Why is "Keeping the zero / tied differences and using the original $n$" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Treating an exact match with the median as a + or - rather than discarding it. How to avoid it: Strike out every difference of exactly 0 and reduce n before computing the binomial probability. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Forgetting to double the tail probability in a two-tailed test" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Treating a two-tailed test like a one-tailed test and comparing a single tail with . How to avoid it: For two-tailed, always compute 2 × P(X x_) before comparing with . Source: 9231 Paper 4 mark schemes

Why is "Using variance $0.5n$ instead of $0.25n$ in the normal approximation" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Confusing the mean np = 0.5n with the variance npq = 0.25n. How to avoid it: Variance is npq = n(0.5)(0.5) = 0.25n; the standard deviation is √(0.25n) = 0.5√(n). Source: 9231 Paper 4 Examiner Reports

Why is "Omitting the continuity correction (or shifting the wrong way)" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Forgetting the binomial is discrete, or adding 0.5 when it should be subtracted. How to avoid it: Always apply 0.5, moving X *towards* the mean 0.5n (subtract when X > 0.5n, add when X < 0.5n). Source: 9231 Paper 4 mark schemes

Why is "Choosing the wrong tail for a one-tailed test" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Not matching the direction of H_1 to the upper or lower tail of X. How to avoid it: If H_1 predicts more + signs, use P(X x); if fewer, use P(X x). Source: 9231 Paper 4 Examiner Reports

Why is "Concluding only 'reject $H_0$' without reference to the population median" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Stopping at the statistical decision and not translating it back to the context. How to avoid it: Finish with a sentence about the population median in the words of the question, e.g. 'evidence that the median exceeds m_0'. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Testing for identity of populations using a paired-sample sign test

Q: Why is "Choosing the wrong tail for a one-tailed test" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Not matching the direction of H_1 to the upper or lower tail of X. How to avoid it: If H_1 predicts more + signs, use P(X x); if fewer, use P(X x). Source: 9231 Paper 4 Examiner Reports

Q: Why is "Concluding only 'reject $H_0$' without reference to the population median" a common mistake in Testing for identity of populations using a paired-sample sign test?

Why it happens: Stopping at the statistical decision and not translating it back to the context. How to avoid it: Finish with a sentence about the population median in the words of the question, e.g. 'evidence that the median exceeds m_0'. Source: 9231 Paper 4 Examiner Reports

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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.

The Sign Test (single and paired) — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 (2026-2027 syllabus)

The simplest non-parametric test. Record only the SIGN of each difference, count the plus signs, and treat that count as a binomial variable with $p = 0.5$ . Master discarding ties, the binomial tail probability, the normal approximation for large $n$ , and the in-context conclusion about the population median.

At a glance

Record the sign of (value − hypothesised median), or of each paired difference. Magnitudes are ignored.
Discard ties / zeros — any difference of exactly $0$ is dropped and $n$ reduced.
Under $H_0$ the number of $+$ signs $\sim \text{B}(n, 0.5)$ .
Compare the binomial tail probability with the significance level (double it for a two-tailed test).
Large $n$ : use the normal approximation $\text{N}(0.5n,\ 0.25n)$ with a continuity correction.

What you’ll learn

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

4.4a — Understand the rationale of the sign test as a test about the population median requiring only the order/sign of the data.
4.4b — Carry out a single-sample or paired sign test, discarding ties and using the $\text{B}(n, 0.5)$ distribution for the number of plus signs.
4.4c — Use the normal approximation to the binomial for large samples and conclude in context about the population median.

The sign test — recording signs and counting

Sign of each (value − median) or pair difference → discard zeros → count + signs → number of + signs ~ B(n, 0.5).

When to use it. A single sample tested against a median $m_0$ , or a paired sample (work with the pair differences). Unlike the Wilcoxon signed-rank test, the sign test needs no assumption of symmetry — it only uses the sign of each difference, so it has the weakest assumptions of all the non-parametric tests in 9231.

The procedure — four steps:

Form the differences. For a single sample, $d_i = x_i - m_0$ . For a paired sample, $d_i$ = (first − second) of each pair.
Record only the sign of each $d_i$ : a $+$ if positive, a $-$ if negative.
Discard ties / zeros. Any $d_i = 0$ is dropped and the sample size $n$ reduced accordingly.
Count the plus signs. Let $X$ = number of $+$ signs. Under $H_0$ (median $= m_0$ ), each difference is equally likely to be $+$ or $-$ , so $X \sim \text{B}(n,\ 0.5).$

Each value is compared with the median m₀ = 12; the exact 12 is discarded, leaving n = 3 signs to count.

