OCR A Level Mathematics A: Hypothesis testing: language, binomial tests and the sample mean (H240) – Revision Notes

These notes condense sections 2.05a to 2.05e of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 2.05a to 2.05c are Stage 1, common with AS. Sections 2.05d and 2.05e are A Level only. For full explanations and worked examples read the study guide.

Links: OCR Mathematics A hub, printable checklist, practice questions and free diagnostics.

Vocabulary (2.05a)

Term	Meaning
H₀	Null hypothesis: one stated value of the parameter, e.g. p = 0.2
H₁	Alternative: p < p₀, p > p₀ (one-tail) or p ≠ p₀ (two-tail)
Test statistic	The sample quantity you judge: X (a count) or X̄ (a mean)
Significance level	Probability chosen beforehand as the threshold for “unlikely”
Critical region	Outcomes for which H₀ is rejected
Critical value	Where the critical region starts
Acceptance region	Everything outside the critical region
p-value	Chance, if H₀ holds, of a sample as extreme as yours or more so

Hypotheses are about population parameters. Define the symbol: “p is the probability that a randomly chosen seed germinates”.

Two equivalent decisions

Route	Reject H₀ when
p-value	p-value ≤ significance level
Critical region	observed value is in the critical region

Two-tail: either compare each tail with half the level, or compare double the tail probability with the whole level.

Method in steps: binomial test (2.05b)

1. Define p. State H₀: p = p₀ and H₁.
2. Under H₀, X ~ B(n, p₀).
3. Find the p-value or the critical region.
4. Compare with the level.
5. Conclude in context, without certainty.

H₁	p-value
p > p₀	P(X ≥ x) = 1 − P(X ≤ x − 1)
p < p₀	P(X ≤ x)
p ≠ p₀	2 × the smaller tail probability

Critical region. Choose the largest region whose probability does not exceed the level (the level, or half of it in each tail). The probability of the region is the actual significance level.

Worked reminder. X ~ B(25, 0.3), H₁: p ≠ 0.3, 5% level. Lower tail: P(X ≤ 2) = 0.0090 ≤ 0.025 but P(X ≤ 3) = 0.0332 > 0.025. Upper tail: P(X ≥ 13) = 0.0175 ≤ 0.025 but P(X ≥ 12) = 0.0442 > 0.025. Critical region: X ≤ 2 or X ≥ 13. Actual level: 0.0090 + 0.0175 = 0.0264.

Conclusions (2.05a, 2.05b)

Outcome	Good wording
Reject	The data support rejecting H₀ at the 5% level; the proportion has probably increased.
Not reject	The data do not support rejecting H₀ at the 5% level; no reason to think the proportion has changed.

Do not write “H₀ is accepted”, “H₀ is true”, “this proves” or a bare “reject H₀” with no context.

Errors (2.05c)

Decision	H₀ really true	H₀ really false
Reject H₀	Type I error	Right decision
Keep H₀	Right decision	Type II error

• P(Type I) = P(test statistic in the critical region | H₀ true). It is the actual significance level, never above the stated level.
• P(Type II) = P(test statistic not in the critical region | the stated true value). It needs a particular true parameter value.
• The further the true value from p₀ or μ₀, the smaller P(Type II).
• Sample data are used to make an inference about a population, so any conclusion may be wrong.

Sample mean (2.05d, A Level only)

If X ~ N(μ, σ²) then X̄ ~ N(μ, σ²/n). The standard deviation of X̄ is σ/√n.

Reminder: X ~ N(80, 10²), n = 16. Then X̄ ~ N(80, 2.5²) and P(X̄ < 78) = P(Z < −0.8) = 0.2119, whereas P(X < 78) = P(Z < −0.2) = 0.4207.

Method in steps: normal mean test (2.05e, A Level only)

1. Define μ. State H₀: μ = μ₀ and H₁.
2. Under H₀, X̄ ~ N(μ₀, σ²/n).
3. Standardise: z = (x̄ − μ₀)/(σ/√n), or find the critical region for X̄.
4. Compare with the critical z-value or find the p-value.
5. Conclude in context.

Level	One-tail z	Two-tail z
10%	1.282	1.645
5%	1.645	1.960
1%	2.326	2.576

Worked reminder. H₀: μ = 60, H₁: μ < 60, σ = 5, n = 25, 5% level. Under H₀, X̄ ~ N(60, 1²). The critical region is X̄ < 60 − 1.645 × 1 = 58.36 (2 d.p.). A sample mean of 58.1 is in the region, so reject H₀ and conclude that the mean is likely to be below 60.

