Question

If we have \[{b_{xy}} = 0.4\] and \[{b_{yx}} = 1.6\], then the angle \[\theta \] between two regression lines is
A. \[{\tan ^{ - 1}}(0.09)\]
B. \[{\tan ^{ - 1}}(0.18)\]
C. \[{\tan ^{ - 1}}(0.36)\]
D. \[{\tan ^{ - 1}}(0.72)\]

Answer 1

Hint: We know that, A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. We often use a regression line to predict the value of y for a given value of x.
At first, we will find the slope of the lines by using the geometric means of the coefficient of regression formula, we will get the final answer.

Formula used: Using the formulas
\[{m_1} = \dfrac{1}{r}\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\] and \[{m_2} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
\[\tan \theta = \pm \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}\]

Complete step-by-step solution:
It is given that; \[{b_{xy}} = 0.4\] and \[{b_{yx}} = 1.6\].
We have to find the value of angle \[\theta \] between two regression lines.
We know that, the geometric means of the coefficient of regression \[r = \sqrt {{b_{xy}} \times {b_{yx}}} \]
Substitute the values we get,
$\Rightarrow$\[r = \sqrt {1.6 \times 0.4} \]
Simplifying we get,
$\Rightarrow$\[r = \sqrt {0.64} = 0.8\]
Again, we know that, \[{b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
Simplifying we get,
$\Rightarrow$\[\dfrac{{{\sigma _y}}}{{{\sigma _x}}} = \dfrac{{{b_{yx}}}}{r}\]
Substitute the values we get,
$\Rightarrow$\[\dfrac{{{\sigma _y}}}{{{\sigma _x}}} = \dfrac{{1.6}}{{0.8}} = 2\]
We know that, the slope of first line \[{m_1} = \dfrac{1}{r}\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
Substitute the values we get,
$\Rightarrow$\[{m_1} = \dfrac{1}{{0.8}}(2) = 2.5\]
Also, the slope of second line \[{m_2} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
Substitute the values we get,
$\Rightarrow$\[{m_2} = 0.8 \times 2 = 1.6\]
We know that, if the slope of two lines are \[{m_1}\]and \[{m_2}\] respectively, then the angle between them is,
$\Rightarrow$\[\theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}\]
Substitute the value of \[{m_1}\]and \[{m_2}\] in the above formula we get,
$\Rightarrow$\[\tan \theta = \dfrac{{2.5 - 1.6}}{{1 + (2.5)(1.6)}}\]
Simplifying the values, we get,
$\Rightarrow$\[\tan \theta = \dfrac{{0.9}}{{1 + 4}}\]
Simplifying the values again, we get,
$\Rightarrow$\[\tan \theta = 0.18\]
So, \[\theta = {\tan ^{ - 1}}(0.18)\]
Hence, the angle \[\theta \] between two regression lines is \[\theta = {\tan ^{ - 1}}(0.18)\].

Hence, the correct option is B) \[{\tan ^{ - 1}}(0.18)\].

Note: Regression coefficient of y on x is \[{b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
Regression coefficient of x on y is \[{b_{xy}} = r\dfrac{{{\sigma _x}}}{{{\sigma _y}}}\]
\[{m_1}\] be the slope of the line of regression of y on x =\[{b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
\[{m_2}\] be the slope of the line of regression of x on y =\[\dfrac{1}{{{b_{xy}}}} = \dfrac{1}{r}.\dfrac{{{\sigma _y}}}{{{\sigma _x}}}\]
So, we get,
\[\tan \theta = \pm \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}} = \pm \dfrac{{(1 - {r^2}){\sigma _x}{\sigma _y}}}{{r({\sigma _x}^2 + {\sigma _y}^2)}}\]