Why is "Ranking the signed differences instead of the absolute differences" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Skipping the |d| column and ranking d directly, so negatives are ranked as 'smallest'. How to avoid it: Always build a separate |d| column and rank from that; re-attach the signs only afterwards. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Keeping zero differences and using the original $n$" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Treating a zero like any other value rather than discarding it. How to avoid it: Strike out every d = 0 and reduce n before ranking — then use that reduced n for the critical value. Source: 9231 Paper 4 mark schemes

Why is "Giving tied values consecutive ranks instead of the averaged rank" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Ranking each value as it appears without checking for equal values. How to avoid it: Whenever values are equal, share the average of the positions they occupy; verify with ranks = N(N+1)/2. Source: 9231 Paper 4 Examiner Reports

Why is "Summing the ranks of the larger sample (or ranking samples separately)" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Forgetting that W is defined on the smaller sample, or not pooling before ranking. How to avoid it: Pool all data into one ordered list, rank once, then add only the smaller sample's ranks. Source: 9231 Paper 4 Examiner Reports

Why is "Using only the lower critical value in a two-tailed test" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Forgetting the symmetric upper critical value m(m+n+1) - W_L (rank-sum) or comparing only one rank sum. How to avoid it: For two-tailed rank-sum, reject if W W_L OR W upper value; for signed-rank, T = (W_+, W_-) already covers both tails. Source: 9231 Paper 4 mark schemes

Why is "Concluding only 'reject $H_0$' without reference to the population" a common mistake in Wilcoxon signed-rank test and the Wilcoxon rank-sum test?

Why it happens: Stopping at the statistical decision and not translating it back to the context. How to avoid it: Finish every test with a sentence about the population median / the two populations in the words of the question. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Wilcoxon signed-rank test and the Wilcoxon rank-sum test

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Detailed Study Notes

Detailed notes on Non-parametric tests for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Wilcoxon Signed-rank and Rank-sum Tests — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 (2026-2027 syllabus)

Two rank-based tests. The Wilcoxon signed-rank test handles a single or paired sample (symmetric population); the rank-sum / Mann–Whitney test handles two independent samples. Master the ranking discipline, the test statistic, the MF19 critical values, and the normal approximation for large samples.

At a glance

Signed-rank test — single/paired sample. Rank the ABSOLUTE differences, sum the $+$ and $-$ ranks; statistic $T = \min(W_+, W_-)$ .
Rank the |differences|, average ties, DISCARD zeros — three rules that decide most marks.
Rank-sum / Mann–Whitney — two independent samples. Pool, rank, sum the ranks of the SMALLER sample → $W$ .
Critical values come from MF19 (Wilcoxon tables). Compare your statistic with the tabled value.
Large samples: use the normal approximation with the stated mean and variance of the statistic.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

4.4d — Carry out the Wilcoxon signed-rank test (single-sample and paired) including the correct ranking of absolute differences and the rejection rule.
4.4e — Carry out the Wilcoxon rank-sum (Mann–Whitney) test for two independent samples using the rank sum of the smaller sample.
4.4f — Use the normal approximation to the test statistic for large samples, quoting the appropriate mean and variance, and conclude in context.

The Wilcoxon signed-rank test — single and paired sample

Differences → discard zeros → rank the absolute values (average ties) → sum + and − ranks → T = smaller of the two.

When to use it. A single sample tested against a median $m_0$ , or a paired sample (work with the differences). The population (of values or differences) must be symmetric about its median.

The procedure — five disciplined steps:

Differences. For a single sample, compute $d_i = x_i - m_0$ . For a paired sample, $d_i =$ (pair difference).
Discard zeros. Any $d_i = 0$ is dropped, and $n$ is reduced accordingly.
Rank the ABSOLUTE differences $|d_i|$ from smallest ( $1$ ) to largest. Tied values share the average rank (e.g. two values tied for ranks 3 and 4 both get $3.5$ ).
Re-attach the signs and sum the ranks: $W_+ =$ sum of ranks of positive $d_i$ , $W_- =$ sum of ranks of negative $d_i$ .
Test statistic $T = \min(W_+, W_-)$ . A small $T$ is evidence against $H_0$ .

Check: $W_+ + W_- = \dfrac{n(n+1)}{2}$ (the sum of all ranks $1$ to $n$ ) — a fast way to catch arithmetic slips.

The two |d| = 5 values tie for ranks 3 and 4, so each gets 3.5. T is the smaller rank sum.

Cambridge tip. The two killers are: (i) ranking the signed differences instead of the absolute differences, and (ii) forgetting to discard zeros. Build the table with a column for $|d|$ — it prevents both errors.

Rank $|d|$ , average ties, discard zeros, reduce $n$ .
$T = \min(W_+, W_-)$ ; small $T$ → reject $H_0$ .
Check $W_+ + W_- = \frac{n(n+1)}{2}$ .

