Why is "Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$" a common mistake in Pooled estimate of a population variance?

Why it happens: Treating the pooled formula like a simple total-over-count average, forgetting that each sample loses one degree of freedom. How to avoid it: Memorise the denominator as (n_1-1)+(n_2-1) = n_1+n_2-2. It matches the degrees of freedom of the resulting t-test. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Taking the simple average $\frac{s_1^2 + s_2^2}{2}$ when $n_1 \neq n_2$" a common mistake in Pooled estimate of a population variance?

Why it happens: Assuming both samples deserve equal weight; the simple average happens to work only when sample sizes are equal. How to avoid it: Always weight by the degrees of freedom (n_1-1) and (n_2-1). Only collapse to a simple average when you have checked n_1 = n_2. Source: 9231 Paper 4 mark schemes

Why is "Pooling standard deviations $s_1, s_2$ directly instead of variances" a common mistake in Pooled estimate of a population variance?

Why it happens: The question gives s_1 and s_2 (standard deviations) and the student plugs them in without squaring. How to avoid it: The formula uses s_1^2 and s_2^2. Square any standard deviation first; pool variances, never standard deviations. Source: 9231 Paper 4 Examiner Reports

Why is "Using the divisor-$n$ (biased) variance from raw data instead of $s^2$" a common mistake in Pooled estimate of a population variance?

Why it happens: Computing 1/n( x^2 - ( x)^2/n) — the population-style formula — for each sample. How to avoid it: For an unbiased estimate use the (n-1) divisor, or pool the sums of squares directly so the only division is by n_1+n_2-2. Source: 9231 Paper 4 Examiner Reports

Why is "Rounding $s_1^2$ and $s_2^2$ to 3 s.f. before pooling" a common mistake in Pooled estimate of a population variance?

Why it happens: Rounding each intermediate value out of habit, which then propagates into the final answer. How to avoid it: Carry full precision (ideally pool the exact sums of squares) and round ONLY the final s_p^2 to 3 s.f. Source: 9231 Paper 4 — accuracy-mark losses

Why is "Reporting an $s_p^2$ that does not lie between $s_1^2$ and $s_2^2$" a common mistake in Pooled estimate of a population variance?

Why it happens: An arithmetic slip in the weighting goes unnoticed because no sanity check is done. How to avoid it: Every pooled value MUST sit between the two sample variances. If it doesn't, re-check the weights and arithmetic before moving on. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Pooled estimate of a population variance

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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.

Pooled Estimate of a Population Variance — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 Further Probability & Statistics (2026-2027 syllabus)

When two samples are believed to come from populations with the same (unknown) variance, you can combine their information into one stronger estimate. The pooled variance $s_p^2$ is a weighted average of the two sample variances — and it is the engine that powers the two-sample $t$ -test.

At a glance

Pooled variance $s_p^2 = \dfrac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$ — a single best estimate of one common population variance $\sigma^2$ .
Weighted average: the weights are the degrees of freedom $(n_1-1)$ and $(n_2-1)$ , so $s_p^2$ ALWAYS lies between $s_1^2$ and $s_2^2$ .
Equal-variance assumption: pooling is only valid when both populations are assumed to share the same variance $\sigma^2$ .
From raw data: compute each $s^2 = \dfrac{1}{n-1}\left(\sum x^2 - \dfrac{(\sum x)^2}{n}\right)$ first, then pool.
Where it leads: $s_p^2$ feeds the two-sample $t$ -test (subtopic 15.3) with $n_1 + n_2 - 2$ degrees of freedom.

What you’ll learn

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

4.2a — Calculate a pooled estimate of a population variance from two samples, given either summary statistics or raw data.
4.2b — Understand that pooling assumes the two populations share a common variance $\sigma^2$ , and recognise the degrees of freedom $n_1 + n_2 - 2$ .
4.2c — Recognise when a question requires a pooled estimate (in preparation for a two-sample $t$ -test in 4.2/4.3) rather than two separate variance estimates.

Why pool — combining two samples into one estimate

If both samples come from populations with the same variance, merging their information gives a sharper estimate.

The situation. You have two independent samples — say the test scores of students taught by Method A and by Method B. You believe the spread of scores is the same in both populations (the methods might shift the average, but not the variability). In symbols: both populations have the same variance $\sigma^2$ , even if their means differ.

The problem. Sample A gives an unbiased estimate $s_1^2$ of $\sigma^2$ , and sample B gives another, $s_2^2$ . They will rarely be equal — random sampling makes them wobble. Which do you trust? Neither alone: a larger sample carries more information, so it deserves more weight.

