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Question: If we have \[{b_{xy}} = 0.4\] and \[{b_{yx}} = 1.6\], then the angle \[\theta \] between two regress...

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

Explanation

Solution

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
m1=1rσyσx{m_1} = \dfrac{1}{r}\dfrac{{{\sigma _y}}}{{{\sigma _x}}} and m2=rσyσx{m_2} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}
tan⁡θ=±m2−m11+m1m2\tan \theta = \pm \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}

Complete step-by-step solution:
It is given that; bxy=0.4{b_{xy}} = 0.4 and byx=1.6{b_{yx}} = 1.6.
We have to find the value of angle θ\theta between two regression lines.
We know that, the geometric means of the coefficient of regression r=bxy×byxr = \sqrt {{b_{xy}} \times {b_{yx}}}
Substitute the values we get,
\Rightarrow$$$r = \sqrt {1.6 \times 0.4} $$ Simplifying we get, \Rightarrowr = \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 m1=1rσyσx{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}$$ respectively, then the angle between them is, $\Rightarrow\theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}Substitutethevalueof Substitute the value of{m_1}andand {m_2} in the above formula we get, $\Rightarrow$$$\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$$$\tan \theta = 0.18So, So,\theta = {\tan ^{ - 1}}(0.18)Hence,theangle Hence, the angle\theta betweentworegressionlinesisbetween two regression lines is\theta = {\tan ^{ - 1}}(0.18)$$.

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

Note: Regression coefficient of y on x is byx=rσyσx{b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}
Regression coefficient of x on y is bxy=rσxσy{b_{xy}} = r\dfrac{{{\sigma _x}}}{{{\sigma _y}}}
m1{m_1} be the slope of the line of regression of y on x =byx=rσyσx{b_{yx}} = r\dfrac{{{\sigma _y}}}{{{\sigma _x}}}
m2{m_2} be the slope of the line of regression of x on y =1bxy=1r.σyσx\dfrac{1}{{{b_{xy}}}} = \dfrac{1}{r}.\dfrac{{{\sigma _y}}}{{{\sigma _x}}}
So, we get,
tan⁡θ=±m2−m11+m1m2=±(1−r2)σxσyr(σx2+σy2)\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)}}