See the full worked example for wilcoxon signed-rank test and the wilcoxon rank-sum test →

The Wilcoxon rank-sum (Mann–Whitney) test — two independent samples

Pool both samples, rank everything, sum the ranks of the smaller sample → W.

When to use it. Two independent samples (sizes $m \le n$ ), non-normal, testing whether the populations are identical.

The procedure:

Pool all $m + n$ observations into one ordered list.
Rank them all together, $1$ (smallest) to $m+n$ (largest), averaging ties as before.
Sum the ranks of the SMALLER sample (size $m$ ). Call this rank sum $W$ (sometimes $R_s$ ).
Compare $W$ with the MF19 rank-sum critical value for sample sizes $(m, n)$ at the chosen level.

The MF19 table gives a lower critical value $W_L$ (and, by symmetry, an upper one). A rank sum that is too small or too large signals a difference.

Mann–Whitney form. Some courses convert $W$ to the Mann–Whitney statistic $U = W - \tfrac{1}{2}m(m+1),$ but reading $W$ directly against the MF19 Wilcoxon rank-sum table is fully acceptable in 9231 — just match your $W$ to the sample sizes used in the table.

All seven values are ranked together; the rank sum of the smaller sample is W = 10, compared with the MF19 critical value for (3, 4).

Cambridge tip. Always take the rank sum of the smaller sample, and pool both samples before ranking. Ranking each sample separately is a guaranteed zero.

Pool both samples, rank together, average ties.
$W$ = rank sum of the smaller sample.
Compare $W$ with the MF19 rank-sum critical value for $(m, n)$ .

See the full worked example for wilcoxon signed-rank test and the wilcoxon rank-sum test →

Large samples — the normal approximation

For large n, each statistic is approximately normal. Standardise with the tabulated mean and variance.

When the sample is too large for the MF19 tables, each test statistic is approximately normally distributed under $H_0$ , and you standardise to a $z$ -value.

Signed-rank test (after discarding zeros, $n$ differences). Under $H_0$ : $E(T) = \frac{n(n+1)}{4}, \qquad \operatorname{Var}(T) = \frac{n(n+1)(2n+1)}{24}$

Rank-sum / Mann–Whitney test (smaller sample size $m$ , larger $n$ ). Under $H_0$ , the rank sum $W$ of the smaller sample has: $E(W) = \frac{m(m+n+1)}{2}, \qquad \operatorname{Var}(W) = \frac{mn(m+n+1)}{12}$

Standardise and compare with the normal critical value: $z = \frac{(\text{statistic}) - E}{\sqrt{\operatorname{Var}}}$

(A continuity correction of $\pm 0.5$ may be applied — move the statistic $0.5$ towards the mean.)

Then conclude in context. Compare $|z|$ (or $z$ ) with the critical value (e.g. $1.96$ two-tailed at 5%, $1.645$ one-tailed). If significant, reject $H_0$ and state the conclusion about the population median / the populations, not just "reject $H_0$ ".

Standardise the statistic to z and reject H₀ if it falls in a tail beyond the critical value.

Cambridge tip. Quote the correct mean/variance pair for the statistic you are using — mixing up the signed-rank and rank-sum formulae is a classic large-sample slip.

Signed-rank: $E(T)=\frac{n(n+1)}{4}$ , $\operatorname{Var}(T)=\frac{n(n+1)(2n+1)}{24}$ .
Rank-sum: $E(W)=\frac{m(m+n+1)}{2}$ , $\operatorname{Var}(W)=\frac{mn(m+n+1)}{12}$ .
Standardise to $z$ , optionally continuity-correct, conclude in context.

See the full worked example for wilcoxon signed-rank test and the wilcoxon rank-sum test →

Quick recap

Signed-rank: differences → discard zeros → rank $|d|$ (average ties) → $T = \min(W_+, W_-)$ .
Rank-sum: pool, rank together, sum the ranks of the SMALLER sample → $W$ .
Critical values from the MF19 Wilcoxon tables; small statistic → reject.
Large $n$ : signed-rank $E(T)=\frac{n(n+1)}{4}$ , $\operatorname{Var}(T)=\frac{n(n+1)(2n+1)}{24}$ .
Large $n$ : rank-sum $E(W)=\frac{m(m+n+1)}{2}$ , $\operatorname{Var}(W)=\frac{mn(m+n+1)}{12}$ .
Standardise, compare with the normal critical value, conclude about the median / populations.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$T = \min(W_+, W_-)$ and $W_+ + W_- = \frac{n(n+1)}{2}$ .
Rank the |differences|, average ties, discard zeros.
Rank-sum $W$ = rank sum of the smaller sample (pool then rank).
Signed-rank normal: $E(T)=\frac{n(n+1)}{4}$ , $\operatorname{Var}(T)=\frac{n(n+1)(2n+1)}{24}$ .
Rank-sum normal: $E(W)=\frac{m(m+n+1)}{2}$ , $\operatorname{Var}(W)=\frac{mn(m+n+1)}{12}$ .