The fix — pool them. The pooled estimate $s_p^2$ blends both into one number, weighting each by its degrees of freedom:

$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$

This is the single best (unbiased) estimate of the common variance $\sigma^2$ , using all the data.

Two samples, each estimating the same population variance σ², are merged into one pooled estimate s_p² that uses all the data.

Cambridge tip. Pooling is only legitimate when you may assume the two populations share a variance. The exam will usually state this ("assume equal population variances") — if it does not, you cannot pool.

Pool only when both populations are assumed to have the same variance $\sigma^2$ .
Larger samples carry more weight — the weights are the degrees of freedom.
$s_p^2$ uses all the data, so it beats either $s_1^2$ or $s_2^2$ alone.

See the full worked example for pooled estimate of a population variance →

The pooled-variance formula and its weighted-average meaning

s_p² is the total of the two 'sums of squares' divided by the total degrees of freedom.

Write the pooled estimate carefully — it is a weighted average, not an ordinary average:

$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{(n_1 - 1) + (n_2 - 1)} = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$

Read it as "total spread ÷ total degrees of freedom". The quantity $(n-1)s^2$ equals $\sum(x - \bar{x})^2$ , the sum of squared deviations for that sample. So the numerator is just the combined sum of squares from both samples, and the denominator is the combined degrees of freedom.

The weighted-average sanity check. Because $s_p^2$ is a weighted average of $s_1^2$ and $s_2^2$ , it must lie between them:

$\min(s_1^2, s_2^2) \;\leq\; s_p^2 \;\leq\; \max(s_1^2, s_2^2)$

If your answer falls outside that range, you have made an arithmetic slip — check immediately.

The pooled value always sits between the two sample variances, pulled towards the one with more degrees of freedom.

Cambridge tip. Do NOT just average $s_1^2$ and $s_2^2$ unless $n_1 = n_2$ . When the sample sizes differ, the simple average is wrong; the degrees-of-freedom weights are essential.

$(n-1)s^2 = \sum(x-\bar{x})^2$ is the sum of squares — add the two, divide by $n_1+n_2-2$ .
$s_p^2$ always lies between $s_1^2$ and $s_2^2$ — a free error check.
Only when $n_1 = n_2$ does pooling reduce to the simple average.

See the full worked example for pooled estimate of a population variance →

Computing s_p² from raw data or summary statistics

First turn raw data into each sample variance, then pool — and watch the (n−1) divisor.

Exam data arrives in two forms. Be ready for both.

Form 1 — summary statistics already given. If the question states $n_1, s_1^2, n_2, s_2^2$ directly, substitute straight into the pooled formula. (Read carefully: a question may give the biased divisor- $n$ variance or the standard deviation $s$ , not $s^2$ . Convert first.)

Form 2 — raw data or $\sum x$ , $\sum x^2$ . Compute each sample's unbiased variance using the computational formula:

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{\left(\sum x\right)^2}{n}\right)$

The bracket $\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ is exactly the sum of squares $\sum(x-\bar{x})^2$ , so you can pool without dividing by $(n-1)$ at all:

$s_p^2 = \frac{\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)_1 + \left(\sum x^2 - \frac{(\sum x)^2}{n}\right)_2}{n_1 + n_2 - 2}$

That is the slickest route: add the two sums of squares on top, divide by $n_1 + n_2 - 2$ on the bottom. One division, no rounding mid-way.

Worked sketch. Sample 1: $n_1 = 6$ , $\sum x = 48$ , $\sum x^2 = 402$ . Sum of squares $= 402 - \frac{48^2}{6} = 402 - 384 = 18$ . Sample 2: $n_2 = 5$ , $\sum y = 35$ , $\sum y^2 = 261$ . Sum of squares $= 261 - \frac{35^2}{5} = 261 - 245 = 16$ . Then $s_p^2 = \dfrac{18 + 16}{6 + 5 - 2} = \dfrac{34}{9} = 3.78$ (3 s.f.).

Cambridge tip. Keep full precision (do not round the two sample variances to 3 s.f. and then pool) — round only the final $s_p^2$ . Premature rounding is a classic accuracy-mark loss.

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ for each sample — the unbiased estimator.
Slick route: add both sums of squares, divide once by $n_1+n_2-2$ .
Round only at the end; carry full precision through the calculation.

See the full worked example for pooled estimate of a population variance →

Where s_p² is used — preview of the two-sample t-test

The pooled variance is the standard-error ingredient of the two-sample t-test, on n1+n2−2 degrees of freedom.

