WJEC WJEC A2 style-style practice: A machine should produce rods of mean length 50\,cm with standard deviation 2\,cm. A sample of 16 rods has mean 51\,cm. Test at the 5 per cent level whether the mean length has increased.

Use the distribution of the sample mean, which is Normal with the same mean and a standard deviation reduced by the square root of the sample size. H_0\!: μ = 50, H_1\!: μ > 50 (one-tailed). Under H_0, the sample mean X N\!(50, 2^216), so its standard deviation is 2sqrt(16) = 0.5. Standardise the observed mean: z = 51 - 500.5 = 2. Critical z for 5 per cent one tail is 1.6449. Since 2 > 1.6449, reject H_0. There is evidence at the 5 per cent level that the mean length has increased. Markers reward the standard error σsqrt(n), the z-statistic of 2, the comparison with 1.6449, and a conclusion in context.

WalesMathsSyllabus dot point

How do we test for correlation, and test a claim about the mean of a Normal distribution?

Hypothesis testing for a correlation coefficient, and hypothesis testing for the mean of a Normal distribution using the distribution of the sample mean.

A focused answer to WJEC A2 Unit 4 hypothesis testing, covering testing a correlation coefficient against a critical value and testing the mean of a Normal distribution using the distribution of the sample mean.

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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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

WJEC wants you to carry out a hypothesis test for a correlation coefficient (comparing the sample $r$ with a tabulated critical value) and a hypothesis test for the mean of a Normal distribution (using the fact that the sample mean is itself Normally distributed with a smaller standard deviation). Both extend the AS binomial test to new contexts and need the same disciplined structure.

The answer

Testing a correlation coefficient

The product moment correlation coefficient $r$ measures the strength and direction of linear association in a sample, between $-1$ and $1$ . A hypothesis test asks whether the underlying population correlation $\rho$ is non-zero.

The critical value depends on both $n$ and the significance level, and is read from a supplied table, not calculated.

Testing a Normal mean

The crucial fact is the distribution of the sample mean: averaging reduces variability.

Test a claim about a Normal mean

A drink should contain $330\,\text{ml}$ , Normally distributed with $\sigma = 4\,\text{ml}$ . A sample of $25$ cans has mean $328\,\text{ml}$ . Test at the 5 per cent level whether the mean has fallen.

step State the hypotheses

$H_0\!: \mu = 330$ , $H_1\!: \mu < 330$ (one-tailed).

step State the distribution of the sample mean

$\bar{X} \sim N\!\left(330, \dfrac{4^2}{25}\right)$ , so the standard error is $\dfrac{4}{\sqrt{25}} = 0.8$ .

step Compute the test statistic

$z = \dfrac{328 - 330}{0.8} = \dfrac{-2}{0.8} = -2.5$ .

step Compare and conclude

The critical value for the lower 5 per cent tail is $-1.6449$ . Since $-2.5 < -1.6449$ , reject $H_0$ : there is evidence the mean has fallen below $330\,\text{ml}$ .

Examples in context

Example 1. A weak but significant correlation. With a large sample of $100$ pairs, even a modest $r = 0.25$ can exceed the critical value and lead to rejecting $H_0$ . Significance is not the same as a strong relationship: a large sample detects small effects, so always report the size of $r$ alongside the test conclusion.

Example 2. Why bigger samples sharpen a mean test. Doubling the sample size from $16$ to $64$ halves the standard error from $\dfrac{\sigma}{4}$ to $\dfrac{\sigma}{8}$ , so the same difference of sample mean from $\mu_0$ gives a z-statistic twice as large. Larger samples make a genuine shift in the mean easier to detect.

Try this

Q1. State the null hypothesis for a test of whether two variables are correlated. [1 mark]

Cue. $H_0\!: \rho = 0$ (no correlation in the population).

Q2. A sample of $36$ from $N(\mu, 6^2)$ is used to test the mean. Find the standard error of the sample mean. [2 marks]

Cue. $\dfrac{\sigma}{\sqrt{n}} = \dfrac{6}{\sqrt{36}} = 1$ .

Q3. A sample $r = 0.55$ is compared with a critical value of $0.60$ in a one-tailed test. State the conclusion. [2 marks]

Cue. $0.55 < 0.60$ , so do not reject $H_0$ : no significant evidence of correlation.

Exam-style practice questions

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

WJEC A2 style5 marksA sample of 12 pairs gives a product moment correlation coefficient of

r = 0.62

. Test at the 5 per cent level whether there is positive correlation, given the critical value is

0.4973

Show worked answer →

Set up a one-tailed correlation test and compare the sample r with the critical value.

Let $\rho$ be the population correlation coefficient. $H_0\!: \rho = 0$ , $H_1\!: \rho > 0$ (one-tailed, positive).

The critical value for $n = 12$ at 5 per cent (one tail) is $0.4973$ .

The sample $r = 0.62 > 0.4973$ , so the result lies in the critical region.

Reject $H_0$ : there is evidence at the 5 per cent level of positive correlation between the variables.

Markers reward stating the hypotheses in terms of $\rho$ , comparing $r$ with the critical value, and a conclusion in context. Testing $r$ against $0$ rather than the tabulated critical value is the standard error.

WJEC A2 style6 marksA machine should produce rods of mean length

50\,\text{cm}

with standard deviation

2\,\text{cm}

. A sample of 16 rods has mean

51\,\text{cm}

. Test at the 5 per cent level whether the mean length has increased.

Show worked answer →

Use the distribution of the sample mean, which is Normal with the same mean and a standard deviation reduced by the square root of the sample size.

$H_0\!: \mu = 50$ , $H_1\!: \mu > 50$ (one-tailed).

Under $H_0$ , the sample mean $\bar{X} \sim N\!\left(50, \dfrac{2^2}{16}\right)$ , so its standard deviation is $\dfrac{2}{\sqrt{16}} = 0.5$ .

Standardise the observed mean: $z = \dfrac{51 - 50}{0.5} = 2$ .

Critical z for 5 per cent one tail is $1.6449$ . Since $2 > 1.6449$ , reject $H_0$ .

There is evidence at the 5 per cent level that the mean length has increased. Markers reward the standard error $\dfrac{\sigma}{\sqrt{n}}$ , the z-statistic of $2$ , the comparison with $1.6449$ , and a conclusion in context.

Related dot points

Sources & how we know this

WJEC GCE AS/A Level Mathematics specification (from 2017) — WJEC (2017)