z = 2.0000, p = 0.0455 — significant. The exact binomial test gives 0.0569, which is not. Same data, opposite verdicts at 0.05.
60 of 100 = 60.0000% against a hypothesised 50.0000%
z = 2.00000, p = 0.045500
The exact binomial test gives p = 0.056888. They disagree by enough to matter here — read the exact one.
z (score test)
2.00000
null p in the SE
p (score test)
0.045500
two-sided
p (exact binomial)
0.056888
no approximation
Observed proportion
60.0000%
60 of 100
z (Wald variant)
2.04124
observed p in the SE
p (Wald variant)
0.041227
the common error
Gap in p
0.004273
from one substitution
Same verdict at 0.05?
Yes
both agree
The shaded area is the p-value: the chance of a result at least this far from 50.0000% in either direction, if the null were true. Both tails are shaded because the test is two-sided; a one-sided test would halve this and requires deciding the direction beforehand.
The standard error uses the NULL proportion, not the observed one. Under the null hypothesis the true proportion IS p₀, so the variance of the estimate is p₀(1 − p₀)/n — there is no reason to estimate a quantity the hypothesis already specifies. Substituting p̂ gives a Wald test instead, and on this data the two differ by 0.004273 in the p-value.
The score test and the Wilson interval are one procedure. The set of hypothesised proportions this test fails to reject at 0.05 is exactly the 95% Wilson interval, to six decimal places. If a p-value and a confidence interval ever disagree about significance, they were not built from the same test.
What this tool shows
60 heads in 100 tosses gives p = 0.0455 by the z-test and p = 0.0569 by the exact binomial test. Significant and not significant, on identical data, at the conventional threshold. The tool runs both, plus the Wald variant that comes from putting the observed proportion in the standard error where the null proportion belongs.
The score z-test, with the null proportion in the standard error
The exact two-sided binomial test, summed rather than approximated
The Wald variant, and how far its p-value differs from the score test’s
A warning when the smaller cell is too small for the approximation
The null distribution drawn with both tails shaded
The identity linking this test to the Wilson confidence interval
Score and Wald Exact binomial Small-n flagged Tails plotted
The standard error uses the null proportion, not the observed one.
Updated 12 September 2026 · Works in any browser, no installation
The test asks whether an observed proportion is far enough from a hypothesised one to be surprising if the hypothesis were true. z = (p̂ − p₀) ÷ √(p₀(1 − p₀)/n), and the p-value is the area in both tails beyond it. Where the counts are small the exact binomial test is the one to read, and this tool computes it alongside.
At a glance
Formula shown
z = (p̂ − p₀) / √(p₀(1 − p₀)/n), where p₀ is the hypothesised proportion and p̂ = x/n the observed one. The variance in the denominator comes from the NULL hypothesis, which specifies it — that makes this a score test. Substituting p̂ there gives the Wald variant, a different statistic with a different p-value. The exact test sums the binomial probability of every outcome at least as improbable as the observed one, which needs no approximation and no minimum sample size.
Scenario support
Testing a conversion rate against a target, a defect rate against a specification, a pass rate against a historical baseline, a coin or die for fairness, or any single proportion against a value fixed in advance.
Educational estimate
Planning support from the values you enter — not professional advice.
The standard error uses the null proportion
This is the detail two implementations of the same test disagree about, and the reasoning settles it cleanly.
Under the null hypothesis, the true proportion IS p₀. So the variance of p̂ is p₀(1 − p₀)/n — a known quantity, because the hypothesis specified it. There is nothing to estimate.
Substituting p̂ estimates something the hypothesis already told you. That gives the Wald test, which is a legitimate procedure with worse small-sample behaviour and no advantage here.
On 60 of 100 against 0.5 the two give z = 2.0000 and z = 2.0412. Small, and enough to move a p-value across a threshold on data that is closer to the boundary.
The score version is also the one that matches the confidence interval. Its non-rejection region is exactly the Wilson interval; the Wald test’s corresponds to the Wald interval, whose coverage failures are documented on that page.
So a p-value and an interval that disagree about significance were not built from the same test — which is worth knowing, because most software reports a score test and a Wald interval side by side without saying so.
When the approximation and the exact test disagree
The z-test approximates a discrete binomial distribution with a continuous normal one. The exact test does not approximate anything, and the gap between them is computable.
60 of 100 against 0.5: z-test 0.0455, exact 0.0569. One crosses the conventional threshold and the other does not, on a sample of 100 that no rule of thumb would call small.
9 of 15 against 0.5: z-test 0.4386, exact 0.6072. A 39% gap. Both are far from significant, so the verdict is unchanged — but the approximation consistently errs in the direction of overstating the evidence.
The approximation stays poor near 0 and 1 however large n is. Three defects in 400 against a 0.5% standard gives 0.4784 against the exact 0.4580. A large n does not rescue a proportion close to a boundary, which is the case people least expect.
The exact test is not always preferable. Because binomial p-values are discrete it is conservative — its realised error rate sits below the nominal level rather than at it — so it trades a guarantee for some power. That is the same trade the sign test makes.
The tool computes both up to n = 20,000 and says when they disagree materially. Above that the approximation is reliable and the enumeration is no longer worth its cost.
The rule of thumb, and what it misses
The usual condition is that both np₀ and n(1 − p₀) should exceed 5, or 10 in stricter treatments. It is a reasonable filter and it is not sufficient.
The tool checks the observed cells rather than the expected ones and flags when either falls below 10, because the binomial’s skewness — which is what the normal approximation fails to capture — depends on where the proportion sits.
