Why is "Dividing the corrected sum of squares by $n$ instead of $n-1$" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Confusing the population-variance formula (÷n) with the unbiased estimate (÷n-1), or pressing the \sigma_n calculator key. How to avoid it: For a test you ALWAYS need the unbiased estimate: s^2 = 1/n-1( x^2 - ( x)^2/n). On a calculator use the \sigma_n-1 / s_x button. Source: 9231 Paper 4 Examiner Reports — recurring comment

Why is "Using the $5\%$ table column for a two-tailed test (should be $2.5\%$)" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Reading the significance level (5\%) straight into the table without splitting it between the two tails. How to avoid it: Two-tailed at 5\% means 2.5\% in each tail — look up the 2.5\% column. Say 'two tails, halve the level' before reading MF19. Source: 9231 Paper 4 mark schemes

Why is "Concluding only 'reject $H_0$' with no statement in context" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Treating the test as a pure calculation and stopping at the comparison. How to avoid it: Always finish with a sentence naming the variable and direction: 'There is (in)sufficient evidence at the X\% level that [the mean …] has [increased/changed/…].' Source: 9231 Paper 4 Examiner Reports

Why is "Using a $z$-test when the variance is estimated from a small sample" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Defaulting to the normal distribution; not noticing that ^2 is unknown. How to avoid it: Check the conditions: unknown variance + small sample + normal population ⇒ t-test with = n-1. Only use z if is given. Source: 9231 Paper 4 Examiner Reports

Why is "Comparing a negative $t$ with a positive critical value in a lower-tail test" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Forgetting that for H_1: t_crit. How to avoid it: For a lower-tail test reject when t < -(critical value). Sketch the curve and shade the lower tail to fix the sign. Source: 9231 Paper 4 — one-tailed test questions

Why is "Rounding $s$, $s^2$ or the standard error too early, distorting $t$" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Writing down a 3 s.f. value and re-entering it, accumulating rounding error before the final division. How to avoid it: Carry full calculator precision through s, s/√(n) and t; only round the final t (to 3 s.f.) and never round before comparing with a near-boundary critical value. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Hypothesis Testing for a Population Mean using the t-Test

Q: Why is "Rounding $s$, $s^2$ or the standard error too early, distorting $t$" a common mistake in Hypothesis Testing for a Population Mean using the t-Test?

Why it happens: Writing down a 3 s.f. value and re-entering it, accumulating rounding error before the final division. How to avoid it: Carry full calculator precision through s, s/√(n) and t; only round the final t (to 3 s.f.) and never round before comparing with a near-boundary critical value. Source: 9231 Paper 4 Examiner Reports

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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 Testing for a Population Mean using the t-Test — Cambridge International AS & A Level Further Mathematics 9231 Paper 4 (2026-2027 syllabus)

When the sample is small and the population variance is unknown, you cannot use a $z$ -test. Instead you estimate the variance from the data, form $t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}$ , and compare it with a critical value from the $t$ -distribution on $\nu = n-1$ degrees of freedom. Master the unbiased variance estimate and the five-step test structure.

At a glance

Use the $t$ -test when the population is normal, the sample is small, and $\sigma^2$ is unknown (estimated from the data).
Unbiased variance estimate $s^2 = \dfrac{1}{n-1}\left(\sum x^2 - \dfrac{(\sum x)^2}{n}\right)$ — the divisor is $n-1$ , not $n$ .
Test statistic $t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}$ with degrees of freedom $\nu = n-1$ .
Critical values come from the $t$ -distribution tables in MF19 — state the df and whether the test is one- or two-tailed.
Always conclude in context — name the variable and the direction (e.g. 'evidence the mean lifetime has increased'), never just 'reject $H_0$ '.

What you’ll learn

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

4.2a — Formulate hypotheses and carry out a test for the mean of a normal distribution with unknown variance using a small sample, applying the $t$ -distribution.
4.2b — Calculate the unbiased estimate $s^2$ of the population variance from sample data and use it to form the test statistic $t = \frac{\bar{x}-\mu}{s/\sqrt{n}}$ on $\nu = n-1$ degrees of freedom.
4.2c — Identify whether a one-tailed or two-tailed test is required, read the correct critical value from the $t$ -tables, and state a conclusion in the context of the question.

When to reach for the t-test (not the z-test)

Small sample + normal population + unknown variance = t-test. Otherwise reconsider.

The key decision. Before testing a mean you must choose the right distribution. Three conditions point to the $t$ -test:

The population is normally distributed (stated, or reasonable to assume).
The population variance $\sigma^2$ is unknown — you only have sample data.
The sample is small (typically $n < 30$ ), so you cannot lean on the Central Limit Theorem to rescue a $z$ -test.

When $\sigma^2$ is unknown you must estimate it from the sample. That estimate, $s^2$ , is itself a random quantity, so it adds extra uncertainty. The $t$ -distribution accounts for this: it looks like the standard normal but with fatter tails, and it depends on a single parameter, the degrees of freedom $\nu = n-1$ .

