t-test vs z-test: when to use each

The quick answer: t-test vs z-test

Here is the whole decision in one line: test a proportion with a z-test, and test a mean with a t-test. That rule covers nearly every inference question you will meet in AP Statistics, so if you remember nothing else, remember that.

The reason comes down to a single quantity, the population standard deviation, written $\sigma$ (the Greek letter sigma). When you test a mean, $\sigma$ is unknown, so you replace it with the sample standard deviation $s$ computed from your own data. Estimating $\sigma$ adds extra uncertainty, and the t distribution is built to absorb it. When you test a proportion, there is no separate standard deviation to estimate, so the normal (z) distribution applies directly.

Put another way: data summarized as a sample proportion $\hat{p}$ (p-hat, the sample proportion) use z, and data summarized as a sample mean $\bar{x}$ (x-bar, the sample mean) use t.

When to use a t-test

Use a t-test whenever the parameter you care about is a population mean $\mu$ (mu, the population mean) and you do not know the true population standard deviation $\sigma$. That last condition is the deciding factor, and in practice it is almost always met, because knowing $\sigma$ exactly would mean you already understand the spread of the whole population you are trying to study.

Since you cannot use $\sigma$, you estimate the spread from the sample and build the standard error from $s$:

where $s$ is the sample standard deviation and $n$ is the sample size. Plugging an estimate into the denominator of the test statistic makes that statistic bounce around more than a clean z-score would, and the t distribution with $n - 1$ degrees of freedom is the exact correction for that extra bounce.

There is one textbook exception. If $\sigma$ were known, the statistic $(\bar{x} - \mu_0)/(\sigma/\sqrt{n})$ would follow a standard normal distribution and you could use z for a mean. Real datasets rarely hand you $\sigma$, so means are tested with t.

The difference between the t and z distribution

The difference between the t and z distribution is shape. Both curves are symmetric and bell-shaped, and both are centered at 0, but the t distribution has heavier tails, meaning more of its area sits far from the center. Those heavier tails produce larger critical values, so a t-based confidence interval comes out a little wider than a z-based one at the same confidence level. That extra width is the honest cost of estimating $\sigma$ instead of knowing it.

How heavy the tails are depends on the degrees of freedom, which equal $n - 1$ for a one-sample t procedure. A small sample has few degrees of freedom and the heaviest tails. As $n$ grows, $s$ becomes a steadier estimate of $\sigma$, the degrees of freedom rise, and the t curve tightens toward the standard normal curve. With infinite degrees of freedom the two are identical.

By the time $n$ reaches the low hundreds the gap is tiny, which is why t works as a safe default for means at any sample size. See standard error vs standard deviation if the $s/\sqrt{n}$ step is still fuzzy.

z-test or t-test: a side-by-side comparison

This table lines up the two tests on the features that actually decide which one you run. In the standard error row, $p_0$ is the proportion assumed under the null hypothesis.

Two-sample versions follow the same rule

The same logic extends to comparing two groups. To compare two proportions (for example, the approval rate in two cities), you run a two-sample z-test. Under the null hypothesis that the two proportions are equal, you pool the data into a combined proportion $\hat{p}_c = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2}$ (p-hat-c, the combined sample proportion) and build the standard error $\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$. To compare two means (for example, average test scores under two teaching methods), you run a two-sample t-test, with standard error $\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ and no pooling, since each group's spread is estimated from its own sample.

The tie-breaker never changes. Proportions stay with z because their standard deviation is fixed by the proportion itself, and means stay with t because each group's spread is estimated from its own sample. For the finer points of comparing proportions, see one-proportion vs two-proportion z-test.

How t-tests and z-tests map to the AP exam

On the Fall 2026 AP Statistics course, the split is clean. Unit 3, Inference for Categorical Data: Proportions, uses z procedures for one proportion and for the difference between two proportions. Unit 4, Inference for Quantitative Data: Means, uses t procedures for one mean and for the difference between two means, because the population standard deviation is unknown and the sample standard deviation is used instead.

Formulas and tables are provided during both sections of the exam, so you do not have to recall critical values from memory, but you are expected to choose the correct procedure yourself. When you are deciding, ask first whether the response is categorical or quantitative, then apply the rule.

You can check your work with the one-sample t-test calculator or the proportion z-test calculator, and see the full menu of options in which statistical test to use.