Independent-Samples t-Test

Tests whether the means of two independent groups differ on a continuous outcome — e.g., do male and female employees differ in job satisfaction?

Data requirements

Assumptions

• Independence of observations — participants in one group are unrelated to those in the other, and no participant appears twice.
• Normality — the outcome is approximately normally distributed within each group (check Shapiro–Wilk, histograms, Q–Q plots); with n ≥ 30 per group the test is robust to moderate violations.
• Homogeneity of variance — the two groups have similar variances (check Levene's test); if violated, report the Welch correction.
• The outcome is continuous (interval/ratio); the grouping variable is genuinely dichotomous, not an artificially split continuous variable.

Hypotheses

The concept

The test computes a t statistic: the difference between the two sample means divided by the standard error of that difference. Intuitively, it asks how many 'standard errors of the difference' apart the two means are. If the groups truly came from the same population, t values near zero are common and large values are rare.

The resulting t is compared against the t distribution with (n₁ + n₂ − 2) degrees of freedom to obtain a p-value — the probability of observing a difference this large (or larger) if H₀ were true. Because statistical significance says nothing about practical importance, always pair the p-value with an effect size: Cohen's d expresses the mean difference in standard-deviation units (≈ 0.20 small, 0.50 medium, 0.80 large).

Worked example

A researcher asks whether work-from-home (n = 62) and in-office (n = 58) employees differ in job satisfaction, measured with a 5-item Likert-type scale (1–5, averaged). Mean satisfaction is 3.92 (SD = 0.61) for remote employees and 3.64 (SD = 0.66) for office employees.

Levene's test is non-significant (variances equal), normality is acceptable, so the standard independent t-test applies. The result: t(118) = 2.41, p = .017, d = 0.44 — remote employees report significantly higher satisfaction, a small-to-medium effect.

How to run it

# outcome: satisfaction (numeric), group: workmode (factor: Remote/Office)
library(effectsize)

# 1. Descriptives by group
aggregate(satisfaction ~ workmode, data = df, FUN = function(x) c(M = mean(x), SD = sd(x)))

# 2. Check assumptions
shapiro.test(df$satisfaction[df$workmode == "Remote"])
shapiro.test(df$satisfaction[df$workmode == "Office"])
car::leveneTest(satisfaction ~ workmode, data = df)

# 3. The test (var.equal = TRUE for classic Student's t)
t.test(satisfaction ~ workmode, data = df, var.equal = TRUE)

# 4. Effect size
cohens_d(satisfaction ~ workmode, data = df)

import pandas as pd
from scipy import stats
import pingouin as pg   # pip install pingouin

remote = df.loc[df["workmode"] == "Remote", "satisfaction"]
office = df.loc[df["workmode"] == "Office", "satisfaction"]

# 1. Descriptives
print(df.groupby("workmode")["satisfaction"].agg(["mean", "std", "count"]))

# 2. Assumptions
print(stats.shapiro(remote), stats.shapiro(office))
print(stats.levene(remote, office))

# 3. Test + Cohen's d in one call
print(pg.ttest(remote, office, correction=False))  # correction=True -> Welch

Interpreting the output

• t and df — the test statistic and degrees of freedom (n₁ + n₂ − 2); report both.
• p-value — if p < .05 (or your chosen α), the group means differ significantly; if p ≥ .05, you failed to find a difference (which is not proof of equality).
• Mean difference and its 95% CI — a CI that excludes 0 corresponds to a significant result and shows the plausible range of the true difference.
• Cohen's d — the size of the difference in SD units; a significant but tiny d may be practically meaningless in large samples.
• Direction — look at the group means to state which group scored higher; the sign of t depends only on coding order.

APA-style reporting

Common mistakes

Related methods