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

This guide follows the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. It covers sections 2.05a to 2.05e of Statistical Hypothesis Testing. Sections 2.05a to 2.05c are Stage 1 content, common with AS. Sections 2.05d and 2.05e, the sample mean and the test for a normal mean, are A Level only. Hypothesis testing for correlation (2.05f and 2.05g) has its own pages, so it is not repeated here.

Links: OCR Mathematics A hub, printable checklist, revision notes, practice questions, OCR statistics overview and free diagnostics.

What this unit covers

Section	Skill	Stage
2.05a	Use the language: null and alternative hypothesis, significance level, test statistic, one-tail and two-tail test, critical value, critical region, acceptance region, p-value	Stage 1
2.05b	Carry out a test for the proportion in a binomial distribution and interpret it in context	Stage 1
2.05c	See a sample as evidence about a population, and know that the significance level is the chance of wrongly rejecting a true null hypothesis	Stage 1
2.05d	Know that a sample mean is a random variable, and that X̄ ~ N(μ, σ²/n) when X ~ N(μ, σ²)	A Level only
2.05e	Test the mean of a normal distribution when the variance is known, given or assumed	A Level only

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

2.05a: the language

• Null hypothesis H₀: the assumption you test. It names one value of the parameter, such as p = 0.15.
• Alternative hypothesis H₁: what you accept as likely if the data cast doubt on H₀. It is p > 0.15, p < 0.15 or p ≠ 0.15.
• One-tail test: H₁ has a single direction (< or >). Two-tail test: H₁ uses ≠.
• Test statistic: the quantity from the sample that you judge, such as the number of successes X, or the sample mean X̄.
• Significance level: the probability, set before you look at the data, that is small enough to count as surprising. Typical levels are 10%, 5% and 1%.
• Critical region: every outcome that would make you reject H₀. Its boundary is the critical value, and the outcomes left over form the acceptance region.
• p-value: how likely it is, if H₀ holds, to get a sample as extreme as yours or more so.

Define every symbol in the hypotheses, and give the parameter in the population, never the sample value. A good statement is: H₀: p = 0.15, H₁: p > 0.15, where p is the probability that a randomly chosen plug is faulty.

Two ways to decide, always the same decision:

1. Compare the p-value with the significance level. If p-value ≤ level, reject H₀.
2. Build the critical region, then check whether your observed value falls inside it.

For a two-tail test, each tail gets half the significance level (or double the one-tail probability to get the p-value).

Writing the conclusion. The test cannot prove anything, so the wording must be tentative and in context. If H₀ is rejected, write something like: “At the 5% level the data support rejecting H₀, so the proportion of faulty plugs has probably increased.” If it is not rejected: “At the 5% level the data do not support rejecting H₀, so there is no reason to think the proportion has changed.” Avoid “H₀ is true”, “accept H₀” and “the claim is proved”.

2.05b: binomial tests

The test statistic is X, the number of successes in a sample of n independent trials. Under H₀, X ~ B(n, p₀).

Method

1. Define p and state H₀ and H₁.
2. State the distribution of X under H₀.
3. Find the p-value, or the critical region, using binomial cumulative probabilities.
4. Compare with the significance level.
5. Conclude in context.

Example 1: upper tail, p-value method

A supplier says 15% of its plugs are faulty. A buyer thinks the rate is higher and tests 22 plugs at the 5% level. Seven are faulty.

H₀: p = 0.15    H₁: p > 0.15    p = probability a plug is faulty
Under H₀, X ~ B(22, 0.15)
P(X ≥ 7) = 1 − P(X ≤ 6) = 0.0368
0.0368 < 0.05

Reject H₀. At the 5% level the evidence points to a proportion of faulty plugs above 0.15.

At the 1% level the decision changes: 0.0368 > 0.01, so there is no evidence to reject H₀. The same data can give different conclusions at different levels, which is why you set the level first.

Example 2: lower tail, critical region method

A club says 60% of members renew. A new treasurer suspects fewer do. Sample size 30, 5% level. Find the critical region.

H₀: p = 0.6    H₁: p < 0.6    X ~ B(30, 0.6)
P(X ≤ 13) = 0.0481 ≤ 0.05
P(X ≤ 14) = 0.0971 > 0.05

The critical region is X ≤ 13. The critical value is 13. The probability of falling in the critical region when H₀ is true is 0.0481. This is the actual significance level. It is below 5%, never above it, because you take the largest region whose probability does not exceed the level (the specification states this).

If 12 members renew, 12 is in the region (P(X ≤ 12) = 0.0212), so reject H₀. If 14 renew, it is not, so there is no evidence of a fall.

Example 3: two-tail test

A café claims 45% of customers order a hot drink. A sample of 20 is taken to see whether the proportion differs, at the 10% level. Each tail gets 5%.

H₀: p = 0.45    H₁: p ≠ 0.45    X ~ B(20, 0.45)
Lower: P(X ≤ 4) = 0.0189 ≤ 0.05,  P(X ≤ 5) = 0.0553 > 0.05
Upper: P(X ≥ 14) = 0.0214 ≤ 0.05, P(X ≥ 13) = 0.0580 > 0.05

Critical region: X ≤ 4 or X ≥ 14. Actual significance level: 0.0189 + 0.0214 = 0.0403, about 4.03%.

