US vs Japanese cars, pooled t (NIST)
- Input
- n 249, x̄ 20,14458, s 6,41470 × n 79, x̄ 30,48101, s 6,10771
- Expected output
- t = −12,6206, gl = 326, crítico ±1,9673
Rejects H0 at 5%.
two sample t test calculator
Type the mean, standard deviation and size of each group, or paste the raw data, and the "t, 2 means" tab does the rest. The default is Welch, which does not require equal variances; the pooled t is one click away. The result brings the written hypothesis, the p-value, the rejection region on the chart and a sentence ready for your report.
Rejects H0 at 5%.
p of the order of 1.6e-25.
Fails to reject H0; d = 0.46.
95% CI of the difference: −4.10 to 15.50.
Same data in the paired t: p = 0.0028.
The p-value is the probability of observing a statistic as extreme as your sample, or more, assuming H0 is true. It is not the probability that H0 is true. With p = 0.0794 and α = 0.05, the data are not "unlikely enough" to reject H0; with p = 0.0028, they are. Very small p-values appear in scientific notation: 2.5e-7 means 0.00000025.
When in doubt, Welch. It loses almost no power when variances are equal and protects the p-value when they are not. The pooled version is only justified with clearly equal variances and similar sizes.
Yes. Switch to "Raw data" and paste each group into its box, using new lines, semicolons, commas or spaces and decimal commas. The calculator computes n, mean and standard deviation and also runs the Levene test.
In "Advanced options", enter the difference under H0 (δ₀). With δ₀ = 5 and one-sided right, for example, the alternative hypothesis becomes μ₁ − μ₂ > 5.
10 values read (separator: comma).
10 values read (separator: comma).
Interpretation
There is not enough statistical evidence, at significance level α = 0.05, that the difference between the population means (group 1 − group 2) is different from 0 (observed difference: -1.58; p-value = 0.079394). Effect size: Cohen d = -0.8322 (large). Failing to reject H0 does not prove it true: the data simply do not provide enough evidence against it, which can also reflect a small sample.
| Group | Size (n) | Mean | s |
|---|---|---|---|
| Group 1 | 10 | 0.75 | 1.789 |
| Group 2 | 10 | 2.33 | 2.0022 |
Difference between means (group 1 − group 2): -1.58. Standard error: 0.8491.
F test of variances: F = 0.7983, p = 0.742720. No sign of different variances.
Levene (Brown-Forsythe): F = 0.2482, p = 0.624374.
SE = √(s₁²/n₁ + s₂²/n₂) = √(3.2006/10 + 4.009/10) = 0.8491t = ((x̄₁ − x̄₂) − δ₀) / SE = ((0.75 − 2.33) − 0) / 0.8491 = -1.8608df = (s₁²/n₁ + s₂²/n₂)² / [ (s₁²/n₁)²/(n₁−1) + (s₂²/n₂)²/(n₂−1) ] = 17.7765p = 2 · P(T ≥ |t|) = 2 · P(T ≥ 1.8608) = 0.079394tcrit = ±2.1028CI 95% = [-3.3655; 0.2055]Two independent samples, each approximately normal (or large). Welch does not require equal variances; the pooled t does. If the same subjects were measured twice, use the paired t.