This subtopic is about forming $s_p^2$ correctly. Here is why the syllabus cares — a short preview of subtopic 15.3.

The two-sample $t$ -test. To test whether two population means differ ( $H_0: \mu_1 = \mu_2$ ) when the populations are normal with the same unknown variance, the test statistic is

$t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$

compared against the $t$ -distribution with $\nu = n_1 + n_2 - 2$ degrees of freedom. The pooled standard deviation $s_p = \sqrt{s_p^2}$ sits right in the denominator, and the degrees of freedom are exactly the denominator of the pooling formula. So getting $s_p^2$ right is the foundation of the whole test.

Why "same variance" is essential. The pooled formula assumes a single common $\sigma^2$ . If the two populations clearly have different variances, pooling them is meaningless — a different (Welch-type) procedure is needed, which is outside this syllabus. Cambridge questions will tell you to assume equal variances; your job is to spot that instruction and pool accordingly.

Cambridge tip. Note the chain: pooled variance $s_p^2 \Rightarrow$ pooled standard deviation $s_p = \sqrt{s_p^2} \Rightarrow$ standard error $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ . In 15.3 you will need $s_p$ (the square root), so don't stop at $s_p^2$ when the question continues into a test.

$s_p^2$ is the variance ingredient of the two-sample $t$ -test statistic.
Degrees of freedom for that test are $n_1 + n_2 - 2$ — the pooling denominator.
Pooling requires equal population variances; unequal variances need a different method (out of syllabus).

See the full worked example for pooled estimate of a population variance →

Quick recap

$s_p^2 = \dfrac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$ — the weighted average of the two sample variances.
Pool only when both populations are assumed to share a common variance $\sigma^2$ .
$s_p^2$ always lies between $s_1^2$ and $s_2^2$ — use this to check your arithmetic.
From raw data: each $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ , or add the two sums of squares and divide once.
Carry full precision; round only the final $s_p^2$ to 3 s.f.
The pooled estimate feeds the two-sample $t$ -test on $n_1 + n_2 - 2$ degrees of freedom.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$s_p^2 = \dfrac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$ — the pooled estimate of a common variance.
$(n-1)s^2 = \sum x^2 - \frac{(\sum x)^2}{n} = \sum(x-\bar{x})^2$ — the sum of squares.
$s_p^2$ lies between $s_1^2$ and $s_2^2$ (weighted-average check).
Degrees of freedom $= n_1 + n_2 - 2$ for the two-sample $t$ -test that uses $s_p^2$ .
Equal-variance assumption is required before pooling.

How it’s examined

Subtopic 15.2 almost always appears as the FIRST step of a Paper 4 two-sample $t$ -test question (worth 1-3 of the question's marks). Candidates are given two samples (as summary statistics or as raw data) and must form $s_p^2$ correctly before testing the difference of means in 15.3. The standard mark scheme awards M1 for a correct pooled-variance expression and A1 for the value (3 s.f.). Marks are lost by (i) using $n_1 + n_2$ instead of $n_1 + n_2 - 2$ , (ii) simple-averaging the two variances when $n_1 \neq n_2$ , and (iii) confusing $s^2$ with the divisor- $n$ variance from raw data. Because the rest of the question depends on $s_p^2$ , an early slip cascades — so the weighted-average between-check is high value.

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

Take this whole topic with you

Step-by-step worked examples — Pooled estimate of a population variance

Step-by-step solutions to past-paper-style questions on pooled estimate of a population variance, written exactly the way a tutor would explain them at the board.

Question type:

