Significance Tests for a Mean
- State hypotheses and compute a one-sample t test statistic
- Find a p-value using the t-distribution and make a decision
- Connect the confidence interval and two-sided test
The one-sample t test
To test a claim about μ, set H₀: μ = μ₀ against Hₐ (μ <, >, or ≠ μ₀). The t statistic measures how many estimated standard errors x-bar lies from μ₀, and we find the p-value from the t-distribution with n − 1 degrees of freedom. The same conditions as the t-interval apply: Random, 10%, and Normal/Large Sample.
Interval–test duality
A two-sided test at level α and a (1 − α) confidence interval agree: if the hypothesized μ₀ lies outside the confidence interval, you would reject H₀ at level α; if μ₀ lies inside, you fail to reject. This link lets a confidence interval double as a two-sided test, and it is a favorite AP connection to explain.
A manufacturer claims mean fill = 500 mL. An SRS of 36 bottles gives x-bar = 496 mL, s = 12 mL. Test Hₐ: μ ≠ 500 at α = 0.05.
- 1.Hypotheses: H₀: μ = 500 vs. Hₐ: μ ≠ 500. Conditions: random; n = 36 ≥ 30 so CLT covers shape; < 10% of production.
- 2.Standard error: s/sqrt(n) = 12/sqrt(36) = 12/6 = 2.
- 3.t = (496 − 500)/2 = −4/2 = −2.0, with df = 35.
- 4.Two-sided p-value = 2·P(t₃₅ < −2.0) ≈ 2(0.027) ≈ 0.053, which is greater than 0.05.
A 95% confidence interval for μ is (48, 54). For a two-sided test of H₀: μ = 50 at α = 0.05, what is the decision?
A large p-value means "not enough evidence," not "H₀ is true." Failing to reject H₀ that μ = 500 does not prove the mean is exactly 500 — it only means the data are consistent with that value. Never state a conclusion as proof of the null.
A one-sample t test gives t = 2.8 with df = 20 for a one-sided (upper) alternative. The p-value is the:
Answer the 2 checkpoints as you read.
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