Why $\nu = n-1$ ? Computing $s^2$ uses the sample mean $\bar{x}$ , which "uses up" one piece of information from the $n$ data values. Only $n-1$ values are then free to vary — hence $n-1$ degrees of freedom.

As $\nu$ grows, the $t$ -distribution gets closer to the standard normal $N(0,1)$ .

The t-distribution (red) has more probability in the tails than the standard normal (blue), so its critical values are larger — a small sample demands stronger evidence.

Cambridge tip. If a question gives you the population variance, or says "it is known that the standard deviation is …", that is a $z$ -test, not a $t$ -test. The $t$ -test is specifically for an unknown variance estimated from the data.

Use $t$ when: normal population, small sample, $\sigma^2$ unknown.
The $t$ -distribution has fatter tails than $N(0,1)$ to absorb the extra uncertainty in $s^2$ .
Degrees of freedom $\nu = n-1$ because $\bar{x}$ uses up one value.

See the full worked example for hypothesis testing for a population mean using the t-test →

Estimating the variance — the unbiased s²

Divide the corrected sum of squares by n − 1, never by n.

Because $\sigma^2$ is unknown, you estimate it. The unbiased estimate of the population variance is:

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

The bracket $\sum x^2 - \dfrac{(\sum x)^2}{n}$ is called the corrected sum of squares, $S_{xx}$ . Dividing by $n-1$ (not $n$ ) is what makes the estimate unbiased — on average it equals the true $\sigma^2$ .

Two common data formats:

Raw totals given: you are handed $n$ , $\sum x$ and $\sum x^2$ . Plug straight in.
Coded data: sometimes the data are coded as $y = x - a$ for some constant $a$ . Variance is unaffected by the shift, so $s^2$ computed on $y$ equals $s^2$ for $x$ ; only the mean shifts back by $+a$ .

Worked mini-calculation. Suppose $n = 8$ , $\sum x = 612$ , $\sum x^2 = 47\,000$ . Then:

$\bar{x} = \frac{612}{8} = 76.5$ $s^2 = \frac{1}{7}\left(47000 - \frac{612^2}{8}\right) = \frac{1}{7}\left(47000 - 46818\right) = \frac{182}{7} = 26.0$

So $s = \sqrt{26.0} = 5.10$ (3 s.f.).

The corrected sum of squares divided by n − 1 gives the unbiased variance estimate s², the building block of the test statistic.

Cambridge tip. Your calculator's $\sigma_{n-1}$ (or $s_x$ ) button gives exactly this $s$ . Make sure you read off the $n-1$ version, not the $\sigma_n$ population value — that error silently inflates your $t$ .

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ — divisor is $n-1$ .
The corrected sum of squares $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}$ .
On a calculator use $\sigma_{n-1}$ / $s_x$ , never $\sigma_n$ .

See the full worked example for hypothesis testing for a population mean using the t-test →

The five-step test structure

Hypotheses → test statistic → critical value → compare → conclude in context.

Every $t$ -test for a mean follows the same five steps. Lay them out clearly — examiners award method marks for the structure even before the final decision.

Step 1 — State the hypotheses. Let $\mu$ be the population mean. Always: $H_0: \mu = \mu_0$ The alternative depends on the claim being tested:

"has changed / is different" → two-tailed $H_1: \mu \neq \mu_0$
"has increased / is greater" → one-tailed $H_1: \mu > \mu_0$
"has decreased / is less" → one-tailed $H_1: \mu < \mu_0$

Step 2 — Compute the test statistic. $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}, \qquad \nu = n-1$ (quote to 3 s.f.; keep full precision in working).

Step 3 — Find the critical value from the $t$ -tables in MF19. You need the degrees of freedom $\nu = n-1$ and the significance level in the correct tail:

One-tailed at $5\%$ → use the $5\%$ column.
Two-tailed at $5\%$ → split into $2.5\%$ each tail → use the $2.5\%$ column.

Step 4 — Compare. Reject $H_0$ if the test statistic falls in the critical (rejection) region:

One-tailed upper ( $H_1:\mu>\mu_0$ ): reject if $t >$ critical value.
One-tailed lower ( $H_1:\mu<\mu_0$ ): reject if $t < -(\text{critical value})$ .
Two-tailed: reject if $|t| >$ critical value.

Step 5 — Conclude IN CONTEXT. Translate the decision back into the wording of the problem. Not "reject $H_0$ " but, for example, "there is sufficient evidence at the $5\%$ level that the mean lifetime has increased."

For a one-tailed upper test at 5%, the rejection region is the upper 5% tail. If the test statistic exceeds the critical value, reject H₀ and conclude in context.

Cambridge tip. For a two-tailed test it is the most common slip to look up the $5\%$ column instead of $2.5\%$ . "Two-tailed at $5\%$ means $2.5\%$ in each tail" — say it out loud before reading the table.

