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OCR A-level Mathematics A (H240) · 2.05 Statistical Hypothesis Testing
Mini-Lesson

Hypothesis testing

OCR section 2.05 puts a claim on trial. You set up H₀ and H₁, compute a test statistic, compare it with a critical value or p-value, and reach a conclusion in context. Tests: the binomial proportion, the Normal mean, and correlation.

Hypotheses & p-values Binomial test Normal mean & correlation H₀ is on trial — and it is innocent until proven guilty

The final mark is always for the conclusion in context — never just 'reject H₀'. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Hypothesis testing · language

The language of a test

  • Null hypothesis H₀ — the claim being tested; assumed true until the evidence says otherwise. It always states a specific value, e.g. p = 0.25 or μ = 50.
  • Alternative hypothesis H₁ — what you suspect instead, e.g. p > 0.25 (one-tailed) or p ≠ 0.25 (two-tailed).
  • Significance level α — how much evidence you demand. Typically 5% or 1%.
  • p-value — the probability of getting a result at least as extreme as the one observed, assuming H₀ is true.
  • Critical region — the set of test-statistic values that lead you to reject H₀.
p-value < significance level  ⇒  REJECT H₀otherwise there is insufficient evidence to reject H₀ — which is NOT the same as proving H₀ true

Two-tailed tests split the significance level. A two-tailed test at 5% puts 2.5% in each tail. Forgetting to halve it is the single most common error in this topic.

Quick check

What is a p-value?

?In a hypothesis test, the p-value is:
Hypothesis testing · binomial

Testing a binomial proportion

Worked example

A treatment is claimed to work on 25% of patients. A researcher believes it works on more. In a trial of 20 patients, 9 improve. Test at the 5% level.

H₀: p = 0.25    H₁: p > 0.25   (one-tailed)

Under H₀, X ~ B(20, 0.25).

p-value = P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.9591 = 0.0409.

0.0409 < 0.05, so reject H₀.

Conclusion in context: there is significant evidence at the 5% level that the treatment works on more than 25% of patients.

Note the direction. H₁ says ‘more’, so the extreme results are the large ones — hence P(X ≥ 9), not P(X = 9) and not P(X ≤ 9). If H₁ were p < 0.25 you would compute P(X ≤ 9) instead.

Calculate

Your turn — the binomial p-value

1Under H₀, X ~ B(20, 0.25) and H₁ is p > 0.25. You observe x = 9. Given P(X ≤ 8) = 0.9591, find the p-value = P(X ≥ 9), to 4 decimal places.
Hint: P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.9591.
Quick check

The conclusion

?The p-value is 0.0409 and the test is at the 5% significance level. What do you conclude?
Hypothesis testing · critical regions

Critical regions and actual significance

Instead of a p-value, you can find the critical region in advance — the values of the test statistic that would make you reject H₀.

Because the binomial is discrete, you can rarely hit the significance level exactly. The actual significance level is the true probability of landing in the critical region when H₀ is true.

Worked example — X ~ B(10, 0.5), two-tailed test at 5%

2.5% is wanted in each tail. Under H₀, each outcome has probability 10Cr/1024.

Lower tail: P(X ≤ 1) = (1 + 10)/1024 = 11/1024 = 0.01074.  (P(X ≤ 2) = 56/1024 = 0.0547, which is too big — so the lower critical region is X ≤ 1.)

Upper tail: by symmetry, P(X ≥ 9) = 0.01074.

Actual significance level = 0.01074 + 0.01074 = 0.0215, i.e. 2.15% — noticeably less than the nominal 5%.

Why it matters: the discreteness makes the test more conservative than advertised. The actual significance level is the real probability of a Type I error.

