Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

Here n=6 ,

$${\boldsymbol{m}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{2}$$

$$\boldsymbol{\mathrm{\therefore }} R = 1 - \frac{6}{n\left(n^2-1\right)}\ \left\{\ \sum{d^2_i+\ \frac{1}{12}\ (}m^3_1-\ m_1\right)\}$$ $$ = 1 - \frac{6}{6\left(6^2-1\right)}\ \left\{\ 7.5+\ \frac{1}{12}\left(2^3-2\right)\right\}$$ $$ = 1 - \frac{1}{35}\ \left\{\ 7.5+\frac{1}{12}*6\right\}$$ $$= 0.7714$$

$\boldsymbol{\mathrm{\therefore }}$ The value of rank correlation coefficient = 0.7714