coefficient, meant by $r$ . It is at times called Pearson's correlation coefficient after its originator and is a proportion of straight affiliation. On the off chance that a bended line is expected to communicate the relationship, other and more convoluted proportions of the correlation should be utilized.