Why is "Running a two-sample test on paired data (or vice versa), and using the wrong degrees of freedom" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Not checking whether the two sets of readings are linked to the same subjects before choosing the test. How to avoid it: Ask: 'are the measurements matched, item by item?' If yes, paired ( = n-1); if independent, two-sample ( = n_1+n_2-2). Differing sample sizes rule out a paired test. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$ in the pooled variance" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Confusing the pooled-variance denominator with a simple sample-size total. How to avoid it: The denominator is the degrees of freedom n_1 + n_2 - 2 — the same number you use for the t-table. Source: 9231 Paper 4 mark schemes

Why is "Putting $s_p^2$ (instead of $s_p$) in the test-statistic denominator" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Forgetting that the standard error needs the standard deviation, not the variance. How to avoid it: After finding s_p^2, take its square root: the denominator is s_p√(1/n_1+1/n_2). Source: 9231 Paper 4 Examiner Reports

Why is "Using a one-tailed critical value for a two-tailed test (or vice versa)" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Not splitting the significance level between the two tails for a two-tailed alternative. How to avoid it: Two-tailed at 5% uses the 2.5% column (e.g. t_10 = 2.228); one-tailed at 5% uses the 5% column (t_10 = 1.812). Decide the tail from H_1 first. Source: 9231 Paper 4 Examiner Reports

Why is "Writing only 'reject $H_0$' / 'accept $H_0$' with no contextual conclusion" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Treating the comparison as the end of the question and forgetting the final A1 is for context. How to avoid it: Always finish with a sentence in the problem's own words and at the stated significance level — e.g. 'sufficient evidence at the 5% level that the mean yield increased'. Source: 9231 Paper 4 Examiner Reports — most frequent lost mark

Why is "Pooling variances (or using the $t$-test) when the population variances are known" a common mistake in Formulating hypotheses concerning the difference of population?

Why it happens: Applying the two-sample t procedure by reflex without checking what information is given. How to avoid it: If ^2 values are stated (or samples are large), use the normal z-test with √(\sigma_1^2/n_1 + \sigma_2^2/n_2) — no pooling, no degrees of freedom. Source: 9231 Paper 4 mark schemes

Cambridge International A Levels Further Mathematics (9231)

Formulating hypotheses concerning the difference of population

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Detailed Study Notes

Detailed notes on Inference using Normal and t-distributions for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Hypothesis Tests for a Difference of Population Means — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 (2026-2027 syllabus)

Two samples, one question: are the population means really different? Learn to choose between the two-sample pooled $t$ -test, the paired $t$ -test and the known-variance $z$ -test, then run all five steps and — crucially — state your conclusion in context.

At a glance

Two-sample pooled $t$ -test compares two independent samples when the (unknown) population variances are assumed equal: $\nu = n_1 + n_2 - 2$ .
Paired $t$ -test is for matched data (e.g. before/after on the same subjects) — work with the differences $d$ and use $\nu = n - 1$ .
Choosing the test is itself examined: paired when the data are naturally linked, two-sample when the samples are independent.
Known variance or large samples $\Rightarrow$ use the normal ( $z$ ) distribution instead of $t$ .
Conclusion in context is compulsory — never just "reject $H_0$ ".

What you’ll learn

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

4.2 — Formulate hypotheses concerning the difference of population means, and apply a hypothesis test using a two-sample $t$ -test (pooled variance) for independent samples from normal populations with equal but unknown variance.
4.2 — Apply a paired-sample $t$ -test, recognising when matched (paired) data require analysis of the differences with $\nu = n - 1$ degrees of freedom.
4.2 — Use the normal distribution for a difference of means when the population variances are known (or the samples are large), and state every conclusion in the context of the question.

The two-sample (pooled) t-test for independent samples

Two independent samples, equal but unknown variances — pool the variances and use ν = n₁ + n₂ − 2.

The setting. You have two separate, independent samples — say the yields of crops on field A and on field B — drawn from two normal populations with means $\mu_1$ and $\mu_2$ . You want to test whether the population means differ. Because the population variances are unknown but assumed equal, you combine ("pool") the two sample variances into one best estimate.

The pooled variance. This is the single most error-prone formula, so write it out carefully: $s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$

It is a weighted average of the two sample variances, weighted by their degrees of freedom $(n-1)$ .

The test statistic. $t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}, \qquad \nu = n_1 + n_2 - 2$

Under $H_0$ you normally assume $\mu_1 - \mu_2 = 0$ , so the bracket $(\mu_1 - \mu_2)$ vanishes — but write it in, because some questions test a specified difference.

The five-step structure (use it every single time):

State $H_0: \mu_1 = \mu_2$ and $H_1$ (one- or two-tailed).
State the significance level and identify the test.
Compute $s_p^2$ , then the test statistic $t$ .
Compare with the critical value from the $t$ -table (MF19) at $\nu = n_1 + n_2 - 2$ .
Conclude in context.

