Find the correlation coefficient

Question

In studying the relation between the two variables $x$ and $y$ , if the equation of the regression line of $y$ on $x$ was $$y=0.421x+0.67$$ and the equation of the regression line of $x$ on $y$ was $$x=1.58y+3.9$$ \Find\ \ The linear correlation coefficient between $x$ and $y$ My solution is $$r= \pm\sqrt{0.421\times 1.58}= \pm0.8155$$ Does my solution correct or i would not take the negative value into account ?

Answer 1

You are right, but you can find the sign too.

The correlation coefficient of $x$ and $y$ is $\dfrac{\operatorname{cov}(x,y)}{\sqrt{\operatorname{var}(x)\operatorname{var}(y)}}$.

The slope in the regression $y=ax+b$ is given by $a=\dfrac{\operatorname{cov}(x,y)}{\operatorname{var}(x)}$.

Likewise, the slope in the regression $x=a'y+b'$ is given by $a'=\dfrac{\operatorname{cov}(x,y)}{\operatorname{var}(y)}$.

Hence the correlation coefficient is $\pm\sqrt{aa'}$. But it's also of the same sign as $a$ (or $a'$), hence positive here.

Answer 2

The two least-squares line are:

$$y=0.421x+0.67 \\ x=1.58y+3.9$$

You should have $$ \frac{y-\nu}{\tau} = \rho\left( \frac{x-\mu} \sigma \right) \\ \frac{x-\mu}\sigma = \rho\left( \frac{y-\nu} \tau \right) $$ where

• $\mu$ is the average $x$-value,
• $\nu$ is the average $y$-value,
• $\sigma$ is the standard deviation of the $x$-values,
• $\tau$ is the standard deviation of the $y$-values,
• $\rho$ is the correlation.

Thus \begin{align} & \frac{\rho\tau} \sigma = 0.421, \\[10pt] & \frac{\rho\sigma} \tau = 1.58. \end{align} Multiplying left sides and right sides, you get $\rho^2 = 0.421\times 1.58.$

But notice also that $\rho$ must be positive since the slopes are positive.