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 Spearman’s rank correlation formula.
For \(n\) observations, Spearman’s rank correlation coefficient is
\[ \rho = 1-\frac{6\sum d_i^2}{n(n^2-1)} \] where \(d_i\) denotes the difference in ranks for the \(i\)-th contestant.
Here,
\[ n=10 \] and the initially computed value is
\[ \rho=0.6 \]

Step 2: Find the originally used value of \(\sum d_i^2\).
Using the formula,
\[ 0.6 = 1-\frac{6\sum d_i^2}{10(10^2-1)} \] \[ 0.6 = 1-\frac{6\sum d_i^2}{10(99)} \] \[ 0.4 = \frac{6\sum d_i^2}{990} \] \[ 6\sum d_i^2 = 396 \] \[ \sum d_i^2 = 66 \]

Step 3: Correct the value of \(\sum d_i^2\).
One difference was incorrectly taken as
\[ 2 \] instead of the correct value
\[ 5 \] So, the correction in \(\sum d_i^2\) is
\[ 5^2-2^2 = 25-4 = 21 \] Therefore, the corrected value is
\[ \sum d_i^2=66+21=87 \]

Step 4: Compute the corrected rank correlation coefficient.
\[ \rho = 1-\frac{6(87)}{10(99)} \] \[ = 1-\frac{522}{990} \] \[ = 1-0.527272\ldots \] \[ = 0.472727\ldots \] Rounded to two decimal places,
\[ \rho\approx 0.47 \]

Step 5: Final conclusion.
Hence, the corrected value of Spearman’s rank correlation coefficient is
\[ \boxed{0.47} \]