How it’s examined

A full Wilcoxon question in Paper 4 typically carries 7-10 marks: marks for the correct ranking table (with averaged ties and discarded zeros), for the statistic, for reading the MF19 critical value at the right tail, and for an in-context conclusion. Large-sample parts test the standardisation with the correct mean and variance. The heaviest mark losses are ranking signed instead of absolute differences, mishandling ties, and concluding only 'reject $H_0$ ' without reference to the population median.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 4); 9231/41, 9231/42, 9231/43 Further Probability & Statistics — past papers, mark schemes and the MF19 formula booklet. Last reviewed 2026-06-23.

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Step-by-step worked examples — Wilcoxon signed-rank test and the Wilcoxon rank-sum test

Step-by-step solutions to past-paper-style questions on wilcoxon signed-rank test and the wilcoxon rank-sum test, written exactly the way a tutor would explain them at the board.

Question type:

1Wilcoxon signed-rank test for a single sample median
Getting startedMulti-step problem• signed-rank, single-sample
Question
The reaction times (seconds) of 8 people are $0.7,\ 1.1,\ 0.9,\ 1.4,\ 1.0,\ 0.6,\ 1.3,\ 0.95$ . Test, at the 5% level, whether the population median differs from $1.0$ s using the Wilcoxon signed-rank test. (MF19 lower critical value for $n=8$ , two-tailed 5% is $3$ .)
Step-by-step solution
1. Step 1
  State hypotheses. $H_0$ : population median $m = 1.0$ ; $H_1$ : $m \neq 1.0$ (two-tailed).
2. Step 2
  Compute differences $d_i = x_i - 1.0$ : $-0.3,\ 0.1,\ -0.1,\ 0.4,\ 0.0,\ -0.4,\ 0.3,\ -0.05$ . The value $1.0$ gives $d = 0$ — discard it, so $n = 7$ .
3. Step 3
  Rank the absolute differences $|d|$ : $0.05,\ 0.1,\ 0.1,\ 0.3,\ 0.3,\ 0.4,\ 0.4$ . The two $0.1$ tie for ranks 2,3 → $2.5$ each; the two $0.3$ tie for ranks 4,5 → $4.5$ each; the two $0.4$ tie for ranks 6,7 → $6.5$ each.
4. Step 4
  Sum the signed ranks. Positive $d$ : $0.1\,(2.5),\ 0.4\,(6.5),\ 0.3\,(4.5)$ . Negative $d$ : $0.3\,(4.5),\ 0.1\,(2.5),\ 0.4\,(6.5),\ 0.05\,(1)$ .
  $W_+ = 2.5 + 6.5 + 4.5 = 13.5, \qquad W_- = 4.5 + 2.5 + 6.5 + 1 = 14.5$
5. Step 5
  Test statistic and check. $T = \min(W_+, W_-) = 13.5$ . Check $W_+ + W_- = 28 = \frac{7 \times 8}{2}$ ✓.
  $T = \min(13.5, 14.5) = 13.5$
6. Step 6
  Compare and conclude. Critical value $= 3$ . Since $T = 13.5 > 3$ , do not reject $H_0$ .
Answer
$T = 13.5 > 3$ , so there is insufficient evidence at the 5% level that the population median reaction time differs from $1.0$ s.
Examiner tip
Discarding the exact $1.0$ (giving $n=7$ , not $8$ ) and averaging the three pairs of ties are the marks most often lost. Always finish with a sentence about the population median, not just 'accept $H_0$ '.
2Wilcoxon rank-sum test for two independent samples
Getting startedMulti-step problem• rank-sum, two-sample
Question
Sample A (size 4): $12,\ 15,\ 9,\ 20$ . Sample B (size 5): $11,\ 14,\ 8,\ 18,\ 13$ . Test at the 5% level (two-tailed) whether the populations have the same distribution, using the Wilcoxon rank-sum test. (MF19 lower critical value for $(m,n)=(4,5)$ , two-tailed 5% is $12$ .)
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : the two population distributions are identical; $H_1$ : they differ (two-tailed).
2. Step 2
  Pool and rank all 9 values. Ordered: $8(1),\ 9(2),\ 11(3),\ 12(4),\ 13(5),\ 14(6),\ 15(7),\ 18(8),\ 20(9)$ . No ties.
3. Step 3
  Sum the ranks of the SMALLER sample (A, size $m = 4$ ): values $12,15,9,20$ have ranks $4,7,2,9$ .
  $W = 4 + 7 + 2 + 9 = 22$
4. Step 4
  Compare with MF19. Lower critical value $= 12$ ; the upper value (by symmetry) is $m(m+n+1) - 12 = 4(10) - 12 = 28$ . Reject if $W \le 12$ or $W \ge 28$ .
  $12 < W = 22 < 28$
5. Step 5
  Conclude. $W = 22$ lies inside the acceptance region, so do not reject $H_0$ .
Answer
$W = 22$ , which is not in the critical region ( $\le 12$ or $\ge 28$ ); there is insufficient evidence at the 5% level that the two populations differ.
Examiner tip
Always take the rank sum of the smaller sample, and pool before ranking. Reading only the lower tail (forgetting the symmetric upper value $28$ ) costs the conclusion mark in two-tailed questions.
3Paired signed-rank test on 'before and after' data
Building confidenceMulti-step problem• signed-rank, paired
Question
Eight athletes record times (s) before and after training. The differences (before − after) are $1.2,\ 0.5,\ -0.3,\ 0.8,\ 0.0,\ 1.5,\ 0.4,\ 0.9$ . Test at the 5% level whether training reduces times (one-tailed). MF19 one-tailed 5% critical value for $n=7$ is $3$ .
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : median difference $= 0$ ; $H_1$ : median difference $> 0$ (times reduced, so before − after positive). One-tailed.