1Pooled variance from two sample variances
Getting startedDirect calculation• pooled-variance, summary-statistics
Question
Two samples are drawn from populations assumed to have the same variance. Sample 1 has $n_1 = 8$ and variance $s_1^2 = 12.4$ ; sample 2 has $n_2 = 10$ and variance $s_2^2 = 9.6$ . Find the pooled estimate of the population variance.
Step-by-step solution
1. Step 1
  State the pooled formula. Weights are the degrees of freedom $(n_1-1)$ and $(n_2-1)$ .
  $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$
2. Step 2
  Substitute $n_1=8$ , $s_1^2=12.4$ , $n_2=10$ , $s_2^2=9.6$ .
  $s_p^2 = \frac{(7)(12.4) + (9)(9.6)}{8 + 10 - 2} = \frac{86.8 + 86.4}{16}$
3. Step 3
  Evaluate.
  $s_p^2 = \frac{173.2}{16} = 10.825$
Answer
$s_p^2 = 10.825 \approx 10.8$ (3 s.f.).
Examiner tip
Check: $10.8$ lies between $s_2^2 = 9.6$ and $s_1^2 = 12.4$ , as a weighted average must. Note the denominator is $n_1+n_2-2 = 16$ , NOT $18$ .
2Equal sample sizes — pooling reduces to a simple average
Getting startedDirect calculation• pooled-variance, weighted-average
Question
Two samples each of size $12$ come from populations with a common variance. Their sample variances are $s_1^2 = 15.2$ and $s_2^2 = 18.4$ . Find the pooled estimate $s_p^2$ .
Step-by-step solution
1. Step 1
  Apply the formula with $n_1 = n_2 = 12$ , so each weight is $11$ .
  $s_p^2 = \frac{(11)(15.2) + (11)(18.4)}{12 + 12 - 2}$
2. Step 2
  Simplify. With equal weights the $11$ cancels, leaving the simple average.
  $s_p^2 = \frac{11(15.2 + 18.4)}{22} = \frac{15.2 + 18.4}{2} = \frac{33.6}{2}$
3. Step 3
  Evaluate.
  $s_p^2 = 16.8$
Answer
$s_p^2 = 16.8$ .
Examiner tip
Only when $n_1 = n_2$ does the pooled estimate equal the simple mean of the two variances. When sample sizes differ, you MUST use the degrees-of-freedom weights.
3Pooled variance from raw data (sums)
Building confidenceMulti-step problem• pooled-variance, raw-data, sum-of-squares
Question
Sample A: $n = 6$ , $\sum x = 48$ , $\sum x^2 = 402$ . Sample B: $n = 5$ , $\sum y = 35$ , $\sum y^2 = 261$ . Assuming a common population variance, find the pooled estimate.
Step-by-step solution
1. Step 1
  Sum of squares for A using $\sum x^2 - \frac{(\sum x)^2}{n}$ .
  $\text{SS}_A = 402 - \frac{48^2}{6} = 402 - 384 = 18$
2. Step 2
  Sum of squares for B the same way.
  $\text{SS}_B = 261 - \frac{35^2}{5} = 261 - 245 = 16$
3. Step 3
  Pool directly: add the sums of squares, divide by $n_1 + n_2 - 2 = 9$ .
  $s_p^2 = \frac{\text{SS}_A + \text{SS}_B}{n_1 + n_2 - 2} = \frac{18 + 16}{6 + 5 - 2} = \frac{34}{9}$
4. Step 4
  Evaluate.
  $s_p^2 = 3.777\ldots$
Answer
$s_p^2 = \dfrac{34}{9} = 3.78$ (3 s.f.).
Examiner tip
Adding the two sums of squares and dividing ONCE avoids rounding errors. You never had to compute $s_1^2$ or $s_2^2$ separately — the cleanest route in the exam.
4Pooled variance when one sample gives a standard deviation
Building confidenceMulti-step problem• pooled-variance, standard-deviation, conversion
Question
Sample 1: $n_1 = 15$ , sample variance $s_1^2 = 23.6$ . Sample 2: $n_2 = 10$ , sample standard deviation $s_2 = 4.347$ (so $s_2^2 = 18.9$ to 3 s.f.). Find the pooled estimate $s_p^2$ .
Step-by-step solution
1. Step 1
  Convert the standard deviation to a variance before pooling. $s_2^2 = 4.347^2 = 18.9$ .
  $s_2^2 = (4.347)^2 = 18.896\ldots \approx 18.9$
2. Step 2
  Apply the pooled formula.
  $s_p^2 = \frac{(14)(23.6) + (9)(18.9)}{15 + 10 - 2}$
3. Step 3
  Evaluate the numerator and denominator.
  $s_p^2 = \frac{330.4 + 170.1}{23} = \frac{500.5}{23}$
4. Step 4
  Compute.
  $s_p^2 = 21.76\ldots$
Answer
$s_p^2 = 21.8$ (3 s.f.).
Examiner tip
The check works: $21.8$ lies between $18.9$ and $23.6$ . The most common slip is pooling $s_2 = 4.347$ directly without squaring it first — always pool variances, never standard deviations.
5Pooled variance from two raw data sets
StretchMulti-step problem• pooled-variance, raw-data, full-calculation
Question
The masses (g) of two batches of components are recorded. Batch X: $4, 6, 7, 9, 9$ . Batch Y: $5, 8, 8, 11$ . Assuming the two populations share a variance, find the pooled estimate $s_p^2$ .
Step-by-step solution
1. Step 1
  Batch X: $n_X = 5$ , $\sum x = 4+6+7+9+9 = 35$ , $\sum x^2 = 16+36+49+81+81 = 263$ .
  $\text{SS}_X = 263 - \frac{35^2}{5} = 263 - 245 = 18$
2. Step 2
  Batch Y: $n_Y = 4$ , $\sum y = 5+8+8+11 = 32$ , $\sum y^2 = 25+64+64+121 = 274$ .
  $\text{SS}_Y = 274 - \frac{32^2}{4} = 274 - 256 = 18$
3. Step 3
  Pool: add the sums of squares, divide by $n_X + n_Y - 2 = 7$ .
  $s_p^2 = \frac{18 + 18}{5 + 4 - 2} = \frac{36}{7}$
4. Step 4
  Evaluate.
  $s_p^2 = 5.142\ldots$
Answer
$s_p^2 = \dfrac{36}{7} = 5.14$ (3 s.f.).
Examiner tip
Lay out $\sum x$ and $\sum x^2$ for each batch in a clear table — this is where careless arithmetic loses marks. Both sums of squares being $18$ here is a coincidence, not a rule.
6Pool, then build the two-sample standard error
StretchMulti-step problem• pooled-variance, standard-error, t-test-preview
Question
Two samples have $n_1 = 10$ , $s_1^2 = 5.2$ and $n_2 = 14$ , $s_2^2 = 6.8$ , from populations with a common variance. (a) Find the pooled estimate $s_p^2$ . (b) Hence find the standard error $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ used in a two-sample $t$ -test, and state its degrees of freedom.
Step-by-step solution
1. Step 1
  (a) Pool the variances.
  $s_p^2 = \frac{(9)(5.2) + (13)(6.8)}{10 + 14 - 2} = \frac{46.8 + 88.4}{22} = \frac{135.2}{22} = 6.1454\ldots$
2. Step 2
  (b) Pooled standard deviation — take the square root of $s_p^2$ (keep full precision).
  $s_p = \sqrt{6.1454\ldots} = 2.4790\ldots$
3. Step 3
  Standard error of the difference of means.
  $s_p\sqrt{\tfrac{1}{10}+\tfrac{1}{14}} = 2.4790\sqrt{0.1+0.07142\ldots} = 2.4790\sqrt{0.17142\ldots}$
4. Step 4
  Evaluate, and state degrees of freedom $= n_1 + n_2 - 2$ .
  $s_p\sqrt{\tfrac{1}{10}+\tfrac{1}{14}} = 2.4790 \times 0.41403 = 1.0264\ldots, \quad \nu = 22$
Answer
(a) $s_p^2 = 6.15$ (3 s.f.). (b) Standard error $= 1.03$ (3 s.f.), on $\nu = 22$ degrees of freedom.
Examiner tip
The pooled standard deviation $s_p = \sqrt{s_p^2}$ (not $s_p^2$ itself) goes into the standard error. The degrees of freedom $n_1 + n_2 - 2 = 22$ are exactly the pooling denominator — this is the link into subtopic 15.3.

