t-Distribution

Why estimating variability changes the reference distribution

When population SD σ is known, a normally distributed sample mean can be standardized with σ/√n. Usually σ is unknown and is replaced by sample SD s. The denominator now varies from sample to sample too.

For independent observations from a normal population:

$T=\frac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}$

The t distribution accounts for this extra uncertainty. It is symmetric around zero and has heavier tails than standard normal. Its shape depends on degrees of freedom, not on the units of the observations.

Degrees of freedom depend on the design

Procedure	Relevant df
One-sample mean	n − 1
Paired mean difference	Number of independent pairs − 1
Pooled independent two-sample test	n₁ + n₂ − 2
Welch independent two-sample test	Welch–Satterthwaite df, often fractional
Simple-regression slope test	n − 2

Do not use n − 1 for every statistic with a t reference distribution. Degrees of freedom reflect what was estimated in the procedure.

As df grows, t approaches standard normal. That is why using t with a large sample usually changes the numerical cutoff little. It does not justify switching to z automatically because n passed 30.

Read critical values by tail convention

For a two-sided 95% interval or a two-sided test at α = 0.05, leave 0.025 in each tail. The positive cutoff is the cumulative 97.5th percentile.

df	Two-sided 5% positive cutoff
4	2.7764
9	2.2622
15	2.1314
24	2.0639
60	2.0003
Standard normal limit	1.9600

For a one-sided test at α = 0.05, leave 0.05 in the relevant tail instead. For df = 15, the positive upper-tail cutoff is 1.7531, not 2.1314.

The interactive t-table lets you select tails. Printed upper-tail tables may label columns with 0.025 or 0.05; cumulative tables may label them 0.975 or 0.95. These can describe the same percentile under different conventions.

Worked interval example

A sample has mean 20, sample SD 4, and n = 16. Use df = 15 and t* = 2.1314 for a 95% interval:

$SE=4/\sqrt{16}=1,\qquad CI=20\pm2.1314=[17.8686,22.1314]$

Using z = 1.9600 would give [18.0400, 21.9600], which is narrower and ignores the exact small-sample reference distribution under the normal model.

The t correction handles estimated variance; it does not handle selection bias, severe outliers, or dependence. Check that the procedure fits the design.

What heavier tails do and do not imply

At small df, a large standardized statistic is more common under t than under normal, so a more extreme cutoff is needed for the same tail probability. This is uncertainty in the statistic’s reference distribution, not a claim that every underlying population is heavy-tailed.

For df = 1, the t distribution is Cauchy and has no finite mean or variance. For df > 1 its mean is zero, and for df > 2 its variance is df/(df − 2). Symmetry alone does not guarantee finite moments.

Practice

A paired dataset has 12 complete independent pairs. Which df and tail column do you use for a two-sided test at α = 0.05?

Show the explanation

Use df = 11, because the analysis uses 12 differences, not 24 independent values. In an upper-tail table, use probability 0.025. In a cumulative table, use 0.975. The positive cutoff is about 2.2010.

Reference and next steps

NIST’s t-distribution reference describes the density and degrees-of-freedom dependence. Continue with t-tests and confidence intervals.