T-Test Calculator
Choose the test, enter means, standard deviations and sample sizes, and see whether the result is significant at your chosen alpha.
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Formula
- One-sample and paired:
t = (mean − μ0) ÷ (s ÷ √n),df = n − 1(for paired, use the differences) - Welch two-sample:
t = (x̄1 − x̄2 − μ0) ÷ √(s1²/n1 + s2²/n2) - Welch-Satterthwaite:
df = (a + b)² ÷ (a² ÷ (n1 − 1) + b² ÷ (n2 − 1)), witha = s1²/n1,b = s2²/n2 - Two-tailed
p = 2 × P(T ≤ −|t|); reject H0 whenp < α
Worked example
Welch test, group 1: mean 82, s 10, n 30; group 2: mean 76, s 12, n 25.
- a = 100 ÷ 30 = 3.3333, b = 144 ÷ 25 = 5.76, standard error = √9.0933 = 3.0155
- t = 6 ÷ 3.0155 = 1.9897
- df = 9.0933² ÷ (3.3333² ÷ 29 + 5.76² ÷ 24) = 46.83
- p = 0.0525, so at α = 0.05 the difference is not significant (critical t = 2.012).
Frequently asked questions
How is the one-sample t statistic calculated?
t = (x̄ − μ0) ÷ (s ÷ √n) with n − 1 degrees of freedom. For x̄ = 52, μ0 = 50, s = 6 and n = 25, t = 2 ÷ 1.2 = 1.667, df = 24 and p = 0.1086, so not significant at 0.05.
What is the Welch-Satterthwaite degrees of freedom?
For two samples with possibly unequal variances, df = (a + b)² ÷ (a² ÷ (n1 − 1) + b² ÷ (n2 − 1)), where a = s1² ÷ n1 and b = s2² ÷ n2. For 82 (s 10, n 30) against 76 (s 12, n 25), t = 1.990, df = 46.83 and p = 0.0525.
How do I run a paired t-test from summary statistics?
Take the difference within each pair, then enter the mean and standard deviation of those differences and the number of pairs. A mean difference of 1.2, standard deviation 2.5 and 16 pairs gives t = 1.92, df = 15, p = 0.0741.
What does the p-value mean?
It is the probability of a t value at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true. Compare it with α to decide, and consider the effect size as well. The test assumes independent observations and approximately normal data, or a reasonably large sample.