Question

Ranks obtained by 5 students in two tests are as follows:

Then the Spearman's rank correlation coefficient \( (\rho) \) is:

• A. \( 0.6 \)
• B. \( 0.4 \)
• C. \( 0.2 \)
• D. \( 0 \)

Hint:

Always ensure \( \sum d = 0 \) as a quick check before squaring. If the sum of differences is not zero, there is a calculation error in the individual differences.

Answer 1

The Correct Option is A

Concept: Spearman's rank correlation coefficient measures the strength of association between two ranked variables.

• Formula: \( \rho = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} \).

• \( d \): Difference between the ranks of each pair.

• \( n \): Number of pairs of observations.

Step 1: Calculate the differences \( d \) and their squares \( d^2 \).
Let \( X \) be Test-1 and \( Y \) be Test-2. Pairs (X, Y): (1, 2), (3, 1), (2, 3), (4, 5), (5, 4). \[ d_1 = 1-2 = -1 \implies d_1^2 = 1 \] \[ d_2 = 3-1 = 2 \implies d_2^2 = 4 \] \[ d_3 = 2-3 = -1 \implies d_3^2 = 1 \] \[ d_4 = 4-5 = -1 \implies d_4^2 = 1 \] \[ d_5 = 5-4 = 1 \implies d_5^2 = 1 \]

Step 2: Calculate the sum \( \sum d^2 \) and identify \( n \).
Summing the squared differences: \[ \sum d^2 = 1 + 4 + 1 + 1 + 1 = 8 \] Number of students \( n = 5 \).

Step 3: Apply the formula for \( \rho \).
\[ \rho = 1 - \frac{6(8)}{5(5^2 - 1)} = 1 - \frac{48}{5(24)} \] \[ \rho = 1 - \frac{48}{120} = 1 - 0.4 = 0.6 \] Final Answer: (A)