Student-t Distribution and t-Test

Let $Z\sim\mathcal{N}(0,1)$ and $V\sim\chi^2_\nu$ be independent. Then $T = \frac{Z}{\sqrt{V/\nu}}\sim t_\nu.$

$Z$ is the source of unit-variance Normal noise. $V/\nu$ is an empirical estimate of unit variance (since $\mathbb{E}[V] = \nu$); it converges to $1$ as $\nu\to\infty$. Dividing $Z$ by a noisy estimate of the scale inflates the tails by a polynomial amount. The thicker the noise (smaller $\nu$), the heavier the tails of $T$.

The sample mean of an i.i.d. Normal sample has numerator $\sqrt n(\bar X - \mu)/\sigma$ that is standard Normal, and denominator $S/\sigma$ where $S^2/\sigma^2$ is $\chi^2_{n-1}/(n-1)$ (i.e., a Chi-squared divided by its degrees of freedom). Independence of $\bar X$ and $S^2$ for Normal samples (see normal distribution) is what makes the standardized statistic exactly $t_{n-1}$ rather than an arbitrary ratio.

Independence of $Z$ and $V$ is essential. In the t-test the relevant $Z$ is $\sqrt n(\bar X - \mu)/\sigma$ and the relevant $V$ is $(n-1)S^2/\sigma^2$; their independence is a consequence of Cochran's theorem applied to Normal samples. For non-Normal samples, $\bar X$ and $S^2$ are asymptotically uncorrelated but not independent, so the t-distribution is exact only under Normality. Outside Normality, the test is asymptotic, and its accuracy in moderate samples depends on tail weight and skewness.

To test $H_0: \mu = \mu_0$ against $H_1: \mu\ne\mu_0$, use $T = \frac{\bar X_n - \mu_0}{S/\sqrt n},\qquad S^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i - \bar X_n)^2.$ Under $H_0$, $T\sim t_{n-1}$ exactly. The two-sided test rejects at level $\alpha$ when $|T| > t_{n-1,1-\alpha/2}$, the $(1-\alpha/2)$ quantile of $t_{n-1}$. The one-sided tests against $\mu > \mu_0$ or $\mu < \mu_0$ use the corresponding tail.

$T$ is a standardized sample mean. Under the null, the numerator has standard error $\sigma/\sqrt n$, so $\sqrt n(\bar X - \mu_0)/\sigma$ is standard Normal. The denominator divides by the sample estimate $S$ instead of the true $\sigma$. The ratio is Normal divided by the root of a normalized Chi-squared, which is exactly Student-t with $n - 1$ degrees of freedom.

The one-sample t-test is the basic parametric test for a sample mean. It is the test you reach for when you want to know whether a sample mean differs from a fixed reference value. The 95% confidence interval $\bar X_n\pm t_{n-1, 0.975}\cdot S/\sqrt n$ is the inverted test region. Both the test and the interval are exact under Normality and asymptotically valid (with size $\to\alpha$ and correct coverage) under any distribution with finite variance, by the central limit theorem combined with the asymptotic equivalence of $t_{n-1}$ and the standard Normal as $n\to\infty$.

The exact $t_{n-1}$ distribution requires Normal data. With heavy-tailed data, the t-statistic has heavier tails than $t_{n-1}$ predicts, and rejection rates are inflated above the nominal level. With skewed data and small samples, the test is biased in the direction of the longer tail. Permutation tests (see permutation tests) are the distribution-free alternative.

To test $H_0: \mu_X = \mu_Y$, define the pooled variance $S_p^2 = \frac{(n_X - 1)S_X^2 + (n_Y - 1)S_Y^2}{n_X + n_Y - 2}$ and the statistic $T = \frac{\bar X - \bar Y}{S_p\sqrt{1/n_X + 1/n_Y}}.$ Under $H_0$ and common variance, $T\sim t_{n_X + n_Y - 2}$ exactly.

The pooled variance $S_p^2$ averages the two sample variances, weighted by their degrees of freedom. Under the common-variance assumption, $(n_X + n_Y - 2)S_p^2/\sigma^2\sim\chi^2_{n_X+n_Y-2}$, and the difference of sample means $\bar X - \bar Y$ is independent of $S_p^2$. The standardized difference is therefore Normal-over-root-Chi-squared with the pooled degrees of freedom.

This is the canonical parametric test for "did treatment A change the mean compared to treatment B" under the simplifying assumption that the two groups share a common variance. It is the test the original Gosset paper introduced under the "Student" pseudonym (Biometrika, 1908). The same procedure gives a confidence interval for $\mu_X - \mu_Y$ by inversion.

The common-variance assumption matters. With unequal variances and unequal sample sizes, the pooled t-test has the wrong size: rejection rates can be much higher or lower than nominal, depending on which group has more variance and more data. The Welch t-test below is the right replacement and is the default in modern software.

The Welch statistic is $T_W = \frac{\bar X - \bar Y}{\sqrt{S_X^2/n_X + S_Y^2/n_Y}},$ and is approximately $t_\nu$ under the null with Welch-Satterthwaite degrees of freedom $\nu = \frac{(S_X^2/n_X + S_Y^2/n_Y)^2}{(S_X^2/n_X)^2/(n_X-1) + (S_Y^2/n_Y)^2/(n_Y-1)}.$

The denominator uses each group's own variance estimate. The price for not pooling is that the denominator is not a multiple of a single Chi-squared, so the statistic is not exactly t-distributed. The Welch-Satterthwaite approximation matches the first two moments of the denominator squared to those of a scaled Chi-squared, and the resulting degrees of freedom is the moment-matched degrees of freedom. The approximation is accurate even at moderate sample sizes when the variance ratio is far from one.

Welch's test is the default two-sample t-test in R (t.test(...) without var.equal = TRUE), SciPy (stats.ttest_ind(..., equal_var=False)), and most modern statistical software. Use it unless you have a specific reason to believe the variances are equal. The cost over the equal-variance pooled test is a fractional reduction in degrees of freedom, which is negligible at moderate sample sizes.

Welch's test still assumes Normal data within each group, although the approximation degrades more gracefully under non-Normality than the exact pooled test. For heavy-tailed or strongly skewed data, prefer a permutation test or a rank-based test (Mann-Whitney). The Welch-Satterthwaite degrees of freedom can be a non-integer; software interpolates the Chi-squared CDF.

To test $H_0: \delta = 0$ for pairs $(X_i, Y_i)$ with $D_i = X_i - Y_i$, compute $T = \frac{\bar D_n}{S_D/\sqrt n},\qquad S_D^2 = \frac{1}{n-1}\sum_{i=1}^n(D_i - \bar D_n)^2.$ Under $H_0$ and Normality of the differences, $T\sim t_{n-1}$ exactly.

A paired sample reduces to a one-sample t-test on the within-pair differences. The pairing eliminates between-subject variability and increases power compared to a two-sample test on the raw values, provided the pairs are genuinely linked (same subject before and after, matched pairs, twins).

Pre-versus-post designs, twin studies, matched-pair clinical trials, and within-subject crossover trials all use the paired t-test. The power gain over a two-sample test is substantial when the within-pair correlation is high; the test exploits that correlation by subtracting out the shared subject-level baseline. Ignoring pairing and using a two-sample test on the raw values gives a correct level (the test is still valid) but lower power.

The Normality assumption on the differences is what is required, not Normality of each marginal. Two heavily skewed marginals can have nearly Normal differences if the pairing captures the right shared structure; conversely, two Normal marginals can have non-Normal differences if the pairing is loose. Plot the differences before assuming Normality; if heavy-tailed, use a sign test or Wilcoxon signed-rank test.