The binomial is symmetric only at p = 0.5. Away from it the distribution is skewed, and the normal approximation places the two tails symmetrically when the truth is not. That is the source of the error, and it is largest exactly where the rule of thumb is most likely to be satisfied by a large n.
A continuity correction helps and is not applied here. Subtracting 1/(2n) from the absolute difference before dividing makes the approximation closer to the exact result, at the cost of a second convention to disclose. With the exact test computed alongside, correcting an approximation is the wrong tool for the job.
The practical rule is simpler than the arithmetic one: if the two p-values on this page disagree about anything you care about, read the exact one.
One-sided and two-sided, and when it is a choice
The tool reports two-sided p-values. Halving one for a one-sided test is arithmetically trivial and methodologically loaded.
A one-sided test is legitimate when the direction was fixed in advance and a result in the other direction would be treated exactly like no effect — not as an interesting finding, not as grounds for further investigation. That condition is stricter than it sounds.
Choosing the direction after seeing the data doubles your error rate while appearing to halve the p-value. It is the cheapest available form of p-hacking and the hardest to detect from a write-up.
Two-sided is the honest default for exactly the reason it is the harder test: it makes no claim about what you would have concluded from a surprise in the other direction.
The plot shades both tails so the two-sidedness is visible rather than implied by a number. A one-sided test would shade one, and the difference in the shaded area is the difference in the p-value.
What the p-value is not
Three misreadings, all common enough to be worth naming, and one consequence.
It is not the probability the null hypothesis is true. It is the probability of data at least this extreme IF the null were true — the other direction of the conditional, which is the same reversal the conditional probability calculator exists to prevent.
It is not a measure of effect size. A trivial departure from p₀ produces a tiny p-value on a large enough sample. The observed proportion and its confidence interval say how far from p₀ the truth might be; the p-value says only whether the sample could distinguish them.
A non-significant result is not evidence of no difference. It is a failure to detect one, which at small n is the overwhelmingly likely outcome whatever the truth. Showing a difference is SMALL requires an equivalence test, which is a different procedure.
And the threshold is a convention, not a property of the world. p = 0.0455 and p = 0.0569 — the two figures on the coin preset — describe almost identical evidence, and only one of them is conventionally reportable as a finding.
Reporting a proportion test
Four things, and the second is the one that most often goes missing.
Report the counts, not just the percentage. “60% (p = 0.046)” loses the sample size, which is half the information. “60 of 100” carries it.
Report the interval beside the p-value. The interval says what the proportion might be; the p-value says only whether p₀ is inside it. Where they come from the same test, the interval is strictly more informative.
Say which test. Score, Wald and exact give different p-values, and readers routinely assume whichever their own software produces.
State the hypothesised value and where it came from. A proportion tested against a target chosen after seeing the data is not a test of anything — and p₀ is the one input to this procedure that nothing in the data can check.
Method. The z-statistic uses the null proportion in the standard error, which makes it a score test, and the Wald variant is computed alongside so the difference between the two conventions is visible rather than assumed away. The exact two-sided p-value sums the binomial probability of every outcome no more likely than the observed one, computed in log space so it stays accurate to n = 20,000. The suite verifies the identity that ties this test to the Wilson interval — every Wilson bound returns p = 0.05 from this test, checked on all 39 interior outcomes at n = 40 — and confirms the score test is the exact procedure whose inversion produces that interval. That engine is verified on every change against 75 assertions. The count and the per-case breakdown are published on the formula verification page.
Related calculators
Where this goes next:
Wilson Score IntervalWilson, Wald, Agresti-Coull and Clopper-Pearson on one set of counts, with the EXACT coverage each delivers at your sample size — a "95%" Wald interval covers the truth 80.85% of the time at n = 30, p = 0.1.
Two-Proportion Z-TestCompares two rates with the pooled z-test and a Newcombe interval for the difference, and shows that z² equals the 2×2 chi-square statistic exactly — so the “z-test or chi-square” question has no content.
Binomial DistributionExact binomial probabilities at any n — including thousands, where a factorial overflows — with the normal approximation beside them and its error measured, which is 0.6% at the centre and 261% in the tail.
p-valueA p-value from a t or z statistic, one- or two-tailed — with a panel that holds an effect fixed and grows the sample, so you can watch significance appear from nothing but n.
Confidence IntervalIntervals for a mean or a proportion using t at every sample size and Wilson rather than the textbook Wald formula — with both methods shown, because Wald returns [0,0] at zero successes.
Sample SizeResponses needed for a target margin of error, with the finite-population correction and a table of the whole cost curve — because n scales with 1/margin², so the last point of precision costs more than the first ten.
An educational tool. A p-value describes how surprising the data would be under a hypothesis you supplied — it cannot tell you whether the sample was representative, whether the hypothesised value was reasonable, or how many other tests were run before this one.
Published the score z-test with the exact binomial test and the Wald variant computed alongside. On 60 of 100 against 0.5 the z-test gives p = 0.0455 and the exact test 0.0569 - opposite verdicts at the conventional threshold, on identical data.
The standard error uses the NULL proportion, which is what makes this a score test. Substituting the observed proportion gives z = 2.0412 instead of 2.0000 on that same data, and it is a different procedure rather than a rounding difference.
The exact p-value is summed in log space so it stays accurate to n = 20,000, and the tool says when the two disagree materially rather than leaving it to be noticed.
Measured the approximation's failure near the boundaries: 3 defects in 400 against a 0.5% standard gives 0.4784 by z and 0.4580 exactly - a large n does not rescue a proportion close to zero.
The score test's non-rejection region is exactly the Wilson interval, verified on all 39 interior outcomes at n = 40.
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