Paper 3 · Quantitative Aptitude — CA Foundation 2025 Q94 Solution · Spearman's Rank Correlation | ConceptFirst

CA Foundation 2025 · Paper 3 · Quantitative AptitudeQ94 · 1 mark↻ Appears in 4 of 6 yearsOfficial key verified

Compute the rank correlation coefficient from the following data:

Figure for question 94

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Spearman's Rank Correlation

Spearman's rank correlation coefficient measures the strength and direction of association between two variables using their ranks rather than their raw values. It is the natural tool for qualitative or ordinal data — judges' rankings, preference orders, or figures too imprecise to measure exactly.

Rank both series, find how far apart the paired ranks are, and turn the sum of squared differences into a coefficient between $-1$ and $+1$ .

How it works

Each item receives a rank in the first series and a rank in the second. If the two rankings agree closely, the paired ranks stay near each other and the differences $d$ are small, pushing $R$ toward $+1$ . If they disagree, the differences grow and $R$ falls toward $-1$ .

R = 1 - \dfrac{6\sum d^{2}}{n(n^{2}-1)}

Interpreting and adjusting

$R = +1$ : perfect agreement in ranking.
$R = -1$ : perfect disagreement (exact reversal).
$R = 0$ : no rank association.
Tied ranks: give tied items the average of the ranks they occupy, and add a correction of $\dfrac{1}{12}(m^{3} - m)$ to $\sum d^{2}$ for each group of $m$ tied items.

Quick example

Five students are ranked by two judges:

Student	Judge X rank	Judge Y rank	$d$	$d^{2}$
A	1	2	$-1$	1
B	2	1	$1$	1
C	3	4	$-1$	1
D	4	3	$1$	1
E	5	5	$0$	0

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$r=1-\dfrac{6\Sigma d^2}{n(n^2-1)}=1-\dfrac{30}{990}\approx0.97$ .

Step 1 — Write the rank-correlation formula

r=1-\frac{6\,\Sigma d^{2}}{n(n^{2}-1)}

Step 2 — Substitute $n=10,\ \Sigma d^2=5$

n(n^2-1)=10\times(100-1)=990

r=1-\frac{6\times5}{990}=1-\frac{30}{990}=1-0.0303=0.9697\approx0.97

Why the other options are wrong: $0.95,0.96,0.99$ come from rounding slips or a wrong $n(n^2-1)$ . …

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