Five steps: hypotheses, statistic, critical value, compare, conclude in context.
Two-tailed at $5\%$ uses the $2.5\%$ column (split the level between two tails).
Method marks come from the structure — lay all five steps out clearly.

See the full worked example for hypothesis testing for a population mean using the t-test →

One-tailed vs two-tailed — reading the question

The wording of the claim fixes H₁ and which column of the tables you use.

Choosing the correct tail is a comprehension skill as much as a statistical one. The signal is in the verb of the claim.

Two-tailed — the claim is about a change in either direction:

"Has the mean changed?"
"Is the mean different from $\mu_0$ ?"
"Test whether $\mu \neq \mu_0$ ."

$\Rightarrow H_1: \mu \neq \mu_0$ , and at the $5\%$ level you use the $2.5\%$ critical value (with $\pm$ ).

One-tailed — the claim points in a specific direction:

"Has the mean increased?" $\Rightarrow H_1: \mu > \mu_0$
"Is the new method faster (smaller time)?" $\Rightarrow H_1: \mu < \mu_0$

$\Rightarrow$ use the $5\%$ critical value in that single tail.

Watch the direction of the rejection region. For $H_1:\mu<\mu_0$ the test statistic will be negative, so you compare $t$ with the negative critical value: reject if $t < -t_{\text{crit}}$ .

Significance levels other than 5%. The same logic scales: a two-tailed test at $1\%$ uses the $0.5\%$ column; a one-tailed test at $1\%$ uses the $1\%$ column. MF19 tabulates $10\%, 5\%, 2.5\%, 1\%, 0.5\%$ (and more) — pick the one matching your tail probability.

Cambridge tip. Underline the key verb in the question ("increased", "changed", "differs") before you write $H_1$ . A correct $H_1$ guarantees you reach for the right column; a wrong tail loses both the critical-value mark and the conclusion mark.

'Changed / different' ⇒ two-tailed ( $H_1:\mu\neq\mu_0$ , use $2.5\%$ at the $5\%$ level).
'Increased / greater' ⇒ one-tailed upper; 'decreased / less' ⇒ one-tailed lower.
For a lower-tail test compare with the negative critical value.

See the full worked example for hypothesis testing for a population mean using the t-test →

Quick recap

Use the $t$ -test for a normal population, small sample, unknown $\sigma^2$ .
Unbiased estimate $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ — divide by $n-1$ .
Test statistic $t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}$ on $\nu = n-1$ degrees of freedom.
Five steps: hypotheses → statistic → critical value (MF19) → compare → conclude in context.
Two-tailed at $5\%$ ⇒ use the $2.5\%$ critical value; one-tailed ⇒ the $5\%$ value.
Every conclusion must name the variable and direction — never just 'reject $H_0$ '.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Unbiased variance $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ .
Test statistic $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ , $\;\nu = n-1$ .
Conditions: normal population, small sample, $\sigma^2$ unknown.
Tail rule: two-tailed at $5\%$ → $2.5\%$ column; one-tailed at $5\%$ → $5\%$ column.
Conclusion template: 'There is (in)sufficient evidence at the $X\%$ level that [variable] has [direction].'

How it’s examined

Topic 4.2 appears regularly in Paper 4 (Further Probability & Statistics), usually as a 6-9 mark structured question. You are typically given either raw data or the summary totals $\sum x$ and $\sum x^2$ , asked to compute the unbiased $s^2$ , form $t$ , and carry out the test. Marks are lost almost entirely through: dividing the sum of squares by $n$ instead of $n-1$ ; using the wrong tail (the $5\%$ vs $2.5\%$ table column); and giving a bare 'reject $H_0$ ' with no contextual conclusion. Because the $t$ -tables are provided in MF19, the high-value preparation is fluent variance computation and disciplined, context-rich conclusions.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 4 Further Probability & Statistics); MF19 List of Formulae and Statistical Tables — t-distribution percentage points. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Hypothesis Testing for a Population Mean using the t-Test

Step-by-step solutions to past-paper-style questions on hypothesis testing for a population mean using the t-test, written exactly the way a tutor would explain them at the board.

Question type:

