Paired t-Test

Overview

The paired t-test compares two numerical measurements that are linked in meaningful pairs. In R, it is implemented by t.test(..., paired = TRUE) [3].

Common examples include:

• before-and-after measurements from the same participant;
• matched case–control pairs;
• measurements from two eyes or limbs when one pair is retained per person;
• two laboratory methods applied to the same specimen;
• and crossover-period measurements from the same participant under a simple design.

The analysis reduces each pair to one difference and tests whether the population mean difference equals a prespecified value, usually zero.

The independent unit is the pair, not the individual measurement.

Why this matters

A paired design can be more precise than an independent design because each participant acts as their own comparison. Stable between-person differences are removed from the paired contrast.

However, pairing helps only when:

• the pairs are correctly linked;
• the paired measurements are meaningfully comparable;
• and the within-pair correlation is represented.

Analysing paired observations as independent discards information and usually gives an incorrect standard error.

The normality assumption applies to the distribution of paired differences, not separately to the before and after measurements. General medical-statistics principles for paired analysis and reporting are discussed by Altman [4].

When to use this method

Use a paired t-test when:

• there are exactly two linked numerical measurements per independent pair;
• pairing is known and correct;
• the mean paired difference is the estimand;
• pairs are independent of other pairs;
• and the difference distribution is sufficiently compatible with the procedure.

Do not use it for:

• two groups containing different participants;
• three or more repeated time points;
• incomplete pairs analysed without a missing-data plan;
• clustered pairs;
• or a randomised trial when the treatment effect requires a between-group comparison.

Research question and variables

The estimand is:

$$\mu_D=E(D)$$

where $D$ is follow-up minus baseline systolic blood pressure.

The hypotheses are:

$$H_0:\mu_D=0$$ $$H_1:\mu_D\ne0$$

The direction must be stated. Reversing the order changes the sign but not the two-sided p-value.

Data structure

Wide paired-data structure

simulated data

participant_id	sbp_baseline	sbp_followup	difference
P001	134	132	−2
P002	152	152	0
P003	152	139	−13
P004	178	168	−10
P005	153	139	−14

Note: Difference = follow-up minus baseline. The complete dataset contains 30 pairs.

The two variables must be aligned by participant identifier. Sorting one column independently destroys the pairing.

Statistical assumptions

The baseline and follow-up distributions do not each need to be normal. The important distribution is the set of within-pair differences.

How the method works

Let $\bar{d}$ be the sample mean of the $n$ paired differences and $s_d$ their sample SD.

The 95% CI is:

$$\bar{d} \pm t_{0.975,n-1} \frac{s_d}{\sqrt{n}}$$

Worked example

Reproducible R analysis

paired_bp <- data.frame(
  participant_id = sprintf("P%03d", 1:30),
  sbp_baseline = c(
    134, 152, 152, 178, 153, 143, 169, 139, 135, 159,
    162, 151, 131, 140, 158, 156, 152, 140, 142, 150,
    155, 163, 165, 146, 151, 129, 138, 142, 150, 152
  ),
  sbp_followup = c(
    132, 152, 139, 168, 139, 137, 176, 131, 125, 142,
    151, 138, 126, 125, 144, 157, 155, 143, 131, 137,
    151, 164, 164, 135, 154, 124, 124, 128, 143, 138
  )
)

paired_bp$difference <- paired_bp$sbp_followup - paired_bp$sbp_baseline

fit <- t.test(
  paired_bp$sbp_followup,
  paired_bp$sbp_baseline,
  paired = TRUE,
  alternative = "two.sided",
  conf.level = 0.95
)

fit

R console output

	Paired t-test

data:  paired_bp$sbp_followup and paired_bp$sbp_baseline
t = -5.7575, df = 29, p-value = 3.104e-06
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 -9.667289 -4.599378
sample estimates:
mean difference
      -7.133333

Cleaned output code

result <- data.frame(
  pairs = nrow(paired_bp),
  baseline_mean = mean(paired_bp$sbp_baseline),
  baseline_sd = sd(paired_bp$sbp_baseline),
  followup_mean = mean(paired_bp$sbp_followup),
  followup_sd = sd(paired_bp$sbp_followup),
  mean_difference = mean(paired_bp$difference),
  difference_sd = sd(paired_bp$difference),
  ci_lower = fit$conf.int[1],
  ci_upper = fit$conf.int[2],
  t = unname(fit$statistic),
  df = unname(fit$parameter),
  p_value = fit$p.value,
  dz = mean(paired_bp$difference) / sd(paired_bp$difference)
)

round(result, 3)

Cleaned R output

 pairs baseline_mean baseline_sd followup_mean followup_sd mean_difference difference_sd ci_lower ci_upper      t df p_value     dz
    30       149.567      11.524       142.433      14.105          -7.133         6.786   -9.667   -4.599 -5.758 29 3.10e-06 -1.051

For reporting, write $p<0.001$, not p=0.000.

Interpreting the output

Presenting the results

Follow-up minus baseline mean difference −7.1 mmHg (95% CI −9.7 to −4.6); paired $t(29)=-5.76$, $p<0.001$.

Suggested reporting

Paired analysis is not automatically a causal before-and-after analysis MMed extension

Pairing addresses statistical dependence. It does not create a valid counterfactual.

In an uncontrolled before-and-after study, observed change can reflect:

• natural history;
• regression to the mean;
• maturation;
• seasonal change;
• concurrent treatment;
• or measurement drift.

Causal interpretation requires an appropriate comparison design and assumptions.

Common mistakes

Knowledge check

What is the independent analytical unit in a paired t-test?

Answer and reasoning

The pair. Each pair produces one difference, and differences from separate pairs are assumed independent.

Which distribution should be examined for the normality-related assumption?

Answer and reasoning

The distribution of the paired differences, not the baseline and follow-up distributions separately.

Why does reversing baseline and follow-up not change the two-sided p-value?

Answer and reasoning

It changes every difference’s sign and reverses the test statistic and confidence interval, but the absolute statistic remains the same.

Does the significant decrease prove the medication review caused the improvement?

Answer and reasoning

No. Pairing controls person-level stability but does not control secular change, regression to the mean or co-interventions.

What method is needed for baseline, month 3 and month 6 measurements?

Answer and reasoning

A repeated-measurement method such as repeated-measures ANOVA, a mixed-effects model or GEE, depending on the estimand and assumptions.

Key take-home messages

References

1. Student. The probable error of a mean. Biometrika. 1908;6(1):1–25. https://doi.org/10.1093/biomet/6.1.1
2. Bland JM, Altman DG. Comparisons against baseline within randomised groups are often used and can be highly misleading. Trials. 2011;12:264. https://doi.org/10.1186/1745-6215-12-264
3. R Core Team. Student’s t-test. R documentation, package stats version 4.6.1. Vienna: R Foundation for Statistical Computing; 2026. https://search.r-project.org/R/refmans/stats/html/t.test.html
4. Altman DG. Practical statistics for medical research. London: Chapman and Hall; 1991.
5. Lakens D. Calculating and reporting effect sizes to facilitate cumulative science: a practical primer for t-tests and ANOVAs. Front Psychol. 2013;4:863. https://doi.org/10.3389/fpsyg.2013.00863