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EnglandMathsSyllabus dot point

How do you use sample data to test a claim about a population, and decide whether the evidence is strong enough?

Setting up null and alternative hypotheses, the significance level, one-tailed and two-tailed tests, hypothesis tests for a binomial proportion and for a normal mean, critical regions, and interpreting the conclusion in context.

A focused answer to the AQA A-Level Mathematics hypothesis testing content, covering null and alternative hypotheses, significance levels, one and two-tailed tests for a binomial proportion and a normal mean, critical regions, and stating conclusions in context.

Generated by Claude Opus 4.811 min answer

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

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  1. What this dot point is asking
  2. Setting up a test
  3. The significance level and errors
  4. Binomial proportion test
  5. Critical regions
  6. Test for a normal mean
  7. A reliable five-step layout

What this dot point is asking

AQA wants you to set up null and alternative hypotheses, choose a significance level, run one-tailed and two-tailed tests for a binomial proportion and for the mean of a normal distribution with known variance, find critical regions, and state a clear conclusion in context. This is a Paper 3 staple and rewards a disciplined layout: hypotheses, model, calculation, comparison, conclusion.

Setting up a test

Choose the direction from the wording. "Has the rate fallen" is one-tailed lower; "has the rate changed" is two-tailed. State H0H_0 and H1H_1 in terms of the population parameter, not the sample statistic.

The significance level and errors

The significance level α\alpha, often 55 percent or 11 percent, is the threshold below which you reject the null hypothesis. It is precisely the probability of a Type I error: rejecting H0H_0 when it is actually true. For a two-tailed test you split α\alpha between the two tails, so each tail carries α2\frac{\alpha}{2}.

Binomial proportion test

For a test on a proportion, model the count XX as binomial under H0H_0 and compute the probability of a result as extreme as, or more extreme than, the one observed.

Critical regions

The critical region is the set of values of the test statistic that would lead you to reject H0H_0. For the binomial you find the smallest (or largest) count whose tail probability is within α\alpha; for the normal mean you compare the standardised statistic with the critical z value. Stating the critical region in advance lets you simply check whether the observed value lies inside it.

Test for a normal mean

For a sample of size nn from a normal distribution with known standard deviation σ\sigma, the sample mean under H0H_0 satisfies XˉN(μ0,σ2n)\bar{X} \sim N\left(\mu_0, \frac{\sigma^2}{n}\right).

A reliable five-step layout

Examiners award method marks for a clear structure, so use the same skeleton every time. First, define the parameter and state H0H_0 and H1H_1 in terms of it. Second, state the distribution of the test statistic under H0H_0 (the binomial B(n,p0)B(n, p_0), or the sample mean N(μ0,σ2/n)N(\mu_0, \sigma^2 / n)). Third, decide the significance level and whether the test is one- or two-tailed, halving the level across the tails if two-tailed. Fourth, compute either the probability of the observed result or more extreme, or the standardised statistic and the critical value. Fifth, compare and write a conclusion in the context of the original claim.

A subtle point is the difference between the two equivalent approaches. The probability (p-value) approach compares P(observed or more extreme)P(\text{observed or more extreme}) with α\alpha; the critical-region approach finds the boundary in advance and checks whether the observation falls beyond it. Both must reach the same decision, and AQA accepts either, but mixing them (comparing a probability with a z value, say) loses marks. Choose one and apply it consistently.

Exam-style practice questions

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

AQA 20186 marksPaper 3, Section B. A seed supplier claims that 7070 percent of its seeds germinate. A gardener plants 2525 seeds and finds that 1313 germinate. Test, at the 55 percent significance level, whether the true germination rate is lower than claimed. State your hypotheses and conclusion clearly in context.
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Let pp be the germination probability and XX the number germinating, XB(25,p)X \sim B(25, p). Hypotheses: H0:p=0.7H_0: p = 0.7 against H1:p<0.7H_1: p < 0.7 (one-tailed, lower). Under H0H_0, find P(X13)=0.0442P(X \le 13) = 0.0442 from the cumulative binomial. Since 0.0442<0.050.0442 < 0.05, reject H0H_0. There is sufficient evidence at the 55 percent level that the germination rate is lower than the claimed 7070 percent. Markers reward correctly stated hypotheses in terms of pp, comparing P(X13)P(X \le 13) (not P(X=13)P(X = 13)) with 0.050.05, and a contextual conclusion rather than a bare "reject".

AQA 20227 marksPaper 3, Section B. A machine fills bottles whose contents are normally distributed with mean 500500 ml and known standard deviation 44 ml. After maintenance, a sample of 1616 bottles has mean 502.3502.3 ml. Test, at the 55 percent significance level, whether the mean has changed. (a) State the hypotheses and the distribution of the sample mean under the null. (b) Carry out the test using the standardised test statistic. (c) State the conclusion in context.
Show worked answer →

Hypotheses: H0:μ=500H_0: \mu = 500 against H1:μ500H_1: \mu \ne 500 (two-tailed). Under H0H_0 the sample mean XˉN(500,4216)=N(500,1)\bar{X} \sim N\left(500, \frac{4^2}{16}\right) = N(500, 1), so the standard error is 11. The test statistic is z=502.35001=2.3z = \frac{502.3 - 500}{1} = 2.3. For a two-tailed test at 55 percent the critical values are ±1.96\pm 1.96. Since 2.3>1.962.3 > 1.96, the result is in the critical region, so reject H0H_0. There is evidence at the 55 percent level that the mean fill has changed from 500500 ml. Markers reward the correct standard error σn\frac{\sigma}{\sqrt{n}}, splitting the 55 percent across two tails to get ±1.96\pm 1.96, and a contextual conclusion.

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