For a two-tailed test at 5%, each tail holds 2.5%. Reject $H_0$ if your test statistic falls in either red region.

Cambridge tip. The standard error in the denominator uses $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ — that is $s_p$ (the square root), not $s_p^2$ . Compute $s_p^2$ first, then take its square root before dividing.

Pool the variances: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ .
Test statistic uses $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ with $\nu = n_1+n_2-2$ .
Assumptions: independent samples, normal populations, equal variances.

See the full worked example for formulating hypotheses concerning the difference of population →

The paired (matched-pairs) t-test

Matched data — collapse each pair to a difference d = x − y, then run a one-sample t-test on the differences with ν = n − 1.

When data come in pairs. Sometimes the two measurements are naturally linked: the same student's score before and after a course, or the left and right tyre of the same car. Here the two samples are NOT independent — a high "before" tends to go with a high "after". Treating them as independent would throw away exactly the information that makes the test powerful.

The key move. Reduce each pair $(x, y)$ to a single difference $d = x - y$ . You now have one sample of differences $d_1, d_2, \ldots, d_n$ , and you run an ordinary one-sample $t$ -test on them.

The test statistic. $t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}, \qquad \nu = n - 1$

where $\bar{d}$ is the mean of the differences, $s_d$ is the sample standard deviation of the differences, and $n$ is the number of pairs. Under $H_0$ you usually take $\mu_d = 0$ (no difference).

Critical distinction — the degrees of freedom. For the paired test $\nu = n - 1$ , where $n$ is the number of pairs. This is not $n_1 + n_2 - 2$ . Mixing these up is the classic paired-vs-two-sample error.

Five matched pairs become five differences. From here it is just a one-sample $t$-test on $d$ with $\nu = n - 1 = 4$.

Cambridge tip. When you compute $s_d$ on the differences, use the unbiased (divide-by- $(n-1)$ ) form. If the question gives $\sum d$ and $\sum d^2$ , then $s_d^2 = \frac{1}{n-1}\left(\sum d^2 - \frac{(\sum d)^2}{n}\right)$ .

Reduce each pair to $d = x - y$ , then run a one-sample test on $d$ .
Test statistic $t = \frac{\bar{d} - \mu_d}{s_d/\sqrt{n}}$ with $\nu = n - 1$ ( $n$ = number of pairs).
Use the differences' own standard deviation $s_d$ , not the original samples'.

See the full worked example for formulating hypotheses concerning the difference of population →

Choosing the right test (and using the normal when variance is known)

Paired vs two-sample depends on whether the data are matched; switch to z when variances are known or samples are large.

Paired or two-sample? Ask one question: are the two sets of measurements naturally linked, item by item?

Yes, matched (same subject measured twice; left/right of the same object; before/after) $\Rightarrow$ paired $t$ -test on the differences.
No, independent (two separate groups of different individuals) $\Rightarrow$ two-sample pooled $t$ -test.

A quick tell: if the two samples must have the same size $n_1 = n_2$ because the data come as pairs, you are almost certainly looking at a paired test. If the sample sizes differ, it cannot be paired.

Using the normal distribution instead of $t$ . The $t$ -distribution is needed only because the population variances are unknown and estimated from small samples. Switch to the normal ( $z$ ) distribution when:

the population variances $\sigma_1^2, \sigma_2^2$ are known, or
the samples are large (so $s^2 \approx \sigma^2$ and $t \approx z$ ).

For known variances and independent samples the statistic is $z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}$

Notice there is no pooling here — each population keeps its own variance — and the critical value comes from the normal table ( $z = 1.645$ for a 5% one-tailed test, $1.96$ for 2.5%), with no degrees of freedom.

The five-step structure (identical for $z$ and $t$ ):

State $H_0$ and $H_1$ .
State the level and name the test.
Compute the test statistic.
Compare with the critical value.
Conclude in context.

First ask if the data are paired. If not, use the pooled $t$-test — unless the variances are known (or $n$ is large), in which case use the normal $z$-test.

Cambridge tip. A one-line justification of your choice ("the data are paired because each plot was measured before and after treatment") often earns an explicit mark. Don't skip it.

Paired = matched data; two-sample = independent groups.
Known variances (or large $n$ ) $\Rightarrow$ normal $z$ -test, no pooling, no df.
Always justify the test choice and conclude in context.

