What is the difference between a one-sample t-test and a z-test?

Use a z-test when the population standard deviation \sigma is known or when n \geq 30. Use a one-sample t-test when you have only the sample standard deviation s and a small sample. The t-distribution has heavier tails than the normal distribution to account for extra uncertainty when estimating the population standard deviation from a sample.

Why does the formula use n - 1 instead of n for calculating sample standard deviation?

The denominator n - 1 appears because calculating s uses the sample mean \bar{x}, which consumes one "degree of freedom." You have only n - 1 independent pieces of information left after computing the mean. This adjustment (called Bessel's correction) makes s an unbiased estimator of \sigma.

What does a negative t-value mean?

A negative t-value simply means the sample mean is below the hypothesized mean \mu_0. A positive t-value means it is above. When deciding significance, use the absolute value |t|, since what matters is distance from \mu_0 in either direction (unless you are performing a one-tailed test, in which case direction matters).

How do I know if my computed t-value is significant?

Compare your |t| to the critical value from a t-table using degrees of freedom df = n - 1 and your chosen significance level (typically \alpha = 0.05 for two-tailed tests). If |t| exceeds the critical value, reject the null hypothesis; otherwise, fail to reject it.

One-Sample t Test

A one-sample t test compares a sample mean to a hypothesized population mean when the population standard deviation is unknown.

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

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What each symbol means

What One-Sample t Test takes

Symbol	Meaning
$t$	The test statistic that you calculate; compare its absolute value to a critical value from a t-table to decide whether to reject the null hypothesis.
$\bar{x}$	The sample mean, calculated as the sum of all observations divided by n, measured in the same units as the original data.
$\mu_0$	The hypothesized population mean you are testing against, specified in the null hypothesis and written in the same units as the data.
$s$	The sample standard deviation measuring the spread of observations around the sample mean; using it instead of population $\sigma$ is why the t-distribution is needed.
$n$	The sample size, or number of observations; larger samples yield smaller standard errors and more stable estimates of the population mean.

When to use it

Use this when you have a single sample and want to test whether its mean differs from a claimed or hypothesized population value.

Level

Usually taught in: Statistics · Appears on: AP Statistics

Worked examples

1. Testing resting heart rate against a claimed value

Problem

An athlete's coach claims the athlete's resting heart rate is 60 bpm. Over 5 days, measurements are 58, 62, 61, 59, and 65 bpm. Calculate the t-statistic to test whether the true mean rate is 60.

$\bar{x} = \frac{58 + 62 + 61 + 59 + 65}{5} = \frac{305}{5} = 61$
Sum all five measurements and divide by $n = 5$ to find the sample mean.
$s = \sqrt{\frac{(-3)^2 + 1^2 + 0^2 + (-2)^2 + 4^2}{5-1}} = \sqrt{\frac{30}{4}} \approx 2.738$
Find each value's deviation from the mean (-3, 1, 0, -2, 4), square them, sum to 30, divide by $n - 1 = 4$ , then take the square root.
$\frac{s}{\sqrt{n}} = \frac{2.738}{\sqrt{5}} \approx 1.225$
Divide the sample standard deviation by $\sqrt{n}$ to get the standard error, which scales variability by sample size.
$t = \frac{61 - 60}{1.225} \approx 0.816$
Subtract the hypothesized mean $\mu_0 = 60$ from the sample mean and divide by the standard error to get the test statistic.

Answer: $t \approx 0.816$

We use a t-test here because we have a small sample (n=5) and know only the sample standard deviation, not the population value. The positive t-value shows the sample mean is slightly above the hypothesized 60 bpm, scaled by the variability in our measurements.

2. Testing mean session time on a website

Problem

A website claims users spend an average of 5 minutes per session. Seven randomly selected sessions are timed: 4.2, 5.8, 4.9, 6.1, 4.5, 5.3, and 4.8 minutes. Calculate t.

$\bar{x} = \frac{4.2 + 5.8 + 4.9 + 6.1 + 4.5 + 5.3 + 4.8}{7} = \frac{35.6}{7} \approx 5.086$
Add all seven measurements and divide by $n = 7$ ; decimal precision matters here.
$s = \sqrt{\frac{0.785 + 0.510 + 0.035 + 1.028 + 0.343 + 0.046 + 0.082}{6}} \approx \sqrt{0.472} \approx 0.687$
The squared deviations (from 5.086) sum to approximately 2.829; divide by $n - 1 = 6$ and take the square root for the standard deviation.
$\frac{s}{\sqrt{n}} = \frac{0.687}{\sqrt{7}} \approx \frac{0.687}{2.646} \approx 0.260$
Standard error accounts for the fact that a sample of 7 gives us a more precise estimate of the population mean than a single observation would.
$t = \frac{5.086 - 5}{0.260} \approx 0.331$
The sample mean is only slightly above the hypothesized value and well within the observed variability, giving a small t-statistic.

Answer: $t \approx 0.331$

With decimal data and a medium sample size, we see that even though the sample mean (5.086) exceeds the claimed 5 minutes, the t-statistic is small because this difference is small relative to the variability seen in the sample.