1Choosing the right test
Getting startedIdentify & classify• t-test, conditions
Question
A machine is meant to fill bottles to a mean of $500$ ml. A quality inspector takes a random sample of $9$ bottles, assumes the volumes are normally distributed, and computes the sample mean and an estimate of the variance from the data. Should the inspector use a $z$ -test or a $t$ -test, and how many degrees of freedom apply?
Step-by-step solution
1. Step 1
  Check the population. The volumes are assumed normally distributed — good, a $t$ -test requires (or assumes) normality.
2. Step 2
  Check the variance. The population variance is NOT given; it is estimated from the sample. This rules out a $z$ -test.
3. Step 3
  Check the sample size. $n = 9$ is small, so the Central Limit Theorem cannot rescue a normal approximation. Use the $t$ -test with $\nu = n - 1$ .
  $\nu = 9 - 1 = 8$
Answer
Use a $t$ -test (normal population, small sample, unknown variance) on $\nu = 8$ degrees of freedom.
Examiner tip
The deciding factor is the unknown, estimated variance. If $\sigma$ had been given, it would be a $z$ -test even with $n=9$ .
2Computing the unbiased variance estimate
Getting startedDirect calculation• variance, unbiased-estimate
Question
A sample of $n = 8$ observations gives $\sum x = 96$ and $\sum x^2 = 1228$ . Find the sample mean $\bar{x}$ and the unbiased estimate $s^2$ of the population variance.
Step-by-step solution
1. Step 1
  Sample mean.
  $\bar{x} = \frac{\sum x}{n} = \frac{96}{8} = 12$
2. Step 2
  Corrected sum of squares $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n}$ .
  $S_{xx} = 1228 - \frac{96^2}{8} = 1228 - 1152 = 76$
3. Step 3
  Divide by $n-1$ (this is what makes it unbiased).
  $s^2 = \frac{76}{8-1} = \frac{76}{7} = 10.857\ldots$
Answer
$\bar{x} = 12$ and $s^2 = \dfrac{76}{7} = 10.9$ (3 s.f.).
Examiner tip
Dividing by $n = 8$ instead of $n - 1 = 7$ gives $9.5$ — a wrong, biased value that derails the whole test. The $n-1$ divisor is non-negotiable.
3A complete one-tailed test from raw totals
Building confidenceMulti-step problem• t-test, one-tailed, full-test
Question
A supplier claims the mean breaking strength of its cables is $14$ kN. A buyer suspects the true mean is higher. A random sample of $10$ cables gives $\sum x = 152$ and $\sum x^2 = 2330.8$ (in kN). Test the supplier's claim against the buyer's suspicion at the $5\%$ significance level. Assume strength is normally distributed.
Step-by-step solution
1. Step 1
  State hypotheses. 'Higher' is directional ⇒ one-tailed upper.
  $H_0: \mu = 14, \qquad H_1: \mu > 14$
2. Step 2
  Sample statistics. Mean and unbiased variance.
  $\bar{x} = \frac{152}{10} = 15.2, \quad s^2 = \frac{1}{9}\left(2330.8 - \frac{152^2}{10}\right) = \frac{20.4}{9} = 2.2667$
3. Step 3
  Test statistic with $\nu = n - 1 = 9$ .
  $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{15.2 - 14}{\sqrt{2.2667}/\sqrt{10}} = \frac{1.2}{0.47610} = 2.521$
4. Step 4
  Critical value from MF19: $\nu = 9$ , one-tailed $5\%$ gives $t_{9}(5\%) = 1.833$ . Compare.
  $2.521 > 1.833 \;\Rightarrow\; \text{reject } H_0$
5. Step 5
  Conclude in context.
Answer
$t = 2.52 > 1.833$ , so reject $H_0$ . There is sufficient evidence at the $5\%$ level that the mean breaking strength of the cables is greater than $14$ kN.
Examiner tip
Quote $t$ to 3 s.f. ( $2.52$ ) but keep full precision while computing. The contextual conclusion (about breaking strength) is required for the final mark.
4A two-tailed test — watch the table column
Building confidenceMulti-step problem• t-test, two-tailed, full-test
Question
A chemical process is designed to give a mean yield of $62$ g. After a maintenance change, a manager wants to know whether the mean yield has changed. A random sample of $12$ runs gives $\bar{x} = 63.8$ g and $s = 3.0$ g (the unbiased estimate). Test, at the $5\%$ level, whether the mean yield has changed. Assume normality.
Step-by-step solution
1. Step 1
  State hypotheses. 'Changed' is non-directional ⇒ two-tailed.
  $H_0: \mu = 62, \qquad H_1: \mu \neq 62$
2. Step 2
  Test statistic with $\nu = 12 - 1 = 11$ .
  $t = \frac{63.8 - 62}{3.0/\sqrt{12}} = \frac{1.8}{0.86603} = 2.078$
3. Step 3
  Critical value. Two-tailed at $5\%$ ⇒ $2.5\%$ in each tail. From MF19, $t_{11}(2.5\%) = 2.201$ .
  $\text{Critical region: } |t| > 2.201$
4. Step 4
  Compare.
  $|2.078| = 2.078 < 2.201 \;\Rightarrow\; \text{do not reject } H_0$
5. Step 5
  Conclude in context.
Answer
$|t| = 2.08 < 2.201$ , so do not reject $H_0$ . There is insufficient evidence at the $5\%$ level that the mean yield of the process has changed.
Examiner tip
Had you wrongly used $t_{11}(5\%) = 1.796$ you would have rejected $H_0$ — the classic two-tailed trap. 'Changed' demands the $2.5\%$ column.
5Building the test from a raw data list
StretchMulti-step problem• t-test, raw-data, lower-tail
Question
A new diet is claimed to reduce mean daily calorie intake below the previous mean of $2100$ kcal. Eight volunteers record intakes (kcal): $2050, 1980, 2010, 1950, 2120, 1990, 2030, 1970$ . Test the claim at the $5\%$ level, assuming intake is normally distributed.
Step-by-step solution
1. Step 1
  Hypotheses. 'Reduce / below' is directional ⇒ one-tailed lower.
  $H_0: \mu = 2100, \qquad H_1: \mu < 2100$
2. Step 2
  Totals. $n = 8$ ; $\sum x = 16100$ ; $\sum x^2 = 32\,421\,800$ .
  $\bar{x} = \frac{16100}{8} = 2012.5$
3. Step 3
  Unbiased variance.
  $s^2 = \frac{1}{7}\left(32421800 - \frac{16100^2}{8}\right) = \frac{1}{7}\left(32421800 - 32401250\right) = \frac{20550}{7} = 2935.7$
4. Step 4
  Test statistic, $\nu = 7$ .
  $t = \frac{2012.5 - 2100}{\sqrt{2935.7}/\sqrt{8}} = \frac{-87.5}{19.158} = -4.568$
5. Step 5
  Critical value. Lower-tail $5\%$ , $\nu = 7$ : $t_{7}(5\%) = 1.895$ , so reject if $t < -1.895$ . Compare and conclude.
  $-4.568 < -1.895 \;\Rightarrow\; \text{reject } H_0$
Answer
$t = -4.57 < -1.895$ , so reject $H_0$ . There is sufficient evidence at the $5\%$ level that the diet reduces the mean daily calorie intake below $2100$ kcal.
Examiner tip
For a lower-tail test the statistic is negative and you compare with the negative critical value. A frequent slip is comparing $|t|$ with $+1.895$ and muddling the direction.
6Working backwards — the critical sample mean
StretchShow that / prove• t-test, critical-value, inverse
Question
A random sample of $15$ items from a normal population gives an unbiased variance estimate $s^2 = 4.84$ . The hypotheses are $H_0: \mu = 20$ against $H_1: \mu > 20$ , tested at the $5\%$ level. Find the smallest value of the sample mean $\bar{x}$ for which $H_0$ would be rejected.
Step-by-step solution
1. Step 1
  Critical value. $\nu = 15 - 1 = 14$ , one-tailed $5\%$ : $t_{14}(5\%) = 1.761$ .
2. Step 2
  Reject boundary. $H_0$ is rejected when the test statistic exceeds $1.761$ .
  $\frac{\bar{x} - 20}{s/\sqrt{n}} > 1.761$
3. Step 3
  Standard error. $s = \sqrt{4.84} = 2.2$ , so $s/\sqrt{n} = 2.2/\sqrt{15}$ .
  $\frac{s}{\sqrt{n}} = \frac{2.2}{3.8730} = 0.56803$
4. Step 4
  Solve for $\bar{x}$ at the boundary.
  $\bar{x} > 20 + 1.761 \times 0.56803 = 20 + 1.0003 = 21.000$
Answer
$H_0$ is rejected when $\bar{x} > 21.0$ (3 s.f.); the smallest such mean is approximately $21.0$ .
Examiner tip
This 'inverse' style rearranges the test inequality. Keep the standard error $s/\sqrt{n}$ to full precision, then multiply by the critical value — rounding $0.568$ too early shifts the boundary.