See the full worked example for formulating hypotheses concerning the difference of population →

Quick recap

Two-sample pooled $t$ : $t = \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{1/n_1+1/n_2}}$ , $\nu = n_1+n_2-2$ .
Pooled variance $s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$ — a df-weighted average.
Paired $t$ : work with $d = x - y$ , $t = \frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$ , $\nu = n - 1$ .
Choose paired for matched data, two-sample for independent samples.
Known variance (or large $n$ ) $\Rightarrow$ normal $z$ -test (no pooling, no df).
Every conclusion must be stated in the context of the question.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Pooled variance $s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$ .
Two-sample $t$ statistic $t = \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{1/n_1+1/n_2}}$ , $\nu = n_1+n_2-2$ .
Paired $t$ statistic $t = \frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$ , $\nu = n - 1$ .
Known-variance $z$ $z = \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$ .
Five steps: hypotheses, level + test, statistic, compare, conclude in context.

How it’s examined

Topic 4.2 is a staple of Paper 4 (typically 6-10 marks). The question gives two sets of data (or summary statistics $\sum x$ , $\sum x^2$ ) and asks you to test for a difference of means. The first mark-scheme hurdle is choosing the correct test — examiner reports repeatedly penalise candidates who run a two-sample test on paired data, or who forget to pool the variances. The pooled variance and the $\nu = n_1+n_2-2$ vs $\nu = n-1$ distinction are the most common arithmetic and conceptual slips. The final A1 mark is almost always reserved for a conclusion stated in context — "there is insufficient evidence at the 5% level that the diet reduces mean weight" — so a bare "do not reject $H_0$ " loses it. Critical values come from the MF19 $t$ - and normal tables; state the table, df and tail you used.

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

Take this whole topic with you

Step-by-step worked examples — Formulating hypotheses concerning the difference of population

Step-by-step solutions to past-paper-style questions on formulating hypotheses concerning the difference of population, written exactly the way a tutor would explain them at the board.

Question type:

