Question

If we have $b_{xy}=0.4$ and $b_{yx}=1.6$, then the angle $\theta$ between two regression lines is

• A. ${\tan }^{-1}\left(0.09\right)$
• B. ${\tan }^{-1}\left(0.18\right)$
• C. ${\tan }^{-1}\left(0.36\right)$
• D. ${\tan }^{-1}\left(0.72\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(0.18\right)$

Given, $b_{xy}=0.4,b_{yx}=1.6$

We have to find $\theta$.

$r=\sqrt{1.6\times 0.4}=\sqrt{0.64}=0.8$

We know that $b_{yx}=r\frac{{\sigma }_y}{{\sigma }_x}$

$\Rightarrow \frac{{\sigma }_y}{{\sigma }_x}=\frac{b_{yx}}{r}=\frac{1.6}{0.8}=2$

We know that $m_1=\frac{1}{r}\frac{{\sigma }_y}{{\sigma }_x}$

$\Rightarrow m_1=\frac{1}{0.8}\left(2\right)=2.5$

Also $m_2=r\frac{{\sigma }_y}{{\sigma }_x}=0.8\times 2=1.6$

Thus $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$

$=\frac{2.5-1.6}{1+\left(2.5\right)\left(1.6\right)}$

$=\frac{0.9}{1+4}$

$=\frac{0.9}{5}$

Therefore, $\tan \theta =0.18$

$\Rightarrow \theta ={\tan }^{-1}\left(0.18\right)$