Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- A full test has four parts: hypotheses, method and conditions, calculations, conclusion.
- The test statistic t says how many standard errors x̄ is from μ₀.
- If H₀ is true and the conditions hold, t follows a t-distribution with n − 1 degrees of freedom.
- The p-value is the probability, assuming H₀ is true, of a test statistic at least as extreme as the observed one, in the direction of Hₐ.
- The conclusion is about the parameter (μ or μd) and the population, never about x̄.
Key relationships
| Item | One-sample mean | Matched pairs |
|---|---|---|
| Test statistic | t = (x̄ − μ₀) / (s / √n) | t = (x̄d − 0) / (sd / √n) |
| n | sample size | number of pairs (differences) |
| Degrees of freedom | n − 1 | n − 1 |
| Hₐ with > | p-value = P(T ≥ t) | same |
| Hₐ with < | p-value = P(T ≤ t) | same |
| Hₐ with ≠ | p-value = 2 × P(T ≥ |t|) | same |
| Decision | p-value ≤ α: reject H₀; p-value > α: fail to reject H₀ | same |
Assumptions and conventions
- σ is unknown, so the standard error uses s and the p-value uses a t-distribution.
- The conditions from Topic 4.4 must hold: random sample or randomized experiment; n ≤ 10% of N when sampling without replacement; normal population, n ≥ 30, or no strong skew or outliers in the sample (of differences).
- A t-table gives a range for the p-value. If your df is missing, use the next smaller df.
- If no α is given, use 0.05 and say so.
Mistakes to avoid
- Using z or the normal curve instead of the t-distribution.
- df = n, or counting both measurements in each pair.
- Wrong tail or not doubling for a two-sided test.
- “The p-value is the probability that H₀ is true.” It is calculated assuming H₀ is true.
- “Accept H₀” or “proves”. Use “fail to reject” and “convincing evidence”.
- No linkage: compare the p-value with α, using both numbers.
- Concluding about x̄ instead of the true mean of the population.
Quick self-check
- A random sample of 16 gives x̄ = 81.5 and s = 6. Find t for H₀: μ = 78. (t = 3.5 / (6 / 4) ≈ 2.33, df = 15)
- For Hₐ: μ > μ₀, t = 2.05 and n = 25. Find the p-value. (P(T ≥ 2.05) with df = 24 ≈ 0.0257)
- A test gives p-value = 0.08 with α = 0.05. What is the decision? (Fail to reject H₀: there is not convincing evidence for Hₐ)
Next: practice questions.