1Forming H₀ and H₁ for a difference of means
Getting startedIdentify & classify• hypotheses, difference-of-means
Question
A researcher believes a new fertiliser gives a higher mean crop yield ( $\mu_1$ ) than the standard fertiliser ( $\mu_2$ ). Write down suitable null and alternative hypotheses, and state whether the test is one- or two-tailed.
Step-by-step solution
1. Step 1
  Null hypothesis — no difference. The default position is that the two means are equal.
  $H_0: \mu_1 = \mu_2 \quad (\text{equivalently } \mu_1 - \mu_2 = 0)$
2. Step 2
  Alternative hypothesis — read the claim. "Higher" is directional, so $\mu_1$ is claimed to exceed $\mu_2$ .
  $H_1: \mu_1 > \mu_2$
3. Step 3
  Decide the tail. A directional ('>') alternative means a one-tailed test, with the whole significance level in the upper tail.
Answer
$H_0: \mu_1 = \mu_2$ ; $\;H_1: \mu_1 > \mu_2$ ; one-tailed test.
Examiner tip
Hypotheses must be about the population means $\mu$ , never the sample means $\bar{x}$ . The word 'higher'/'increase'/'reduce' signals a one-tailed test; 'differ'/'change' signals two-tailed.
2Deciding: paired or two-sample?
Getting startedIdentify & classify• choosing-test, paired, two-sample
Question
For each scenario, state whether a paired or a two-sample $t$ -test is appropriate. (a) Twelve runners each record a 5 km time before and after a six-week training plan. (b) The lifetimes of 10 bulbs from factory A and 8 bulbs from factory B are compared.
Step-by-step solution
1. Step 1
  (a) Are the data matched? Yes — each runner provides a before and an after time, so the two readings are linked to the same person.
  $\text{paired } t\text{-test, analyse } d = \text{before} - \text{after}$
2. Step 2
  (b) Are the samples independent? Yes — different bulbs, different factories, and the sample sizes even differ ( $10 \ne 8$ ), so they cannot be paired.
  $\text{two-sample (pooled) } t\text{-test}$
Answer
(a) Paired $t$ -test ( $\nu = 11$ ). (b) Two-sample pooled $t$ -test ( $\nu = 16$ ).
Examiner tip
If sample sizes differ the test cannot be paired. A clear one-line reason ('same subject measured twice' / 'two independent groups') is usually worth a mark.
3Full two-sample pooled t-test
Building confidenceMulti-step problem• two-sample, pooled-variance, five-steps
Question
Independent random samples of plant heights (cm) give: sample 1, $n_1 = 6$ , $\bar{x}_1 = 24.0$ , $s_1^2 = 5.0$ ; sample 2, $n_2 = 6$ , $\bar{x}_2 = 21.0$ , $s_2^2 = 4.0$ . Test at the 5% level whether the population means differ. Assume normal populations with equal variance.
Step-by-step solution
1. Step 1
  Hypotheses (two-tailed — 'differ').
  $H_0: \mu_1 = \mu_2, \qquad H_1: \mu_1 \neq \mu_2$
2. Step 2
  Pool the variances.
  $s_p^2 = \frac{(6-1)(5.0) + (6-1)(4.0)}{6 + 6 - 2} = \frac{25 + 20}{10} = 4.5$
3. Step 3
  Test statistic. Here $s_p = \sqrt{4.5} = 2.1213\ldots$
  $t = \frac{(24.0 - 21.0) - 0}{\sqrt{4.5}\,\sqrt{\tfrac{1}{6}+\tfrac{1}{6}}} = \frac{3}{2.1213 \times 0.5774} = \frac{3}{1.2247} = 2.449$
4. Step 4
  Critical value from the $t$ -table, $\nu = n_1+n_2-2 = 10$ , two-tailed 5% (so 2.5% in each tail): $t_{10}(2.5\%) = 2.228$ .
  $2.449 > 2.228$
5. Step 5
  Conclude in context.
Answer
$t = 2.45 > 2.228$ , so reject $H_0$ : there is sufficient evidence at the 5% level that the two population mean heights differ.
Examiner tip
Quote $s_p = \sqrt{4.5}$ (not $4.5$ ) in the denominator. With $\nu = 10$ a two-tailed 5% test uses the 2.5% critical value $2.228$ — using the one-tailed $1.812$ here would be wrong.
4Full paired t-test
Building confidenceMulti-step problem• paired, differences, five-steps
Question
Seven students sit a test before and after a revision class. The differences $d = (\text{after}) - (\text{before})$ give $\sum d = 21$ and $\sum d^2 = 91$ . Test at the 5% level whether the class increases the mean mark. Assume the differences are normally distributed.
Step-by-step solution
1. Step 1
  Hypotheses on the mean difference (one-tailed — 'increases').
  $H_0: \mu_d = 0, \qquad H_1: \mu_d > 0$
2. Step 2
  Mean of differences. $n = 7$ pairs.
  $\bar{d} = \frac{\sum d}{n} = \frac{21}{7} = 3$
3. Step 3
  Sample variance of the differences (divide by $n-1$ ).
  $s_d^2 = \frac{1}{n-1}\left(\sum d^2 - \frac{(\sum d)^2}{n}\right) = \frac{1}{6}\left(91 - \frac{21^2}{7}\right) = \frac{91 - 63}{6} = \frac{28}{6} = 4.6667$
4. Step 4
  Test statistic, $s_d = \sqrt{4.6667} = 2.1602$ , $\nu = n - 1 = 6$ .
  $t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{3}{2.1602/\sqrt{7}} = \frac{3}{0.8165} = 3.674$
5. Step 5
  Compare and conclude. Critical value $t_6(5\%) = 1.943$ (one-tailed). Since $3.674 > 1.943$ , reject $H_0$ .
Answer
$t = 3.67 > 1.943$ , so reject $H_0$ : there is sufficient evidence at the 5% level that the revision class increases the mean mark.
Examiner tip
Degrees of freedom are $n - 1 = 6$ (number of pairs minus 1), NOT $n_1+n_2-2$ . The standard error is $s_d/\sqrt{n}$ — the one-sample form applied to the differences.
5Known-variance normal (z) test
StretchMulti-step problem• known-variance, z-test, normal
Question
Two machines fill bags. From past records the filling weights are normal with known variances $\sigma_1^2 = 9$ and $\sigma_2^2 = 16$ (g²). Samples give $n_1 = 40$ , $\bar{x}_1 = 502$ g and $n_2 = 50$ , $\bar{x}_2 = 500$ g. Test at the 5% level whether machine 1 fills to a higher mean weight.
Step-by-step solution
1. Step 1
  Hypotheses (one-tailed — 'higher').
  $H_0: \mu_1 = \mu_2, \qquad H_1: \mu_1 > \mu_2$
2. Step 2
  Variances are known $\Rightarrow$ use the normal $z$ -test (no pooling — each population keeps its own $\sigma^2$ ).
  $\text{SE} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{9}{40} + \frac{16}{50}} = \sqrt{0.225 + 0.32} = \sqrt{0.545}$
3. Step 3
  Test statistic. $\sqrt{0.545} = 0.73824$ .
  $z = \frac{(502 - 500) - 0}{\sqrt{0.545}} = \frac{2}{0.73824} = 2.709$
4. Step 4
  Compare with the normal critical value $z(5\%) = 1.645$ (one-tailed).
  $2.709 > 1.645$
5. Step 5
  Conclude in context.
Answer
$z = 2.71 > 1.645$ , so reject $H_0$ : there is sufficient evidence at the 5% level that machine 1 fills to a higher mean weight.
Examiner tip
With variances known, use $z$ not $t$ — there is no pooling and no degrees of freedom. Each $\sigma^2$ stays attached to its own sample size.
6Testing a specified difference (μ₁ − μ₂ ≠ 0)
StretchMulti-step problem• two-sample, specified-difference, pooled-variance
Question
A manufacturer claims brand A lasts on average more than 5 hours longer than brand B. Independent samples give: A, $n_1 = 8$ , $\bar{x}_1 = 30$ h, $s_1^2 = 6$ ; B, $n_2 = 6$ , $\bar{x}_2 = 23$ h, $s_2^2 = 4$ . Test the claim at the 5% level (assume normal, equal variances).
Step-by-step solution
1. Step 1
  Hypotheses with a specified difference of 5. The claim is $\mu_1 - \mu_2 > 5$ .
  $H_0: \mu_1 - \mu_2 = 5, \qquad H_1: \mu_1 - \mu_2 > 5$
2. Step 2
  Pool the variances, $\nu = 8 + 6 - 2 = 12$ .
  $s_p^2 = \frac{(8-1)(6) + (6-1)(4)}{8+6-2} = \frac{42 + 20}{12} = \frac{62}{12} = 5.1667$
3. Step 3
  Test statistic — keep the $(\mu_1-\mu_2) = 5$ term in. $s_p = \sqrt{5.1667} = 2.2730$ ; $\sqrt{\tfrac18+\tfrac16} = \sqrt{0.29167} = 0.54006$ .
  $t = \frac{(30 - 23) - 5}{2.2730 \times 0.54006} = \frac{2}{1.2276} = 1.629$
4. Step 4
  Compare with $t_{12}(5\%) = 1.782$ (one-tailed).
  $1.629 < 1.782$
5. Step 5
  Conclude in context.
Answer
$t = 1.63 < 1.782$ , so do not reject $H_0$ : there is insufficient evidence at the 5% level that brand A lasts on average more than 5 hours longer than brand B.
Examiner tip
The hypothesised difference is 5, so the numerator is $(\bar{x}_1-\bar{x}_2) - 5 = 2$ , NOT $7$ . Forgetting to subtract the specified difference is the trap here.

