Paired t-Test Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Paired t-Test.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

A hypothesis test for the mean difference in a paired (matched) data design, where each subject provides two related measurements. The test analyzes the differences d_i = x_{1i} - x_{2i} as a single sample.

You want to know if a tutoring program improves math scores. Instead of comparing two separate groups, you test the SAME students before and after tutoring. Each student is their own control. By looking at the difference (after - before) for each student, you eliminate individual variation and focus purely on the change.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The paired t-test reduces the problem to a one-sample t-test on the differences. By pairing, you control for subject-to-subject variation, making it easier to detect a treatment effect.

Common stuck point: Students use a two-sample t-test when they should use a paired t-test. The giveaway: if the same subjects appear in both groups, or subjects are deliberately matched, it's paired.

Sense of Study hint: When comparing two related measurements on the same subjects, use a paired t-test. First, compute the difference d_i = x_{\text{after}} - x_{\text{before}} for each pair. Then calculate \bar{d} and s_d, and compute t = \bar{d}/(s_d/\sqrt{n}). Finally, compare t to the critical value or find the p-value with n-1 degrees of freedom.

Worked Examples

Example 1

medium
Students' scores before and after tutoring: Before: \{70, 65, 80, 75, 60\}, After: \{75, 70, 85, 80, 70\}. Conduct a paired t-test at \alpha=0.05 to test if tutoring improved scores.

Solution

  1. 1
    Differences d_i = \text{After} - \text{Before}: 5, 5, 5, 5, 10
  2. 2
    \bar{d} = (5+5+5+5+10)/5 = 30/5 = 6; s_d = \sqrt{\frac{\sum(d_i-\bar{d})^2}{n-1}} = \sqrt{\frac{4+1+1+1+16}{4}} = \sqrt{4.25} \approx 2.06
  3. 3
    t-statistic: t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{6}{2.06/\sqrt{5}} = \frac{6}{0.921} \approx 6.51
  4. 4
    df=4; t^*_{0.05, 4} = 2.132 (one-tailed); 6.51 > 2.132 → Reject H_0

Answer

t \approx 6.51 > 2.132. Reject H_0. Tutoring significantly improved scores.
A paired t-test analyzes differences between matched pairs (before-after, or twins). By working with differences, we eliminate between-subject variability, increasing power. The test becomes a one-sample t-test on the differences.

Example 2

hard
Explain when to use a paired t-test vs. a two-sample t-test. If shoe comfort is measured on the same subjects wearing Brand A and Brand B, which test is appropriate and why?

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Five paired differences are: \{2, -1, 3, 0, 4\}. Calculate \bar{d} and s_d, then set up the t-test statistic formula.

Example 2

hard
A paired t-test for blood pressure before and after medication: \bar{d}=-8 mmHg, s_d=5 mmHg, n=16. Construct a 95% CI for the true mean difference and interpret.
← Back to Paired t-Test Formula Explained Practice Mode

Background Knowledge

These ideas may be useful before you work through the harder examples.

hypothesis testingconfidence intervalsampling distributionmeanstandard deviation