Model Answers — Pooled estimate of a population variance

High-scoring sample answers for pooled estimate of a population variance on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Two random samples are taken from normal populations with a common variance. Sample A: $n = 9$ , variance $6.25$ . Sample B: $n = 6$ , variance $4.84$ . Calculate the pooled estimate of the common population variance.
Model answer
Pooled-variance formula: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$

Substitute $n_1 = 9$ , $s_1^2 = 6.25$ , $n_2 = 6$ , $s_2^2 = 4.84$ : $s_p^2 = \frac{(8)(6.25) + (5)(4.84)}{9 + 6 - 2} = \frac{50 + 24.2}{13} = \frac{74.2}{13}$

Evaluate: $s_p^2 = 5.7077\ldots = 5.71 \;(\text{3 s.f.})$
Why this scores
M1 for a correct pooled expression with denominator $n_1+n_2-2 = 13$ , A1 for $5.71$ . Check: $5.71$ lies between $4.84$ and $6.25$ . A common error is dividing by $15$ instead of $13$ .
Question 2
3 marks
In a study, sample 1 has $n_1 = 20$ with variance $8.1$ and sample 2 has $n_2 = 16$ with variance $9.7$ . Assuming equal population variances, find the pooled estimate $s_p^2$ , giving your answer to 3 significant figures.
Model answer
Pooled estimate: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2} = \frac{(19)(8.1) + (15)(9.7)}{20 + 16 - 2}$

