We focus on the difference $\bar{x}_1 - \bar{x}_2$.

Sampling Distribution of $\bar{x}_1 - \bar{x}_2$

When conditions are met:

• Mean: $\mu_{\bar{x}_1 - \bar{x}_2} = \mu_1 - \mu_2$
• Standard error: $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
• Distribution: Approximately $t$ with complicated degrees of freedom (calculator handles this)

Two-Sample t-Interval for $\mu_1 - \mu_2$

Confidence Interval: $(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE$

where $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ and $t^*$ is the critical $t$-value based on the degrees of freedom.

Degrees of Freedom (Welch's adjustment): $df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}$

(Most calculators compute this automatically; always use the more conservative estimate if doing by hand.)

Two-Sample t-Test for $\mu_1 = \mu_2$

Null Hypothesis: $H_0: \mu_1 = \mu_2$ (or equivalently, $\mu_1 - \mu_2 = 0$)

Test Statistic: $t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{SE}$

Compare to the $t$-distribution with the appropriate df to find the p-value.

Alternative Hypotheses:

• Two-tailed: $H_a: \mu_1 \neq \mu_2$
• One-tailed: $H_a: \mu_1 > \mu_2$ or

Conditions for Two-Sample t-Procedures

All methods require:

Robustness of t-Tests

t-procedures are robust to departures from Normality, especially for large samples and balanced designs ($n_1 \approx n_2$). However:

• With very small samples ($n < 15$) and skewed data, be cautious.
• With very large samples ($n > 100$), slight departures from Normality are not a concern.

Worked Example 1: Two-Sample t-Interval

Two high schools compare math proficiency test scores:

• School A: $n_1 = 25$, $\bar{x}_1 = 78$, $s_1 = 8$
• School B: $n_2 = 30$, $\bar{x}_2 = 75$,

Construct a 95% CI for $\mu_A - \mu_B$.

Standard error: $SE = \sqrt{\frac{8^2}{25} + \frac{10^2}{30}} = \sqrt{\frac{64}{25} + \frac{100}{30}} = \sqrt{2.56 + 3.333} = \sqrt{5.893} \approx 2.427$

Degrees of freedom (conservative estimate: use $n - 1 = 24$ from smaller sample): Using $df = 24$ and 95% confidence, $t^* \approx 2.064$.

CI: $(78 - 75) \pm 2.064(2.427) = 3 \pm 5.008 = (-2.008, 8.008)$

We are 95% confident that School A's mean exceeds School B's by between −2.0 and +8.0 points.

Worked Example 2: Two-Sample t-Test

Test $H_0: \mu_1 = \mu_2$ vs. $H_a: \mu_1 \neq \mu_2$ at $\alpha = 0.05$ using the data above.

Test statistic: $t = \frac{3 - 0}{2.427} \approx 1.236$

p-value (two-tailed with $df \approx 24$): $p\text{-value} = 2 \times P(T > 1.236) \approx 2(0.114) \approx 0.228$

Since $p\text{-value} = 0.228 > 0.05$, we fail to reject $H_0$. Insufficient evidence that the school means differ.

Common Pitfalls

⚠️ Confusing SE with $s$: $SE = \sqrt{s_1^2/n_1 + s_2^2/n_2}$ is the standard error of $\bar{x}_1 - \bar{x}_2$. Do not use the sample standard deviations directly as if they were the population SDs.

⚠️ Forgetting Independence: Two-sample t-tests require independent samples. If the samples are paired (same individuals measured twice), use a paired t-test instead.

⚠️ Misinterpreting CI: A 95% CI does not mean there is a 95% probability that $\mu_1 - \mu_2$ is in the interval for this specific data. Rather, the method has 95% long-run success rate.

Calculator Tip

💡 TI-84 / TI-Nspire: Use 2-SampTInt for confidence intervals and 2-SampTTest for hypothesis tests. Enter the summary statistics ($\bar{x}_1, s_1, n_1, \bar{x}_2, s_2, n_2$) and the alternative hypothesis. The calculator computes df and the interval or test automatically (with or without assuming equal variances).

Test statistic: $t = \frac{(6.2 - 7.4) - 0}{0.440} = \frac{-1.2}{0.440} \approx -2.727$

Degrees of freedom (conservative: $df = 14$): For one-tailed, $H_a: \mu_1 < \mu_2$: $P(T < -2.727 \mid df = 14) \approx 0.008$

Since $p\text{-value} \approx 0.008 < 0.05$, we reject $H_0$. Significant evidence that the drug increases sleep compared to placebo.

Degrees of freedom (approximate, using Welch's formula or conservative $df = 49$): For $df = 49$ (or even $\infty$ for large samples) and 99% confidence, $t^* \approx 2.681$.

CI: $(100 - 102) \pm 2.681(2.717) = -2 \pm 7.286 = (-9.286, 5.286)$

Interpretation: At 99% confidence, the true difference $\mu_A - \mu_B$ is between −9.286 and +5.286. Since 0 is in the interval, there is no significant difference between the products at the 0.01 level. The observed 2-unit difference in samples could easily be due to random variation.