Model Answers — Formulating hypotheses concerning the difference of population

High-scoring sample answers for formulating hypotheses concerning the difference of population on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
A psychologist wants to test whether a sleep app changes the mean number of hours slept. Let $\mu_1$ and $\mu_2$ be the population mean hours with and without the app. Write down the hypotheses, state the type of test, and explain whether a paired or two-sample design would be more appropriate if the same people are measured with and without the app.
Model answer
Hypotheses (about population means): $H_0: \mu_1 = \mu_2 \qquad H_1: \mu_1 \neq \mu_2$

The word "changes" is non-directional, so this is a two-tailed test.

Design. If the same people are measured both with and without the app, each person provides a matched pair of readings. The two sets of measurements are therefore not independent, so a paired $t$ -test (analysing the differences $d$ ) is appropriate, with $\nu = n - 1$ degrees of freedom.
Why this scores
B1 for $H_0$ , B1 for $H_1$ , B1 for stating two-tailed, B1 for identifying the paired design with a valid reason. Hypotheses must use $\mu$ , not $\bar{x}$ .
Question 2
4 marks
Two independent samples give $n_1 = 10$ , $s_1^2 = 8$ and $n_2 = 6$ , $s_2^2 = 3$ . (a) Calculate the pooled estimate of the common variance. (b) State the number of degrees of freedom that would be used in a two-sample $t$ -test.
Model answer
(a) The pooled variance is a degrees-of-freedom-weighted average: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2} = \frac{(9)(8) + (5)(3)}{10 + 6 - 2}$ $s_p^2 = \frac{72 + 15}{14} = \frac{87}{14} = 6.214 \;(3\text{ s.f.})$

(b) Degrees of freedom: $\nu = n_1 + n_2 - 2 = 10 + 6 - 2 = 14$
Why this scores
(a) M1 for correct substitution into the pooled-variance formula, A1 for $6.21$ . (b) B1 for $\nu = 14$ . A common slip is dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$ .
Question 3
7 marks
Independent random samples of the masses (kg) of fish from two lakes give: Lake A, $n_1 = 8$ , $\bar{x}_1 = 2.6$ , $s_1^2 = 0.20$ ; Lake B, $n_2 = 6$ , $\bar{x}_2 = 2.3$ , $s_2^2 = 0.14$ . Assuming normal populations with equal variance, test at the 5% level whether the population mean masses differ.
Model answer
Step 1 — Hypotheses (two-tailed): $H_0: \mu_1 = \mu_2 \qquad H_1: \mu_1 \neq \mu_2$

Step 2 — Test and level. Two-sample (pooled) $t$ -test at the 5% significance level.

