Two groups, two variances, added

Take independent random samples of sizes $n_1$ and $n_2$ from two populations with means $\mu_1, \mu_2$ and standard deviations $\sigma_1, \sigma_2$. Across all such pairs of samples,

$$\mu_{\bar{x}_1 - \bar{x}_2} = \mu_1 - \mu_2.$$

The difference of two unbiased estimators is unbiased for the difference of the parameters.

The order of subtraction is chosen once and then kept. Comparing two teaching methods, with method A giving $\bar{x}_1 = 78.4$ from 30 students and method B giving $\bar{x}_2 = 71.2$ from 28, the difference $\bar{x}_1 - \bar{x}_2 = 7.2$ says A's mean runs 7.2 points higher. Reversing the order gives $-7.2$, the same fact stated the other way, and a report that never states the order leaves a reader unable to tell which.

Because the samples are independent,

$$\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}.$$

The plus sign survives the minus sign in the difference: each group contributes its own uncertainty to the comparison. If variances subtracted, comparing two equally variable groups would produce a difference with no variability at all, which is plainly false.

With $\sigma$ unknown, as always, the sample standard deviations stand in and the result is the standard error:

$$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{8.1^2}{30} + \frac{9.6^2}{28}} = \sqrt{2.187 + 3.291} \approx 2.341.$$

Three wrong routes give plausible-looking numbers. Subtracting the variances gives $2.187 - 3.291 = -1.104$, a negative radicand. Adding the standard errors gives $1.479 + 1.814 = 3.293$, too large by 40%. And pooling the two samples into one group of 58 discards the two-group structure entirely.

Note also that the two contributions are $\frac{s^2}{n}$ for each group separately, so a small group dominates: with $n_2 = 5$ and $s_2 = 9.6$, that group alone would contribute $\frac{92.16}{5} = 18.4$ and swamp anything the larger group brought.

With two estimated standard deviations, the exact distribution of the standardized difference is not a t distribution at all, only close to one, and the matching degrees of freedom come out fractional. Two answers are accepted:

1. Technology's value. A calculator or software reports a fractional $df$, computed from the two sample sizes and standard deviations. For this study it is about 53.0.
2. The conservative value. Use $df = \min(n_1, n_2) - 1$, which here is $28 - 1 = 27$. It is smaller than the true value, so critical values are slightly larger and intervals slightly wider: safe, and easy to do by hand.

What is not acceptable is $n_1 + n_2 - 1$ or $n_1 + n_2 - 2$ chosen by habit, or a df borrowed from a one-sample setting. The conservative rule and technology's value will rarely change a conclusion, and either has to be stated so a reader knows which was used.

Each condition from the one-sample case applies to each group, and one new condition applies once:

1. Random: both samples random, or subjects randomly assigned to the two treatments.
2. Independence between the groups: no individual appears in both. Before-and-after measurements on the same people are paired, and the variances-add derivation does not apply to them.
3. 10%: each sample at most a tenth of its own population, checked separately.
4. Normal or large enough: for each group, $n \ge 30$, a stated normal population, or a plot showing no strong skew and no outliers. Both groups must clear it.

The shape condition is checked group by group, so a study with $n_1 = 40$ and $n_2 = 9$ still needs a plot of the second group. And with the counts large enough, the sampling distribution of $\bar{x}_1 - \bar{x}_2$ is approximately normal, which is what makes the t procedures of the next four topics work.