Model Answers — Hypothesis Testing for a Population Mean using the t-Test

High-scoring sample answers for hypothesis testing for a population mean using the t-test on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
A random sample of $6$ measurements gives $\sum x = 84$ and $\sum x^2 = 1186$ . Calculate the unbiased estimate of the population variance, $s^2$ .
Model answer
Corrected sum of squares: $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 1186 - \frac{84^2}{6} = 1186 - 1176 = 10$

Divide by $n - 1 = 5$ : $s^2 = \frac{10}{5} = 2$
Why this scores
M1 for the corrected sum of squares $S_{xx}$ , M1 for dividing by $n-1$ , A1 for $s^2 = 2$ . Dividing by $n = 6$ (giving $1.67$ ) loses both the method and accuracy marks.
Question 2
3 marks
Explain why a $t$ -test, rather than a $z$ -test, is appropriate when testing the mean of a normal population from a small sample whose variance is estimated from the data. State the degrees of freedom for a sample of size $n$ .
Model answer
Because the population variance $\sigma^2$ is unknown and must be estimated from the sample by $s^2$ , the standardised statistic does not follow the standard normal distribution. The extra uncertainty introduced by estimating the variance is captured by the $t$ -distribution, which has heavier tails than $N(0,1)$ .

A $z$ -test would only be valid if $\sigma^2$ were known (or the sample large enough for the estimate to be effectively exact).