Type II reminder. In the test above, suppose the true mean is 57.5. Then X̄ ~ N(57.5, 1²) and a Type II error means X̄ ≥ 58.355. P(Z ≥ 0.855) = 0.1963, so about a 20% chance of missing the change.

The specification excludes tests for the mean of a non-normal distribution, the normal approximation to the binomial, and estimation of population parameters.

Must-know distinctions

• H₀ and H₁: H₀ always has “=”. The direction of H₁ comes from the wording of the question, not the data.
• Significance level and p-value: the level is chosen before; the p-value comes from the data.
• Critical value and test statistic: the critical value is a boundary; the test statistic is the sample result.
• Type I and Type II: Type I is wrongly rejecting a true H₀. Type II is failing to reject a false H₀.
• σ and σ/√n: use σ for individual values, σ/√n for sample means.

Quick self-test

1. A retailer says 20% of items are returned. A buyer suspects more are. Write H₀ and H₁, defining p.
2. X ~ B(12, 0.2), H₁: p > 0.2, 5% level. Find the critical region and its actual significance level.
3. X ~ B(10, 0.5), H₁: p ≠ 0.5, 5% level. Find the critical region and its actual size.
4. A test has H₀: p = 0.2 against H₁: p > 0.2 with n = 10 and critical region X ≥ 5. Find P(Type I error).
5. For the test in question 4, find P(Type II error) when the true value of p is 0.5.
6. A statistician writes “H₀ is accepted, so the machine is working properly”. Give a better conclusion for a non-rejection.
7. X ~ N(80, 10²), n = 16. Write the distribution of X̄ and find P(X̄ < 78).
8. H₀: μ = 200, H₁: μ > 200, σ = 12, n = 36, x̄ = 203.5. Find z and the p-value.
9. For the test in question 8, decide at the 5% level and write a conclusion.
10. A two-tail test on a normal mean gives z = −2.4. Find the p-value and decide at the 1% level.

Answers

1. H₀: p = 0.2, H₁: p > 0.2, where p is the probability that an item is returned.
2. P(X ≥ 5) = 0.0726 > 0.05 and P(X ≥ 6) = 0.0194 ≤ 0.05. Critical region X ≥ 6, actual significance level 0.0194.
3. P(X ≤ 1) = 0.0107 ≤ 0.025 and P(X ≤ 2) = 0.0547 > 0.025; by symmetry P(X ≥ 9) = 0.0107. Critical region X ≤ 1 or X ≥ 9, actual level 0.0215.
4. P(X ≥ 5 | p = 0.2) = 0.0328.
5. P(X ≤ 4 | p = 0.5) = 0.377.
6. “The data do not support rejecting H₀ at the stated level, so nothing suggests the machine is faulty.”
7. X̄ ~ N(80, 2.5²). P(X̄ < 78) = P(Z < −0.8) = 0.2119.
8. Standard deviation of X̄ = 12/6 = 2. z = 3.5/2 = 1.75. p-value = P(Z > 1.75) = 0.0401.
9. 1.75 > 1.645, and 0.0401 < 0.05, so reject H₀: at the 5% level the mean is probably above 200.
10. p-value = 2 × P(Z < −2.4) = 0.0164. Since 0.0164 > 0.01 (and |z| = 2.4 < 2.576), the data do not support rejecting H₀ at the 1% level.

Where marks are usually lost

• Writing hypotheses with the sample result, for example H₀: p = 0.4 when 0.4 is the sample proportion.
• Leaving out the definition of the parameter, or defining it as a sample quantity.
• Using P(X = x) or P(X > x) when the test needs P(X ≥ x), so the boundary value is wrong by one.
• Not stating the distribution of X or X̄ under H₀.
• Comparing a one-tail probability with the full level in a two-tail test.
• Dividing by σ instead of σ/√n for a sample mean.
• Finishing with “reject H₀” or “accept H₀” and no sentence in context, or a conclusion that sounds certain.
• Treating a test that does not reject H₀ as proof that H₀ is true.
• Giving P(Type II) without using the true parameter value stated in the question.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. This page covers section 2.05, parts a to e.