When is a z-test valid for testing a mean instead of a t-test?

## Same skeleton, different denominator Every test in this family is built the same way: ``` statistic = (observed estimate - value under the null) / (standard error of the estimate) ``` Testing a battery's mean life against a manufacturer's claim of 40 hours, the numerator is `xbar - 40`. What separates a z-test from a t-test is entirely the denominator. - **z-test for a mean:** the denominator is `sigma / sqrt(n)`, where `sigma` is the *population* standard deviation, treated as a known constant. Under the null, and given normal data (or a large enough sample), the statistic follows the standard normal distribution. - **t-test for a mean:** the denominator is `s / sqrt(n)`, where `s` is the *sample* standard deviation computed from the same data. Under the null, the statistic follows a t distribution with `n - 1` degrees of freedom. ## Why estimating the denominator changes the distribution If `sigma` is known, only the numerator varies from sample to sample, and the ratio is normal. If you substitute `s`, the denominator wobbles too. Sometimes `s` happens to come out small, and a modest deviation in the numerator is then divided by a small number, producing an unusually large statistic. That double source of randomness makes extreme values more common than the normal distribution predicts — the t distribution's heavier tails are exactly the price of not knowing `sigma`. Using a normal critical value while estimating `sigma` from a small sample therefore rejects too often: the test's real false-positive rate exceeds its nominal level. ## When is sigma actually known? Almost never for a mean measured on a fresh sample. Textbook cases where the claim is defensible: - A measuring instrument with a manufacturer-certified precision, where the measurement error is a documented property of the device rather than something you estimate. - A long-running, stable process with years of historical data, where the process standard deviation is known far more precisely than anything the current small sample could tell you. - A simulation or a designed sampling scheme where you set the noise level yourself. Even then, "known" is a modelling assumption: the historical standard deviation is still an estimate, just one based on so much data that its own uncertainty is negligible. ## The case where z is genuinely right: proportions For a proportion, the variance is a function of the mean. A binary outcome with success probability `p` has variance `p * (1 - p)`, so under a *specific* null hypothesis such as `p = 0.5`, the standard error under the null is completely determined: ``` SE_0 = sqrt(p0 * (1 - p0) / n) ``` Nothing is estimated, so a normal reference distribution is appropriate and the test is legitimately a z-test. This is why textbooks introduce z-tests for proportions and t-tests for means: it is not an arbitrary convention, it reflects which parameters the null pins down. The caveat is that the normal approximation to the binomial needs enough data — the usual rough guideline is at least about 10 expected successes and 10 expected failures. ## How much does it matter in practice? Less than the emphasis in a first course suggests, once `n` is large. The two-sided 5% critical value is 1.96 for the normal. For a t distribution it is about 2.78 at 4 degrees of freedom, about 2.09 at 20, about 2.01 at 50, and about 1.97 at 200. So: - With a handful of observations, using 1.96 instead of the correct t value materially inflates the false-positive rate. - With 200 observations, the difference between 1.972 and 1.960 changes essentially no decision, which is why analysts of large datasets often speak loosely of "the z-score" even when they computed a t statistic. Because the t-test converges to the z-test as `n` grows, and never the other way round, **t is the safe default**. You lose nothing by using it when `n` is large, and you are protected when it is small. ## A worked framing A manufacturer claims a mean battery life of 40 hours. You test 12 units and observe a sample mean of 37.5 hours with a sample standard deviation of 4 hours. Because `s` came from those 12 units, the statistic is ``` t = (37.5 - 40) / (4 / sqrt(12)) = -2.5 / 1.1547 = -2.165, df = 11 ``` and it is compared against the t distribution with 11 degrees of freedom, not the normal. Had the manufacturer instead published a certified process standard deviation of 4 hours, the same arithmetic would give a z of -2.165 compared against the normal — same number, different reference distribution, and a smaller p-value. ## What interviewers listen for Two things. First, that you locate the difference in the denominator — known `sigma` versus estimated `s` — rather than in the sample size, which is a consequence rather than the rule. Second, that you know why the rule exists: the estimated denominator adds randomness, the t distribution absorbs it through its degrees of freedom, and the correction vanishes as `n` grows.