Perform parametric tests comparing means. Conduct one-sample t-tests against a hypothesized population mean, or two-sample tests (Welch's) comparing independent group means.
Last updated: March 2026
Mean of your sample data
Null hypothesis value (H₀)
Sample standard deviation
Number of observations
Std Error (SE)
1.5205
p-value (two-tailed)
1.1076
A t-test is a parametric hypothesis test comparing one or more sample means to a hypothesized value or to each other. Unlike the z-test, t-tests use the t-distribution, which has heavier tails and accounts for uncertainty in estimating the population standard deviation from sample data.
t = (x̄ − μ₀) / (s / √n)
or
t = (x̄₁ − x̄₂) / SE
The test produces a t-statistic and degrees of freedom (df), from which you compute a p-value. If p < α (typically 0.05), you reject the null hypothesis, indicating the means differ significantly.
One-Sample: Tests whether a sample mean differs from a hypothesized population mean. Example: testing if a drug changes blood pressure.
Two-Sample (Welch's): Tests whether two independent sample means differ. Welch's version does not assume equal population variances, making it more robust.
Select One-Sample if testing a single sample against a hypothesized mean. Select Two-Sample if comparing two independent groups.
For one-sample: enter x̄ (sample mean), μ₀ (hypothesized mean), s (sample SD), and n. For two-sample: enter both groups' means, SDs, and sizes.
A larger |t| indicates stronger evidence against the null hypothesis. The sign indicates direction: positive if x̄ {'>'} μ₀, negative if x̄ {'<'} μ₀.
One-sample: df = n − 1. Two-sample (Welch's): df is computed using the Welch-Satterthwaite equation to adjust for unequal variances.
Compare p-value to your significance level α (commonly 0.05). If p-value {'<'} α, reject H₀; the means differ significantly.
Scenario: A pharmacist tests a new low-sodium medication to verify it meets FDA standards. The label claims μ₀ = 100 mg sodium per dose. A random sample of 20 doses shows x̄ = 102.5 mg with s = 6.8 mg.
One-Sample t-Test:
H₀: μ = 100 mg (meets label claim)
H₁: μ ≠ 100 mg (does not meet claim)
t = (102.5 − 100) / (6.8 / √20) = 1.6408
df = 20 − 1 = 19
p-value (two-tailed) ≈ 0.1177
Interpretation: Since p-value (0.1177) > 0.05, we fail to reject H₀. There is insufficient evidence that the medication differs from the label claim at the 5% significance level.
Two-Group Example: Compare blood pressure reduction in two treatment groups: Group 1 (x̄₁ = 88.3 mmHg, s₁ = 5.2, n₁ = 18) vs. Group 2 (x̄₂ = 84.9 mmHg, s₂ = 4.7, n₂ = 22). Welch's test would compute t ≈ 2.24 with df ≈ 37, yielding p-value ≈ 0.0314, indicating the groups differ significantly.