The regression lines of y on x and of x on y for some data have the correlation r = 0.6, with σx =…

Step 1. The slopes. $m_1 = 0.6 \times \dfrac{8}{6} = 0.8$. Also $b_{xy} = 0.6 \times \dfrac{6}{8} = 0.45$, so $m_2 = \dfrac{1}{0.45} = 2.222$.

Step 2. The tangent. $\tan\theta = \dfrac{2.222 - 0.8}{1 + 0.8 \times 2.222} = \dfrac{1.422}{2.778} = 0.512$.

Step 3. The angle. $\theta = \tan^{-1} 0.512 = 27.1^\circ$.

Step 4. The general formula. With $m_1 = \dfrac{r\sigma_y}{\sigma_x}$ and $m_2 = \dfrac{\sigma_y}{r\sigma_x}$, $m_2 - m_1 = \dfrac{\sigma_y}{\sigma_x}\cdot\dfrac{1 - r^2}{r}$ and $1 + m_1 m_2 = 1 + \dfrac{\sigma_y^2}{\sigma_x^2} = \dfrac{\sigma_x^2 + \sigma_y^2}{\sigma_x^2}$. Dividing, $\tan\theta = \dfrac{1 - r^2}{\lvert r \rvert}\cdot\dfrac{\sigma_x\sigma_y}{\sigma_x^2 + \sigma_y^2}$. Here $\dfrac{0.64}{0.6} \times \dfrac{48}{100} = 0.512$.