Step 3 — Pooled variance: $s_p^2 = \frac{(7)(0.20) + (5)(0.14)}{8 + 6 - 2} = \frac{1.40 + 0.70}{12} = \frac{2.10}{12} = 0.175$

Test statistic ( $s_p = \sqrt{0.175} = 0.41833$ ): $t = \frac{(2.6 - 2.3) - 0}{0.41833\sqrt{\tfrac{1}{8}+\tfrac{1}{6}}} = \frac{0.3}{0.41833 \times 0.54006} = \frac{0.3}{0.22594} = 1.328$

Step 4 — Critical value. $\nu = n_1 + n_2 - 2 = 12$ ; two-tailed 5% gives $t_{12}(2.5\%) = 2.179$ . $1.328 < 2.179$

Step 5 — Conclusion in context. Do not reject $H_0$ : there is insufficient evidence at the 5% level that the population mean masses of fish in the two lakes differ.
Why this scores
M1 A1 for the pooled variance $0.175$ ; M1 for the test-statistic structure, A1 for $t = 1.33$ ; B1 for the critical value $2.179$ ( $\nu=12$ , two-tailed); M1 for the comparison; A1 for the contextual conclusion. The final A1 is lost by writing only 'accept $H_0$ '.
Question 4
7 marks
The cholesterol levels of 9 patients are measured before and after a treatment. The differences $d = (\text{before}) - (\text{after})$ have $\sum d = 13.5$ and $\sum d^2 = 31.5$ . Assuming the differences are normally distributed, test at the 5% level whether the treatment reduces the mean cholesterol level.
Model answer
Step 1 — Hypotheses. A reduction means $d = \text{before} - \text{after} > 0$ : $H_0: \mu_d = 0 \qquad H_1: \mu_d > 0$

Step 2 — Test and level. Paired $t$ -test at the 5% significance level ( $n = 9$ pairs).

Step 3 — Sample statistics of the differences: $\bar{d} = \frac{\sum d}{n} = \frac{13.5}{9} = 1.5$ $s_d^2 = \frac{1}{n-1}\left(\sum d^2 - \frac{(\sum d)^2}{n}\right) = \frac{1}{8}\left(31.5 - \frac{13.5^2}{9}\right) = \frac{31.5 - 20.25}{8} = \frac{11.25}{8} = 1.40625$

Test statistic ( $s_d = \sqrt{1.40625} = 1.18585$ , $\nu = n - 1 = 8$ ): $t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{1.5}{1.18585/3} = \frac{1.5}{0.39528} = 3.795$

Step 4 — Critical value. $t_8(5\%) = 1.860$ (one-tailed). $3.795 > 1.860$

Step 5 — Conclusion in context. Reject $H_0$ : there is sufficient evidence at the 5% level that the treatment reduces the mean cholesterol level.
Why this scores
M1 A1 for $\bar{d} = 1.5$ and $s_d^2 = 1.40625$ ; M1 for the test statistic with $s_d/\sqrt{n}$ , A1 for $t = 3.80$ ; B1 for $t_8(5\%) = 1.860$ (note $\nu = 8$ , NOT 16); M1 comparison; A1 contextual conclusion. Using $\nu = n_1+n_2-2$ here is the standard error.
Question 5
7 marks
The breaking strengths of ropes from two suppliers are normally distributed with known standard deviations $\sigma_1 = 12$ N and $\sigma_2 = 15$ N. Samples give $n_1 = 50$ , $\bar{x}_1 = 305$ N and $n_2 = 60$ , $\bar{x}_2 = 298$ N. Test at the 2.5% level whether supplier 1's ropes have a greater mean breaking strength.
Model answer
Step 1 — Hypotheses (one-tailed — 'greater'): $H_0: \mu_1 = \mu_2 \qquad H_1: \mu_1 > \mu_2$

Step 2 — Test and level. The population variances are known, so use the normal ( $z$ ) test at the 2.5% significance level (no pooling, no degrees of freedom).

Step 3 — Standard error and test statistic. $\sigma_1^2 = 144$ , $\sigma_2^2 = 225$ : $\text{SE} = \sqrt{\frac{144}{50} + \frac{225}{60}} = \sqrt{2.88 + 3.75} = \sqrt{6.63} = 2.5749$ $z = \frac{(305 - 298) - 0}{2.5749} = \frac{7}{2.5749} = 2.718$

Step 4 — Critical value. $z(2.5\%) = 1.96$ (one-tailed at 2.5%). $2.718 > 1.96$

Step 5 — Conclusion in context. Reject $H_0$ : there is sufficient evidence at the 2.5% level that supplier 1's ropes have a greater mean breaking strength.
Why this scores
M1 for the SE with separate $\sigma^2/n$ terms (no pooling), A1 for $\text{SE} = 2.57$ ; M1 A1 for $z = 2.72$ ; B1 for the critical value $1.96$ (a one-tailed 2.5% test); M1 comparison; A1 contextual conclusion. Squaring the standard deviations first ( $\sigma^2 = 144, 225$ ) is essential.
Question 6
9 marks
A teacher records the marks of two independent groups. Group X ( $n_1 = 5$ ): 62, 65, 70, 68, 70. Group Y ( $n_2 = 7$ ): 58, 60, 63, 59, 61, 64, 62. Assuming normal populations with equal variance, test at the 5% level whether the two population mean marks differ. (You may use $\sum x_X = 335$ , $\sum x_X^2 = 22513$ , $\sum x_Y = 427$ , $\sum x_Y^2 = 26075$ .)
Model answer
Step 1 — Hypotheses (two-tailed): $H_0: \mu_X = \mu_Y \qquad H_1: \mu_X \neq \mu_Y$