Cambridge tip. The sign test throws away the size of each difference and keeps only its direction — which is exactly why it works without a symmetry assumption, but also why it is less powerful than the signed-rank test. Mention this trade-off if a question asks you to justify the choice of test.

Differences → signs only → discard zeros → reduce $n$ .
$X$ = number of $+$ signs $\sim \text{B}(n, 0.5)$ under $H_0$ .
Weakest assumptions: no normality, no symmetry required.

See the full worked example for testing for identity of populations using a paired-sample sign test →

The binomial tail probability and the decision

Compute the probability of a result at least as extreme under B(n, 0.5); compare with α (double it for two-tailed).

Having counted $X =$ number of $+$ signs, decide whether $X$ is surprisingly far from the expected $0.5n$ under $H_0$ .

One-tailed test. Suppose $H_1$ predicts more $+$ signs than $-$ signs and you observed $X = x$ . Compute the upper tail $P(X \ge x) = \sum_{k=x}^{n} \binom{n}{k}(0.5)^n.$ If this probability $\le \alpha$ (the significance level), reject $H_0$ . If $H_1$ predicts fewer $+$ signs, use the lower tail $P(X \le x)$ .

Two-tailed test. Work with whichever count is smaller (the number of $+$ or $-$ signs). Find the probability of a result at least that extreme in one tail, then double it before comparing with $\alpha$ : $2 \times P(X \le x_{\min}) \le \alpha \;\Rightarrow\; \text{reject } H_0.$

Reading the tables. The MF19 cumulative binomial tables give $P(X \le x)$ for $p = 0.5$ directly — read the value, or compute by hand for small $n$ . By symmetry $P(X \ge x) = P(X \le n - x)$ .

Cambridge tip. State the distribution $X \sim \text{B}(n, 0.5)$ explicitly, then write the exact probability you are comparing with $\alpha$ . Examiners award the method mark for the correct tail and (for two-tailed) for doubling it — even a small arithmetic slip then keeps most of the marks.

One-tailed: compare $P(X \ge x)$ or $P(X \le x)$ with $\alpha$ .
Two-tailed: use the smaller count and double the one-tail probability.
$P(X \ge x) = P(X \le n - x)$ by the symmetry of $\text{B}(n, 0.5)$ .

See the full worked example for testing for identity of populations using a paired-sample sign test →

Large samples and choosing the sign test

For large n the count of + signs is approximately N(0.5n, 0.25n); standardise with a continuity correction.

Normal approximation. When $n$ is large, the binomial $\text{B}(n, 0.5)$ is approximately normal: $X \approx \text{N}\big(0.5n,\ 0.25n\big),$ since $np = 0.5n$ and $npq = n(0.5)(0.5) = 0.25n$ . Standardise with a continuity correction of $\pm 0.5$ (move $X$ towards the mean): $z = \frac{(X \pm 0.5) - 0.5n}{\sqrt{0.25n}}.$ Compare $z$ with the normal critical value ( $1.645$ one-tailed at 5%, $1.96$ two-tailed at 5%).

For large n, the count of + signs is approximately N(0.5n, 0.25n); reject H₀ if z falls in the tail.

When the sign test is preferred. Choose the sign test when the data are only ordinal, when nothing can be assumed about the shape of the distribution, or when you have only the direction of each difference (not its size). Its weak assumptions make it robust, but discarding the magnitudes makes it less powerful than the Wilcoxon signed-rank test when symmetry can be assumed.

Concluding in context. Always translate the decision into a statement about the population median: e.g. "there is evidence at the 5% level that the population median exceeds $m_0$ ", rather than a bare "reject $H_0$ ".

Cambridge tip. The variance is $0.25n$ , not $0.5n$ — a frequent slip in the large-sample part. And the continuity correction shifts $X$ towards the mean $0.5n$ , so subtract $0.5$ when $X > 0.5n$ and add $0.5$ when $X < 0.5n$ .