The degrees of freedom are: $\nu = n - 1$ because the sample mean $\bar{x}$ used in computing $s^2$ removes one degree of freedom.
Why this scores
B1 for 'variance unknown / estimated', B1 for naming the $t$ -distribution (heavier tails), B1 for $\nu = n - 1$ . A bare 'small sample' without the unknown-variance reason is incomplete.
Question 3
6 marks
A manufacturer states that the mean lifetime of its batteries is $40$ hours. A consumer group believes the true mean is less. A random sample of $10$ batteries gives a mean lifetime of $38.2$ hours with an unbiased variance estimate $s^2 = 6.25$ hours². Carry out a suitable test at the $5\%$ significance level. (Lifetimes are normally distributed.)
Model answer
Step 1 — Hypotheses. 'Less' is directional, so a one-tailed (lower) test: $H_0: \mu = 40, \qquad H_1: \mu < 40$

Step 2 — Test statistic. With $\bar{x} = 38.2$ , $s = \sqrt{6.25} = 2.5$ , $n = 10$ , $\nu = 9$ : $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{38.2 - 40}{2.5/\sqrt{10}} = \frac{-1.8}{0.79057} = -2.277$

Step 3 — Critical value. One-tailed $5\%$ on $\nu = 9$ : $t_{9}(5\%) = 1.833$ . The rejection region is $t < -1.833$ .

Step 4 — Compare. $-2.277 < -1.833 \;\Rightarrow\; \text{reject } H_0$

Step 5 — Conclusion. There is sufficient evidence at the $5\%$ level that the mean lifetime of the batteries is less than $40$ hours, supporting the consumer group's belief.
Why this scores
B1 hypotheses; M1 A1 for the test statistic $-2.28$ ; B1 for the critical value $1.833$ with the correct (negative) region; M1 for a valid comparison; A1 for a conclusion in context. 'Reject $H_0$ ' alone does not earn the final A1.
Question 4
7 marks
The mean mass of a certain component is supposed to be $250$ g. A random sample of $8$ components has masses (g): $248, 252, 247, 255, 249, 246, 253, 250$ . Test, at the $5\%$ level, whether the mean mass differs from $250$ g. Assume masses are normally distributed.
Model answer
Step 1 — Hypotheses. 'Differs' is non-directional, so two-tailed: $H_0: \mu = 250, \qquad H_1: \mu \neq 250$

Step 2 — Sample statistics. $n = 8$ , $\sum x = 2000$ , $\sum x^2 = 500\,068$ . $\bar{x} = \frac{2000}{8} = 250, \qquad s^2 = \frac{1}{7}\left(500068 - \frac{2000^2}{8}\right) = \frac{68}{7} = 9.714$

Step 3 — Test statistic, $\nu = 7$ : $t = \frac{250 - 250}{\sqrt{9.714}/\sqrt{8}} = \frac{0}{1.1023} = 0$

Step 4 — Critical value. Two-tailed at $5\%$ ⇒ $2.5\%$ each tail: $t_{7}(2.5\%) = 2.365$ . Reject if $|t| > 2.365$ .

Step 5 — Compare and conclude. $|0| = 0 < 2.365 \;\Rightarrow\; \text{do not reject } H_0$ There is insufficient evidence at the $5\%$ level that the mean mass of the components differs from $250$ g.
Why this scores
B1 hypotheses; M1 for $s^2$ (with $n-1$ ); M1 A1 for $t = 0$ ; B1 for $t_{7}(2.5\%) = 2.365$ (the $2.5\%$ column for a two-tailed test); M1 A1 for the contextual conclusion. The neat $\bar{x} = 250$ makes $t = 0$ — a sanity-check that the sample mean equals the hypothesised value.
Question 5
7 marks
A psychologist claims that a training programme increases the mean score on a memory task above the established mean of $72$ . A random sample of $12$ trained participants gives $\sum x = 906$ and $\sum x^2 = 68\,610$ . Test the claim at the $5\%$ significance level, stating your hypotheses and conclusion clearly. (Scores are normally distributed.)
Model answer
Step 1 — Hypotheses. 'Increases / above' is directional ⇒ one-tailed upper: $H_0: \mu = 72, \qquad H_1: \mu > 72$

Step 2 — Sample statistics, $n = 12$ : $\bar{x} = \frac{906}{12} = 75.5$ $s^2 = \frac{1}{11}\left(68610 - \frac{906^2}{12}\right) = \frac{1}{11}\left(68610 - 68403\right) = \frac{207}{11} = 18.818$

Step 3 — Test statistic, $\nu = 11$ : $t = \frac{75.5 - 72}{\sqrt{18.818}/\sqrt{12}} = \frac{3.5}{1.2524} = 2.795$

Step 4 — Critical value. One-tailed $5\%$ , $\nu = 11$ : $t_{11}(5\%) = 1.796$ . Reject if $t > 1.796$ .