2. Step 2
  Discard the zero. One difference is $0.0$ → discard, leaving $n = 7$ differences.
3. Step 3
  Rank $|d|$ : $0.3(1),\ 0.4(2),\ 0.5(3),\ 0.8(4),\ 0.9(5),\ 1.2(6),\ 1.5(7)$ . No ties.
4. Step 4
  Sum signed ranks. Only $-0.3$ is negative (rank 1); the rest are positive.
  $W_- = 1, \qquad W_+ = 2+3+4+5+6+7 = 27$
5. Step 5
  Statistic. For a one-tailed test that the difference is positive, the relevant statistic is the smaller sum, $T = W_- = 1$ . Check $1 + 27 = 28 = \frac{7\times 8}{2}$ ✓.
  $T = \min(W_+, W_-) = 1$
6. Step 6
  Compare. $T = 1 \le 3$ , so reject $H_0$ .
Answer
$T = 1 \le 3$ , so reject $H_0$ : there is evidence at the 5% level that training reduces the athletes' times.
Examiner tip
In a one-tailed test confirm the direction of the evidence matches $H_1$ ( $W_+$ large, $W_-$ small here). Discarding the zero to give $n=7$ is essential — keeping $n=8$ uses the wrong critical value.
4Rank-sum test with tied observations
Building confidenceMulti-step problem• rank-sum, ties
Question
Group X (size 4): $5,\ 7,\ 7,\ 10$ . Group Y (size 5): $6,\ 7,\ 9,\ 11,\ 12$ . Find the rank sum $W$ of the smaller group, handling the ties correctly.
Step-by-step solution
1. Step 1
  Pool and order all 9 values: $5,\ 6,\ 7,\ 7,\ 7,\ 9,\ 10,\ 11,\ 12$ .
2. Step 2
  Assign ranks, averaging ties. $5(1),\ 6(2)$ ; the three $7$ s occupy positions 3,4,5 → average rank $\frac{3+4+5}{3} = 4$ each; then $9(6),\ 10(7),\ 11(8),\ 12(9)$ .
3. Step 3
  Identify which $7$ s belong to X. X has two $7$ s and the values $5$ and $10$ : ranks $1,\ 4,\ 4,\ 7$ (the two X-sevens each take the averaged rank $4$ ).
  $W = 1 + 4 + 4 + 7 = 16$
4. Step 4
  Sanity check. Total of all ranks $= \frac{9\times 10}{2} = 45$ . Y's rank sum $= 2 + 4 + 6 + 8 + 9 = 29$ , and $16 + 29 = 45$ ✓.
Answer
$W = 16$ (rank sum of the smaller group X), with the three tied $7$ s each given the averaged rank $4$ .
Examiner tip
All identical values share one averaged rank regardless of which group they sit in. Splitting the tie (e.g. giving X-sevens ranks 3 and 5) is a guaranteed lost mark — the check $W_X + W_Y = \frac{N(N+1)}{2}$ catches it.
5Large-sample normal approximation for the signed-rank statistic
StretchMulti-step problem• signed-rank, normal-approximation
Question
A Wilcoxon signed-rank test on $n = 30$ differences (no zeros) gives $T = 140$ . Test at the 5% level (two-tailed) whether the population median is as hypothesised, using the normal approximation with a continuity correction.
Step-by-step solution
1. Step 1
  Mean and variance of $T$ under $H_0$ .
  $E(T) = \frac{n(n+1)}{4} = \frac{30 \times 31}{4} = 232.5$
2. Step 2
  Variance.
  $\operatorname{Var}(T) = \frac{n(n+1)(2n+1)}{24} = \frac{30 \times 31 \times 61}{24} = 2363.75$
3. Step 3
  Continuity correction. $T = 140$ is below the mean, so move it $0.5$ up towards the mean: use $140.5$ .
  $z = \frac{140.5 - 232.5}{\sqrt{2363.75}} = \frac{-92}{48.62} = -1.892$
4. Step 4
  Compare with the critical value. Two-tailed 5% critical value is $\pm 1.96$ . Here $|z| = 1.892 < 1.96$ .
5. Step 5
  Conclude. $|z| < 1.96$ , so do not reject $H_0$ at the 5% level.
Answer
$z = -1.89$ (3 s.f.); since $|z| = 1.89 < 1.96$ , there is insufficient evidence at the 5% level to reject $H_0$ .
Examiner tip
Quote the signed-rank pair $E(T)=\frac{n(n+1)}{4}$ , $\operatorname{Var}(T)=\frac{n(n+1)(2n+1)}{24}$ — mixing in the rank-sum formulae is the classic error. The continuity correction shifts $T$ towards the mean.
6Large-sample normal approximation for the rank-sum statistic
StretchMulti-step problem• rank-sum, normal-approximation
Question
Two independent samples have sizes $m = 12$ (smaller) and $n = 15$ . The rank sum of the smaller sample is $W = 120$ . Test at the 5% level (two-tailed) whether the populations differ, using the normal approximation.
Step-by-step solution
1. Step 1
  Mean of $W$ under $H_0$ .
  $E(W) = \frac{m(m+n+1)}{2} = \frac{12(12+15+1)}{2} = \frac{12 \times 28}{2} = 168$
2. Step 2
  Variance of $W$ .
  $\operatorname{Var}(W) = \frac{mn(m+n+1)}{12} = \frac{12 \times 15 \times 28}{12} = 420$
3. Step 3
  Standardise (continuity correction: $W = 120$ is below the mean, use $120.5$ ).
  $z = \frac{120.5 - 168}{\sqrt{420}} = \frac{-47.5}{20.49} = -2.318$
4. Step 4
  Compare. Two-tailed 5% critical value $\pm 1.96$ . Here $|z| = 2.318 > 1.96$ .
5. Step 5
  Conclude. Reject $H_0$ .
Answer
$z = -2.32$ (3 s.f.); since $|z| = 2.32 > 1.96$ , reject $H_0$ : there is evidence at the 5% level that the two populations differ.
Examiner tip
Use the rank-sum pair $E(W)=\frac{m(m+n+1)}{2}$ , $\operatorname{Var}(W)=\frac{mn(m+n+1)}{12}$ with $m$ the SMALLER sample size. A common slip is putting the larger size as $m$ .

