Question

The coefficient of rank correlation of the marks obtained by 10 students in two particular subjects was found to be 0.5. It was later discovered that the difference in ranks in two subjects obtained by one of the students was wrongly taken as 3 instead of 7. What should be the correct value of coefficient of rank correlation.

Answer 1

Given, r_k = 0.5
N = 10

Now,
$$\begin{gathered} r_k=1-\frac{6{\mathrm{\Sigma D}}^2}{{\mathrm{N}}^3-\mathrm{N}} \\ \mathrm{Substituting}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{formula}\mathrm{we}\mathrm{get}, \\ 0.5=1-\frac{6{\mathrm{\Sigma D}}^2}{1000-10} \\ \mathrm{or},495=990-6{\mathrm{\Sigma D}}^2 \\ \mathrm{or},\Sigma D^2=\frac{495}{6} \\ \mathrm{or},\mathrm{\Sigma }{D}^2=82.5 \\ \mathrm{Wrong}\mathrm{\Sigma }{D}^2=82.5 \\ \mathrm{Correct}\Sigma D^2=82.5-{\left(3\right)}^2+{\left(7\right)}^2 \\ =122.5 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{correct}\mathrm{\Sigma }{D}^2\mathrm{in}\mathrm{the}\mathrm{formula}\mathrm{for}\mathrm{correlation}coefficient: \\ \mathrm{Correct}\mathrm{Rank}\mathrm{Correlation}\mathrm{Coefficient}(\mathit{}{r}_{\mathit{k}})=1-\frac{6\times 122.5}{1000-10} \\ \Rightarrow \mathit{}{r}_k=0.257 \end{gathered}$$

Hence, correct coefficient of rank correlation is 0.257.