Student sleep data, paired
- Input
- A: 0,7 −1,6 −0,2 −1,2 −0,1 3,4 3,7 0,8 0,0 2,0; B: 1,9 0,8 1,1 0,1 −0,1 4,4 5,5 1,6 4,6 3,4
- Expected output
- t = −4,0621, gl = 9, p = 0,0028
Rejects H0 at 5%.
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paired t test calculator
Paste measurement A (before) and measurement B (after) in the same order, one box for each, and the "Paired t" tab subtracts pair by pair and tests the mean of the differences. If you already have the mean and standard deviation of the differences, switch to summary statistics and type just those two numbers and the number of pairs.
Rejects H0 at 5%.
Does not contain zero.
Fails to reject H0: pairing made the difference.
dz = 0.59 (medium).
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.
Ask whether each value in one group has a natural partner in the other: the same subject, the same site measured twice, twins, matched pairs. If so, paired. If the groups are different people with no correspondence, independent.
The paired test requires the same number of values in both lists, in the same subject order. The calculator warns when sizes differ. If someone has missing data, remove the whole pair.
It is the effect size of the paired test: the mean of the differences divided by the standard deviation of the differences. It uses the same Cohen ranges as d: 0.2 small, 0.5 medium and 0.8 large.
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.