When to Use Z-Test vs T-Test vs Chi-Square

These methods differ in the question and reference model

A z-test, t-test, and chi-square test are not interchangeable ways to obtain a p-value. Start with the parameter and design.

Method	Typical target	Key distinction
Mean z-test	A population mean	Population SD genuinely known; suitable normal model for the mean
Mean t-test	A population mean or mean difference	Variability estimated from observations
Proportion z-test	A success probability or difference	Binomial-based normal approximation with suitable counts
Categorical chi-square test	Category probabilities or association	Observed versus expected counts

This table describes common uses, not every procedure with these reference distributions.

Z versus t for a mean

With known σ, a one-sample statistic is z = (x̄ − μ₀)/(σ/√n). With σ unknown and estimated by s, the corresponding t statistic is (x̄ − μ₀)/(s/√n), with df = n − 1 under the usual one-sample model.

The fact that n is 30, 100, or 1,000 does not make a sample SD a known population SD. At large df, t and z have nearly the same cutoffs, which explains the classroom approximation without turning it into a selection rule.

For x̄ = 498, s = 4, n = 16, and μ₀ = 500, the t statistic is −2 with 15 df and two-sided p ≈ 0.0639. Using a standard normal reference instead would give p ≈ 0.0455, changing a threshold decision by using a different model.

Different t-tests for different designs

One-sample tests compare a mean to a specified value. Welch’s independent two-sample test compares unrelated groups without assuming equal variances. Paired tests analyze differences within matched units.

For pairing, n counts complete independent pairs. A before-and-after t-test can detect mean change but does not by itself establish that an intervention caused it. Read t-tests for raw data and interpretation.

Proportions use binomial information

For binary outcomes, the outcome is a success count out of a number of trials. A proportion z-test uses a normal approximation to a binomial-based statistic; it is not a known-σ mean test merely because it uses z.

Check expected success and failure counts under the relevant model and the independence of units. Small or rare-event data may call for exact binomial or other suitable procedures. The one-sample tests lesson works through the distinction.

Chi-square analyzes category counts

A goodness-of-fit test compares one distribution with specified category proportions. An independence test compares observed counts with counts expected if row and column classifications are independent.

For independence, E = row total × column total / grand total, and df = (r − 1)(c − 1). The statistic Σ(O − E)²/E is nonnegative, and categorical tests use its upper tail. Adequate expected counts and independent observations are essential for the approximation.

For a 2×2 table, an uncorrected two-proportion z-test and an uncorrected Pearson chi-square test correspond through χ² = z². Continuity corrections and exact methods can break numerical equivalence because they are different calculations.

What if there are several means?

Classical one-way ANOVA compares several independent group means through an F ratio, with its own variance and error assumptions. An overall significant result says not all means are equal, not which pairs differ. Unequal variances or repeated measurements can require different procedures.

Use the reference with the actual question

Open the test-selection guide, then read the method’s worked lesson. Table choice should follow model choice: z-table, t-table, chi-square table, or F-table. None of these references fixes a biased sample or dependence in the observations.