Paired t Tests
In the previous lesson, we compared two independent groups.
In many real-world situations, however, measurements are naturally linked or paired. In such cases, using an independent t test would be incorrect.
The paired t test is designed specifically for these dependent observations.
What Does “Paired” Mean?
Data is considered paired when:
- Each observation in one sample is matched with one in the other
- The same subject is measured twice
- Measurements are taken before and after a treatment
Examples include:
- Weight before and after a diet
- Blood pressure before and after medication
- Test scores before and after training
Key Idea Behind the Paired t Test
Instead of comparing two separate means, the paired t test focuses on the difference within each pair.
The test then determines whether the average difference is significantly different from zero.
When Do We Use a Paired t Test?
A paired t test is appropriate when:
- Data comes in pairs
- The differences are approximately normally distributed
- The pairs are randomly selected
Setting Up the Hypotheses
Let d represent the difference between paired observations.
| Hypothesis | Statement |
|---|---|
| H₀ | μd = 0 (no average difference) |
| H₁ | μd ≠ 0 (average difference exists) |
The Paired t Test Statistic
The test statistic is computed using the differences:
t = ( d̄ − 0 ) ÷ ( sd / √n )
- d̄ = mean of the differences
- sd = standard deviation of the differences
- n = number of pairs
Deep Numerical Example (Step-by-Step)
A fitness program measures participants’ weights before and after a 6-week program.
| Participant | Before (kg) | After (kg) | Difference (Before − After) |
|---|---|---|---|
| 1 | 80 | 76 | 4 |
| 2 | 72 | 70 | 2 |
| 3 | 90 | 85 | 5 |
| 4 | 68 | 66 | 2 |
- Mean difference (d̄) = 3.25
- Standard deviation of differences (sd) = 1.26
- n = 4
- α = 0.05
Step 1: State the Hypotheses
H₀: μd = 0
H₁: μd ≠ 0
Step 2: Calculate the t Statistic
t = 3.25 ÷ (1.26 / √4)
t = 3.25 ÷ 0.63 ≈ 5.16
Step 3: Make the Decision
Degrees of freedom = n − 1 = 3
At α = 0.05, the critical t value ≈ 3.18
Since 5.16 > 3.18:
Decision: Reject the null hypothesis
Interpretation in Plain English
There is strong statistical evidence that the fitness program produced a real change in weight.
Paired vs Independent t Test
| Paired t Test | Independent t Test |
|---|---|
| Same subjects measured twice | Different subjects in each group |
| Analyzes differences | Analyzes separate means |
| More powerful for matched data | Used when no pairing exists |
Common Mistakes to Avoid
- Using independent t test for paired data
- Forgetting to compute differences
- Assuming pairing when none exists
- Ignoring outliers in differences
Quick Check
What is tested in a paired t test?
Practice Quiz
Question 1:
When should a paired t test be used?
Question 2:
What is the null hypothesis in a paired t test?
Question 3:
Why is pairing useful?
Mini Practice
A training program measures employee efficiency scores before and after training.
- Before: 68, 70, 72, 74
- After: 74, 76, 78, 80
Test whether the training improved efficiency at α = 0.05.
What’s Next
In the next lesson, we will study Correlation (Pearson and Spearman), which examines relationships between numerical variables.