# 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:

$\left( a \right){\text{ Correlation coefficient between x and y}}$

\[\left( b \right){\text{ }}\sigma _y^2{\text{ if }}\sigma _x^2 = 4\]

Answer

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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

**Complete step-by-step answer:**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.

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