Two pairs of students are tied in the Statistics marks, so each tied pair must be given the average rank, and a correction term added before applying Spearman's formula. …

Step 1 — Rank the Statistics marks (X), highest = rank 1, averaging tied ranks. Sorted descending: D(90), A(85), B(85), C(78), F(72), G(72), E(65). A and B tie for 2nd/3rd $\Rightarrow$ both get $\frac{2+3}{2}=2.5$. F and G tie for 5th/6th $\Rightarrow$ both get $\frac{5+6}{2}=5.5$.

Step 2 — Rank the Accountancy marks (Y), no ties. Sorted descending: D(88), B(80), A(75), C(70), G(68), F(65), E(60) $\Rightarrow$ ranks 1–7 respectively, with no averaging needed.

Step 3 — Tabulate $R_x$, $R_y$, $d$, $d^2$ by student.

As a cross-check, computing $\rho$ WITHOUT applying the tie correction (i.e. using the uncorrected $\sum d^2=1.0$ alone) would give $\rho=1-\frac{6(1.0)}{336}\approx0.982$ — a noticeably different, incorre …

Assigning consecutive ranks (2 and 3) to tied values instead of the average rank 2.5 to both; forgetting to add the correction term for EVERY tied g …