Calculate

Your turn — actual significance level

2For X ~ B(10, 0.5), the critical region for a two-tailed test is X ≤ 1 or X ≥ 9. Given P(X ≤ 1) = 0.01074 and P(X ≥ 9) = 0.01074, find the actual significance level, to 4 decimal places.
Hint: add the two tails: 0.01074 + 0.01074.
Hypothesis testing · normal mean

Testing a Normal mean

If X ~ N(μ, σ²), then the sample mean of n observations has a smaller spread:

X̄ ~ N( μ , σ²/n )   ⇒   z = (x̄ − μ) ÷ (σ/√n)σ/√n is the standard error — averaging reduces the variability by a factor of √n
Worked example

X ~ N(μ, 6²). A sample of n = 25 has mean x̄ = 52.4. Test H₀: μ = 50 against H₁: μ > 50 at the 5% level.

Standard error = σ/√n = 6/√25 = 6/5 = 1.2.

z = (52.4 − 50) / 1.2 = 2.4 / 1.2 = 2.

Method 1 — critical value: for a one-tailed 5% test, zcrit = 1.645. Since 2 > 1.645, reject H₀.

Method 2 — p-value: P(Z > 2) = 1 − Φ(2) = 1 − 0.97725 = 0.0228. Since 0.0228 < 0.05, reject H₀. Both agree. ✓

In context: there is significant evidence at the 5% level that the mean has increased above 50.

The critical values to know: one-tailed 5% ⇒ 1.645; one-tailed 1% ⇒ 2.326; two-tailed 5% ⇒ ±1.96; two-tailed 1% ⇒ ±2.576.

Calculate

Your turn — the test statistic

3X ~ N(μ, 6²). A sample of n = 25 gives a mean of 52.4. Testing H₀: μ = 50, find the test statistic z.
z =
Hint: standard error = 6/√25 = 1.2. Then z = (52.4 − 50) ÷ 1.2.
Calculate

Your turn — the p-value

4Continuing: H₁ is μ > 50 and z = 2. Given Φ(2) = 0.97725, find the p-value P(Z > 2), to 4 decimal places.
Hint: one-tailed, upper tail: p = 1 − Φ(2) = 1 − 0.97725.
Calculate

Your turn — the critical value

5For a two-tailed test at the 5% significance level using a Normal test statistic, what is the positive critical value of z?
z =
Hint: two-tailed at 5% means 2.5% in each tail, so you need the z with 0.975 below it.
Quick check

Two-tailed at 5%

?A two-tailed test is carried out at the 5% significance level. How is the 5% distributed?
Hypothesis testing · correlation and errors

Correlation tests and Type I errors

Testing correlation. You can test whether a sample correlation coefficient r is strong enough to be evidence of a real population correlation ρ.

  • H₀: ρ = 0 (no correlation in the population). H₁: ρ > 0, ρ < 0 or ρ ≠ 0.
  • Compare your |r| with the critical value from OCR’s table for the given n and significance level.
  • If |r| exceeds the critical value, reject H₀ — there is evidence of correlation.

The critical value depends on n. With a small sample, even a fairly big r can arise by chance, so the bar is high. With a large sample, a modest r can be significant. This is why the sample size must always be quoted.

Type I error. Rejecting H₀ when it is actually true — a false positive. Its probability is the (actual) significance level. Choosing 1% instead of 5% makes a Type I error less likely, but makes it harder to detect a real effect.

Quick check

Type I error

?What is a Type I error?
Sort it

Reject or not?

Tap the evidence, then tap the correct decision.

❌ Reject H₀

✅ Do not reject H₀

❓ Not enough information

Match it

The language of testing

Tap a term on the left, then its definition.

Term
Definition
Recap

The big ideas to know

Set up: H₀ states a specific value; H₁ is what you suspect; state the significance level

Decide: p-value < α ⇒ reject H₀; or compare the test statistic with the critical value

Two-tailed: split α — 5% two-tailed means 2.5% per tail, zcrit = ±1.96

Binomial test: X ~ B(n, p) under H₀; p-value = P(X ≥ x) or P(X ≤ x), matching H₁

Actual significance: the true tail probability of the discrete critical region — usually less than α

Normal mean: X̄ ~ N(μ, σ²/n), so z = (x̄ − μ)/(σ/√n)

Correlation: compare |r| with the critical value for that n; H₀ is ρ = 0

Type I error: rejecting a true H₀; its probability is the significance level

That is OCR 2.05 — and the last mark is always the conclusion in context. Press Finish to see your score.

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