Carrying Out a Test for a Mean · Mise en œuvre d'un test pour une moyenne
Do not double the wrong tail
- For t=−2 with 15 degrees of freedom, the left tail is about 0.032. The right tail is about 0.968.
- A lower-mean alternative uses 0.032. A two-sided alternative uses twice the smaller tail beyond |t|, about 0.064, not twice 0.968.
Find the p-value
- Convert the $t$ statistic to a p-value using the $t$-distribution at $df = n-1$. One-sided $H_a$: the tail area beyond $t$ in that direction.
- Two-sided $H_a$: double the smaller tail beyond $|t|$. The p-value counts results at least as extreme in the direction or directions specified by $H_a$, assuming $H_0$.
Using rounded t left-tail area 0.032 at t = -2 and df = 15, match the alternative to its p-value.
The right tail is the complement. The symmetric two-sided test combines both equally extreme smaller tails.
Compare to alpha
- Compare the p-value to the significance level $\alpha$ (usually $0.05$). p $\le \alpha$ → the result is statistically significant.
- p $> \alpha$ → not significant. Set $\alpha$ before running the test.
Reject or fail to reject
- p $\le \alpha$ → reject $H_0$ (strong enough evidence against it). p $> \alpha$ → fail to reject $H_0$ (not enough evidence).
- Never "accept $H_0$" — only "fail to reject." Same decision rule as any significance test.
With p-value = 0.032 and α = 0.05, the decision is...
0.032 ≤ 0.05 → reject H0.
The p-value for a t-test is read from...
Use the t-distribution with the correct df.
For a symmetric two-sided t-test, the p-value is twice the smaller one-tail area.
Use twice the area beyond |t| in the correct t distribution. Doubling the larger tail is incorrect.
Failing to reject H0 proves the population mean equals μ0.
It only means insufficient evidence against H0.
The conclusion of a mean test should be stated...
Tie the conclusion to μ, in context, with units.
Conclude in context
- Write the conclusion about the population mean, in context, with units. "There is convincing evidence the mean fill is below $500$ mL" (if rejected).
- Or "there is not convincing evidence..." (if not). Failing to reject does not · non prove $H_0$ true.
Get the p-value from the $t$-distribution, not the normal — the correct $df = n-1$ matters, especially for small $n$ where $t$ and · et $z$ differ most. And match the tail(s) to $H_a$: the specified tail for one-sided, twice the smaller tail for a symmetric two-sided test. As always, a large p-value means "insufficient evidence," never "proof of $H_0$."
The lower-sided test rejects at 0.05 while the two-sided test does not. Which are justified?
Preselection preserves the intended test meaning. The distinction does not justify post-hoc direction changes.
From 7.4: $t = -2$ with $df = 15$, one-sided $H_a: \mu < 500$, $\alpha = 0.05$.
- p-value: area left of $t = -2$ at $df = 15 \approx 0.032$.
- Compare: $0.032 \le 0.05$ → reject $H_0$.
- Conclude: convincing evidence the true mean fill is below $500$ mL.
Carry the reasoning to a new case
- At alpha=0.05 the left-sided test rejects and the two-sided test does not.
- This difference follows the preselected alternative, not a licence to change it after seeing data.
Using the given left-tail area 0.032, find the two-sided p-value.
For the symmetric t distribution, double the tail beyond |t|: 2×0.032=0.064.
Find the p-value from the $t$ statistic at $df = n-1$ (twice the smaller tail for two-sided), compare to $\alpha$: reject $H_0$ if p $\le \alpha$, else fail to reject. State the conclusion about the population mean in context — and remember failing to reject never proves $H_0$.