Now suppose 4 customers order a hot drink. Using the p-value, 2 × P(X ≤ 4) = 2 × 0.0189 = 0.0377 (3 s.f.), less than 0.10. Reject H₀. The critical region gives the same decision, because 4 is in it.

2.05c: inference and errors

A sample is used to infer something about a whole population, so the conclusion carries risk. Two things can go wrong.

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

• P(Type I error) is the probability of being in the critical region when H₀ is true. For a discrete test this is the actual significance level, which is at most the stated level.
• P(Type II error) is the probability of not being in the critical region when H₀ is false. It needs a specific true value of the parameter.

Example 4: finding both

Use the critical region in Example 3 (X ≤ 4 or X ≥ 14, n = 20).

• P(Type I) = 0.0403.
• Suppose the true proportion is 0.7. Then X ~ B(20, 0.7) and a Type II error means 5 ≤ X ≤ 13:

P(5 ≤ X ≤ 13) = P(X ≤ 13) − P(X ≤ 4) = 0.392

So there is about a 39% chance that the test misses the fact that the proportion is not 0.45. If the true proportion is 0.3 the chance is 0.762: the nearer the truth is to p₀, the harder it is to detect.

In context, a Type I error here means concluding the café’s claim is wrong when it is correct. A Type II error means keeping the claim when it is wrong.

2.05d: the sample mean (A Level only)

Take a sample of n values from a population. The sample mean X̄ changes from sample to sample, so it is a random variable.

If X ~ N(μ, σ²), then X̄ ~ N(μ, σ²/n).

The mean stays μ, but the variance shrinks, so the standard deviation of X̄ is σ/√n. Bigger samples give means that cluster more tightly around μ.

Example 5

Journey times X (minutes) are N(48, 6²). Take n = 9.

X̄ ~ N(48, 6²/9) = N(48, 2²)
P(X̄ > 50) = P(Z > 1) = 0.1587
P(X > 50) = P(Z > 1/3) = 0.3694
P(46 < X̄ < 50) = 0.6827

2.05e: the test for a normal mean (A Level only)

You assume the population is normal and that σ is known, given or assumed. Under H₀: μ = μ₀, X̄ ~ N(μ₀, σ²/n). Standardise with

z = (x̄ − μ₀) / (σ/√n)

and compare with the standard normal critical values (the specification says tables of percentage points are provided in the assessment). The values to know are:

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

Example 6: one-tail

Delivery times have standard deviation 2.8 minutes. A firm claims a mean of 34 minutes. A rival believes it is longer. A sample of 40 deliveries has mean 34.9 minutes. Test at the 5% level.

H₀: μ = 34    H₁: μ > 34    μ = population mean delivery time (minutes)
Under H₀, X̄ ~ N(34, 2.8²/40), standard deviation 0.4427
z = (34.9 − 34)/0.4427 = 2.033
p-value = P(Z > 2.033) = 0.021

2.033 > 1.645, and 0.021 < 0.05, so reject H₀. There is evidence at the 5% level that the mean delivery time is more than 34 minutes.

The critical region in terms of the mean is X̄ ≥ 34 + 1.645 × 0.4427 = 34.73 (2 d.p.). At the 1% level the boundary is 35.03, so 34.9 would not be in the region, and there would be no evidence to reject H₀.

Example 7: two-tail

A machine is set to fill bottles with mean 330 ml. The fill has standard deviation 3.5 ml. A sample of 25 bottles gives mean 328.9 ml. Test at the 5% level whether the mean has changed.

H₀: μ = 330    H₁: μ ≠ 330
Under H₀, X̄ ~ N(330, 3.5²/25), standard deviation 0.7
z = (328.9 − 330)/0.7 = −1.571
p-value = 2 × P(Z < −1.571) = 0.116

|z| = 1.571 < 1.960, and 0.116 > 0.05, so the data do not support rejecting H₀ at the 5% level. Nothing suggests the mean fill has changed.

The critical region is X̄ < 328.63 or X̄ > 331.37, to 2 d.p.

Example 8: Type II error for a mean

Use Example 6. The 5% critical value for the mean is 34.728. Suppose the true mean is 35.2, with the same standard deviation. Then X̄ ~ N(35.2, 0.4427²), and

P(Type II) = P(X̄ < 34.728) = P(Z < −1.066) = 0.143

Common errors

• Writing H₀: p = 7/22 or H₀: μ = 34.9. Hypotheses are about the population parameter, never the sample result.
• Using P(X = 7) instead of P(X ≥ 7) for an upper-tail test. Write it as 1 minus the cumulative probability up to k − 1.
• Forgetting to halve the level (or double the probability) in a two-tail test.
• Dividing by σ rather than σ/√n when standardising a sample mean.
• Concluding “the machine has changed” with certainty, or writing “accept H₀”.

Next steps

Condense this with the revision notes, then try the practice questions. The free diagnostics show which topics need more work.

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.