Large $n$ : $X \approx \text{N}(0.5n, 0.25n)$ , continuity-correct, standardise to $z$ .
Prefer the sign test for ordinal data or when only the sign is known.
Conclude about the population median, in the words of the question.

See the full worked example for testing for identity of populations using a paired-sample sign test →

Quick recap

Sign test: record the sign of (value − median) or each pair difference; discard zeros.
Number of $+$ signs $X \sim \text{B}(n, 0.5)$ under $H_0$ .
One-tailed: compare the relevant tail $P(X \ge x)$ or $P(X \le x)$ with $\alpha$ .
Two-tailed: take the smaller count, double the one-tail probability, compare with $\alpha$ .
Large $n$ : $X \approx \text{N}(0.5n, 0.25n)$ with a continuity correction.
Weakest assumptions of the non-parametric tests; conclude about the population median in context.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Under $H_0$ : number of $+$ signs $\sim \text{B}(n, 0.5)$ .
Discard zeros / ties and reduce $n$ .
Two-tailed: use the smaller count and DOUBLE the one-tail probability.
Large $n$ : $X \approx \text{N}(0.5n,\ 0.25n)$ — variance $0.25n$ .
Continuity correction: move $X$ by $0.5$ towards the mean $0.5n$ .

How it’s examined

A sign test in Paper 4 typically carries 5-8 marks: marks for stating $H_0$ / $H_1$ about the median, for the signs table with ties discarded, for stating $X \sim \text{B}(n, 0.5)$ , for the correct tail probability (doubled if two-tailed), and for an in-context conclusion. Large-sample parts test the $\text{N}(0.5n, 0.25n)$ approximation with a continuity correction. The heaviest losses are keeping the zero differences, forgetting to double for a two-tailed test, using variance $0.5n$ instead of $0.25n$ , 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.

Take this whole topic with you

Step-by-step worked examples — Testing for identity of populations using a paired-sample sign test

Step-by-step solutions to past-paper-style questions on testing for identity of populations using a paired-sample sign test, written exactly the way a tutor would explain them at the board.

Question type:

1One-tailed sign test for a single-sample median
Getting startedMulti-step problem• sign-test, single-sample, one-tailed
Question
A manufacturer claims the median lifetime of a component is at least 1000 hours. A sample of 10 components lasts $1020,\ 980,\ 1050,\ 970,\ 1100,\ 960,\ 1010,\ 1000,\ 1080,\ 940$ hours. Test at the 5% level whether the median is less than 1000 hours.
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : median $m = 1000$ ; $H_1$ : $m < 1000$ (one-tailed).
2. Step 2
  Signs of $(x - 1000)$ : $+,-,+,-,+,-,+,0,+,-$ . The exact $1000$ gives $0$ — discard it, so $n = 9$ .
3. Step 3
  Count. $+$ signs $= 5$ , $-$ signs $= 4$ . Since $H_1$ is 'median lower', evidence is few $+$ signs; the relevant tail is the lower one. Let $X \sim \text{B}(9, 0.5)$ .
4. Step 4
  Tail probability. $P(X \le 5)$ — but $X = 5 > 4.5$ is on the wrong side, so the data do not even point towards $H_1$ .
  $P(X \le 5) = 0.7461 > 0.05$
5. Step 5
  Conclude. The probability far exceeds 5% (and the count favours $H_0$ ), so do not reject $H_0$ .
Answer
Do not reject $H_0$ : there is no evidence at the 5% level that the median lifetime is less than 1000 hours.
Examiner tip
Discarding the exact $1000$ (giving $n = 9$ ) is essential. With more $+$ than $-$ signs the data already contradict $H_1$ , so the conclusion follows immediately — but you must still state the distribution $\text{B}(9, 0.5)$ .
2Paired sign test on 'before and after' data
Getting startedMulti-step problem• sign-test, paired, one-tailed
Question
Eight people record their pulse rate before and after a coffee. The changes (after − before) are $+4,\ +2,\ -1,\ +5,\ 0,\ +3,\ +1,\ -2$ . Test at the 5% level whether coffee increases pulse rate (one-tailed). $X \sim \text{B}(n, 0.5)$ .
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : median change $= 0$ ; $H_1$ : median change $> 0$ (pulse increases). One-tailed.
2. Step 2
  Signs. $+,+,-,+,0,+,+,-$ . Discard the $0$ , so $n = 7$ .
3. Step 3
  Count. $+$ signs $= 5$ , $-$ signs $= 2$ . Let $X$ = number of $+$ signs, $X \sim \text{B}(7, 0.5)$ . Evidence for $H_1$ is many $+$ signs, so use the upper tail.
4. Step 4
  Upper-tail probability.
  $P(X \ge 5) = 1 - P(X \le 4) = 1 - 0.7734 = 0.2266$
5. Step 5
  Compare. $0.2266 > 0.05$ , so do not reject $H_0$ .
Answer
$P(X \ge 5) = 0.227 > 0.05$ , so there is insufficient evidence at the 5% level that coffee increases pulse rate.
Examiner tip
For a paired test use the pair differences exactly like a single sample. State $X \sim \text{B}(7, 0.5)$ and the correct (upper) tail; $P(X \ge 5) = P(X \le 2) = 0.2266$ by symmetry.
3Two-tailed sign test (double the tail probability)
Building confidenceMulti-step problem• sign-test, two-tailed
Question
A die is suspected of being unfair on the median score. Twelve throws give differences from the claimed median $3.5$ with signs $+,+,+,-,+,+,-,+,+,+,-,+$ . Test at the 5% level (two-tailed) whether the median differs from 3.5. $X \sim \text{B}(12, 0.5)$ .
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : median $= 3.5$ ; $H_1$ : median $\neq 3.5$ (two-tailed). No zeros, so $n = 12$ .
2. Step 2
  Count signs. $+$ signs $= 9$ , $-$ signs $= 3$ . Work with the smaller count, $3$ . Let $X \sim \text{B}(12, 0.5)$ .
3. Step 3
  One-tail probability for the smaller count.
  $P(X \le 3) = 0.0730$
4. Step 4
  Double it for a two-tailed test.
  $2 \times 0.0730 = 0.1460$
5. Step 5
  Compare. $0.1460 > 0.05$ , so do not reject $H_0$ .
Answer
$2P(X \le 3) = 0.146 > 0.05$ , so there is insufficient evidence at the 5% level that the median differs from 3.5.
Examiner tip
The two-tailed test compares $2 \times P(X \le x_{\min})$ with $\alpha$ . Forgetting to double is the most common error here — it would wrongly give $0.073 > 0.05$ but with a borderline that misleads students into the wrong reasoning.
4Justifying the choice of the sign test
Building confidenceWord problem• sign-test, choice-of-test
Question
Judges rank 9 cakes as 'better' or 'worse' than a standard recipe, recording only the direction (not by how much), with 7 'better' and 2 'worse'. Explain why a sign test is appropriate and test at the 5% level whether the new recipe is preferred (one-tailed).
Step-by-step solution
1. Step 1
  Why the sign test. Only the direction of each comparison is known (no measured difference), so a Wilcoxon signed-rank test (which needs the magnitudes and a symmetry assumption) cannot be used. The sign test uses only the signs — its weakest-assumption nature fits ordinal data exactly.
2. Step 2
  Hypotheses. $H_0$ : $P(\text{better}) = 0.5$ (no preference); $H_1$ : $P(\text{better}) > 0.5$ . Let $X$ = number 'better', $X \sim \text{B}(9, 0.5)$ , $X = 7$ .
3. Step 3
  Upper-tail probability.
  $P(X \ge 7) = 1 - P(X \le 6) = 1 - 0.9102 = 0.0898$
4. Step 4
  Compare. $0.0898 > 0.05$ , so do not reject $H_0$ .
Answer
The sign test is appropriate because only the direction of each judgement is available. $P(X \ge 7) = 0.0898 > 0.05$ , so there is insufficient evidence at the 5% level that the new recipe is preferred.
Examiner tip
The 'why this test' mark is earned by linking the ordinal / direction-only data to the sign test's weak assumptions. $P(X \ge 7) = P(X \le 2) = 0.0898$ by symmetry.
5Large-sample sign test using the normal approximation
StretchMulti-step problem• sign-test, normal-approximation
Question
In a survey of 80 households (no ties), 50 report higher spending than the claimed median and 30 report lower. Test at the 5% level (two-tailed) whether the median spending differs from the claimed value, using a normal approximation with a continuity correction.
Step-by-step solution
1. Step 1
  Hypotheses. $H_0$ : median $= m_0$ ; $H_1$ : median $\neq m_0$ (two-tailed). $X$ = number of $+$ signs $= 50$ , $n = 80$ .
2. Step 2
  Distribution under $H_0$ . $X \sim \text{B}(80, 0.5)$ , approximated by
  $X \approx \text{N}(0.5 \times 80,\ 0.25 \times 80) = \text{N}(40, 20)$
3. Step 3
  Continuity correction. $X = 50 > 40$ , so move $0.5$ down: use $49.5$ .
  $z = \frac{49.5 - 40}{\sqrt{20}} = \frac{9.5}{4.472} = 2.124$
4. Step 4
  Compare. Two-tailed 5% critical value $\pm 1.96$ . Here $|z| = 2.124 > 1.96$ .
5. Step 5
  Conclude. Reject $H_0$ .
Answer
$z = 2.12$ (3 s.f.); since $|z| = 2.12 > 1.96$ , reject $H_0$ : there is evidence at the 5% level that the median spending differs from the claimed value.
Examiner tip
Use variance $0.25n = 20$ , not $0.5n$ . The continuity correction moves $X$ towards the mean ( $50 \to 49.5$ ). Quote the conclusion about the population median.
6Finding the critical region for a sign test
StretchMulti-step problem• sign-test, critical-region
Question
A one-tailed sign test ( $H_1$ : median $> m_0$ ) is to be carried out on $n = 10$ observations at the 5% level. Find the critical region for $X$ , the number of $+$ signs, and state the actual significance level.
Step-by-step solution
1. Step 1
  Distribution. Under $H_0$ , $X \sim \text{B}(10, 0.5)$ . Reject for large $X$ , so find the smallest $c$ with $P(X \ge c) \le 0.05$ .
2. Step 2
  Tabulate the upper tail.
  $P(X \ge 8) = 1 - 0.9453 = 0.0547, \qquad P(X \ge 9) = 1 - 0.9893 = 0.0107$
3. Step 3
  Choose $c$ . $P(X \ge 8) = 0.0547 > 0.05$ but $P(X \ge 9) = 0.0107 \le 0.05$ , so the critical region is $X \ge 9$ .
4. Step 4
  Actual significance level.
  $P(X \ge 9) = 0.0107 \approx 1.07\%$
Answer
Critical region: $X \ge 9$ (i.e. $9$ or $10$ plus signs). Actual significance level $= 0.0107$ (1.07%).
Examiner tip
The discrete binomial means the actual level (1.07%) is below the nominal 5%. Choosing $X \ge 8$ would over-shoot 5% ( $0.0547$ ) — the critical region must keep the tail probability $\le \alpha$ .

