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

A Student's t-test table is a reference chart that lists critical t-values organized by degrees of freedom (rows) and significance levels (columns). You compare your calculated t-statistic against these critical values to decide whether to reject the null hypothesis.

The table provides quantiles tα,νt_{\alpha, \nu} of the Student's t-distribution, where α\alpha is the tail probability (significance level for a one-tailed test, or half the significance level for a two-tailed test) and ν\nu is the number of degrees of freedom. A test statistic exceeding the tabled critical value in absolute terms leads to rejection of H0H_0 at the corresponding significance level.

Key Formula

t=xˉ−μ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
Where:
  • xˉ\bar{x} = Sample mean
  • μ0\mu_0 = Hypothesized population mean under H₀
  • ss = Sample standard deviation
  • nn = Sample size

How It Works

First, calculate your degrees of freedom, which for a one-sample or paired t-test equals n−1n - 1. Next, choose your significance level α\alpha (commonly 0.05) and decide whether you need a one-tailed or two-tailed test. For a two-tailed test at α=0.05\alpha = 0.05, look under the column labeled 0.025 (since 0.05 is split across both tails). Find the row matching your degrees of freedom and read the critical value. If your computed ∣t∣|t| exceeds that critical value, you reject the null hypothesis.

Worked Example

Problem: A sample of 16 students has a mean test score of 78 and a sample standard deviation of 8. Test whether the population mean differs from 74 at the 0.05 significance level (two-tailed).
Compute the t-statistic: Plug values into the t formula.
t=78−748/16=42=2.0t = \frac{78 - 74}{8 / \sqrt{16}} = \frac{4}{2} = 2.0
Find degrees of freedom: Degrees of freedom equals n − 1.
ν=16−1=15\nu = 16 - 1 = 15
Look up the critical value: For a two-tailed test at α = 0.05, use the column for α/2 = 0.025 with ν = 15. The t-table gives a critical value of approximately 2.131.
t0.025, 15=2.131t_{0.025,\,15} = 2.131
Compare and decide: Since |t| = 2.0 < 2.131, you do not reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude the population mean differs from 74.
Answer: Fail to reject H₀. The computed t-value of 2.0 does not exceed the critical value of 2.131 at the 0.05 significance level with 15 degrees of freedom.

Why It Matters

The t-table is essential in introductory statistics courses whenever you test hypotheses or construct confidence intervals with small samples and unknown population variance. In practice, many software tools compute exact p-values, but reading a t-table builds your understanding of the relationship between degrees of freedom, tail area, and critical values.

Common Mistakes

Mistake: Using the one-tailed column value for a two-tailed test (or vice versa).
Correction: For a two-tailed test at α = 0.05, look under the 0.025 column since the rejection region is split into both tails. For a one-tailed test at α = 0.05, use the 0.05 column directly.