What is hypothesis testing?

Choosing a one- or two-tailed test wrongly. "Less than" or "more than" is one-tailed; "different from" or "changed" is two-tailed, which splits the significance level between both ends.

For X Po(4), write down the mean and variance. [1 mark]

Northern IrelandFurther MathsSyllabus dot point

How do you set up and carry out a hypothesis test, and use continuous and Poisson distributions?

The Poisson distribution as a model, the normal distribution and standardising, the Central Limit Theorem for the distribution of the sample mean, and hypothesis testing including null and alternative hypotheses, significance levels and conclusions.

A CCEA A2 Further Maths Statistics answer on the Poisson distribution, the normal distribution and standardising, the Central Limit Theorem for the sample mean, and the structure of a hypothesis test with null and alternative hypotheses, significance levels and conclusions.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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Try this

What this dot point is asking

CCEA wants you to use the Poisson distribution as a model, work with the normal distribution (standardising to $z$ ), apply the Central Limit Theorem to the distribution of the sample mean, and carry out a hypothesis test with clearly stated hypotheses, a significance level, a test statistic and a conclusion in context.

The answer

The Poisson distribution

The normal distribution and standardising

The Central Limit Theorem

Hypothesis testing

A hypothesis test on a mean

A supplier claims the mean lifetime of a component is $1200$ hours with standard deviation $50$ hours. A sample of $25$ components has mean $1180$ hours. Test at the $5\%$ level whether the true mean is less than claimed.

step State the hypotheses

H_0: \mu = 1200, \qquad H_1: \mu < 1200 \quad (\text{one-tailed}).

step Find the distribution of the sample mean

By the CLT, $\bar{X} \sim N\!\left(1200, \dfrac{50^2}{25}\right) = N(1200, 100)$ , so the standard error is $\sqrt{100} = 10$ .

step Standardise and compare

z = \frac{1180 - 1200}{10} = -2.0.

The one-tailed

5\%

critical value is

-1.645

. Since

-2.0 < -1.645

, the statistic is in the critical region.

step Conclude in context

Reject $H_0$ : there is significant evidence at the $5\%$ level that the mean lifetime is less than $1200$ hours.

Examples in context

Example 1. Modelling rare events. The number of flaws per metre of cable, or accidents per week at a junction, is modelled by the Poisson distribution. Engineers use $\text{Po}(\lambda)$ to predict how often a threshold number of faults will occur and to plan inspections.

Example 2. Quality assurance. A factory samples products and tests whether the mean has drifted from target. The CLT lets them use a normal test on the sample mean even when individual items are not normally distributed, which is the statistical backbone of process control.

Try this

Q1. For $X \sim \text{Po}(4)$ , write down the mean and variance. [1 mark]

Cue. Both equal $4$ .

Q2. A sample of $100$ has population standard deviation $20$ . Find the standard error of the sample mean. [1 mark]

Cue. $\frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{100}} = 2$ .

Q3. A test of "the mean has changed" is one-tailed or two-tailed? [1 mark]

Cue. Two-tailed.

Exam-style practice questions

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

CCEA A2 20206 marksThe number of calls to a helpline follows a Poisson distribution with mean

3

per hour. Find the probability of exactly

2

calls in an hour, and the probability of more than

1

call in an hour.

Show worked answer →

Let $X$ be the number of calls per hour, $X \sim \text{Po}(3)$ , with $P(X = r) = e^{-3}\dfrac{3^r}{r!}$ .

Exactly $2$ calls: $P(X = 2) = e^{-3}\dfrac{3^2}{2!} = e^{-3}\dfrac{9}{2} = 4.5e^{-3} = 0.224.$

More than $1$ call means $P(X > 1) = 1 - P(X = 0) - P(X = 1)$ :

$P(X = 0) = e^{-3} = 0.0498, \qquad P(X = 1) = e^{-3}(3) = 0.149.$

$P(X > 1) = 1 - 0.0498 - 0.149 = 0.801.$

Markers reward the Poisson formula, the exact probability, and using the complement for more than one call.

CCEA A2 20188 marksA machine fills bottles with a mean volume of

500\,\text{ml}

and standard deviation

8\,\text{ml}

. A sample of

64

bottles has a mean of

497\,\text{ml}

. Test at the

5\%

level whether the mean has decreased.

Show worked answer →

State the hypotheses: $H_0: \mu = 500$ against $H_1: \mu < 500$ (a one-tailed test for a decrease).

By the Central Limit Theorem, the sample mean is approximately normal: $\bar{X} \sim N\left(500, \dfrac{8^2}{64}\right) = N(500, 1)$ , so the standard error is $1\,\text{ml}$ .

Standardise the observed mean: $z = \dfrac{497 - 500}{1} = -3.0.$

The critical value for a one-tailed test at $5\%$ is $z = -1.645$ . Since $-3.0 < -1.645$ , the result is in the critical region.

Conclusion: reject $H_0$ . There is significant evidence at the $5\%$ level that the mean volume has decreased.

Markers reward the hypotheses, the standard error from the CLT, the test statistic, comparison with the critical value, and a conclusion in context.

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)