Compute the numerator and denominator: $s_p^2 = \frac{153.9 + 145.5}{34} = \frac{299.4}{34}$

Evaluate: $s_p^2 = 8.8058\ldots = 8.81 \;(\text{3 s.f.})$
Why this scores
M1 for the pooled expression (correct weights $19$ and $15$ ), A1 for $8.81$ . The value lies between $8.1$ and $9.7$ — the weighted average leans towards $8.1$ because sample 1 carries more degrees of freedom.
Question 3
4 marks
Two samples are summarised below, drawn from populations assumed to have the same variance. Sample A: $n = 7$ , $\sum x = 84$ , $\sum x^2 = 1036$ . Sample B: $n = 8$ , $\sum y = 96$ , $\sum y^2 = 1180$ . Calculate the pooled estimate of the population variance.
Model answer
Sum of squares for each sample using $\sum x^2 - \frac{(\sum x)^2}{n}$ : $\text{SS}_A = 1036 - \frac{84^2}{7} = 1036 - 1008 = 28$ $\text{SS}_B = 1180 - \frac{96^2}{8} = 1180 - 1152 = 28$

Pool — add the sums of squares, divide by $n_1 + n_2 - 2$ : $s_p^2 = \frac{\text{SS}_A + \text{SS}_B}{n_1 + n_2 - 2} = \frac{28 + 28}{7 + 8 - 2} = \frac{56}{13}$

Evaluate: $s_p^2 = 4.3076\ldots = 4.31 \;(\text{3 s.f.})$
Why this scores
M1 for at least one correct sum of squares, M1 for pooling with denominator $13$ , A1 for $\frac{56}{13}$ , A1 for $4.31$ . Working with the combined sum of squares ( $56$ ) over $13$ avoids any mid-calculation rounding.
Question 4
4 marks
A researcher records two samples from normal populations with a common variance. Sample 1: $n_1 = 10$ , sample standard deviation $s_1 = 3.2$ . Sample 2: $n_2 = 13$ , sample standard deviation $s_2 = 4.1$ . Find the pooled estimate of the population variance.
Model answer
Convert standard deviations to variances (we pool variances, not standard deviations): $s_1^2 = 3.2^2 = 10.24, \qquad s_2^2 = 4.1^2 = 16.81$

Pooled estimate: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2} = \frac{(9)(10.24) + (12)(16.81)}{10 + 13 - 2}$

Compute: $s_p^2 = \frac{92.16 + 201.72}{21} = \frac{293.88}{21} = 13.994\ldots$

Therefore: $s_p^2 = 14.0 \;(\text{3 s.f.})$
Why this scores
B1 for squaring both standard deviations, M1 for the pooled expression (denominator $21$ ), A1 for $13.99\ldots$ , A1 for $14.0$ . Pooling $s_1 = 3.2$ and $s_2 = 4.1$ directly (forgetting to square) is the headline error here.
Question 5
5 marks
Two independent samples are taken from populations assumed to have the same variance $\sigma^2$ . Sample X: $n_X = 10$ , $\sum x = 150$ , $\sum x^2 = 2284$ . Sample Y: $n_Y = 12$ , $\sum y = 192$ , $\sum y^2 = 3120$ . (a) Show that the pooled estimate of $\sigma^2$ is $4.1$ . (b) State the pooled standard deviation $s_p$ , to 3 significant figures.
Model answer
(a) Sum of squares for each sample. $\text{SS}_X = \sum x^2 - \frac{(\sum x)^2}{n_X} = 2284 - \frac{150^2}{10} = 2284 - 2250 = 34$ $\text{SS}_Y = \sum y^2 - \frac{(\sum y)^2}{n_Y} = 3120 - \frac{192^2}{12} = 3120 - 3072 = 48$

Pool with denominator $n_X + n_Y - 2 = 20$ : $s_p^2 = \frac{\text{SS}_X + \text{SS}_Y}{n_X + n_Y - 2} = \frac{34 + 48}{20} = \frac{82}{20} = 4.1 \quad \text{(as required)}$