Model Answers — Testing for identity of populations using a paired-sample sign test

High-scoring sample answers for testing for identity of populations using a paired-sample sign test on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
6 marks
A teacher claims the median test mark is 60. A sample of 11 students scores $65,\ 58,\ 72,\ 60,\ 55,\ 68,\ 49,\ 70,\ 62,\ 53,\ 66$ . Test at the 5% level (two-tailed) whether the population median differs from 60.
Model answer
Hypotheses. $H_0$ : median $m = 60$ ; $H_1$ : $m \neq 60$ (two-tailed, 5%).

Signs of $(x - 60)$ : $+,-,+,0,-,+,-,+,+,-,+$ . The exact $60$ gives $0$ → discard, so $n = 10$ .

Counts: $+$ signs $= 6$ , $-$ signs $= 4$ . Work with the smaller count, $4$ . Let $X \sim \text{B}(10, 0.5)$ .

One-tail probability: $P(X \le 4) = 0.3770$

Double for two-tailed: $2 \times 0.3770 = 0.7540$

Conclusion. $0.7540 > 0.05$ , so do not reject $H_0$ . There is insufficient evidence at the 5% level that the population median mark differs from 60.
Why this scores
B1 hypotheses about the median, M1 signs with the zero discarded ( $n=10$ ), B1 state $X \sim \text{B}(10, 0.5)$ , M1 $P(X \le 4)$ , M1 double it, A1 in-context conclusion.
Question 2
6 marks
Nine patients' blood pressure is measured before and after a drug. The changes (after − before) are $-8,\ -3,\ +1,\ -5,\ -6,\ 0,\ -2,\ -4,\ +2$ . Test at the 5% level whether the drug lowers blood pressure (one-tailed).
Model answer
Hypotheses. $H_0$ : median change $= 0$ ; $H_1$ : median change $< 0$ (drug lowers pressure). One-tailed, 5%.

