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The t distribution (also called Student’s t distribution) is a symmetric, bell-shaped probability distribution used to describe a standardized sample mean when the population standard deviation has to be estimated from the data. It looks like the standard normal curve but has heavier tails, and its exact shape depends on one number, the degrees of freedom. As that number grows, the t distribution approaches the normal distribution.
What is the t distribution? Definition in one paragraph
The t distribution is a family of curves indexed by a single parameter, ν (nu), the degrees of freedom. Each curve is centred on zero and symmetric. Compared with the standard normal curve, a t curve puts more probability in the tails, so extreme values are less surprising. That extra tail weight is the price of estimating the standard deviation from a small sample instead of knowing it. NIST’s engineering statistics handbook describes it as a standardized distribution with no location or scale parameters when used for testing, and notes that it approaches a normal distribution as ν becomes large. For how degrees of freedom are counted, see degrees of freedom in statistics.
t distribution formula
The probability density function in the NIST handbook is:
f(x) = (1 + x²/ν)-(ν+1)/2 / ( B(0.5, 0.5ν) · √ν )
- x: the value of the t statistic, any real number.
- ν: degrees of freedom, a positive number. For a one-sample t procedure it is N − 1; for a pooled two-sample test it is N1 + N2 − 2.
- B(a, b): the beta function, B(a, b) = ∫01 ta−1(1 − t)b−1 dt. It scales the curve so the total area is 1.
In practice you rarely evaluate this density by hand. You read a critical value from a table or call a function (see the Excel, R and Python section below). What matters is how the summary properties change with ν.
Which moments exist for which degrees of freedom
Because the tails are heavy, some summary measures do not exist for small ν. The NIST handbook gives these conditions:
| Summary measure | Value | Defined when |
|---|---|---|
| Mean | 0 | ν > 1 |
| Standard deviation | √(ν / (ν − 2)) | ν > 2 |
| Skewness | 0 (symmetric) | ν > 3 |
| Kurtosis | Depends on ν | ν > 4 |
The variance is therefore ν / (ν − 2), always larger than 1 and shrinking toward 1 as ν increases. That is why a t curve is wider than the standard normal curve and converges to it.
How much wider? Two-sided 95% critical values by degrees of freedom
This is where the heavy tails matter in practice. The table below uses the NIST critical-value table (column for probability 0.975, which is the two-sided 95% cutoff) and the normal-distribution row for comparison.
| Degrees of freedom (ν) | Two-sided 95% critical value |
|---|---|
| 1 | 12.706 |
| 2 | 4.303 |
| 5 | 2.571 |
| 9 | 2.262 |
| 10 | 2.228 |
| 20 | 2.086 |
| 30 | 2.042 |
| 60 | 2.000 |
| Normal distribution (ν = infinity) | 1.960 |
With 9 degrees of freedom the cutoff is 2.262, about 15% larger than the normal value of 1.960, so a confidence interval built with it is correspondingly wider. By ν = 30 the gap is about 4%, and NIST describes the normal approximation as quite good for ν above 30. To look up other values, see how to find a critical value.
t distribution examples: a worked one-sample case
The data below are ten hypothetical repeated measurements, invented for illustration only: 12, 15, 11, 14, 13, 16, 12, 15, 14, 13. Suppose we want a 95% confidence interval for the population mean and a test of whether the mean differs from 12.
- Sample mean. The sum is 135 and N = 10, so the mean is 135 / 10 = 13.5.
- Deviations from the mean. −1.5, 1.5, −2.5, 0.5, −0.5, 2.5, −1.5, 1.5, 0.5, −0.5.
- Sum of squared deviations. 2.25 + 2.25 + 6.25 + 0.25 + 0.25 + 6.25 + 2.25 + 2.25 + 0.25 + 0.25 = 22.5.
- Sample variance and standard deviation. s² = 22.5 / (10 − 1) = 2.5, so s = 1.581.
- Standard error. s / √N = 1.581 / 3.162 = 0.5.
- Critical value. Degrees of freedom = N − 1 = 9. From the table above, t = 2.262.
- 95% confidence interval. 13.5 ± 2.262 × 0.5 = 13.5 ± 1.131, which gives 12.37 to 14.63. This is the form given in the NIST handbook: the sample mean plus or minus the t critical value times s / √N.
- t statistic against a hypothesised mean of 12. t = (13.5 − 12) / 0.5 = 3.00.
Because |3.00| is larger than 2.262, the two-sided test rejects the hypothesis that the mean is 12 at the 0.05 level; the exact two-sided p-value for t = 3.00 with 9 degrees of freedom is about .015. Had we wrongly used the normal cutoff of 1.960, the interval would have been 13.5 ± 0.98, about 13% too narrow.
