Skip to main content
EnglandMathsSyllabus dot point

How do you use sample evidence to decide whether to accept or reject a claim about a population?

Null and alternative hypotheses, one- and two-tailed tests, significance levels and critical regions, hypothesis tests for a binomial proportion, and for a correlation coefficient and a normal mean.

A focused answer to the Edexcel A-Level Mathematics hypothesis testing content, covering null and alternative hypotheses, one- and two-tailed tests, significance levels and critical regions, tests for a binomial proportion, the correlation coefficient and the mean of a normal distribution.

Generated by Claude Opus 4.811 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Edexcel wants you to set up null and alternative hypotheses, carry out one- and two-tailed tests at a stated significance level, find and use critical regions, test a hypothesis about a binomial proportion pp, test a sample correlation coefficient against zero, and test the mean of a normally distributed variable.

The answer

Hypotheses and significance

A strategy for any hypothesis test

Every test on the paper follows the same five steps. First, define the parameter and state H0H_0 and H1H_1, deciding whether the alternative is one-tailed or two-tailed from the wording. Second, state the distribution of the test statistic assuming H0H_0 is true, such as XB(n,p0)X \sim B(n, p_0). Third, calculate the probability of a result as extreme as, or more extreme than, the one observed (or find the critical region). Fourth, compare with the significance level. Fifth, state the conclusion in the context of the question, not just "reject H0H_0". The wording "greater than" or "less than" signals a one-tailed test, while "changed" or "different from" signals a two-tailed test.

Critical regions

A binomial test

Examples in context

Try this

Q1. State suitable hypotheses for a two-tailed test that a proportion has changed from 0.30.3. [2 marks]

  • Cue. H0:p=0.3H_0: p = 0.3 and H1:p0.3H_1: p \ne 0.3.

Q2. For XB(10,0.2)X \sim B(10, 0.2), H0:p=0.2H_0: p = 0.2, H1:p>0.2H_1: p > 0.2 at 5%5\%, you observe X=5X = 5. Given P(X5)0.0328P(X \ge 5) \approx 0.0328, state the conclusion. [3 marks]

  • Cue. 0.0328<0.050.0328 < 0.05, so reject H0H_0: evidence that pp has increased.

Exam-style practice questions

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

Edexcel 20206 marksA manufacturer claims that at most 10%10\% of its components are faulty. In a random sample of 2525 components, 66 are found to be faulty. Test, at the 5%5\% significance level, whether the proportion of faulty components is greater than 0.100.10.
Show worked answer →

State hypotheses (B1): H0:p=0.10H_0: p = 0.10, H1:p>0.10H_1: p > 0.10, with XB(25,0.10)X \sim B(25, 0.10) where XX is the number faulty.

This is a one-tailed test at 5%5\% (M1). Find P(X6)=1P(X5)P(X \ge 6) = 1 - P(X \le 5) (M1).

From the binomial, P(X5)0.9666P(X \le 5) \approx 0.9666, so P(X6)0.0334P(X \ge 6) \approx 0.0334 (A1).

Since 0.0334<0.050.0334 < 0.05, the result lies in the critical region, so reject H0H_0 (M1).

There is evidence at the 5%5\% level that more than 10%10\% of components are faulty (A1).

Markers reward the hypotheses, the correct tail probability, the comparison, and a contextual conclusion.

Edexcel 20235 marksA coin is tossed 2020 times and lands heads 1515 times. Test at the 5%5\% level whether the coin is biased towards heads, using XB(20,0.5)X \sim B(20, 0.5).
Show worked answer →

State hypotheses (B1): H0:p=0.5H_0: p = 0.5, H1:p>0.5H_1: p > 0.5 (one-tailed).

Find P(X15)=1P(X14)P(X \ge 15) = 1 - P(X \le 14) under XB(20,0.5)X \sim B(20, 0.5) (M1). From tables P(X14)0.9793P(X \le 14) \approx 0.9793, so P(X15)0.0207P(X \ge 15) \approx 0.0207 (A1).

Compare with 5%5\% (M1): 0.0207<0.050.0207 < 0.05, so reject H0H_0 (A1).

There is evidence the coin is biased towards heads.

Markers reward the hypotheses, the tail probability, and the comparison with conclusion in context.

Related dot points

Sources & how we know this