SingaporeMathsSyllabus dot point

How do we test a claim about a population mean using sample evidence?

Carry out a hypothesis test for a population mean, stating hypotheses, computing a test statistic or p-value, and interpreting the conclusion in context

A focused answer to the H2 Mathematics outcome on hypothesis testing. Setting up null and alternative hypotheses, one- and two-tailed tests, the test statistic and p-value, the significance level, and interpreting the conclusion.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
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What this dot point is asking

SEAB wants you to carry out a hypothesis test for a population mean: state the null and alternative hypotheses, decide one- or two-tailed, compute a test statistic and $p$ -value (or compare with a critical value), and interpret the conclusion in context at the stated significance level.

The answer

The hypotheses

The null hypothesis $H_0$ states the value being tested, for example $\mu = \mu_0$ .
The alternative hypothesis $H_1$ states what we suspect: $\mu < \mu_0$ or $\mu > \mu_0$ (one-tailed) or $\mu \neq \mu_0$ (two-tailed).

The direction of $H_1$ comes from the question's wording ("less than", "greater than", or just "different").

The test statistic

Assuming $H_0$ true, and using the CLT, $\bar{X} \sim \mathrm{N}\left(\mu_0, \dfrac{\sigma^2}{n}\right)$ . The standardised test statistic is

z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}.

When the population variance is unknown, use the unbiased sample estimate $s^2$ (for large $n$ the normal model still applies via the CLT).

The p-value and the decision

The $p$ -value is the probability, under $H_0$ , of a sample result at least as extreme as the one observed. The decision rule:

If $p$ -value $<$ significance level, reject $H_0$ .
Otherwise, do not reject $H_0$ (there is insufficient evidence).

For a two-tailed test, compare the two-tailed $p$ -value (or split the significance level between the tails).

The significance level and errors

The significance level (such as $5\%$ ) is the probability of rejecting a true $H_0$ , a Type I error. Choosing a smaller level makes rejection harder, reducing false positives.

Worked example

A teacher claims the mean test score is $60$ . A sample of $36$ students has mean $63$ with population standard deviation $9$ . Test at the $5\%$ level whether the mean differs from $60$ .

Step 1: State the hypotheses

$H_0$ : $\mu = 60$ ; $H_1$ : $\mu \neq 60$ (two-tailed, since "differs").

Step 2: Find the standard error and test statistic

Standard error $= \dfrac{9}{\sqrt{36}} = 1.5$ . Test statistic $z = \dfrac{63 - 60}{1.5} = 2$ .

Step 3: Find the p-value

Two-tailed: $p = 2\,\mathrm{P}(Z > 2) = 2(1 - 0.9772) = 2(0.0228) = 0.0456$ .

Step 4: Compare and decide

$0.0456 < 0.05$ , so reject $H_0$ .

Step 5: Conclude in context

There is significant evidence at the $5\%$ level that the mean test score differs from $60$ .

Examples in context

Example 1. Drug efficacy. A trial tests whether a new treatment lowers mean blood pressure: $H_0$ of no change against $H_1$ of a decrease, with a small significance level chosen to guard against falsely approving an ineffective drug.

Example 2. Manufacturing tolerance. A factory tests whether the mean fill volume differs from the target (two-tailed), rejecting $H_0$ when the sample evidence is extreme enough to act on, the routine statistical quality check.

Try this

Q1. State suitable hypotheses to test whether a mean has increased above $20$ . [2 marks]

Cue. $H_0: \mu = 20$ , $H_1: \mu > 20$ (one-tailed, upper).

Q2. A test gives a $p$ -value of $0.03$ at the $5\%$ level. State the conclusion. [1 mark]

Cue. Since $0.03 < 0.05$ , reject $H_0$ .

Q3. Explain what a Type I error is. [2 marks]

Cue. Rejecting the null hypothesis when it is in fact true; its probability is the significance level.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksA manufacturer claims its bags have mean mass

500\ \text{g}

. A sample of

40

bags has mean

496\ \text{g}

. The population standard deviation is

12\ \text{g}

. Test at the

5\%

level whether the mean mass is less than claimed.

Show worked answer →

$H_0$ : $\mu = 500$ ; $H_1$ : $\mu < 500$ (one-tailed, lower).

Under $H_0$ , $\bar{X} \sim \mathrm{N}\left(500, \dfrac{12^2}{40}\right)$ , standard error $\dfrac{12}{\sqrt{40}} \approx 1.897$ .

Test statistic: $z = \dfrac{496 - 500}{1.897} \approx -2.108$ .

$p$ -value $= \mathrm{P}(Z < -2.108) \approx 0.0175$ . Since $0.0175 < 0.05$ , reject $H_0$ .

There is significant evidence at the $5\%$ level that the mean mass is less than $500\ \text{g}$ .

Markers reward correct hypotheses, the test statistic, the $p$ -value, comparison with $0.05$ , and a contextual conclusion.

Original5 marksExplain the meaning of a

5\%

significance level and the difference between a one-tailed and a two-tailed test.

Show worked answer →

A $5\%$ significance level means we reject $H_0$ if the observed result (or more extreme) would occur with probability less than $0.05$ when $H_0$ is true; it is the probability of wrongly rejecting a true $H_0$ (a Type I error).

A one-tailed test has $H_1$ specifying a direction (for example $\mu < 500$ or $\mu > 500$ ), placing all the rejection region in one tail. A two-tailed test has $H_1: \mu \neq 500$ , splitting the rejection region between both tails ( $2.5\%$ each at the $5\%$ level).

Markers reward defining the significance level as the Type I error probability and correctly distinguishing the one- and two-tailed alternatives.

What this dot point is asking

The answer

The hypotheses

The test statistic

The p-value and the decision

The significance level and errors

Examples in context

Try this

Exam-style practice questions

Related dot points