Model Answers — Wilcoxon signed-rank test and the Wilcoxon rank-sum test

High-scoring sample answers for wilcoxon signed-rank test and the wilcoxon rank-sum test on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
7 marks
The masses (g) of 9 chocolate bars are $48,\ 51,\ 49,\ 53,\ 50,\ 47,\ 52,\ 50,\ 54$ . A bar is supposed to have median mass $50$ g. Test at the 5% level whether the population median differs from $50$ g using the Wilcoxon signed-rank test. (Two-tailed 5% critical value for $n=7$ is $2$ .)
Model answer
Hypotheses. $H_0$ : median $m = 50$ ; $H_1$ : $m \neq 50$ (two-tailed, 5%).

Differences $d = x - 50$ : $-2,\ 1,\ -1,\ 3,\ 0,\ -3,\ 2,\ 0,\ 4$ .

The two $0$ differences are discarded, leaving $n = 7$ .

Rank $|d|$ (average ties):

| $|d|$ | 1 | 1 | 2 | 2 | 3 | 3 | 4 | |---|---|---|---|---|---|---|---| | rank | 1.5 | 1.5 | 3.5 | 3.5 | 5.5 | 5.5 | 7 |

Signed rank sums: $W_+ = 1.5\,(d=1) + 5.5\,(d=3) + 3.5\,(d=2) + 7\,(d=4) = 17.5$ $W_- = 3.5\,(d=-2) + 1.5\,(d=-1) + 5.5\,(d=-3) = 10.5$

Check: $17.5 + 10.5 = 28 = \frac{7 \times 8}{2}$ ✓.

Statistic: $T = \min(17.5, 10.5) = 10.5$ .

Conclusion: critical value $= 2$ . Since $T = 10.5 > 2$ , do not reject $H_0$ . There is insufficient evidence at the 5% level that the population median mass differs from $50$ g.
Why this scores
B1 hypotheses, M1 differences with both zeros discarded ( $n=7$ ), M1 ranking of $|d|$ with averaged ties, A1 $W_+$ and $W_-$ , A1 $T=10.5$ , M1 compare with 2, A1 in-context conclusion.
Question 2
7 marks
Two fertilisers are compared by crop yield. Fertiliser P (5 plots): $22,\ 25,\ 19,\ 28,\ 24$ . Fertiliser Q (6 plots): $21,\ 23,\ 18,\ 27,\ 20,\ 26$ . Test at the 5% level (two-tailed) whether the yields differ, using the Wilcoxon rank-sum test. (MF19 lower critical value for $(5,6)$ two-tailed 5% is $18$ .)
Model answer
Hypotheses. $H_0$ : the yield distributions are identical; $H_1$ : they differ (two-tailed, 5%).

