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Question: Angle between two lines of regression is given by...

Angle between two lines of regression is given by

A

tan⁡−1(byx+1bxy1−bxybyx)\tan^{- 1}\left( \frac{b_{yx} + \frac{1}{b_{xy}}}{1 - \frac{b_{xy}}{b_{yx}}} \right)

B

tan⁡−1(byx−bxy−1byx+bxy)\tan^{- 1}\left( \frac{b_{yx} - b_{xy} - 1}{b_{yx} + b_{xy}} \right)

C

tan⁡−1(bxy−1byx1+bxybyx)\tan^{- 1}\left( \frac{b_{xy} - \frac{1}{b_{yx}}}{1 + \frac{b_{xy}}{b_{yx}}} \right)

D

tan⁡−1(byx−bxy1+byx.bxy)\tan^{- 1}\left( \frac{b_{yx} - b_{xy}}{1 + b_{yx}.b_{xy}} \right)

Answer

tan⁡−1(bxy−1byx1+bxybyx)\tan^{- 1}\left( \frac{b_{xy} - \frac{1}{b_{yx}}}{1 + \frac{b_{xy}}{b_{yx}}} \right)

Explanation

Solution

Equation of regression line of y on x is,

y−yˉ=rσyσx(x−xˉ)y - \bar{y} = r\frac{\sigma_{y}}{\sigma_{x}}\left( x - \bar{x} \right) or y−yˉ=byx(x−xˉ)y - \bar{y} = b_{yx}(x - \bar{x})

m1=m_{1} = Slope of regression line of yy on x=byxx = b_{yx}

Now, equation of regression line of x on y is,

x−xˉ=bxy(y−yˉ)x - \bar{x} = b_{xy}(y - \bar{y})

rxy=20(36)(25)=23=0.66r_{xy} = \frac{20}{\sqrt{(36)(25)}} = \frac{2}{3} = 0.66 slope of regression line of x on y=1bxyy = \frac{1}{b_{xy}}

If angle between them is θ\theta, then

tan⁡θ=±m1−m21+m1m2=byx−1bxy1+byxbxy\tan\theta = \pm \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = \frac{b_{yx} - \frac{1}{b_{xy}}}{1 + \frac{b_{yx}}{b_{xy}}}or tan⁡θ=(bxy−1byx1+bxybyx)\tan\theta = \left( \frac{b_{xy} - \frac{1}{b_{yx}}}{1 + \frac{b_{xy}}{b_{yx}}} \right).