How this instrument works
The one-sample z-test checks whether a sample mean (x-bar) is meaningfully different from a hypothesized population value (mu0), measured in standard errors: z = (x-bar − mu0) / (sigma / sqrt(n)). A z of 0 means the sample mean landed exactly on the hypothesized value; larger positive or negative z-values mean the sample mean sits many standard errors away, which counts as evidence against the hypothesis that the true mean equals mu0.
The defining feature of this test — and what separates it from this site's t-statistic instrument — is that sigma here is treated as a known population standard deviation, fixed in advance from an established process or a large body of prior data, not calculated from the sample currently being tested. Because sigma is assumed known rather than estimated, the z-statistic is compared against the standard normal distribution rather than a t-distribution, since there's no extra estimation uncertainty in the standard error to account for.
In practice, a genuinely known population sigma is the exception rather than the rule — most real studies estimate variability from the very sample they collected, which calls for the t-test instead. The z-test remains useful for processes with a long, stable measurement history where sigma has been pinned down by a large volume of past data, and it's also the standard teaching example before introducing the t-test's extra layer of uncertainty. For very large samples, the t-distribution converges to the normal distribution anyway, so the two tests' results become practically indistinguishable regardless of which one is technically correct.
- Enter your observed average into Sample mean (x-bar).
- Enter the value you're testing against into Hypothesized population mean (mu0).
- Enter the known population value into Known population standard deviation (sigma) — this must be an established figure, not one calculated from your current sample.
- Enter your observation count into Sample size (n).
- Read z-statistic, then compare it against a standard normal (z) critical value, such as 1.96 for a two-tailed test at 5% significance.
- If you only have a standard deviation estimated from your own sample rather than a known population sigma, use this site's t-statistic instrument instead.
Worked example — x-bar = 105 against a hypothesized 100
A process is known, from long production history, to have a population standard deviation of sigma = 15 grams. A new sample of n = 25 units has a mean of x-bar = 105 grams, and the question is whether the process mean has shifted from its 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 z-statistic reads (105 − 100) / 3 = 1.6667. Compared against the standard normal distribution, a two-tailed critical value of 1.96 at 5% significance means this particular shift falls short of being statistically significant — it's plausibly ordinary sampling variation around an unchanged process mean.
Questions
How is this different from the t-statistic calculator on this site?
Both use the identical formula shape, and with matching 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 z = t = 1.6667. The difference is entirely about where the standard deviation comes from: this instrument requires sigma to be a known population value fixed in advance, compared against the standard normal distribution, while the t-statistic instrument estimates s from the sample itself and compares against a t-distribution with heavier tails to account for that added uncertainty.
When is it actually appropriate to treat sigma as 'known'?
When you have a long, stable history of measurements — an established manufacturing process, a well-characterized instrument, or a large prior study — that has pinned down the population standard deviation independently of the current sample. In most real research and quality-control situations, this genuinely known figure isn't available, and the sample's own standard deviation has to stand in, which calls for the t-test instead.
Why does the z-test use the standard normal distribution instead of the t-distribution?
Because sigma is assumed known rather than estimated, there's no additional layer of uncertainty coming from estimating the standard deviation — only the sample mean itself varies from sample to sample. The t-distribution's heavier tails exist specifically to account for that extra estimation uncertainty, so once sigma is genuinely known, the standard normal distribution is the mathematically correct comparison.
Does the z-test give a different answer than the t-test for large samples?
Practically, no. As sample size grows, the t-distribution converges toward the standard normal distribution, so critical values and conclusions from the two tests become nearly identical past roughly n = 30, even though they rest on different assumptions about sigma. The distinction matters most for small samples, where the t-distribution's heavier tails require a more extreme statistic to reach significance.
Is this the one-sample or two-sample z-test?
One-sample — it compares a single sample's mean against one fixed hypothesized value, assuming a known population sigma. Comparing two independent samples' means against each other, whether with known or estimated standard deviations, is a different calculation; this site's two-sample t-test instrument covers the estimated-variance version of that comparison.