(b) Pooled standard deviation: $s_p = \sqrt{4.1} = 2.0248\ldots = 2.02 \;(\text{3 s.f.})$
Why this scores
Part (a): M1 for both sums of squares, M1 for pooling over $20$ , A1 for the shown value $4.1$ (this is a 'Show that', so the $34$ , $48$ and $20$ must all be visible). Part (b): B1 for $s_p = 2.02$ . Quoting only $4.1$ without the two SS values scores no method marks on a 'Show that'.
Question 6
6 marks
Two samples from normal populations with a common variance give $n_1 = 10$ , $\bar{x}_1 = 52.0$ , $s_1^2 = 4.4$ and $n_2 = 12$ , $\bar{x}_2 = 48.5$ , $s_2^2 = 4.0$ . (a) Find the pooled estimate $s_p^2$ . (b) Hence find the standard error of $\bar{x}_1 - \bar{x}_2$ , i.e. $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ , and state the degrees of freedom of the associated $t$ -statistic.
Model answer
(a) Pool the variances. $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2} = \frac{(9)(4.4) + (11)(4.0)}{10 + 12 - 2}$ $s_p^2 = \frac{39.6 + 44.0}{20} = \frac{83.6}{20} = 4.18$

(b) Pooled standard deviation (keep full precision): $s_p = \sqrt{4.18} = 2.04450\ldots$

Standard error of the difference of sample means: $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} = 2.04450\sqrt{\frac{1}{10}+\frac{1}{12}} = 2.04450\sqrt{0.18333\ldots}$ $= 2.04450 \times 0.428174\ldots = 0.87539\ldots = 0.875 \;(\text{3 s.f.})$

Degrees of freedom: $\nu = n_1 + n_2 - 2 = 20$
Why this scores
Part (a): M1 for the pooled expression, A1 for $4.18$ . Part (b): M1 for $s_p = \sqrt{s_p^2}$ (square root, not $s_p^2$ ), M1 for the $\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$ factor, A1 for $0.875$ ; B1 for $\nu = 20$ . The means $\bar{x}_1, \bar{x}_2$ are not needed until the test itself (15.3) — a distractor here.

Key Formulae — Pooled estimate of a population variance

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

Pooled estimate of a common variance
$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$
$n_1, n_2$
the two sample sizes
$s_1^2, s_2^2$
the two (unbiased) sample variances
$s_p^2$
the pooled estimate of the common population variance $\sigma^2$
When to use
Two samples assumed to come from populations with the SAME variance — combine into one estimate before a two-sample $t$ -test.
Unbiased sample variance (computational form)
$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$
$n$
sample size
$\sum x, \sum x^2$
the sum and sum of squares of the data
When to use
Compute each sample's unbiased variance from raw data or from $\sum x$ and $\sum x^2$ , before pooling.
Sum of squared deviations
$\text{SS} = (n-1)s^2 = \sum x^2 - \frac{(\sum x)^2}{n} = \sum (x - \bar{x})^2$
$\text{SS}$
sum of squared deviations about the mean
$\bar{x}$
the sample mean
When to use
The 'slick' pooling route: add the two SS values on top, divide by $n_1+n_2-2$ — one division, no mid-way rounding.
Pooled variance from sums of squares
$s_p^2 = \frac{\text{SS}_1 + \text{SS}_2}{n_1 + n_2 - 2}$
$\text{SS}_1, \text{SS}_2$
the two sums of squared deviations
When to use
When you already have each $\sum x^2 - \frac{(\sum x)^2}{n}$ — pool directly without first forming $s_1^2, s_2^2$ .
Two-sample t-statistic (uses s_p)
$t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}, \quad \nu = n_1 + n_2 - 2$
$\bar{x}_1, \bar{x}_2$
the two sample means
$s_p$
the pooled standard deviation $\sqrt{s_p^2}$
$\nu$
degrees of freedom for the $t$ -distribution
When to use
Preview of subtopic 15.3 — the test that USES the pooled estimate. Note $s_p$ (square root) appears, on $n_1+n_2-2$ df.

Key Definitions and Keywords — Pooled estimate of a population variance

Definitions to memorise and the exact keywords mark schemes credit for pooled estimate of a population variance answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Pooled estimate of variance
Examiner keyword
A single estimate of a common population variance $\sigma^2$ , formed as a weighted average of two sample variances using their degrees of freedom as weights: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ .
Common (equal) variance assumption
Examiner keyword
The assumption that two populations share the same variance $\sigma^2$ (even if their means differ). This assumption is required before pooling and before a two-sample $t$ -test of means.
Unbiased sample variance $s^2$
Examiner keyword
$s^2 = \frac{1}{n-1}\sum(x-\bar{x})^2$ , the estimator of population variance using the divisor $(n-1)$ . It is the variance that must be used when pooling (NOT the divisor- $n$ version).
Degrees of freedom
Examiner keyword
The effective number of independent pieces of information. Each sample contributes $n-1$ ; the pooled estimate and its $t$ -test have $n_1 + n_2 - 2$ degrees of freedom.
Sum of squares (SS)
$\text{SS} = \sum(x-\bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n} = (n-1)s^2$ , the total squared deviation about the mean. Pooling adds the two SS values and divides by $n_1+n_2-2$ .
Weighted average
An average in which each value contributes in proportion to a weight. Here the weights are the degrees of freedom, so $s_p^2$ always lies between $s_1^2$ and $s_2^2$ .

