Significance tests for a mean

A one-sample t test asks whether data give convincing evidence against a claimed mean $\mu_0$. The null hypothesis is $H_0: \mu = \mu_0$. The alternative is $H_a: \mu > \mu_0$, $\mu < \mu_0$ or $\mu \neq \mu_0$, chosen from the question before you look at the data. Hypotheses are always about the parameter $\mu$, never about $\bar{x}$.

Check the same conditions as for an interval: Random, 10%, Normal/Large Sample. Then compute $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ with df = n - 1.

The p-value is the probability, assuming $H_0$ is true, of getting a t statistic at least as extreme as the one observed, in the direction of $H_a$. For $\mu \neq \mu_0$, count both tails.

Compare the p-value with the significance level $\alpha$. If $p < \alpha$, reject $H_0$: there is convincing evidence for $H_a$. If $p \ge \alpha$, fail to reject $H_0$: there is not convincing evidence for $H_a$.

Interpret a p-value by starting with the assumption: if the true mean were $\mu_0$, the probability of getting a sample mean at least as far from $\mu_0$ as ours, in the direction of $H_a$, by chance alone is p.

Justify a conclusion by comparing p with $\alpha$ in a sentence, then stating the decision and what it means in context: because 0.027 is less than 0.05, we reject $H_0$; there is convincing evidence that the true mean fill of all boxes is less than 500 grams.