Developed by Charles Spearman in 1904, Spearman's Rank Correlation Coefficient ($r_s$ or $
ho$) is the premier non-parametric method for evaluating the monotonic association between two variables. Instead of operating on raw numerical values, it evaluates the relative ranks assigned to the observations. It is particularly indispensable when analyzing qualitative attributes that cannot be measured on a physical scale (such as leadership, beauty, musicality, or coding competency), or when data exhibits extreme non-normality and outliers.
Key Formulas for Spearman's $r_s$:
Case I: Untied Ranks (Unique Values): $$r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$$
where $d_i = R_{x_i} - R_{y_i}$ is the difference between the ranks of the $i^{\text{th}}$ individual, and $n$ is total pairs.
Crucial Algebraic Check: The sum of rank differences must always equal zero: $\sum d_i = 0$.
Case II: Tied (Repeated) Ranks: When two or more items share the same value, each is assigned the average (mean) of the ranks they occupy.
For each group of $m$ tied values, add a correction factor $\frac{m(m^2 - 1)}{12}$ to $\sum d^2$:
$$r_s = 1 - \frac{6 \left[\sum d_i^2 + \sum \frac{m(m^2 - 1)}{12}\right]}{n(n^2 - 1)}$$
Range of $r_s$: $-1 \le r_s \le +1$.
• $r_s = +1$: Complete agreement between rankings ($R_x = R_y$).
• $r_s = -1$: Complete disagreement/inversion between rankings ($R_y = n + 1 - R_x$).
• $r_s = 0$: No monotonic relationship between ranks.
Pearson's $r$ vs. Spearman's $r_s$ Comparison
Feature
Karl Pearson's $r$
Spearman's $r_s$
Type of Measure
Parametric (requires metric data)
Non-parametric (ordinal ranks)
Data Applicable
Quantitative continuous measurements
Qualitative attributes or ranked data
Sensitivity to Outliers
Extremely Sensitive
Highly Robust (outliers bounded by rank)
Type of Association
Strictly Linear
Monotonic (linear or non-linear monotonic)
Worked Benchmark Examples
Example 1 (Ranks Assigned Directly - Judges' Evaluation):
Two judges in a classical debate competition ranked 6 participants as follows:
Judge 1 ($R_1$): $1, \quad 2, \quad 3, \quad 4, \quad 5, \quad 6$
Judge 2 ($R_2$): $2, \quad 4, \quad 1, \quad 5, \quad 6, \quad 3$
Compute the rank correlation coefficient and interpret the consensus between the judges.
Graded textbook exercises on Spearman's Rank Correlation Coefficient from CBSE Class 11 Applied Mathematics.
Problem 1Perfect Inversion Proof
If two judges rank 5 competitors in completely opposite order ($R_1: 1, 2, 3, 4, 5$ and $R_2: 5, 4, 3, 2, 1$), show that Spearman's rank correlation coefficient is strictly $-1$.
Ten competitors in a beauty contest are ranked by three judges in the following order:
Ranks by Judge A: $1, 6, 5, 10, 3, 2, 4, 9, 7, 8$
Ranks by Judge B: $3, 5, 8, 4, 7, 10, 2, 1, 6, 9$
Calculate the coefficient of rank correlation between Judge A and Judge B if $\sum d^2 = 214$.
In a ranking of 10 students, a mark of 75 occurs 3 times in Mathematics and a mark of 82 occurs 2 times in Accountancy. Compute the total correction factor to be added to $\sum d^2$.
Explain why Spearman's rank correlation coefficient is equal to $+1$ for the paired data $(1, 1), (2, 8), (3, 27), (4, 64), (5, 125)$, even though the relationship $Y = X^3$ is strictly non-linear.
Show Step-by-Step Solution
• The function $Y = X^3$ is strictly monotonically increasing ($X_1 < X_2 \implies Y_1 < Y_2$).
• Ranking $X$: $1, 2, 3, 4, 5$.
• Ranking $Y$: $1, 2, 3, 4, 5$.
• The ranks assigned to $X$ and $Y$ are identical: $R_x = R_y \implies d_i = 0$ for all $i$.
• Therefore, $r_s = 1 - 0 = \mathbf{+1}$.
• Key Insight: Spearman's $r_s$ measures monotonic agreement, which is invariant under any strictly increasing transformation.
Problem 5Missing Difference Reconstruction
For 8 pairs of ranks, seven of the rank differences $d_i$ were calculated as: $+2, -1, +3, -4, +1, -3, +1$. What must be the eighth difference, and what is the algebraic law used?
Show Step-by-Step Solution
• Algebraic Law: The sum of differences between two sets of identical ranks is always zero: $\sum_{i=1}^n d_i = 0$.
• Sum of the 7 known differences: $2 - 1 + 3 - 4 + 1 - 3 + 1 = -1$.
• Therefore, the 8th difference must be $d_8 = 0 - (-1) = \mathbf{+1}$.
Comprehensive Practice Worksheet: Spearman's Rank Correlation
Examination standard questions classified into 1-Mark, 2-Mark, and 4-Mark sections based on CBSE Applied Mathematics curriculum standards.
Q2: When is Spearman's rank correlation coefficient preferred over Karl Pearson's coefficient?
Show Answer & Scheme
Answer [1 Mark]: When analyzing qualitative data that can only be ranked (e.g. honesty, beauty, athletic skill) or when the dataset contains extreme outliers violating bivariate normality.