3. Testing mean homework hours against a benchmark

Problem

A college claims its students spend an average of 8 hours per week on homework. A sample of 9 students reports: 7, 9, 8, 10, 6, 9, 7, 8, and 11 hours. Calculate the t-statistic to test this claim.

$\bar{x} = \frac{7 + 9 + 8 + 10 + 6 + 9 + 7 + 8 + 11}{9} = \frac{75}{9} \approx 8.333$
Sum the nine values and divide by $n = 9$ to get the sample mean, which is slightly above the claimed 8 hours.
$s = \sqrt{\frac{1.777 + 0.445 + 0.111 + 2.778 + 5.443 + 0.445 + 1.777 + 0.111 + 7.112}{8}} \approx \sqrt{2.5} \approx 1.581$
Square each deviation from 8.333, sum to approximately 20, divide by $n - 1 = 8$ to get variance 2.5, then take the square root.
$\frac{s}{\sqrt{n}} = \frac{1.581}{\sqrt{9}} = \frac{1.581}{3} \approx 0.527$
A sample of 9 provides good precision; note that $\sqrt{9} = 3$ simplifies this calculation.
$t = \frac{8.333 - 8}{0.527} \approx 0.632$
The observed difference of 0.333 hours from the hypothesized mean 8 is scaled by the standard error to produce the test statistic.

Answer: $t \approx 0.632$

This word problem demonstrates how context shapes interpretation: students study slightly longer on average (8.33 hours), but with the variability seen in this sample, that difference is not strong evidence that all students truly differ from the claimed 8 hours. The t-statistic is small relative to its critical value.

Common mistakes

Where One-Sample t Test usually goes wrong

The mistake	Why it is wrong	The fix
Writing $t = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$ using population standard deviation $\sigma$ instead of sample $s$ .	The one-sample t-test formula requires the sample standard deviation $s$ because we do not know the population value $\sigma$ ; if $\sigma$ were known, you would use a z-test instead.	Always use $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ with the sample standard deviation computed from your data.
Omitting $\sqrt{n}$ and writing $t = \frac{\bar{x} - \mu_0}{s}$ .	Forgetting to divide by $\sqrt{n}$ makes the standard error far too large, causing the test statistic to be artificially small and making real differences appear insignificant.	Always include $\sqrt{n}$ in the denominator: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ .
Interpreting the test statistic as "the sample mean is far from $\mu_0$ " when in fact it measures standardized distance.	The sample mean will naturally differ from $\mu_0$ by random chance; the t-statistic accounts for sample variability, so what matters is whether the difference is large relative to that variability.	Compare the computed $t$ to the critical value for your degrees of freedom ( $n-1$ ) and significance level; only then can you decide if the evidence supports rejecting the null hypothesis.

Tips and when to use something else

If the population standard deviation $\sigma$ is known or $n \geq 30$ , use a z-test instead; the t-test is specifically for unknown $\sigma$ and smaller samples.
Degrees of freedom is always $n - 1$ ; this is what you use to find critical values or p-values in a t-table.
A negative t-value means the sample mean is below the hypothesized value; in a two-tailed test, only the absolute value $|t|$ matters when comparing to the critical value.
Statistical significance (a t-value large enough to reject the null) does not mean the effect is practically important; always consider whether the actual difference matters in context.

Related formulas

Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.Standard Error of the MeanThe standard error of the mean measures how much sample means vary around the true population mean; use it for confidence intervals and hypothesis tests.Confidence Interval for a MeanEstimates the range where a population mean likely falls using sample data when the population standard deviation is unknown.Z-ScoreThe Z-Score tells you how many standard deviations a data point is from the mean, revealing whether it is typical or unusual.

Frequently asked questions

What is the difference between a one-sample t-test and a z-test?: Use a z-test when the population standard deviation $\sigma$ is known or when $n \geq 30$ . Use a one-sample t-test when you have only the sample standard deviation $s$ and a small sample. The t-distribution has heavier tails than the normal distribution to account for extra uncertainty when estimating the population standard deviation from a sample.
Why does the formula use $n - 1$ instead of $n$ for calculating sample standard deviation?: The denominator $n - 1$ appears because calculating $s$ uses the sample mean $\bar{x}$ , which consumes one "degree of freedom." You have only $n - 1$ independent pieces of information left after computing the mean. This adjustment (called Bessel's correction) makes $s$ an unbiased estimator of $\sigma$ .
What does a negative t-value mean?: A negative t-value simply means the sample mean is below the hypothesized mean $\mu_0$ . A positive t-value means it is above. When deciding significance, use the absolute value $|t|$ , since what matters is distance from $\mu_0$ in either direction (unless you are performing a one-tailed test, in which case direction matters).
How do I know if my computed t-value is significant?: Compare your $|t|$ to the critical value from a t-table using degrees of freedom $df = n - 1$ and your chosen significance level (typically $\alpha = 0.05$ for two-tailed tests). If $|t|$ exceeds the critical value, reject the null hypothesis; otherwise, fail to reject it.

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Reviewed 2026-09-18