This mini-lesson covers AQA section O: the full language of testing — H₀, H₁, significance level, test statistic, 1-tail and 2-tail tests, critical value, critical region, acceptance region and p-value — then testing a binomial proportion, testing the mean of a Normal distribution, and interpreting a correlation coefficient.
The critical region is where the test statistic is so extreme that we reject H₀. For a two-tailed 5% test, split it: 2.5% in each tail.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. AQA note: you interpret a given correlation coefficient against a given critical value or p-value — calculating r is excluded. Press Start when you're ready.
Section O1 · language
Setting up a test
A hypothesis test uses a sample to decide between two statements about a population.
H₀ — the null hypothesis. Always states equality (p = 0.25, μ = 30). It is assumed true unless the evidence against it is strong.
H₁ — the alternative hypothesis. Uses >, < (one-tailed) or ≠ (two-tailed).
The significance level (5%, 1%…) is the probability of rejecting H₀ when it is actually true — a Type I error.
The critical region is the set of values of the test statistic that lead you to reject H₀; everything else is the acceptance region.
The p-value is P(result at least as extreme as observed | H₀ true). If p-value < significance level → reject H₀.
Two-tailed = split the level. A 5% two-tailed test puts 2.5% in each tail. Forgetting to halve is the single most common error in this topic.
Sort it
H₀ or H₁?
Tap a hypothesis statement, then tap where it belongs.
⚖️ H₀ (null)
➡️ H₁ one-tailed
↔️ H₁ two-tailed
Section O2 · binomial test
Testing a binomial proportion
A seed company claims 20% of its seeds germinate. A gardener plants 20 seeds and 8 germinate. Test at the 5% level whether the germination rate is higher than claimed.
The full method (write all of it in the exam)
1. Model: let X = number that germinate. Under H₀, X ~ B(20, 0.2).
2. Hypotheses: H₀: p = 0.2 H₁: p > 0.2 (one-tailed).
6. Conclude in context: reject H₀. There is significant evidence at the 5% level that more than 20% of the seeds germinate.
Critical-region method (same test, other route): find the smallest c with P(X ≥ c) ≤ 0.05. Since P(X ≥ 8) = 0.0321 ≤ 0.05 but P(X ≥ 7) = 0.0867 > 0.05, the critical region is X ≥ 8, and the actual significance level is 0.0321, i.e. 3.21%.
Calculate
Binomial p-value
1H₀: p = 0.2, H₁: p > 0.2, with X ~ B(20, 0.2). The observed value is x = 8. Calculate the p-value, P(X ≥ 8), to 4 decimal places.
2A test uses X ~ B(20, 0.35) with H₁: p < 0.35 at the 5% level. Given P(X ≤ 3) = 0.0444 and P(X ≤ 4) = 0.1182, state the largest value c for which the critical region is X ≤ c.
Hint: You need P(X ≤ c) ≤ 0.05. P(X ≤ 3) = 0.0444 ≤ 0.05 ✓ but P(X ≤ 4) = 0.1182 > 0.05 ✗.
Quick check
Reaching a conclusion
?In the seed test above, the p-value was 0.0321 and the significance level was 5%. What is the correct conclusion?
Quick check
What the level means
?What exactly does a 5% significance level mean?
Section O3 · normal mean
Testing the mean of a Normal distribution
If X ~ N(μ, σ²) with known variance, then the mean of a random sample of size n satisfies:
X̄ ~ N(μ, σ² / n) ⇒ z = (x̄ − μ) ÷ (σ / √n)the standard error is σ / √n — bigger samples give a tighter distribution of x̄
Worked example
Machine fills bottles with X ~ N(μ, 3²). A sample of n = 25 bottles has mean x̄ = 26.2. Test at the 5% level whether μ > 25.
H₀: μ = 25 H₁: μ > 25 (one-tailed)
Standard error = σ / √n = 3 ÷ √25 = 3 ÷ 5 = 0.6
z = (26.2 − 25) ÷ 0.6 = 1.2 ÷ 0.6 = 2.0
p-value = P(Z > 2) = 1 − 0.9772 = 0.0228 < 0.05 ✓
Conclusion: reject H₀ — significant evidence that the mean fill is more than 25.
Critical values you should know: one-tailed 5% → z = 1.645; one-tailed 1% → z = 2.326; two-tailed 5% → z = ±1.96; two-tailed 1% → z = ±2.576.
Calculate
Test statistic for a mean
3X ~ N(μ, 3²). A sample of n = 25 has mean x̄ = 26.2. Testing H₀: μ = 25, calculate the test statistic z.
4For that test (one-tailed, H₁: μ > 25), the test statistic is z = 2. Calculate the p-value, P(Z > 2), to 4 decimal places.
Hint: P(Z > 2) = 1 − Φ(2) = 1 − 0.97725.
Calculate
Two-tailed critical value
5A two-tailed test at the 5% significance level uses the Normal distribution. State the positive critical value of z, to 2 decimal places.
Hint: Two-tailed → 2.5% in each tail. Find z with P(Z < z) = 0.975.
Quick check
Spot the error
?A student writes: "H₀: μ = 30, H₁: μ ≠ 30, 5% level, so I reject H₀ if z > 1.645." What is wrong?
Match it
The vocabulary of testing
Tap a definition on the left, then its name on the right.
Definition
Term
Quick check
Testing a correlation coefficient
?For a sample of n = 12 bivariate observations, the product moment correlation coefficient is r = 0.62. A one-tailed test of H₀: ρ = 0 against H₁: ρ > 0 at the 5% level has critical value 0.4973. What is the conclusion?
Section O2 · two-tailed
A two-tailed binomial test
A coin is tossed 10 times to test whether it is biased (in either direction) at the 5% level.
Worked example
H₀: p = 0.5 H₁: p ≠ 0.5 → two-tailed, so put 2.5% in each tail.
Why is the actual level below 5%? Because X is discrete — you can only put whole numbers of outcomes in the critical region, so you can rarely hit exactly 2.5% in each tail. The actual level is always at most the nominal one.
Calculate
Actual significance level
6For X ~ B(10, 0.5) the critical region of a two-tailed 5% test is X ≤ 1 or X ≥ 9, and P(X ≤ 1) = P(X ≥ 9) = 0.0107. Calculate the actual significance level, to 4 decimal places.
Hint: Add the two tails: 0.0107 + 0.0107 (more precisely 0.010742 × 2).
Quick check
Splitting the tails
?In a two-tailed test at the 5% significance level, what probability is placed in each tail?