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The t-Distribution

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Why t, not z

When we do inference for a mean, we rarely know the population standard deviation σ, so we estimate it with the sample standard deviation s. That extra estimation adds uncertainty, and the standardized statistic no longer follows the Normal (z) distribution — it follows a t-distribution. Using z when σ is unknown would understate the true variability.

Shape and degrees of freedom

The t-distribution is symmetric and bell-shaped like the Normal but has heavier tails — it spreads out more to account for the added uncertainty from estimating σ. Its exact shape is set by the degrees of freedom (df). For a one-sample procedure, df = n − 1. As df increases (larger samples), the tails thin and t approaches the standard Normal.

One-sample t statistic and df
t = (x-bar − μ₀) / (s / sqrt(n)) · df = n − 1
Same structure as the z statistic, but s replaces σ and the reference distribution is t with n − 1 degrees of freedom. Larger n → t looks more like z.
Worked example

A sample of n = 16 has x-bar = 52, s = 8, testing μ₀ = 50. Find the degrees of freedom and the t statistic.

  1. 1.Degrees of freedom: df = n − 1 = 16 − 1 = 15.
  2. 2.Standard error: s/sqrt(n) = 8/sqrt(16) = 8/4 = 2.
  3. 3.t = (x-bar − μ₀)/(s/sqrt(n)) = (52 − 50)/2 = 2/2 = 1.
Answer: df = 15 and t = 1.0. The sample mean is one estimated standard error above the hypothesized mean, evaluated against a t-distribution with 15 degrees of freedom.
Checkpoint

Why do we use the t-distribution rather than the z-distribution for inference about a population mean?

Tip

Think of t as "z with extra caution." Estimating σ from a small sample means more uncertainty, so t’s tails are fatter and critical values are larger — giving wider intervals. As n grows, that extra caution fades and t converges to z.

Checkpoint

Compared to the standard Normal distribution, a t-distribution with 5 degrees of freedom has:

Answer the 2 checkpoints as you read.

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