Step 2 — Sample means: $\bar{x}_X = \frac{335}{5} = 67, \qquad \bar{x}_Y = \frac{427}{7} = 61$

Step 3 — Sample variances (unbiased, divide by $n-1$ ): $s_X^2 = \frac{1}{4}\left(22513 - \frac{335^2}{5}\right) = \frac{1}{4}(22513 - 22445) = \frac{68}{4} = 17$ $s_Y^2 = \frac{1}{6}\left(26075 - \frac{427^2}{7}\right) = \frac{1}{6}(26075 - 26047) = \frac{28}{6} = 4.6667$

Step 4 — Pooled variance, $\nu = 5 + 7 - 2 = 10$ : $s_p^2 = \frac{(4)(17) + (6)(4.6667)}{10} = \frac{68 + 28}{10} = \frac{96}{10} = 9.6$

Test statistic ( $s_p = \sqrt{9.6} = 3.0984$ ): $t = \frac{(67 - 61) - 0}{3.0984\sqrt{\tfrac{1}{5}+\tfrac{1}{7}}} = \frac{6}{3.0984 \times 0.58554} = \frac{6}{1.8143} = 3.307$

Step 5 — Compare and conclude. $t_{10}(2.5\%) = 2.228$ (two-tailed 5%). Since $3.307 > 2.228$ , reject $H_0$ : there is sufficient evidence at the 5% level that the two population mean marks differ.
Why this scores
M1 A1 for the two sample variances; M1 A1 for $s_p^2 = 9.6$ ; M1 for the test-statistic structure, A1 for $t = 3.31$ ; B1 for $t_{10}(2.5\%) = 2.228$ ; M1 comparison and A1 contextual conclusion. Using $\sum x^2 - n\bar{x}^2$ over $(n-1)$ for the variances is the key intermediate step.

Key Formulae — Formulating hypotheses concerning the difference of population

The formulae you need to memorise for formulating hypotheses concerning the difference of population on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Two-sample (pooled) t-statistic
$t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}, \quad \nu = n_1 + n_2 - 2$
$\bar{x}_1, \bar{x}_2$
the two sample means
$\mu_1 - \mu_2$
hypothesised difference (often 0)
$s_p$
pooled standard deviation $\sqrt{s_p^2}$
$n_1, n_2$
the two sample sizes
When to use
Two independent samples, normal populations, unknown but equal variances. Critical value from the $t$ -table with $\nu = n_1+n_2-2$ .
Pooled estimate of variance
$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$
$s_1^2, s_2^2$
the two unbiased sample variances
$n_1, n_2$
the two sample sizes
When to use
Before any two-sample $t$ -test with assumed equal variances. A degrees-of-freedom-weighted average of the sample variances.
Paired-sample t-statistic
$t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}, \quad \nu = n - 1$
$d = x - y$
the difference within each matched pair
$\bar{d}$
mean of the differences
$s_d$
sample standard deviation of the differences
$n$
the number of pairs
When to use
Matched/paired data (same subjects measured twice). Reduce to differences and run a one-sample test with $\nu = n - 1$ .
Known-variance z-statistic (difference of means)
$z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}$
$\sigma_1^2, \sigma_2^2$
the known population variances
$\bar{x}_1, \bar{x}_2$
the two sample means
$n_1, n_2$
the two sample sizes
When to use
Independent samples with known variances (or large samples). No pooling and no degrees of freedom; critical value from the normal table.
Unbiased sample variance (from summaries)
$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$
$\sum x, \sum x^2$
sum and sum of squares of the data
$n$
the sample size
When to use
To obtain $s_1^2, s_2^2$ (or $s_d^2$ on the differences) from raw data or summary statistics before running the test.

