Two-Sample t-Test Step by Step: Hypotheses, Calculation, and Interpretation With a Worked Example

The independent two-sample t-test determines whether two group means are significantly different from each other. Example: does a new drug lower blood pressure more than a placebo? The test compares the difference between the sample means to the variability within the samples. If the difference is large relative to the variability, the test concludes that the groups are truly different (not just random fluctuation). The formula: t = (x̄₁ - x̄₂) / √(s²p/n₁ + s²p/n₂), where x̄₁ and x̄₂ are the sample means, s²p is the pooled variance, and n₁ and n₂ are the sample sizes. The assumptions: both groups are independent (no subject appears in both), the dependent variable is continuous and approximately normally distributed in each group (or n > 30 per group), and the variances in both groups are approximately equal (check with Levene's test — if variances are unequal, use Welch's t-test instead). Snap a photo of any t-test problem and StatsIQ identifies the test type, checks the assumptions, calculates the test statistic, determines the p-value, and writes the conclusion — step by step with every formula shown.

Research question: Does caffeine improve reaction time? Two groups: caffeine (n=10) and placebo (n=10). Reaction time measured in milliseconds (lower = faster). Data: Caffeine group: x̄₁ = 245 ms, s₁ = 30 ms. Placebo group: x̄₂ = 268 ms, s₂ = 35 ms. **Step 1: State hypotheses.** H₀: μ₁ = μ₂ (no difference in mean reaction time between groups) Hₐ: μ₁ ≠ μ₂ (there is a difference) — this is a two-tailed test. Significance level: α = 0.05. **Step 2: Calculate pooled variance.** s²p = [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁ + n₂ - 2) s²p = [(9)(900) + (9)(1225)] / 18 s²p = [8100 + 11025] / 18 = 19125 / 18 = 1062.5 **Step 3: Calculate the t-statistic.** t = (x̄₁ - x̄₂) / √(s²p/n₁ + s²p/n₂) t = (245 - 268) / √(1062.5/10 + 1062.5/10) t = -23 / √(106.25 + 106.25) t = -23 / √212.5 t = -23 / 14.58 t = -1.578 **Step 4: Degrees of freedom and p-value.** df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18 Using a t-table or calculator with df=18 and t=-1.578 (two-tailed): p ≈ 0.132 **Step 5: Decision and conclusion.** p = 0.132 > α = 0.05. Fail to reject H₀. Conclusion: There is not sufficient evidence at the 0.05 significance level to conclude that caffeine significantly affects reaction time. The 23 ms difference between groups could be due to random variation. StatsIQ shows every one of these steps when you snap a photo of a t-test problem — including the pooled variance calculation that most students find hardest.

The conclusion must include four elements: the decision (reject or fail to reject H₀), the significance level, the test statistic and p-value, and an interpretation in context. Good conclusion for our example: "An independent samples t-test was conducted to compare reaction times between the caffeine group (M = 245, SD = 30) and the placebo group (M = 268, SD = 35). There was no statistically significant difference between the groups, t(18) = -1.578, p = .132. Caffeine did not significantly improve reaction time in this sample." What NOT to say: "We accept the null hypothesis." You never accept H₀ — you either reject it or fail to reject it. Failing to reject means the evidence was not strong enough to conclude a difference, not that you have proven the groups are equal. The distinction matters because a small sample might simply lack the power to detect a real difference. Also do not say: "The result is insignificant." Say "not statistically significant." Insignificant implies the result does not matter. Not statistically significant means the evidence did not reach the threshold — a subtle but important difference. With a larger sample, the same 23 ms difference might reach significance because the standard error would be smaller. Effect size complements the p-value: Cohen's d = (x̄₁ - x̄₂) / sp = -23 / √1062.5 = -23 / 32.6 = -0.71. This is a medium-to-large effect. The practical difference (23 ms faster) might be meaningful even though it is not statistically significant with n=10 per group — the study may have been underpowered.

Paired (dependent) t-test: when the same subjects are measured twice (before/after, or matched pairs). The formula uses the differences within each pair rather than comparing group means: t = d̄ / (sd / √n), where d̄ is the mean of the differences and sd is the standard deviation of the differences. Use when: pre-test/post-test designs, matched-pair experiments, or when each subject serves as their own control. Welch's t-test: when the equal variance assumption is violated (Levene's test p < 0.05 or one SD is more than double the other). Welch's does not pool the variances — it calculates the standard error from each group's variance separately and adjusts the degrees of freedom downward. Most statistical software defaults to Welch's because it performs well even when variances are equal. If your professor does not specify, Welch's is the safer choice. One-sample t-test: comparing a single group mean to a known value (not another group). Example: is this class's average test score significantly different from the national average of 75? t = (x̄ - μ₀) / (s / √n). Use when: you have one sample and a hypothesized population mean. The choice between these three depends on study design: independent groups → independent t-test (or Welch's). Same subjects measured twice → paired t-test. One group vs a known value → one-sample t-test. StatsIQ identifies which variant applies from the problem description and solves accordingly.