SECTION B: 2-Mark Short Answer Problems
Q3: The rank correlation coefficient between two series of 10 observations was calculated as 0.5. Later, it was discovered that the difference in ranks for one pair was wrongly taken as 3 instead of 7. Find the corrected rank correlation coefficient.
Q4: In a ranking of 8 candidates, two candidates share the 4th rank. How should their ranks be assigned under the standard statistical convention?
Show Answer & Scheme
Answer [2 Marks]:
The two tied candidates occupy the 4th and 5th positions. Under statistical convention, each candidate is assigned the arithmetic average of these ranks:
$$\text{Assigned Rank} = \frac{4 + 5}{2} = \mathbf{4.5}$$
The candidate immediately following them receives the 6th rank.
SECTION C: 4-Mark Comprehensive Numerical Drill
Q5: An executive recruiter evaluates the interview performance ($X$) and written aptitude test ($Y$) of 7 candidates:
Interview ($X$): $68, \quad 75, \quad 90, \quad 75, \quad 50, \quad 62, \quad 40$
Written ($Y$): $80, \quad 90, \quad 85, \quad 70, \quad 60, \quad 70, \quad 55$
(i) Assign appropriate ranks to candidates for $X$ and $Y$ incorporating tied rank adjustments.
(ii) Identify the tie correction factors.
(iii) Calculate Spearman's rank correlation coefficient.
Standard timed test: 4 Objective MCQs / Assertion-Reason with instant automatic grading, plus 6 multi-mark subjective problems with CBSE marking rubrics.
PART 1: Objective Section (4 Marks)
Q1: If the rankings of $n$ students by two teachers are identical, the value of Spearman's rank correlation coefficient is:
Solution: (B) When rankings are identical, $d_i = 0$ for all $i$, so $r_s = 1 - 0 = +1$.
Q2: If two values are tied for the 5th and 6th ranks, what rank is assigned to each of them?
Solution: (C) Average of 5 and 6 is $ rac{5+6}{2} = 5.5$.
Q3: The tie correction factor added to $\sum d^2$ for a value that repeats $m$ times is:
Solution: (B) The standard tie correction factor is $\frac{m(m^2 - 1)}{12}$.
Q4 (Assertion-Reason): Assertion (A): Spearman's rank correlation coefficient is more resilient to extreme data outliers than Karl Pearson's correlation coefficient. Reason (R): Converting numerical observations into ranks bounds extreme values within the rank sequence $1$ to $n$, eliminating outlier leverage.
Solution: (A) Both Assertion and Reason are true and Reason provides the precise explanation of why ranking bounds outlier leverage.
PART 2: Subjective Section (Q5 - Q10)
Q52 Marks
The sum of squares of differences in ranks for 10 pairs of observations was found to be 33. Calculate Spearman's rank correlation coefficient.
Why is $\sum_{i=1}^n d_i = 0$ always true in rank correlation? Provide an algebraic justification.
Show Step-by-Step Solution & Marking Scheme
• $\sum d_i = \sum (R_{x_i} - R_{y_i}) = \sum R_{x_i} - \sum R_{y_i}$ [1 Mark].
• Since both $R_x$ and $R_y$ are permutations of the first $n$ natural numbers $1, 2, \dots, n$, their sums are identical:
$$\sum R_x = \sum R_y = \frac{n(n+1)}{2} \implies \sum d_i = \frac{n(n+1)}{2} - \frac{n(n+1)}{2} = \mathbf{0} \quad [1 \text{ Mark}]$$
Q73 Marks
In a painting competition, 6 contestants are evaluated by two judges. Compute $r_s$ for the ranks:
Judge 1: $1, 2, 3, 4, 5, 6$
Judge 2: $6, 5, 4, 3, 2, 1$
Interpret the relationship between the judgments.
Three values in a series of 12 items share the exact same score of 45. Explain how their ranks should be assigned, and calculate the tie correction factor contributed by this group.
Show Step-by-Step Solution & Marking Scheme
• Suppose the three tied values occupy positions $k, k+1, k+2$. Each is assigned the average rank: $\frac{k + (k+1) + (k+2)}{3} = \mathbf{k+1}$ [1.5 Marks].
• Number of repetitions $m = 3$.
• Correction factor:
$$CF = \frac{m(m^2 - 1)}{12} = \frac{3(3^2 - 1)}{12} = \frac{3(8)}{12} = \frac{24}{12} = \mathbf{2} \quad [1.5 \text{ Marks}]$$
Q9 & Q10Case Study (4 Marks Total)
Case Context: Software Engineering Coding Benchmark
A tech hiring panel ranks 5 full-stack engineering candidates based on System Design ($R_1$) and Algorithmic Problem Solving ($R_2$):
Candidate A: $R_1 = 1, R_2 = 2$
Candidate B: $R_1 = 2, R_2 = 1$
Candidate C: $R_1 = 3, R_2 = 4$
Candidate D: $R_1 = 4, R_2 = 3$
Candidate E: $R_1 = 5, R_2 = 5$
Q9 (2 Marks): Calculate Spearman's rank correlation coefficient $r_s$ between System Design and Algorithmic Problem Solving. Q10 (2 Marks): Interpret the result in terms of candidate competencies and explain why rank correlation is appropriate for recruitment evaluations.
Solution to Q10 [2 Marks]:
• $r_s = +0.80$ demonstrates a high positive rank correlation. Candidates who excel in architectural system design generally exhibit strong algorithmic problem-solving ability.
• Rank correlation is ideal in recruitment because candidate evaluations inherently produce ordinal ranking rather than calibrated metric points, removing individual interviewer scoring scale bias.