Testing one proportion against a claim
A claimed rate (half of customers return, 10% of parts are defective, a coin is fair) meets a sample: x successes in n trials. The one-proportion z-test measures the gap between the sample proportion p̂ and the claim p₀, in standard errors. Under the null the standard error is built from p₀ itself, because if the claim were true, p₀(1−p₀) is the variance the sampling would actually have.
A shop believes half its customers come back. Of 120 tracked customers, 64 returned: p̂ = 0.5333. z = (0.5333 − 0.5) ÷ √(0.25 ÷ 120) = 0.7303, two-tailed p = 0.4652. Nowhere near significance. A 53% return rate in 120 customers is exactly the kind of wobble a true 50% produces. The 95% interval for the proportion (0.444 to 0.623) contains 0.5 comfortably and says the same thing with more information.
Check the conditions
The normal approximation behind this test needs np₀ ≥ 10 and n(1−p₀) ≥ 10: expected successes and failures both in double figures (the calculator's steps show both counts for your numbers). Below that, the exact binomial test is the right tool. Random, independent trials are assumed throughout; a biased sample invalidates everything downstream, silently.
"p = 0.47 proves the rate is 50%." It proves nothing of the sort. The test failed to find evidence against 50%, with a sample too small to detect modest gaps. The interval shows the honest range: the truth could be 46% or 60%.
Frequently asked questions
Why does the test use p0 in the standard error but the interval uses p-hat?
The test asks 'if p0 were true, how surprising is this?', so it uses p0's world. The interval estimates the real p from the data, so it uses p-hat's. Different questions, different standard errors.
What if my successes are fewer than 10?
The z approximation is shaky; an exact binomial calculation is the correct instrument for small counts.