This mini-lesson covers Topic 5 of Edexcel A-level Mathematics (9MA0): the language of testing — null and alternative hypotheses, significance level, test statistic, 1-tail and 2-tail tests, critical value, critical region, acceptance region and p-value — then three tests you must be able to carry out: a test for a binomial proportion p, a test for the mean of a normal distribution, and a test for a correlation coefficient ρ.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s−2 unless a question says otherwise. Press Start when you are ready.
Two-tailed rule: at the 5% level, each tail gets 2.5%. Comparing a one-tailed p-value with 0.025 (or doubling it and comparing with 0.05) is the standard method.
Never say "H₀ is true". The correct wording is: "there is sufficient / insufficient evidence to reject H₀ at the 5% level", followed by a sentence in context.
Tap a statement, then tap where it belongs.
Model the count with X ~ B(n, p), assume H₀ is true, then find the probability of the observed value or more extreme.
A coin-flip style claim: p = 0.3. In n = 20 trials there are 10 successes. Test at the 5% level whether p has increased.
H₀: p = 0.3 · H₁: p > 0.3 (one-tailed)
Assume X ~ B(20, 0.3). p-value = P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.9520 = 0.0480.
0.0480 < 0.05, so reject H₀: there is sufficient evidence at the 5% level that the proportion has increased.
X ~ B(10, 0.5) under H₀: p = 0.5, with H₁: p ≠ 0.5, at the 5% level. Observed: 9 successes.
9 is above the expected value np = 5, so use the upper tail: P(X ≥ 9) = (10 + 1) × 0.5¹⁰ = 11 ÷ 1024 = 0.01074.
Two-tailed: compare 0.01074 with 0.025 — or double it, 2 × 0.01074 = 0.0215, and compare with 0.05.
0.0215 < 0.05, so reject H₀: there is evidence the probability is not 0.5.
If X ~ N(μ, σ²), then the sample mean of n observations satisfies:
Note the variance is divided by n: bigger samples give a more reliable mean. Hypotheses are stated in terms of the population mean μ.
X ~ N(μ, 5²). A sample of n = 25 gives a sample mean of x̄ = 52. Test H₀: μ = 50 against H₁: μ > 50 at the 5% level.
Standard error = σ ÷ √n = 5 ÷ √25 = 5 ÷ 5 = 1.
z = (52 − 50) ÷ 1 = 2. p-value = P(Z > 2) = 1 − 0.9772 = 0.0228.
0.0228 < 0.05 → reject H₀. There is evidence at the 5% level that the mean has increased.
Critical value method: for a one-tailed 5% test the critical value is z = 1.6449; since 2 > 1.6449, reject H₀ — the same conclusion.
Tap an item on the left, then its partner on the right.
The product moment correlation coefficient r measures how close the sample's points lie to a straight line, with |r| ≤ 1. We use it to test the population correlation coefficient, ρ (rho).
A sample of n = 12 pairs gives r = 0.62. Test at the 5% level whether there is positive correlation.
H₀: ρ = 0 · H₁: ρ > 0 (one-tailed). The critical value from the table is 0.4973.
r = 0.62 > 0.4973, so the result is in the critical region → reject H₀.
In context: there is sufficient evidence at the 5% level of a positive correlation between the two variables.
You are not asked to calculate r by hand — a calculator (or the question) supplies it. What is examined is the hypotheses, the comparison and the conclusion in context.
Instead of finding a p-value you can find the critical region in advance: the smallest set of extreme values whose total probability is at most the significance level.
Work down from the top until the tail probability exceeds 5%:
P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.9520 = 0.0480 ≤ 0.05 ✔
P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.8867 = 0.1133 > 0.05 ✘
So the critical region is X ≥ 10, and the actual significance level of the test is 0.0480, i.e. 4.80% — not exactly 5%, because X is discrete.
Actual significance level = the true probability of the test statistic falling in the critical region when H₀ is true. For a discrete distribution it is almost never exactly α — this is a favourite exam question, and the answer is a probability, not "5%".
H₀: always an equality about a population parameter (p, μ or ρ = 0) · H₁: >, < (1-tail) or ≠ (2-tail)
Significance level: the probability of incorrectly rejecting H₀; two-tailed ⇒ split it, 2.5% per tail
Binomial test: assume X ~ B(n, p₀), find P(X ≥ x) or P(X ≤ x), compare with α
Normal mean test: X̄ ~ N(μ, σ²/n), so z = (x̄ − μ) ÷ (σ ÷ √n)
Correlation test: H₀: ρ = 0; compare |r| with the given critical value
Conclusion: “sufficient / insufficient evidence to reject H₀ at the 5% level” — then say what it means in context
That is the whole of 9MA0 Topic 5 — Statistical hypothesis testing. Press Finish to see your score.
You have worked through Statistical hypothesis testing for Edexcel A-level Mathematics (9MA0). 🎉
Your stars: 0 / 0
Next: test yourself in the Verify stage.