Signs of the changes: $-,-,+,-,-,0,-,-,+$ . Discard the $0$ , so $n = 8$ .

Counts: $-$ signs $= 6$ , $+$ signs $= 2$ . Evidence for $H_1$ (lower pressure) is few $+$ signs. Let $X$ = number of $+$ signs, $X \sim \text{B}(8, 0.5)$ , $X = 2$ .

Lower-tail probability: $P(X \le 2) = 0.1445$

Conclusion. $0.1445 > 0.05$ , so do not reject $H_0$ . There is insufficient evidence at the 5% level that the drug lowers blood pressure.
Why this scores
B1 hypotheses, M1 signs with zero discarded ( $n=8$ ), B1 $X \sim \text{B}(8, 0.5)$ with $X$ the number of $+$ signs, M1 correct (lower) tail $P(X \le 2)$ , M1 comparison with $0.05$ , A1 in-context conclusion. No doubling — it is one-tailed.
Question 3
7 marks
A coin-tossing game is claimed to be fair (median outcome neutral). In 14 rounds a player's net result is positive in 11 rounds and negative in 3. Test at the 5% level (two-tailed) whether the game is biased. $X \sim \text{B}(14, 0.5)$ .
Model answer
Hypotheses. $H_0$ : $P(\text{positive}) = 0.5$ (fair); $H_1$ : $P(\text{positive}) \neq 0.5$ (biased). Two-tailed, 5%.

Counts: positive $= 11$ , negative $= 3$ , no ties so $n = 14$ . Work with the smaller count, $3$ . Let $X \sim \text{B}(14, 0.5)$ .

One-tail probability: $P(X \le 3) = \binom{14}{0}(0.5)^{14} + \binom{14}{1}(0.5)^{14} + \binom{14}{2}(0.5)^{14} + \binom{14}{3}(0.5)^{14}$ $= (1 + 14 + 91 + 364)(0.5)^{14} = 470 \times \frac{1}{16384} = 0.02869$

Double for two-tailed: $2 \times 0.02869 = 0.0574$

Conclusion. $0.0574 > 0.05$ , so do not reject $H_0$ . There is insufficient evidence at the 5% level that the game is biased.
Why this scores
B1 hypotheses, M1 identify the smaller count $3$ , M1 $P(X \le 3)$ (tables or the binomial sum), A1 $0.0287$ , M1 double to $0.0574$ , M1 compare with $0.05$ , A1 in-context conclusion. The borderline ( $0.0574$ vs $0.05$ ) makes the doubling decisive.
Question 4
7 marks
Twelve consumers state whether they prefer Brand X or Brand Y (no 'no preference' allowed), with 9 preferring X and 3 preferring Y. (a) Explain why a sign test is suitable here rather than a Wilcoxon signed-rank test. (b) Test at the 5% level (two-tailed) whether consumers have a preference.
Model answer
(a) Each consumer gives only a preference (a direction), with no measured difference and no numerical scale. A Wilcoxon signed-rank test needs the magnitudes of differences and assumes a symmetric distribution; neither is available here. The sign test uses only the signs (directions) of the comparisons, so its weak assumptions exactly suit this ordinal, direction-only data.

(b) $H_0$ : $P(\text{prefer X}) = 0.5$ ; $H_1$ : $P(\text{prefer X}) \neq 0.5$ (two-tailed, 5%). Let $X$ = number preferring X, $X \sim \text{B}(12, 0.5)$ . Smaller count is $3$ (prefer Y).

$P(X \le 3) = 0.0730$ $2 \times 0.0730 = 0.1460$

Conclusion. $0.1460 > 0.05$ , so do not reject $H_0$ . There is insufficient evidence at the 5% level that consumers have a preference.
Why this scores
(a) B1 for linking direction-only/ordinal data to the sign test, B1 for noting the signed-rank test needs magnitudes + symmetry. (b) B1 hypotheses, M1 $P(X \le 3) = 0.073$ , M1 double, M1 compare, A1 conclusion.
Question 5
8 marks
In a sample of 100 paired observations (no ties), 62 differences are positive and 38 are negative. 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%).

