Question

If we have $b_{xy}=\frac{2}{3}$ and $b_{yx}=0.5$, then the angle $\theta$ between two regression lines is

• A. ${\tan }^{-1}\left(\frac{5}{2}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{2}\right)$
• C. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{4}{7}\right)$

Answer 1

Answer: (D)

${\tan }^{-1}\left(\frac{4}{7}\right)$

Given $b_{xy}=\frac{2}{3},b_{yx}=0.5$. We have to find $\theta$.

$r=\sqrt{\frac{2}{3}\times 0.5}=\sqrt{\frac{1}{3}}=0.577$

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

$\Rightarrow \frac{{\sigma }_y}{{\sigma }_x}=\frac{b_{yx}}{r}=\frac{0.5}{0.577}=0.866$

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

$\Rightarrow m_1=\frac{1}{0.577}\left(0.866\right)=1.5$

Also $m_2=r\frac{{\sigma }_y}{{\sigma }_x}=0.577\times 0.866=0.5$

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

$=\frac{1.5-0.5}{1+\left(1.5\right)\left(0.5\right)}$

$=\frac{1}{1+0.75}$

$=\frac{1}{1.75}$

$=\frac{100}{175}$

$=\frac{4}{7}$

$\therefore tan\theta =\frac{4}{7}$

$\Rightarrow \theta =ta{n}^{-1}\left(\frac{4}{7}\right)$