One-Sample t-Test Calculator

Understanding the one-sample t-test — concept and principle

A one-sample t-test checks whether the mean of a sample differs from a fixed reference value μ₀, such as a target fill weight, a historical average or a claimed specification. It is used when the population standard deviation is unknown and has to be estimated from the same data, which is the usual situation.

The 1.7 gap is only 1.23 standard errors wide (p = 0.25), too small to rule out chance.

How it is calculated

Compute the sample mean x̄ and the sample standard deviation s (with divisor n − 1).
Standard error: SE = s ÷ √n.
Test statistic: t = (x̄ − μ₀) ÷ SE.
Degrees of freedom: df = n − 1.
Two-tailed p-value = 2 × P(T > |t|) for Student's t distribution with df degrees of freedom.
Decision at α = 0.05: reject H₀ (the true mean equals μ₀) if p < 0.05.

The principle

If the true mean really is μ₀, the sample mean will still wander around it by about one standard error. The t statistic counts how many standard errors the observed mean lies from μ₀. Because s is itself estimated, t follows Student's t distribution rather than the normal; its heavier tails demand stronger evidence when n is small. The p-value is the probability of seeing a t at least this extreme if H₀ were true.

Worked example

Data: 12, 15, 11, 19, 22, 15, 18, 25, 14, 16 (n = 10); μ₀ = 15.
x̄ = 167 ÷ 10 = 16.7; s = 4.372896.
SE = 4.372896 ÷ √10 = 1.382831.
t = (16.7 − 15) ÷ 1.382831 = 1.229361876, with df = 9.
Two-tailed p = 0.250106155.
Since 0.25 > 0.05, the decision is fail to reject H₀.

Each point is the calculator re-run with only “Hypothesised mean μ₀” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Hypothesised mean μ₀	t
3.75	9.364844878
7.5	6.65301721
11.2	3.941189543
15	1.229361876
18.8	-1.482465791
22.5	-4.194293459
30	-9.617948793

How to read the result

Where does your result fall?

The sample mean is 1.7 above 15, but that gap is only about 1.23 standard errors; for 9 df, |t| would need to exceed 2.262 to be significant at the 5% level. A p of 0.25 means a difference this large would arise by chance about one time in four even if the true mean were 15. "Fail to reject" does not prove the mean is 15; it only says the data are not strong enough to rule it out. Report the mean difference and a confidence interval alongside p.

Common mistakes

Reading a non-significant result as proof that H₀ is true.
Using a one-tailed test after seeing the direction of the data.
Applying the test to strongly skewed small samples or to non-independent observations.

Keep in mind: The test assumes independent observations from a roughly normal population; with small n, outliers can dominate. A p-value says nothing about practical importance.

Frequently asked questions

When should I use a z-test instead?

Only when the population standard deviation σ is known in advance, which is rare outside textbook problems.

How do I get a one-tailed p-value?

If the effect is in the predicted direction, halve the two-tailed p (here 0.125), but decide on one-tailed testing before looking at the data.

What if my data are not normal?

For moderate to large n the t-test is fairly robust. For small, skewed samples consider the Wilcoxon signed-rank test.

Sources: Student (1908). The probable error of a mean. Biometrika, 6(1), 1–25. · NIST/SEMATECH e-Handbook of Statistical Methods.

Checked 8 Nov 2026. Standard formulas, computed in your browser with double-precision numbers.

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