Pool and rank all 11 values: $18(1),\ 19(2),\ 20(3),\ 21(4),\ 22(5),\ 23(6),\ 24(7),\ 25(8),\ 26(9),\ 27(10),\ 28(11)$

Rank sum of the smaller sample P (size $m = 5$ ): values $22,25,19,28,24$ → ranks $5,8,2,11,7$ . $W = 5 + 8 + 2 + 11 + 7 = 33$

Critical region. Lower value $= 18$ ; upper value $= m(m+n+1) - 18 = 5(12) - 18 = 42$ . Reject if $W \le 18$ or $W \ge 42$ .

Conclusion. $18 < W = 33 < 42$ , so $W$ is not in the critical region: do not reject $H_0$ . There is insufficient evidence at the 5% level that the two fertilisers give different yields.
Why this scores
B1 hypotheses, M1 pooled ranking, A1 $W=33$ for the smaller sample, M1 forming the symmetric upper critical value $42$ , M1 comparison, A1 in-context conclusion. (One mark for stating the test is two-tailed and using both tails.)
Question 3
8 marks
Ten students sit a test before and after a revision course. The differences (after − before) in marks are $4,\ 2,\ -1,\ 5,\ 3,\ 0,\ 6,\ 1,\ -2,\ 3$ . Test at the 5% level whether the course improves marks (one-tailed). (MF19 one-tailed 5% critical value for $n=9$ is $8$ .)
Model answer
Hypotheses. $H_0$ : median difference $= 0$ ; $H_1$ : median difference $> 0$ (course improves marks). One-tailed, 5%.

Discard the zero, leaving $n = 9$ differences: $4,\ 2,\ -1,\ 5,\ 3,\ 6,\ 1,\ -2,\ 3$ .

Rank $|d|$ (average ties):

| $|d|$ | 1 | 1 | 2 | 2 | 3 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---|---|---|---| | rank | 1.5 | 1.5 | 3.5 | 3.5 | 5.5 | 5.5 | 7 | 8 | 9 |

Signed sums. Negative differences: $-1$ (rank 1.5), $-2$ (rank 3.5). $W_- = 1.5 + 3.5 = 5, \qquad W_+ = 45 - 5 = 40$

Check: $\frac{9 \times 10}{2} = 45 = 40 + 5$ ✓.

Statistic: $T = \min(W_+, W_-) = 5$ . The direction ( $W_+$ large) is consistent with $H_1$ .

Conclusion: critical value $= 8$ . Since $T = 5 \le 8$ , reject $H_0$ . There is evidence at the 5% level that the revision course improves marks.
Why this scores
B1 hypotheses (one-tailed), M1 discard zero → $n=9$ , M1 ranking with averaged ties, A1 $W_-=5$ , A1 $T=5$ , M1 confirm direction matches $H_1$ , M1 compare with 8, A1 in-context conclusion.
Question 4
8 marks
Two machines fill bottles. Machine A (4 bottles): $498,\ 502,\ 500,\ 505$ . Machine B (5 bottles): $500,\ 497,\ 503,\ 500,\ 506$ . Test at the 5% level (two-tailed) whether the machines differ, using the rank-sum test, handling ties correctly. (MF19 lower critical value for $(4,5)$ two-tailed 5% is $12$ .)
Model answer
Hypotheses. $H_0$ : the two fill-volume distributions are identical; $H_1$ : they differ (two-tailed, 5%).

Pool and order all 9 values: $497,\ 498,\ 500,\ 500,\ 500,\ 502,\ 503,\ 505,\ 506$

Ranks (average ties). $497(1),\ 498(2)$ ; the three $500$ s occupy positions 3,4,5 → averaged rank $4$ each; $502(6),\ 503(7),\ 505(8),\ 506(9)$ .

Rank sum of smaller sample A (size $m = 4$ ): values $498,502,500,505$ → ranks $2,\ 6,\ 4,\ 8$ . $W = 2 + 6 + 4 + 8 = 20$

(Check: B's rank sum $= 1 + 4 + 7 + 4 + 9 = 25$ and $20 + 25 = 45 = \frac{9\times 10}{2}$ ✓.)

Critical region. Lower $= 12$ ; upper $= 4(10) - 12 = 28$ . Reject if $W \le 12$ or $W \ge 28$ .

Conclusion. $12 < W = 20 < 28$ , so do not reject $H_0$ . There is insufficient evidence at the 5% level that the machines differ.
Why this scores
B1 hypotheses, M1 pooled ranking, A1 the three $500$ s averaged to rank $4$ , A1 $W=20$ , M1 symmetric upper value $28$ , M1 comparison, A1 conclusion in context. The tie-handling A1 is the discriminating mark.
Question 5
8 marks
A Wilcoxon signed-rank test on $n = 25$ paired differences (no zeros) gives $W_+ = 95$ and $W_- = 230$ . Using a normal approximation with a continuity correction, test at the 5% level (two-tailed) whether the population median difference is zero.
Model answer
Hypotheses. $H_0$ : median difference $= 0$ ; $H_1$ : median difference $\neq 0$ (two-tailed, 5%).

