Paired t-test

Tests the mean of within-pair differences (before/after, matched pairs) with a one-sample t-test on the differences.

d_i = x_i - y_i,\qquad t = \frac{\bar d - \Delta_0}{s_d/\sqrt{n}},\qquad df = n - 1

$d_i$: Difference within pair i (first − second).
$\bar d, s_d$: Mean and SD (n − 1) of the differences.
$n$: Number of pairs.
$\Delta_0$: Mean difference under H₀ (usually 0).

When to use

Each observation in one condition is naturally linked to one in the other: the same subjects measured twice, twins, matched stores, the same route timed by two methods. Pairing removes between-subject variation, which usually makes the test far more sensitive.

Assumptions

Pairs are independent of each other.
The differences are roughly normal (or n is large); the original measurements need not be.

How to interpret

The test is about $\mu_d$ , the population mean difference. Report $\bar d$ with its CI and $d_z = \bar d/s_d$ .

Common mistakes

Analysing paired data with a two-sample test (wrong SE, wrong df).
Pairing data that are not genuinely linked (e.g. sorting both groups and pairing by rank).
Checking normality of each variable instead of the differences.
Mixing up the direction of the differences when stating a one-sided $H_1$ .

Example

Blood pressure before: 140, 135, 150, 142, 138, 145; after: 136, 132, 143, 141, 133, 139. Differences: 4, 3, 7, 1, 5, 6; $\bar d = 4.33$ , $s_d = 2.16$ , SE $= 0.882$ , $t = 4.91$ , df = 5, $p = 0.0044$ ; 95% CI for the mean drop 2.07 to 6.60 mmHg; $d_z = 2.01$ . Treating the same numbers as independent groups (Welch) gives $p = 0.16$ — the pairing is what makes the change detectable.