Let $X$ = number of positive differences $= 62$ , $n = 100$ .

Distribution under $H_0$ : $X \sim \text{B}(100, 0.5)$ , approximated by $X \approx \text{N}(0.5 \times 100,\ 0.25 \times 100) = \text{N}(50, 25)$

Standardise (continuity correction: $X = 62 > 50$ , use $61.5$ ): $z = \frac{61.5 - 50}{\sqrt{25}} = \frac{11.5}{5} = 2.30$

Conclusion. Two-tailed 5% critical value $\pm 1.96$ . Since $|z| = 2.30 > 1.96$ , reject $H_0$ . There is evidence at the 5% level that the population median difference differs from zero.
Why this scores
B1 hypotheses, M1 $\text{N}(50, 25)$ with variance $0.25n = 25$ , M1 continuity correction to $61.5$ , A1 $z = 2.30$ , M1 compare with $1.96$ , A1 conclusion, plus B1 for stating the binomial-to-normal approximation and B1 for the mean $0.5n$ . Variance $0.5n$ instead of $0.25n$ is the classic slip.
Question 6
8 marks
A two-tailed sign test on $n = 15$ observations is carried out at the 5% level for the number of $+$ signs $X \sim \text{B}(15, 0.5)$ . (a) Find the critical region. (b) State the actual significance level of the test.
Model answer
(a) Two-tailed, so split $\alpha = 0.05$ into $0.025$ in each tail. Under $H_0$ , $X \sim \text{B}(15, 0.5)$ (symmetric about $7.5$ ).

Lower tail: find the largest $a$ with $P(X \le a) \le 0.025$ . $P(X \le 3) = 0.0176 \le 0.025, \qquad P(X \le 4) = 0.0592 > 0.025$ So $X \le 3$ .

Upper tail (by symmetry): $X \ge 12$ , since $P(X \ge 12) = P(X \le 3) = 0.0176$ .

Critical region: $X \le 3$ or $X \ge 12$ .

(b) Actual significance level (sum of both tails): $P(X \le 3) + P(X \ge 12) = 2 \times 0.0176 = 0.0352 \approx 3.52\%$
Why this scores
(a) M1 split into $0.025$ per tail, M1 $P(X \le 3) = 0.0176 \le 0.025$ chosen over $P(X \le 4)$ , A1 lower region, B1 symmetric upper region $X \ge 12$ . (b) M1 sum both tails, A1 $0.0352$ . The discrete distribution makes the actual level (3.52%) below the nominal 5%.

Key Formulae — Testing for identity of populations using a paired-sample sign test

The formulae you need to memorise for testing for identity of populations using a paired-sample sign test on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Sign-test distribution
$X \sim \text{B}(n, 0.5)$
$X$
number of $+$ signs after discarding zeros
$n$
number of non-zero differences
When to use
Under $H_0$ (median $= m_0$ ), each difference is equally likely to be $+$ or $-$ , so the count of $+$ signs is binomial with $p = 0.5$ .
One-tailed decision rule
$P(X \ge x) \le \alpha \;\text{ or }\; P(X \le x) \le \alpha$
$x$
observed number of $+$ signs
$\alpha$
significance level
When to use
One-tailed test: use the upper tail if $H_1$ predicts more $+$ signs, the lower tail if it predicts fewer; reject if the tail probability $\le \alpha$ .
Two-tailed decision rule
$2 \times P(X \le x_{\min}) \le \alpha$
$x_{\min}$
the smaller of the $+$ and $-$ counts
$\alpha$
significance level
When to use
Two-tailed test: take the smaller count, find its one-tail probability, double it, and compare with $\alpha$ .
Normal approximation for large $n$
$X \approx \text{N}\big(0.5n,\ 0.25n\big)$
$0.5n$
mean $np$
$0.25n$
variance $npq$ with $p = q = 0.5$
When to use
Large samples: approximate $\text{B}(n, 0.5)$ by a normal distribution; standardise with a $\pm 0.5$ continuity correction.
Standardised statistic with continuity correction
$z = \frac{(X \pm 0.5) - 0.5n}{\sqrt{0.25n}}$
$X$
observed number of $+$ signs
$n$
number of non-zero differences
When to use
Move $X$ by $0.5$ towards the mean $0.5n$ , standardise, and compare $z$ with the normal critical value.