Step 5 — Compare and conclude. $2.795 > 1.796 \;\Rightarrow\; \text{reject } H_0$ There is sufficient evidence at the $5\%$ level that the training programme increases the mean memory-task score above $72$ .
Why this scores
B1 hypotheses; M1 for $s^2$ using $n-1$ (A1 for $18.8$ ); M1 A1 for $t = 2.80$ ; B1 for $t_{11}(5\%) = 1.796$ ; A1 for a contextual conclusion. Carry $s^2$ to full precision into $t$ ; rounding $s^2$ to $18.8$ early is acceptable here but keep $s/\sqrt{n}$ accurate.
Question 6
8 marks
A filling machine should dispense a mean volume of $500$ ml. A random sample of $9$ bottles gives $\sum x = 4527$ and $\sum x^2 = 2\,277\,165$ (ml). (a) Calculate the sample mean and the unbiased estimate of the population variance. (b) Test, at the $5\%$ level, whether the machine is dispensing a mean volume different from $500$ ml. (Volumes are normally distributed.)
Model answer
(a) Sample statistics, $n = 9$ : $\bar{x} = \frac{4527}{9} = 503$ $s^2 = \frac{1}{8}\left(2277165 - \frac{4527^2}{9}\right) = \frac{1}{8}\left(2277165 - 2277081\right) = \frac{84}{8} = 10.5$

(b) Step 1 — Hypotheses. 'Different' ⇒ two-tailed: $H_0: \mu = 500, \qquad H_1: \mu \neq 500$

Step 2 — Test statistic, $\nu = 9 - 1 = 8$ : $t = \frac{503 - 500}{\sqrt{10.5}/\sqrt{9}} = \frac{3}{1.0801} = 2.778$

Step 3 — Critical value. Two-tailed at $5\%$ ⇒ $2.5\%$ each tail: $t_{8}(2.5\%) = 2.306$ . Reject if $|t| > 2.306$ .

Step 4 — Compare and conclude. $|2.778| = 2.778 > 2.306 \;\Rightarrow\; \text{reject } H_0$ There is sufficient evidence at the $5\%$ level that the machine is dispensing a mean volume different from $500$ ml (here, the evidence points to over-filling).
Why this scores
(a) M1 $S_{xx}$ , A1 $s^2 = 10.5$ , B1 $\bar{x} = 503$ . (b) B1 hypotheses; M1 A1 for $t = 2.78$ ; B1 for $t_{8}(2.5\%) = 2.306$ (the $2.5\%$ column — using $1.860$ from the $5\%$ column is the standard error here); A1 for the contextual conclusion. The two-tailed wording must drive the $2.5\%$ lookup.

Key Formulae — Hypothesis Testing for a Population Mean using the t-Test

The formulae you need to memorise for hypothesis testing for a population mean using the t-test on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Unbiased estimate of population variance
$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{\left(\sum x\right)^2}{n}\right)$
$n$
sample size
$\sum x$
sum of the data values
$\sum x^2$
sum of the squares of the data values
$s^2$
unbiased estimate of the population variance $\sigma^2$
When to use
Whenever the population variance is unknown and must be estimated from a sample. The divisor is $n-1$ (not $n$ ).
t-test statistic for a population mean
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
$\bar{x}$
sample mean
$\mu_0$
the mean assumed under $H_0$
$s$
unbiased estimate of the standard deviation, $\sqrt{s^2}$
$n$
sample size
When to use
The core of the test. Follows a $t$ -distribution on $\nu = n-1$ degrees of freedom when $H_0$ is true and the population is normal.
Degrees of freedom
$\nu = n - 1$
$\nu$
degrees of freedom for the $t$ -distribution
$n$
sample size
When to use
Needed to choose the correct row of the $t$ -tables in MF19. One degree of freedom is lost to estimating the mean.
Estimated standard error of the mean
$\text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}}$
$s$
unbiased sample standard deviation
$n$
sample size
When to use
The denominator of the test statistic. Keep it to full precision before dividing; early rounding distorts $t$ .
Corrected sum of squares
$S_{xx} = \sum x^2 - \frac{\left(\sum x\right)^2}{n}$
$S_{xx}$
corrected sum of squares
$\sum x,\ \sum x^2$
data totals
When to use
Intermediate quantity: $s^2 = S_{xx}/(n-1)$ . Useful when totals are given rather than raw data.

Key Definitions and Keywords — Hypothesis Testing for a Population Mean using the t-Test

Definitions to memorise and the exact keywords mark schemes credit for hypothesis testing for a population mean using the t-test answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Null hypothesis $H_0$
Examiner keyword
The default claim being tested, always of the form $\mu = \mu_0$ (the mean equals a stated value). It is assumed true unless the data provide strong evidence against it.
Alternative hypothesis $H_1$
Examiner keyword
The claim accepted if $H_0$ is rejected: $\mu \neq \mu_0$ (two-tailed), $\mu > \mu_0$ or $\mu < \mu_0$ (one-tailed). Its form is dictated by the wording of the question.
Unbiased estimate $s^2$
Examiner keyword
An estimate of the population variance whose expected value equals the true $\sigma^2$ . Achieved by dividing the corrected sum of squares by $n-1$ rather than $n$ .
Degrees of freedom $\nu$
Examiner keyword
The parameter of the $t$ -distribution, $\nu = n-1$ for a one-sample test. It indexes the row of the $t$ -tables used to find the critical value.
Significance level
Examiner keyword
The probability of rejecting a true $H_0$ (a Type I error), commonly $5\%$ or $1\%$ . It fixes the size of the rejection region in the tails of the $t$ -distribution.
Critical (rejection) region
Examiner keyword
The set of test-statistic values that lead to rejecting $H_0$ , e.g. $t > 1.833$ (one-tailed) or $|t| > 2.201$ (two-tailed). Its boundary is the critical value read from MF19.

