Question

In a singing competition, two judges award ranks for each of the \(10\) contestants. Based on the ranks awarded by the two judges, the Spearman’s rank correlation coefficient \(\rho\) for the \(10\) contestants was computed to be \(0.6\). It was found later that the difference in ranks awarded by the two judges for one of the contestants was incorrectly taken as \(2\), instead of the correct value \(5\). If the correct difference is taken into account, then the corrected value of \(\rho\) equals (rounded off to two decimal places).

Hint:

In Spearman’s rank correlation, errors in rank differences affect the term \(\sum d_i^2\). Always correct the squared differences before recomputing \(\rho\).

Answer 1

Correct Answer: 0.47

Step 1: Recall the formula.
$\rho=1-\frac{6\sum d_i^2}{n(n^2-1)}$ with $n=10$, so the denominator is $990$.

Step 2: Recover the old $\sum d^2$.
From $\rho=0.6$, $0.4=\frac{6\sum d^2}{990}$, giving $\sum d^2=66$.

Step 3: Fix the wrong entry.
Replacing the $2$ by $5$ adds $5^2-2^2=21$, so the corrected sum is $66+21=87$.

Step 4: Recompute $\rho$.
$\rho=1-\frac{6(87)}{990}=1-\frac{522}{990}=0.4727\ldots$

Step 5: Round.
\[ \boxed{0.47} \]