Key Definitions and Keywords — Testing for identity of populations using a paired-sample sign test

Definitions to memorise and the exact keywords mark schemes credit for testing for identity of populations using a paired-sample sign test answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Sign test
Examiner keyword
A non-parametric test for the median of a single or paired population that uses only the sign (direction) of each difference from the hypothesised median, ignoring magnitudes.
Number of plus signs $X$
Examiner keyword
The count of positive differences after discarding zeros. Under $H_0$ it follows $\text{B}(n, 0.5)$ and is the test statistic.
Discarding ties / zeros
Examiner keyword
Any observation equal to the hypothesised median (or any zero pair difference) is removed and the sample size $n$ reduced before testing.
Population median
Examiner keyword
The value $m_0$ that the sign test makes inferences about — the conclusion must always refer to it, not merely to $H_0$ .
Weak assumptions
Examiner keyword
The sign test requires no normality and no symmetry — only that each non-zero difference is independently equally likely to be $+$ or $-$ under $H_0$ . This makes it suitable for ordinal data.
Actual significance level
Because the binomial is discrete, the true probability of the critical region is usually below the nominal level; this exact tail probability is the actual significance level.

Common Mistakes and Misconceptions — Testing for identity of populations using a paired-sample sign test

The traps other students keep falling into on testing for identity of populations using a paired-sample sign test questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Keeping the zero / tied differences and using the original $n$
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Treating an exact match with the median as a $+$ or $-$ rather than discarding it.
How to avoid it
Strike out every difference of exactly $0$ and reduce $n$ before computing the binomial probability.
✕Forgetting to double the tail probability in a two-tailed test
9231 Paper 4 mark schemes
Why it happens
Treating a two-tailed test like a one-tailed test and comparing a single tail with $\alpha$ .
How to avoid it
For two-tailed, always compute $2 \times P(X \le x_{\min})$ before comparing with $\alpha$ .
✕Using variance $0.5n$ instead of $0.25n$ in the normal approximation
9231 Paper 4 Examiner Reports
Why it happens
Confusing the mean $np = 0.5n$ with the variance $npq = 0.25n$ .
How to avoid it
Variance is $npq = n(0.5)(0.5) = 0.25n$ ; the standard deviation is $\sqrt{0.25n} = 0.5\sqrt{n}$ .
✕Omitting the continuity correction (or shifting the wrong way)
9231 Paper 4 mark schemes
Why it happens
Forgetting the binomial is discrete, or adding $0.5$ when it should be subtracted.
How to avoid it
Always apply $\pm 0.5$ , moving $X$ towards the mean $0.5n$ (subtract when $X > 0.5n$ , add when $X < 0.5n$ ).
✕Choosing the wrong tail for a one-tailed test
9231 Paper 4 Examiner Reports
Why it happens
Not matching the direction of $H_1$ to the upper or lower tail of $X$ .
How to avoid it
If $H_1$ predicts more $+$ signs, use $P(X \ge x)$ ; if fewer, use $P(X \le x)$ .
✕Concluding only 'reject $H_0$ ' without reference to the population median
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 with a sentence about the population median in the words of the question, e.g. 'evidence that the median exceeds $m_0$ '.

Testing for identity of populations using a paired-sample sign 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

1One-tailed sign test for a single-sample median

2Paired sign test on 'before and after' data

3Two-tailed sign test (double the tail probability)

4Justifying the choice of the sign test

5Large-sample sign test using the normal approximation

6Finding the critical region for a sign test

Sign-test distribution

One-tailed decision rule

Two-tailed decision rule

Normal approximation for large nnn

Standardised statistic with continuity correction

Sign test

Number of plus signs XXX

Discarding ties / zeros

Population median

Weak assumptions

Actual significance level

✕Keeping the zero / tied differences and using the original nnn

✕Forgetting to double the tail probability in a two-tailed test

✕Using variance 0.5n0.5n0.5n instead of 0.25n0.25n0.25n in the normal approximation

✕Omitting the continuity correction (or shifting the wrong way)

✕Choosing the wrong tail for a one-tailed test

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

Normal approximation for large $n$

Number of plus signs $X$

✕Keeping the zero / tied differences and using the original $n$

✕Using variance $0.5n$ instead of $0.25n$ in the normal approximation

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