Common Mistakes and Misconceptions — Hypothesis Testing for a Population Mean using the t-Test

The traps other students keep falling into on hypothesis testing for a population mean using the t-test questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Dividing the corrected sum of squares by $n$ instead of $n-1$
9231 Paper 4 Examiner Reports — recurring comment
Why it happens
Confusing the population-variance formula (÷ $n$ ) with the unbiased estimate (÷ $n-1$ ), or pressing the $\sigma_n$ calculator key.
How to avoid it
For a test you ALWAYS need the unbiased estimate: $s^2 = \frac{1}{n-1}(\sum x^2 - \frac{(\sum x)^2}{n})$ . On a calculator use the $\sigma_{n-1}$ / $s_x$ button.
✕Using the $5\%$ table column for a two-tailed test (should be $2.5\%$ )
9231 Paper 4 mark schemes
Why it happens
Reading the significance level ( $5\%$ ) straight into the table without splitting it between the two tails.
How to avoid it
Two-tailed at $5\%$ means $2.5\%$ in each tail — look up the $2.5\%$ column. Say 'two tails, halve the level' before reading MF19.
✕Concluding only 'reject $H_0$ ' with no statement in context
9231 Paper 4 Examiner Reports
Why it happens
Treating the test as a pure calculation and stopping at the comparison.
How to avoid it
Always finish with a sentence naming the variable and direction: 'There is (in)sufficient evidence at the $X\%$ level that [the mean …] has [increased/changed/…].'
✕Using a $z$ -test when the variance is estimated from a small sample
9231 Paper 4 Examiner Reports
Why it happens
Defaulting to the normal distribution; not noticing that $\sigma^2$ is unknown.
How to avoid it
Check the conditions: unknown variance + small sample + normal population ⇒ $t$ -test with $\nu = n-1$ . Only use $z$ if $\sigma$ is given.
✕Comparing a negative $t$ with a positive critical value in a lower-tail test
9231 Paper 4 — one-tailed test questions
Why it happens
Forgetting that for $H_1: \mu < \mu_0$ the rejection region is $t < -t_{\text{crit}}$ , not $|t| > t_{\text{crit}}$ .
How to avoid it
For a lower-tail test reject when $t < -(\text{critical value})$ . Sketch the curve and shade the lower tail to fix the sign.
✕Rounding $s$ , $s^2$ or the standard error too early, distorting $t$
9231 Paper 4 Examiner Reports
Why it happens
Writing down a 3 s.f. value and re-entering it, accumulating rounding error before the final division.
How to avoid it
Carry full calculator precision through $s$ , $s/\sqrt{n}$ and $t$ ; only round the final $t$ (to 3 s.f.) and never round before comparing with a near-boundary critical value.

Hypothesis Testing for a Population Mean using the t-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

1Choosing the right test

2Computing the unbiased variance estimate

3A complete one-tailed test from raw totals

4A two-tailed test — watch the table column

5Building the test from a raw data list

6Working backwards — the critical sample mean

Unbiased estimate of population variance

t-test statistic for a population mean

Degrees of freedom

Estimated standard error of the mean

Corrected sum of squares

Null hypothesis H0H_0H0​

Alternative hypothesis H1H_1H1​

Unbiased estimate s2s^2s2

Degrees of freedom ν\nuν

Significance level

Critical (rejection) region

✕Dividing the corrected sum of squares by nnn instead of n−1n-1n−1

✕Using the 5%5\%5% table column for a two-tailed test (should be 2.5%2.5\%2.5%)

✕Concluding only 'reject H0H_0H0​' with no statement in context

✕Using a zzz-test when the variance is estimated from a small sample

✕Comparing a negative ttt with a positive critical value in a lower-tail test

✕Rounding sss, s2s^2s2 or the standard error too early, distorting ttt

Null hypothesis $H_0$

Alternative hypothesis $H_1$

Unbiased estimate $s^2$

Degrees of freedom $\nu$

✕Dividing the corrected sum of squares by $n$ instead of $n-1$

✕Using the $5\%$ table column for a two-tailed test (should be $2.5\%$ )

✕Concluding only 'reject $H_0$ ' with no statement in context

✕Using a $z$ -test when the variance is estimated from a small sample

✕Comparing a negative $t$ with a positive critical value in a lower-tail test

✕Rounding $s$ , $s^2$ or the standard error too early, distorting $t$