Test whether the mean difference between paired measurements (before/after, matched pairs) is significantly different from zero.
The paired t-test analyses the differences within pairs rather than comparing two separate groups. Each subject serves as their own control, which removes between-subject variability and increases statistical power. It is equivalent to a one-sample t-test applied to the differences with μ0 = 0.
Typical applications include before-after studies, crossover trials, and comparisons of two methods measured on the same subjects (e.g., two blood pressure devices on the same arm).
A clinician measures systolic BP in n = 15 patients before and after starting an antihypertensive. The mean of the individual differences (Before − After) is d̄ = 8.2 mmHg and the SD of differences is sd = 6.5 mmHg. Is the reduction significant (Two-sided, α = 0.05)? The calculator provides the following:
1 Open the Paired t-Test calculator.
2 Enter Mean Difference = 8.2, SD of Differences = 6.5, n = 15.
3 Select Two-sided.
4 Result: t = 4.886, df = 14, p = 0.0002, 95% CI = (4.60, 11.80).
The medication produces a statistically significant reduction in BP (mean drop 8.2 mmHg, p < 0.001).
A significant result means the mean paired difference is unlikely to be zero. The 95% CI (4.60, 11.80) gives the plausible range for the true mean reduction. The medication reduces SBP by at least ~5 mmHg and up to ~12 mmHg.
Direction of differences: Be consistent about which direction you compute (Before − After vs. After − Before). A positive mean difference means the first measurement is larger.
where \( \bar{d} \) is the mean of paired differences and \( s_d \) is the standard deviation of differences.
Two-sided: \( p = 2 \cdot P(T \leq -|t|) \)
Left-sided: \( p = P(T \leq t) \)
Right-sided: \( p = 1 - P(T \leq t) \)
A cardiologist measures resting heart rate (bpm) in 20 patients before and after starting a beta-blocker. Mean difference = −8.5 bpm, SD of differences = 6.2 bpm.
Result: t = −6.13, df = 19, p < 0.001; 95% CI for mean difference: (−11.4, −5.6).
Interpretation: The beta-blocker significantly reduced resting heart rate by an average of 8.5 bpm.
A teacher measures spelling test scores (out of 50) for 25 students before and after a phonics programme. Mean gain = 3.8 points, SD of gains = 5.0 points.
Result: t = 3.80, df = 24, p < 0.001; 95% CI for mean gain: (1.7, 5.9).
Interpretation: The phonics programme produced a significant mean improvement of 3.8 points (p < 0.001).
Fifteen machines are calibrated, and output precision (mm) is measured before and after. Mean improvement = 0.04 mm, SD = 0.06 mm.
Result: t = 2.58, df = 14, p = 0.022; 95% CI: (0.007, 0.073).
Interpretation: Calibration significantly improved precision by 0.04 mm on average (p = 0.022). The CI confirms the improvement, though it is modest.
A psychologist measures self-esteem scores (scale 10–50) in 30 participants before and after a group therapy course. Mean change = +4.2, SD = 7.5.
Result: t = 3.07, df = 29, p = 0.005; 95% CI: (1.4, 7.0).
Interpretation: Self-esteem scores increased by an average of 4.2 points post-therapy, a significant improvement (p = 0.005).