Two-Sample T-Test

When to use a two-sample t-test

Use when comparing means of two independent groups with unknown population standard deviations.

Step-by-step

Hypotheses: $H_0: \mu_A = \mu_B$, $H_a: \mu_A \neq \mu_B$ (two-tailed).

Step 1 — Compute the test statistic:

$t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} = \frac{85 - 80}{\sqrt{\frac{25}{20} + \frac{36}{25}}} = \frac{5}{\sqrt{1.25 + 1.44}} = \frac{5}{\sqrt{2.69}} = \frac{5}{1.64} = 3.05$

Step 2 — Degrees of freedom (Welch's approximation):

$df \approx \frac{(s_A^2/n_A + s_B^2/n_B)^2}{\frac{(s_A^2/n_A)^2}{n_A-1} + \frac{(s_B^2/n_B)^2}{n_B-1}} = \frac{2.69^2}{\frac{1.5625}{19} + \frac{2.0736}{24}} = \frac{7.24}{0.0822 + 0.0864} = \frac{7.24}{0.1686} \approx 42.9$

Step 3 — Critical value at $\alpha = 0.05$, $df \approx 43$: $t_{\text{crit}} \approx 2.017$.

Step 4 — Decision: $|t| = 3.05 > 2.017$. Reject $H_0$.

$\boxed{\text{The two groups have significantly different means.}}$

Try it in the visualization

Two bell curves for each group are drawn. The overlap (or lack thereof) shows how different the groups are. The t-statistic and rejection region are marked.