written 7.7 years ago by | modified 2.1 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV
Marks: 5M
Year: May 2015
written 7.7 years ago by | modified 2.1 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV
Marks: 5M
Year: May 2015
written 7.7 years ago by |
Let $L_{yx}\ $be the line of regression of y on x.
Let the angle made by $L_{yz}\ $with X-axis be $\mathrm{\theta}$
$\mathrm{\therefore }$ Tangent of the angle made by $L_{yz}$ with X- axis = tan $\mathrm{\theta}$ = 0.6
But , slope of any line with X-axis = tan $\mathrm{\theta}$
However, slope of $L_{yx}=\ b_{yx}$
$\mathrm{\therefore }$ $b_{yx}$ = tan $\mathrm{\theta}$ = 0.6
But $b_{yx}=r\ \frac{{\sigma }_y}{{\sigma }_x}$
$$\mathrm{\therefore } 0.6 = r\ \frac{2{\sigma }_x}{{\sigma }_x} --------(given)$$ $$\mathrm{\therefore } 0.6 = 2 r$$ $$r=0.3$$
$\boldsymbol{\mathrm{\therefore }}$ The correlation coefficient between x and y(r) = 0.3