One-sample t-test for a mean
Hypothesis testingA cereal box claims a mean fill of 50 g. A random sample of 25 boxes has mean 51.2 g and standard deviation 4.0 g. At α = 0.05, is there evidence the true mean differs from 50 g?
- Hypotheses. H₀: μ = 50 vs. Hₐ: μ ≠ 50 (two-tailed — "differs," not "is more").
- Choose the test and check conditions. Population σ is unknown and n = 25, so use a one-sample t-test (random sample; box weights roughly normal or n moderately large).
- Compute the statistic. t = (x̄ − μ₀)/(s/√n) = (51.2 − 50)/(4.0/√25) = 1.2/0.8 = 1.50, with df = 24.
- Compare. The two-tailed critical value at α = 0.05 with df = 24 is ±2.064. Since |1.50| < 2.064 (p ≈ 0.147), the statistic is not in the rejection region.
Fail to reject H₀: at the 5% level, this sample does not give convincing evidence that the true mean fill differs from 50 g.
Common mistake the solver flags: saying "we accept H₀" or "the mean is 50 g." Failing to reject is not proof the claim is true — only that this sample lacks evidence against it.