Question

The angle $\left(\Theta \right)$ between the two lines of regressions $x+4y=3$ & $3x+y=15$ is given by

• A. ${\tan }^{-1}\left(\frac{7}{11}\right)$
• B. ${\tan }^{-1}\left(\frac{4}{11}\right)$
• C. ${\tan }^{-1}\left(\frac{11}{7}\right)$
• D. None of these

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{11}{7}\right)$

$m_1$= slope of line of regression y on x$=b_{yx}=-\frac{1}{4}$
$m_2$= slope of line of regression x on y$=\frac{1}{b_{xy}}=-3$
$\therefore$ $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{b_{yx}-\frac{1}{b_{xy}}}{1+\frac{b_{yx}}{b_{xy}}}=\frac{-\frac{1}{4}+3}{1+\frac{3}{4}}$
$\therefore$ $\tan \theta =\frac{11}{7}$
$\Rightarrow$ $\theta ={\tan }^{-1}\left(\frac{11}{7}\right)$