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

Choosing the wrong statistical test is one of the most common mistakes in data analysis. This guide compares the three tests students encounter most — z-test, t-test, and chi-square — and gives you clear rules for picking the right one.

Quick Decision Table

Question	Test	Example
Is this sample mean different from a known value? (σ known, large n)	Z-test	Is the average height of recruits different from 170 cm?
Is this sample mean different from a known value? (σ unknown or small n)	One-sample t-test	Do students at this school score differently from the national mean?
Are two group means different?	Two-sample t-test	Do men and women differ in reaction time?
Did a treatment change scores?	Paired t-test	Did the training program improve test scores?
Is there a relationship between two categorical variables?	Chi-square test	Is there an association between gender and voting preference?
Does observed data match expected frequencies?	Chi-square goodness of fit	Does a die roll follow a uniform distribution?

The Z-Test

What it does

Compares a sample mean to a known population mean when you know the population standard deviation (σ).

Formula

$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$

When to use it

• Population σ is known (rare in practice)
• Sample size is large (n ≥ 30)
• Data is approximately normal (or n is large enough for CLT)

Real-world example

A factory produces bolts with a known σ = 0.5 mm. You sample 50 bolts and find $\bar{x}$ = 10.2 mm. Are they different from the target μ = 10 mm?

$z = \frac{10.2 - 10}{0.5 / \sqrt{50}} = \frac{0.2}{0.0707} = 2.83$

Look up z = 2.83 in the Z-Table: p = 0.0023 (one-tailed) or 0.0046 (two-tailed). Reject H₀ at α = 0.05.

Tool

Use our Z-Score Calculator to compute z-values and look up probabilities.

The T-Test

What it does

Compares means when the population σ is unknown (the usual case). The t-distribution is wider than the z-distribution, accounting for extra uncertainty.

Three flavors

One-sample t-test:

$t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

Independent two-sample t-test:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2(1/n_1 + 1/n_2)}}$

Paired t-test:

$t = \frac{\bar{d}}{s_d / \sqrt{n}}$

When to use it

• Population σ is unknown (you estimate it from the sample)
• Comparing means (not proportions or frequencies)
• Data is continuous and approximately normal (or n > 30)

Real-world example

You want to know if a new drug lowers blood pressure. You test 20 patients (before and after):

• Mean difference: d̄ = −8.5 mmHg
• SD of differences: s_d = 6.2
• t = −8.5 / (6.2 / √20) = −8.5 / 1.387 = −6.13
• df = 19, critical t = 2.093

Result: |−6.13| > 2.093, so the drug significantly lowers blood pressure.

Tool

Our T-Test Calculator handles all three types with step-by-step output.

See also: How to Perform a Paired T-Test

The Chi-Square Test

What it does

Tests relationships between categorical variables (not means). There are two main types.

Chi-square test of independence

Tests whether two categorical variables are associated.

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Where O = observed count, E = expected count.

Chi-square goodness of fit

Tests whether observed frequencies match expected frequencies.

When to use it

• Your data is categorical (counts/frequencies)
• You have a contingency table (rows × columns)
• Expected frequencies are ≥ 5 in each cell (rule of thumb)

Real-world example

Is there an association between smoking status and lung disease?

	Lung disease	No disease	Total
Smoker	90	60	150
Non-smoker	30	120	150
Total	120	180	300

Expected values: E = (row total × column total) / grand total

• E(Smoker, Disease) = 150 × 120 / 300 = 60

$\chi^2 = \frac{(90-60)^2}{60} + \frac{(60-90)^2}{90} + \frac{(30-60)^2}{60} + \frac{(120-90)^2}{90}$

$= 15 + 10 + 15 + 10 = 50$

With df = (2−1)(2−1) = 1, look up in the Chi-Square Table: critical value at α = 0.05 is 3.841. Since 50 > 3.841, there is a significant association between smoking and lung disease.

Tool

Use the Bartlett’s Test Calculator for variance homogeneity checks or look up critical values in the Chi-Square Table.

Z-Test vs T-Test: What’s the Difference?

Feature	Z-test	T-test
Population σ	Known	Unknown
Sample size	Usually large (n ≥ 30)	Any size
Distribution used	Standard normal (z)	t-distribution (wider tails)
Degrees of freedom	Not applicable	n − 1 (or pooled)
Use in practice	Rare	Very common

Rule of thumb: If you don’t know σ, use a t-test. With large n, the t and z distributions converge, so the results will be nearly identical.

Decision Flowchart

1. What type of data do you have?

   • Categorical (counts) → Chi-square
   • Continuous (measurements) → Go to 2

2. How many groups?

   • One group vs. known value → Go to 3
   • Two groups → Go to 4
   • Three or more groups → ANOVA (not covered here)

3. Do you know the population σ?

   • Yes → Z-test
   • No → One-sample t-test

4. Are the groups independent or paired?

   • Independent → Independent t-test
   • Paired/matched → Paired t-test

What About ANOVA?

If you’re comparing three or more group means, none of these tests work — you need ANOVA (Analysis of Variance). Running multiple t-tests inflates your Type I error rate. ANOVA compares all groups simultaneously using the F-distribution.

Summary

Test	Data type	Compares	Key assumption
Z-test	Continuous	One mean vs. known value	σ known
T-test	Continuous	1 or 2 means	σ unknown
Chi-square	Categorical	Frequencies / proportions	Expected counts ≥ 5

Related Reading

• How to Choose the Right Statistical Test
• Hypothesis Testing: A Beginner’s Complete Guide
• How to Read Statistical Tables
• Degrees of Freedom Reference