SOLVETUTORMATH SOLVER

Instrument MI-07-180 · Statistics

t-statistic Calculator

When you don't know the population's true standard deviation and have to estimate it from your own sample, the t-statistic — not the z-statistic — is the honest way to test a hypothesized mean.

Instrument MI-07-180
Sheet 1 OF 1
Rev A
Verified
Type 07 — Hypothesis Testing SER. 2026-07180

t-statistic

1.6667

t = (x-bar - mu0) / (s / sqrt(n))

The working Every figure verified twice
  1. t = (105 − 100) ⁄ (15 ⁄ √(25)) = 1.6667
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

The one-sample t-statistic tests whether a sample mean (x-bar) is meaningfully different from some hypothesized population value (mu0), measured in standard errors: t = (x-bar − mu0) / (s / sqrt(n)). A t of 0 means your sample mean landed exactly on the hypothesized value; a large positive or negative t means your sample mean sits many standard errors away, which is evidence against the hypothesis that the true mean equals mu0.

What makes this specifically a t-statistic rather than a z-statistic is where the standard deviation comes from: here, s is calculated from your own sample data, not known in advance. Because s is itself an estimate with its own uncertainty, the t-statistic is compared against the t-distribution rather than the standard normal distribution — a distribution with heavier tails that widen the goalposts for statistical significance, accounting for the extra uncertainty of not knowing sigma for certain. The correct t-distribution to compare against has n − 1 degrees of freedom, which this site's degrees of freedom instrument computes directly.

This is nearly always the more realistic test in practice: real studies estimate variability from the sample they collected, rather than knowing a population's true standard deviation ahead of time. As sample size grows, s becomes a more reliable stand-in for sigma and the t-distribution converges toward the normal distribution, so the t-statistic and this site's z-test instrument produce nearly identical results for very large samples — the two calculators differ only in that theoretical assumption about sigma, not in the arithmetic itself.

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
t — the t-statistic · x-bar — sample mean · mu0 — the hypothesized population mean being tested · s — sample standard deviation, estimated from the data · n — sample size. Compare t against a t-distribution with n − 1 degrees of freedom.
  • Enter your observed average into Sample mean (x-bar).
  • Enter the value you're testing against into Hypothesized population mean (mu0) — often a claimed or historical figure.
  • Enter your sample's variability into Sample standard deviation (s) — calculated from your own data, not assumed in advance.
  • Enter your observation count into Sample size (n).
  • Read t-statistic, then compare it against a t-distribution critical value with n − 1 degrees of freedom to judge statistical significance.

Worked example — x-bar = 105 against a hypothesized 100

A quality-control sample of n = 25 units has a mean weight of x-bar = 105 grams and a sample standard deviation of s = 15 grams, and the question is whether the true process mean has drifted from a specified mu0 = 100 grams. Enter 105, 100, 15, and 25 into their respective fields.

The instrument first computes the standard error, 15 / sqrt(25) = 15 / 5 = 3, then t-statistic reads (105 − 100) / 3 = 1.6667. With n − 1 = 24 degrees of freedom, that t-value would be compared against a t-table, or its p-value computed directly, to decide whether a shift this large is likely to be real drift or ordinary sampling noise.

Questions

How is the t-statistic different from the z-test on this site?

Both use the identical formula shape and, with the same numbers, produce the identical result — the worked examples on both pages use x-bar=105, mu0=100, sample size 25, and a spread of 15, giving t = z = 1.6667. The difference is entirely conceptual: this instrument assumes s is estimated from your own sample data, so you compare against a t-distribution, while the z-test assumes sigma is a known population value fixed in advance, so you compare against the standard normal distribution. Real studies almost always estimate variability from data, which is why the t-statistic is the more common real-world tool.

Why does the t-statistic use a t-distribution instead of the normal distribution?

Because s, the sample standard deviation plugged into the formula, is itself an estimate with sampling uncertainty — you don't know the true population sigma for certain. The t-distribution has heavier tails than the normal distribution to account for that extra layer of uncertainty, meaning you need a more extreme t-value to reach the same significance level than you would with a known-sigma z-test.

What degrees of freedom do I use with this t-statistic?

n − 1, where n is your sample size. This site's degrees of freedom instrument computes that figure directly — for the worked example's n = 25, that's df = 24, the row you'd look up in a t-table to find the critical value for your chosen significance level.

Does the t-statistic converge to the z-statistic for large samples?

Yes. As sample size grows, the sample standard deviation s becomes an increasingly reliable estimate of the true population sigma, and the t-distribution itself converges toward the standard normal distribution. Past roughly n = 30, t-test and z-test critical values are usually close enough that the distinction matters little in practice, though the t-test remains the technically correct choice whenever sigma isn't actually known.

Is this the one-sample or two-sample t-test?

One-sample — it tests a single sample's mean against one hypothesized value. To compare the means of two independent samples against each other instead, for example a treatment group versus a control group, use this site's t-test instrument, which computes the two-sample pooled-variance version with its own pooled variance and degrees of freedom.

References