Statistic: $T = \min(W_+, W_-) = 95$ . (Check $95 + 230 = 325 = \frac{25 \times 26}{2}$ ✓.)

Mean and variance under $H_0$ : $E(T) = \frac{n(n+1)}{4} = \frac{25 \times 26}{4} = 162.5$ $\operatorname{Var}(T) = \frac{n(n+1)(2n+1)}{24} = \frac{25 \times 26 \times 51}{24} = 1381.25$

Standardise (continuity correction: $T = 95$ is below the mean, use $95.5$ ): $z = \frac{95.5 - 162.5}{\sqrt{1381.25}} = \frac{-67}{37.166} = -1.803$

Conclusion. Two-tailed 5% critical value $= \pm 1.96$ . Since $|z| = 1.80 < 1.96$ , do not reject $H_0$ . There is insufficient evidence at the 5% level that the population median difference differs from zero.
Why this scores
B1 hypotheses, M1 $T=95$ , M1 $E(T)$ , M1 $\operatorname{Var}(T)$ , M1 continuity correction $95.5$ , A1 $z=-1.80$ , M1 compare with $1.96$ , A1 conclusion. Quoting the rank-sum mean/variance by mistake loses the two formula marks.
Question 6
8 marks
Two independent samples have sizes $m = 10$ (smaller) and $n = 14$ . The rank sum of the smaller sample is $W = 95$ . Using a normal approximation with a continuity correction, test at the 5% level (two-tailed) whether the populations differ.
Model answer
Hypotheses. $H_0$ : the two distributions are identical; $H_1$ : they differ (two-tailed, 5%).

Mean and variance of $W$ under $H_0$ : $E(W) = \frac{m(m+n+1)}{2} = \frac{10(10+14+1)}{2} = \frac{10 \times 25}{2} = 125$ $\operatorname{Var}(W) = \frac{mn(m+n+1)}{12} = \frac{10 \times 14 \times 25}{12} = 291.67$

Standardise (continuity correction: $W = 95$ is below the mean, use $95.5$ ): $z = \frac{95.5 - 125}{\sqrt{291.67}} = \frac{-29.5}{17.078} = -1.727$

Conclusion. Two-tailed 5% critical value $= \pm 1.96$ . Since $|z| = 1.73 < 1.96$ , do not reject $H_0$ . There is insufficient evidence at the 5% level that the two populations differ.
Why this scores
B1 hypotheses, M1 $E(W)$ , M1 $\operatorname{Var}(W)$ with $m$ the smaller size, M1 continuity correction, A1 $z=-1.73$ , M1 compare $1.96$ , A1 conclusion, plus B1 for stating the smaller sample's rank sum is used. Using $m=14$ inverts the mean.

Key Formulae — Wilcoxon signed-rank test and the Wilcoxon rank-sum test

The formulae you need to memorise for wilcoxon signed-rank test and the wilcoxon rank-sum test on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Signed-rank test statistic
$T = \min(W_+, W_-), \qquad W_+ + W_- = \frac{n(n+1)}{2}$
$W_+$
sum of the ranks of the positive differences
$W_-$
sum of the ranks of the negative differences
$n$
number of non-zero differences (zeros discarded)
When to use
Single-sample or paired Wilcoxon test. Rank the absolute differences, then take the smaller signed-rank sum.
Rank-sum statistic
$W = \sum (\text{ranks of the smaller sample})$
$W$
rank sum of the smaller sample (size $m$ )
$m, n$
sizes of the smaller and larger samples, $m \le n$
When to use
Two independent samples: pool, rank together (averaging ties), then sum the ranks of the smaller sample.
Signed-rank — normal approximation
$E(T) = \frac{n(n+1)}{4}, \qquad \operatorname{Var}(T) = \frac{n(n+1)(2n+1)}{24}$
$T$
the signed-rank statistic
$n$
number of non-zero differences
When to use
Large-sample signed-rank test. Standardise $z = \frac{T - E(T)}{\sqrt{\operatorname{Var}(T)}}$ with a $\pm 0.5$ continuity correction.
Rank-sum — normal approximation
$E(W) = \frac{m(m+n+1)}{2}, \qquad \operatorname{Var}(W) = \frac{mn(m+n+1)}{12}$
$W$
rank sum of the smaller sample
$m, n$
sizes of the smaller and larger samples
When to use
Large-sample rank-sum / Mann–Whitney test. Standardise to $z$ with a continuity correction.
Mann–Whitney conversion
$U = W - \frac{1}{2}m(m+1)$
$U$
the Mann–Whitney statistic
$W$
rank sum of the smaller sample
$m$
size of the smaller sample
When to use
Optional — only when a question or table is expressed in $U$ . Reading $W$ directly against the MF19 rank-sum table is fully acceptable.
Standardised statistic with continuity correction
$z = \frac{(\text{statistic} \pm 0.5) - E}{\sqrt{\operatorname{Var}}}$
$E$
mean of the statistic under $H_0$
$\operatorname{Var}$
variance of the statistic under $H_0$
When to use
Move the statistic $0.5$ towards the mean, then compare $z$ with the normal critical value (e.g. $1.96$ two-tailed at 5%).

