Hypotheses are about mu, never x-bar

$$H_0: \mu = \mu_0 \qquad \text{against} \qquad H_a: \mu < \mu_0, \; \mu > \mu_0, \text{ or } \mu \ne \mu_0.$$

A manufacturer claims its batteries last a mean of 128 hours. A consumer group suspects the true mean is lower and tests 25 randomly selected batteries, obtaining $\bar{x} = 124.6$ and $s = 9.2$. The hypotheses are

$$H_0: \mu = 128 \qquad \text{versus} \qquad H_a: \mu < 128,$$

where $\mu$ is the mean lifetime of all batteries of this type.

Three constructions that are not hypotheses:

1. $H_0: \bar{x} = 128$. The sample mean is 124.6, a known number, so there is nothing to test.
2. $H_0: \mu \ge 128$. The null carries equality, because the p-value is computed from one specific sampling distribution.
3. Hypotheses about "the 25 batteries tested". Those 25 have a mean of exactly 124.6.

The alternative's direction comes from the investigation, not the sample. Seeing 124.6 and then choosing $\mu < 128$ puts the whole significance level in a tail the data already selected.

When each individual or matched pair supplies two measurements, subtract first and state the hypotheses about $\mu_d$, the population mean difference:

$$H_0: \mu_d = 0 \qquad \text{against} \qquad H_a: \mu_d > 0, \; < 0, \text{ or } \ne 0.$$

The null of 0 says the mean did not change. For a training program measured before and after, $H_a: \mu_d > 0$ with the order stated as post minus pre says the program is expected to raise scores.

Two things must be written down. The order of subtraction, since it decides the sign of everything downstream. And the fact that the sample size is the number of pairs: twelve students measured twice give $n = 12$ and $df = 11$, not 24 and 23.

Reading the design is the skill this shares with Topic 4.2. Before-and-after on the same subjects, matched pairs, or two measurements on the same unit are paired. Two unrelated groups are the two-sample procedure of Topic 4.9.

Named with the study's numbers:

1. Random: a random sample or randomized experiment. Here, 25 randomly selected batteries.
2. 10%: $25 \le 0.10N$, satisfied for a production run in the thousands.
3. Normal or large enough: $n \ge 30$, a stated normal population, or a plot of the sample showing no strong skew and no outliers. At $n = 25$ the plot is the evidence, so it is drawn and described.

The procedure is a one-sample t test, with $df = n - 1 = 24$. It is t and not z because $\sigma$ is unknown and $s$ stands in, which is true in essentially every real study. A z test for a mean built from the sample's own $s$ is the recurring error of this unit, and it understates the uncertainty by using critical values that are too small.

Written out for the battery study:

1. Parameter: $\mu$ = the mean lifetime of all batteries of this type.
2. Hypotheses: $H_0: \mu = 128$ against $H_a: \mu < 128$.
3. Conditions: one-sample t test; random sample stated; $25 \le 0.10N$; the dotplot of the 25 lifetimes is roughly symmetric with no outliers.
4. Significance level: $\alpha = 0.05$, fixed now rather than after the p-value appears.

Choosing $\alpha$ in advance is what keeps the decision rule from being written to fit the result, and stating the parameter in context is what keeps the conclusion from drifting onto the 25 batteries actually tested.