Student's t-Test — Definition, Formula & Examples

Student's t-test is a statistical method used to determine whether the mean of a sample (or the difference between two sample means) is significantly different from a hypothesized value. It is designed for situations where the sample size is small or the population standard deviation is unknown.

The t-test evaluates whether a test statistic, computed as the ratio of the departure of an estimated parameter from its null-hypothesis value to its standard error, follows a Student's t-distribution under the null hypothesis. The degrees of freedom depend on the sample size and the specific variant of the test (one-sample, independent two-sample, or paired).

Key Formula

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

Where:

$\bar{x}$ = Sample mean
$\mu_0$ = Hypothesized population mean under the null hypothesis
$s$ = Sample standard deviation
$n$ = Sample size

How It Works

You start by stating a null hypothesis, typically that the population mean equals some value or that two population means are equal. Next, you compute the t-statistic from your sample data. You then compare this t-statistic to the critical value from the t-distribution at your chosen significance level, or equivalently, you compute a p-value. If the p-value is less than your significance level (commonly 0.05), you reject the null hypothesis and conclude that the observed difference is statistically significant.

Worked Example

Problem: A manufacturer claims that its light bulbs last 1000 hours on average. You test 16 bulbs and find a sample mean of 980 hours with a sample standard deviation of 40 hours. At a significance level of 0.05, is there evidence that the true mean differs from 1000?

Step 1: State the hypotheses. The null hypothesis is that the mean lifetime equals 1000 hours, and the alternative is that it does not.

H_0: \mu = 1000 \quad H_a: \mu \neq 1000

Step 2: Compute the t-statistic using the one-sample formula.

t = \frac{980 - 1000}{40 / \sqrt{16}} = \frac{-20}{10} = -2.0

Step 3: Find the critical value. With 15 degrees of freedom and a two-tailed test at the 0.05 level, the critical values are approximately ±2.131. Since |−2.0| = 2.0 < 2.131, we fail to reject the null hypothesis.

df = n - 1 = 15, \quad t_{\text{crit}} \approx \pm 2.131

Answer: At the 0.05 significance level, there is not enough evidence to conclude that the true mean bulb lifetime differs from 1000 hours.

Why It Matters

The t-test appears in nearly every introductory statistics course and is one of the first hypothesis tests students learn. It is widely used in clinical trials, quality control, and social science research whenever sample sizes are moderate and the population variance is unknown.

Common Mistakes

Mistake: Using the t-test when the data are heavily skewed or contain extreme outliers in a small sample.

Correction: The t-test assumes the sampling distribution of the mean is approximately normal. With small samples, check for normality or consider a nonparametric alternative like the Wilcoxon test.