Key Definitions and Keywords — Wilcoxon signed-rank test and the Wilcoxon rank-sum test

Definitions to memorise and the exact keywords mark schemes credit for wilcoxon signed-rank test and the wilcoxon rank-sum test answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Wilcoxon signed-rank test
Examiner keyword
A non-parametric test for the median of a single (or paired) symmetric population. Differences from the hypothesised median are ranked by absolute size; the statistic $T$ is the smaller of the positive and negative rank sums.
Wilcoxon rank-sum (Mann–Whitney) test
Examiner keyword
A non-parametric test for whether two independent samples come from identical distributions. All data are pooled and ranked together; the statistic $W$ is the rank sum of the smaller sample.
Averaged (mid-) ranks for ties
Examiner keyword
When two or more observations are equal, each is given the mean of the ranks they would otherwise occupy (e.g. positions 3 and 4 → both $3.5$ ).
Discarding zero differences
Examiner keyword
In the signed-rank test, any difference equal to zero is removed and the sample size $n$ reduced before ranking.
MF19 critical value
Examiner keyword
The tabulated value from the MF19 Wilcoxon tables against which the test statistic is compared. A statistic less than or equal to the lower critical value gives a significant result.
Continuity correction
A $\pm 0.5$ adjustment applied to a discrete statistic before standardising to the continuous normal distribution, moving the statistic towards the mean.

Common Mistakes and Misconceptions — Wilcoxon signed-rank test and the Wilcoxon rank-sum test

The traps other students keep falling into on wilcoxon signed-rank test and the wilcoxon rank-sum test questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Ranking the signed differences instead of the absolute differences
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Skipping the $|d|$ column and ranking $d$ directly, so negatives are ranked as 'smallest'.
How to avoid it
Always build a separate $|d|$ column and rank from that; re-attach the signs only afterwards.
✕Keeping zero differences and using the original $n$
9231 Paper 4 mark schemes
Why it happens
Treating a zero like any other value rather than discarding it.
How to avoid it
Strike out every $d = 0$ and reduce $n$ before ranking — then use that reduced $n$ for the critical value.
✕Giving tied values consecutive ranks instead of the averaged rank
9231 Paper 4 Examiner Reports
Why it happens
Ranking each value as it appears without checking for equal values.
How to avoid it
Whenever values are equal, share the average of the positions they occupy; verify with $\sum \text{ranks} = \frac{N(N+1)}{2}$ .
✕Summing the ranks of the larger sample (or ranking samples separately)
9231 Paper 4 Examiner Reports
Why it happens
Forgetting that $W$ is defined on the smaller sample, or not pooling before ranking.
How to avoid it
Pool all data into one ordered list, rank once, then add only the smaller sample's ranks.
✕Using only the lower critical value in a two-tailed test
9231 Paper 4 mark schemes
Why it happens
Forgetting the symmetric upper critical value $m(m+n+1) - W_L$ (rank-sum) or comparing only one rank sum.
How to avoid it
For two-tailed rank-sum, reject if $W \le W_L$ OR $W \ge$ upper value; for signed-rank, $T = \min(W_+, W_-)$ already covers both tails.
✕Concluding only 'reject $H_0$ ' without reference to the population
9231 Paper 4 Examiner Reports
Why it happens
Stopping at the statistical decision and not translating it back to the context.
How to avoid it
Finish every test with a sentence about the population median / the two populations in the words of the question.

Wilcoxon signed-rank test and the Wilcoxon rank-sum test

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Wilcoxon signed-rank test for a single sample median

2Wilcoxon rank-sum test for two independent samples

3Paired signed-rank test on 'before and after' data

4Rank-sum test with tied observations

5Large-sample normal approximation for the signed-rank statistic

6Large-sample normal approximation for the rank-sum statistic

Signed-rank test statistic

Rank-sum statistic

Signed-rank — normal approximation

Rank-sum — normal approximation

Mann–Whitney conversion

Standardised statistic with continuity correction

Wilcoxon signed-rank test

Wilcoxon rank-sum (Mann–Whitney) test

Averaged (mid-) ranks for ties

Discarding zero differences

MF19 critical value

Continuity correction

✕Ranking the signed differences instead of the absolute differences

✕Keeping zero differences and using the original nnn

✕Giving tied values consecutive ranks instead of the averaged rank

✕Summing the ranks of the larger sample (or ranking samples separately)

✕Using only the lower critical value in a two-tailed test

✕Concluding only 'reject H0H_0H0​' without reference to the population

✕Keeping zero differences and using the original $n$

✕Concluding only 'reject $H_0$ ' without reference to the population