A two-sample example from the NIST handbook
The handbook’s two-sample t-test example compares miles per gallon for US cars (N = 249, mean 20.14458, s = 6.41470) and Japanese cars (N = 79, mean 30.48101, s = 6.10771). With equal variances assumed, the pooled standard deviation is 6.34260, the degrees of freedom are 249 + 79 − 2 = 326, and the test statistic is T = −12.62059. The critical value is 1.9673, and since |T| exceeds it, the hypothesis of equal means is rejected at 0.05. Notice how close 1.9673 is to the normal 1.960 at this sample size.
Interpretation: reading a t statistic
| Situation | Reading (from the NIST decision rule) |
|---|---|
| Two-sided test, |t| greater than the critical value for 1 − α/2 | Reject the null hypothesis at level α. |
| Two-sided test, |t| not greater than that value | Do not reject the null hypothesis at level α. |
| Upper one-sided test, t greater than the critical value for 1 − α | Reject the null hypothesis. |
| Lower one-sided test, t less than the negative of the critical value for 1 − α | Reject the null hypothesis. |
Because the distribution is symmetric, only positive critical values are tabulated. The related guide how to interpret the t-statistic goes further into what the number does and does not tell you.
Assumptions and common mistakes
- Using the normal cutoff for small samples. With 9 degrees of freedom the 1.960 cutoff understates the width of a 95% interval, as the worked example shows.
- Counting degrees of freedom wrongly. One-sample is N − 1; a pooled two-sample test is N1 + N2 − 2; a test without the equal-variance assumption uses the Welch-Satterthwaite approximation instead.
- Treating the t distribution as a description of raw data. It describes a standardized sample statistic, not the spread of individual observations.
- Ignoring the model conditions. The standard t procedures assume independent observations drawn from a roughly normal population; strong skew or outliers in a small sample undermine the result.
- Reading a confidence level as a probability for one interval. NIST notes that a 95% interval does not mean a 95% probability that this particular interval contains the true mean; the 95% belongs to the method over many samples.
To choose between the t-based and normal-based procedures, compare them in z-test vs t-test, and see the t-test guide for the full procedure.
How to report it (APA example)
A common APA-style pattern reports the statistic, its degrees of freedom in parentheses, and the p-value, plus the interval. For the worked example above:
The mean measurement (M = 13.5, SD = 1.58) differed from 12, t(9) = 3.00, p = .015, 95% CI [12.37, 14.63].
Always report the degrees of freedom, because the same t value means different things at different ν.
How to calculate it in Excel, R or Python
Each snippet below returns the two-sided 95% critical value for 9 degrees of freedom (2.262) and the two-sided p-value for t = 3 (about 0.015).
Excel
=T.INV.2T(0.05, 9)
=T.DIST.2T(3, 9)R
qt(0.975, df = 9)
2 * pt(3, df = 9, lower.tail = FALSE)Python (SciPy)
from scipy import stats
stats.t.ppf(0.975, 9)
2 * stats.t.sf(3, 9)FAQ
What is the t distribution?
It is a symmetric, bell-shaped family of distributions, indexed by degrees of freedom, that describes a standardized sample mean when the standard deviation is estimated from the sample. It has heavier tails than the normal curve and approaches it as the degrees of freedom increase.
When should I use the t distribution instead of the normal distribution?
Use it when the population standard deviation is unknown and estimated from the sample, especially when the sample is small. For large samples the two give nearly the same answer; see the discussion of z-scores and the normal distribution in the related guides.
What is the t distribution formula?
The density is f(x) = (1 + x²/ν)-(ν+1)/2 / (B(0.5, 0.5ν) √ν), with ν degrees of freedom. The test statistic itself is usually (sample mean − hypothesised mean) / (s / √N).
What are the mean and variance of the t distribution?
The mean is 0 when ν > 1, and the variance is ν / (ν − 2) when ν > 2.
Why does a confidence interval use a t value?
Because the standard error uses an estimated standard deviation, the interval needs the wider t critical value. See confidence intervals explained.
Sources
- NIST/SEMATECH e-Handbook of Statistical Methods, “Student’s t Distribution” (section 1.3.6.6.4), itl.nist.gov/div898/handbook/eda/section3/eda3664.htm, checked 2026-10-29.
- NIST/SEMATECH e-Handbook, “Critical Values of the Student’s t Distribution” (1.3.6.7.2), itl.nist.gov/div898/handbook/eda/section3/eda3672.htm, checked 2026-10-29.
- NIST/SEMATECH e-Handbook, “Confidence Limits for the Mean” (1.3.5.2), itl.nist.gov/div898/handbook/eda/section3/eda352.htm, checked 2026-10-29.
- NIST/SEMATECH e-Handbook, “Two-Sample t-Test for Equal Means” (1.3.5.3), itl.nist.gov/div898/handbook/eda/section3/eda353.htm, checked 2026-10-29.