Key Definitions and Keywords — Formulating hypotheses concerning the difference of population

Definitions to memorise and the exact keywords mark schemes credit for formulating hypotheses concerning the difference of population answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Pooled variance
Examiner keyword
A single estimate of the common population variance formed as a degrees-of-freedom-weighted average of two sample variances: $s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$ . Used only when the variances are assumed equal.
Two-sample t-test
Examiner keyword
A hypothesis test for the difference of means of two independent samples from normal populations with equal but unknown variance, using $\nu = n_1 + n_2 - 2$ degrees of freedom.
Paired (matched-pairs) t-test
Examiner keyword
A test for the mean difference of matched data: each pair is reduced to a difference $d = x - y$ and a one-sample $t$ -test is run on the differences, with $\nu = n - 1$ ( $n$ = number of pairs).
Independent samples
Examiner keyword
Two samples in which the choice or value of one observation has no bearing on the other — drawn from separate groups of individuals. Required for the two-sample (not paired) test.
Conclusion in context
Examiner keyword
The final statement of a hypothesis test phrased in the language of the problem, e.g. 'there is sufficient evidence at the 5% level that the diet reduces mean weight', rather than a bare 'reject $H_0$ '.
Significance level
The probability $\alpha$ of rejecting a true $H_0$ (a Type I error). It fixes the size of the rejection region(s): all of $\alpha$ in one tail for a one-tailed test, $\alpha/2$ in each tail for a two-tailed test.

Common Mistakes and Misconceptions — Formulating hypotheses concerning the difference of population

The traps other students keep falling into on formulating hypotheses concerning the difference of population questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Running a two-sample test on paired data (or vice versa), and using the wrong degrees of freedom
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Not checking whether the two sets of readings are linked to the same subjects before choosing the test.
How to avoid it
Ask: 'are the measurements matched, item by item?' If yes, paired ( $\nu = n-1$ ); if independent, two-sample ( $\nu = n_1+n_2-2$ ). Differing sample sizes rule out a paired test.
✕Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$ in the pooled variance
9231 Paper 4 mark schemes
Why it happens
Confusing the pooled-variance denominator with a simple sample-size total.
How to avoid it
The denominator is the degrees of freedom $n_1 + n_2 - 2$ — the same number you use for the $t$ -table.
✕Putting $s_p^2$ (instead of $s_p$ ) in the test-statistic denominator
9231 Paper 4 Examiner Reports
Why it happens
Forgetting that the standard error needs the standard deviation, not the variance.
How to avoid it
After finding $s_p^2$ , take its square root: the denominator is $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ .
✕Using a one-tailed critical value for a two-tailed test (or vice versa)
9231 Paper 4 Examiner Reports
Why it happens
Not splitting the significance level between the two tails for a two-tailed alternative.
How to avoid it
Two-tailed at 5% uses the 2.5% column (e.g. $t_{10} = 2.228$ ); one-tailed at 5% uses the 5% column ( $t_{10} = 1.812$ ). Decide the tail from $H_1$ first.
✕Writing only 'reject $H_0$ ' / 'accept $H_0$ ' with no contextual conclusion
9231 Paper 4 Examiner Reports — most frequent lost mark
Why it happens
Treating the comparison as the end of the question and forgetting the final A1 is for context.
How to avoid it
Always finish with a sentence in the problem's own words and at the stated significance level — e.g. 'sufficient evidence at the 5% level that the mean yield increased'.
✕Pooling variances (or using the $t$ -test) when the population variances are known
9231 Paper 4 mark schemes
Why it happens
Applying the two-sample $t$ procedure by reflex without checking what information is given.
How to avoid it
If $\sigma^2$ values are stated (or samples are large), use the normal $z$ -test with $\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}$ — no pooling, no degrees of freedom.

Formulating hypotheses concerning the difference of population

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Forming H₀ and H₁ for a difference of means

2Deciding: paired or two-sample?

3Full two-sample pooled t-test

4Full paired t-test

5Known-variance normal (z) test

6Testing a specified difference (μ₁ − μ₂ ≠ 0)

Two-sample (pooled) t-statistic

Pooled estimate of variance

Paired-sample t-statistic

Known-variance z-statistic (difference of means)

Unbiased sample variance (from summaries)

Pooled variance

Two-sample t-test

Paired (matched-pairs) t-test

Independent samples

Conclusion in context

Significance level

✕Running a two-sample test on paired data (or vice versa), and using the wrong degrees of freedom

✕Dividing by n1+n2n_1 + n_2n1​+n2​ instead of n1+n2−2n_1 + n_2 - 2n1​+n2​−2 in the pooled variance

✕Putting sp2s_p^2sp2​ (instead of sps_psp​) in the test-statistic denominator

✕Using a one-tailed critical value for a two-tailed test (or vice versa)

✕Writing only 'reject H0H_0H0​' / 'accept H0H_0H0​' with no contextual conclusion

✕Pooling variances (or using the ttt-test) when the population variances are known

✕Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$ in the pooled variance

✕Putting $s_p^2$ (instead of $s_p$ ) in the test-statistic denominator

✕Writing only 'reject $H_0$ ' / 'accept $H_0$ ' with no contextual conclusion

✕Pooling variances (or using the $t$ -test) when the population variances are known