Question

Calculate the coefficient of rank correlation from the following data:

$\begin{array}{ccccccccccc}X& 48& 33& 40& 9& 16& 16& 65& 24& 16& 57\\ Y& 13& 13& 24& 6& 15& 4& 20& 9& 6& 19\end{array}$

Answer 1

$\begin{array}{cccccc}X& Y& R_1& R_2& D=R_1-R_2& D^2\\ 48& 13& 8& 5\cdot 5& 2\cdot 5& 6\cdot 25\\ 33& 13& 6& 5\cdot 5& 0\cdot 5& 0\cdot 25\\ 40& 24& 7& 10& -3& 9\cdot 0\\ 9& 6& 1& 2\cdot 5& -1\cdot 5& 2\cdot 25\\ 16& 15& 3& 7& -4& 16\\ 16& 4& 3& 1& 2& 4\\ 65& 20& 10& 9& 1& 1\\ 24& 9& 5& 4& 1& 1\\ 16& 6& 3& 2\cdot 5& 0\cdot 5& 0\cdot 25\\ 57& 19& 9& 8& 1& 1\cdot 0\\ & & & & & \sum D^2=41\cdot 00\end{array}$

Here, $N=10$

There is one item ($16$) in $X$ series repeated three times and two different items ($13$) and ($6$) in $Y$ series repeated two times of each.

i.e., $m_1=3,m_2=2$ and $m_3=2$

$r_k=1-\frac{6\sum D^2+\frac{1}{12}(m_1^3-m_1)+\frac{1}{12}(m_2^3-m_2)+\frac{1}{12}(m_3^3-m_3)}{N^3-N}$

$\Rightarrow r_k=1-\frac{6\left(41\right)+\frac{1}{12}(3^3-3)+\frac{1}{12}(2^3-2)+\frac{1}{12}(2^3-2)}{{10}^3-10}$

$\Rightarrow r_k=1-\frac{6\left(41\right)+2+\frac{1}{2}+\frac{1}{2}}{9990}$

$\Rightarrow r_k=1-\frac{249}{990}$

$\Rightarrow r_k=\frac{741}{990}$

$\therefore r_k=0.748$

Hence, coefficient of rank correlation $=0.748$