Common Mistakes and Misconceptions — Pooled estimate of a population variance

The traps other students keep falling into on pooled estimate of a population variance questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Treating the pooled formula like a simple total-over-count average, forgetting that each sample loses one degree of freedom.
How to avoid it
Memorise the denominator as $(n_1-1)+(n_2-1) = n_1+n_2-2$ . It matches the degrees of freedom of the resulting $t$ -test.
✕Taking the simple average $\frac{s_1^2 + s_2^2}{2}$ when $n_1 \neq n_2$
9231 Paper 4 mark schemes
Why it happens
Assuming both samples deserve equal weight; the simple average happens to work only when sample sizes are equal.
How to avoid it
Always weight by the degrees of freedom $(n_1-1)$ and $(n_2-1)$ . Only collapse to a simple average when you have checked $n_1 = n_2$ .
✕Pooling standard deviations $s_1, s_2$ directly instead of variances
9231 Paper 4 Examiner Reports
Why it happens
The question gives $s_1$ and $s_2$ (standard deviations) and the student plugs them in without squaring.
How to avoid it
The formula uses $s_1^2$ and $s_2^2$ . Square any standard deviation first; pool variances, never standard deviations.
✕Using the divisor- $n$ (biased) variance from raw data instead of $s^2$
9231 Paper 4 Examiner Reports
Why it happens
Computing $\frac{1}{n}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ — the population-style formula — for each sample.
How to avoid it
For an unbiased estimate use the $(n-1)$ divisor, or pool the sums of squares directly so the only division is by $n_1+n_2-2$ .
✕Rounding $s_1^2$ and $s_2^2$ to 3 s.f. before pooling
9231 Paper 4 — accuracy-mark losses
Why it happens
Rounding each intermediate value out of habit, which then propagates into the final answer.
How to avoid it
Carry full precision (ideally pool the exact sums of squares) and round ONLY the final $s_p^2$ to 3 s.f.
✕Reporting an $s_p^2$ that does not lie between $s_1^2$ and $s_2^2$
9231 Paper 4 Examiner Reports
Why it happens
An arithmetic slip in the weighting goes unnoticed because no sanity check is done.
How to avoid it
Every pooled value MUST sit between the two sample variances. If it doesn't, re-check the weights and arithmetic before moving on.

Pooled estimate of a population variance

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Pooled variance from two sample variances

2Equal sample sizes — pooling reduces to a simple average

3Pooled variance from raw data (sums)

4Pooled variance when one sample gives a standard deviation

5Pooled variance from two raw data sets

6Pool, then build the two-sample standard error

Pooled estimate of a common variance

Unbiased sample variance (computational form)

Sum of squared deviations

Pooled variance from sums of squares

Two-sample t-statistic (uses s_p)

Pooled estimate of variance

Common (equal) variance assumption

Unbiased sample variance s2s^2s2

Degrees of freedom

Sum of squares (SS)

Weighted average

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

✕Taking the simple average s12+s222\frac{s_1^2 + s_2^2}{2}2s12​+s22​​ when n1≠n2n_1 \neq n_2n1​=n2​

✕Pooling standard deviations s1,s2s_1, s_2s1​,s2​ directly instead of variances

✕Using the divisor-nnn (biased) variance from raw data instead of s2s^2s2

✕Rounding s12s_1^2s12​ and s22s_2^2s22​ to 3 s.f. before pooling

✕Reporting an sp2s_p^2sp2​ that does not lie between s12s_1^2s12​ and s22s_2^2s22​

Unbiased sample variance $s^2$

✕Dividing by $n_1 + n_2$ instead of $n_1 + n_2 - 2$

✕Taking the simple average $\frac{s_1^2 + s_2^2}{2}$ when $n_1 \neq n_2$

✕Pooling standard deviations $s_1, s_2$ directly instead of variances

✕Using the divisor- $n$ (biased) variance from raw data instead of $s^2$

✕Rounding $s_1^2$ and $s_2^2$ to 3 s.f. before pooling

✕Reporting an $s_p^2$ that does not lie between $s_1^2$ and $s_2^2$