Read the design before the numbers

Before writing a hypothesis, decide which of two situations the study is:

1. Two independent samples. Different individuals in each group, with no link between them. Test $\mu_1 - \mu_2$ with the two-sample procedure.
2. Paired data. The same individuals measured twice, or individuals matched into pairs. Subtract within pairs and test $\mu_d$ with the one-sample procedure of Topic 4.4.

Thirty students taught by method A and twenty-eight different students taught by method B are two independent samples. Twenty runners who each try two shoe models are paired. Twenty-five pairs of twins with one twin in each condition are paired, even though no individual repeats, because the pairing links the groups.

Getting this wrong is not a small error. The paired analysis subtracts first and has $n - 1$ degrees of freedom on the differences; the two-sample analysis adds two variances and uses a different df. They answer the same research question with different machinery, and only one matches the data.

For two independent samples,

$$H_0: \mu_1 = \mu_2 \qquad \text{equivalently} \qquad H_0: \mu_1 - \mu_2 = 0,$$

against $H_a: \mu_1 - \mu_2 > 0$, $< 0$, or $\ne 0$.

For the teaching-methods study, with $\mu_1$ the mean score for all students taught by method A and $\mu_2$ the mean for all taught by method B, a researcher who expects A to do better writes $H_0: \mu_1 = \mu_2$ against $H_a: \mu_1 > \mu_2$.

Three constructions that are not hypotheses:

1. $H_0: \mu_1 = 78.4$ and $\mu_2 = 71.2$. Those values came from the samples.
2. $H_0: \bar{x}_1 = \bar{x}_2$. The sample means are known and differ by 7.2.
3. $H_0: \mu_1 - \mu_2 = 7.2$. The observed difference is the evidence, not the claim.

The null names no numerical value for either mean, only equality, and the alternative's direction comes from the research question rather than from which sample mean came out higher.

1. Random: both samples random, or subjects randomly assigned to the two conditions.
2. Independence between groups: no individual in both, and no matching between them.
3. 10%: each sample at most a tenth of its own population.
4. Normal or large enough, per group: $n \ge 30$, a stated normal population, or a plot with no strong skew and no outliers, assessed for each group separately.

The procedure is a two-sample t test, and the degrees of freedom are either technology's fractional value or the conservative $\min(n_1, n_2) - 1$, stated either way. As always, a condition is checked by writing the study's numbers next to it, not by naming it.

For the teaching-methods study:

1. Parameters: $\mu_1$ = mean score for all students taught by method A; $\mu_2$ = the same for method B.
2. Hypotheses: $H_0: \mu_1 = \mu_2$ against $H_a: \mu_1 > \mu_2$.
3. Conditions: two-sample t test; random assignment stated; groups independent; each sample under 10% of its population; group A has $n = 30$, and group B's boxplot of 28 scores is roughly symmetric with no outliers.
4. Significance level: $\alpha = 0.05$, fixed in advance.

The design also decides what the eventual conclusion may claim. Random assignment to the two methods licenses a causal statement; two intact classes license only a statement that the means differ, with class composition as an alternative explanation. The test statistic is identical in both cases, so the difference lives entirely in the final sentence.