The conclusion has to show its comparison

$$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}, \qquad df = n - 1.$$

For the battery study, $\bar{x} = 124.6$, $s = 9.2$, $n = 25$, and $\mu_0 = 128$:

$$t = \frac{124.6 - 128}{1.84} \approx -1.85, \qquad df = 24.$$

Against $H_a: \mu < 128$ the p-value is the area to the left of $-1.85$ in the $t_{24}$ distribution, about $0.0385$. Against $H_a: \mu \ne 128$ it would be about $0.077$, twice as large.

Two mechanical points. The tail is chosen by the alternative, exactly as in Unit 3. And the p-value must be read from the t distribution at $df = 24$, not from the normal table: the normal area beyond $-1.85$ is about 0.032, which is about a sixth smaller and makes the evidence look stronger than it is.

For paired data the same formula runs on the differences: $t = \frac{\bar{x}_d - 0}{s_d/\sqrt{n}}$ with $df$ equal to the number of pairs minus one.

At $\alpha = 0.05$ with the one-sided p-value of 0.0385:

"Since the p-value of 0.0385 is less than $\alpha = 0.05$, we reject $H_0$. There is convincing evidence that the mean lifetime of all batteries of this type is less than 128 hours."

Three parts, and a missing one costs credit:

1. The explicit comparison. "Reject $H_0$" with no p-value stated beside $\alpha$ gives a reader nothing to verify. The comparison also has to be arithmetically right: $0.0385 < 0.05$ rejects, while $0.077 > 0.05$ does not.
2. The decision. Reject or fail to reject, never accept.
3. The context. A sentence about "the null hypothesis" that never mentions battery lifetime or the population is not a conclusion.

Note the consistency with Topic 4.3: the two-sided p-value of 0.077 exceeds 0.05, and the 95% interval $(120.8, 128.4)$ contains 128. A two-sided test at $\alpha$ and an interval at $1 - \alpha$ agree, and the one-sided test rejects because it concentrates all of $\alpha$ in the lower tail.

Had the p-value been 0.31, the report would read: there is not convincing evidence that the mean lifetime is less than 128 hours. Not that the mean is 128, not that the manufacturer's claim is correct, and not that the batteries meet the specification.

The reason is the same as always. Many values besides 128 would produce an unsurprising sample, and with $n = 25$ the test has limited power to detect a shortfall of a few hours. A non-significant result reported alongside its sample size lets a reader judge what the study could have detected.

t procedures are robust: they perform close to their stated confidence and error rates even when the population is not exactly normal. What that means in practice depends on $n$:

1. $n < 15$: use t only if the data look roughly symmetric with no outliers. A single wild value at this size dominates both $\bar{x}$ and $s$.
2. $15 \le n < 30$: mild skew is acceptable; strong skew or outliers are not.
3. $n \ge 30$: t is safe for almost any shape short of extreme outliers.

Two opposite failures follow from ignoring this. Running a t test on $n = 8$ with one wild outlier and reporting the p-value without comment treats a number as evidence when the procedure's assumptions have failed. And refusing inference at $n = 45$ over mild skew discards a perfectly valid analysis.

What is never acceptable is deleting an outlier to make the procedure apply. An unusual value is investigated and reported: if it is a recording error it can be corrected, and if it is real it belongs in the analysis, with the results reported both with and without it if its influence is large.