Paired t-Test: A Step-by-Step Guide
What Is a Paired t-Test?
A paired t-test is a statistical method used to determine whether there is a significant difference between two sets of related observations. It’s especially useful when the same subjects are measured under two different conditions.
Common Use Cases
- Test scores before and after a preparatory course
- Comparing two treatment methods on the same group
- Measurements by different instruments on the same subjects
Test Scenario
Question:
Does taking a preparatory course improve scores on standardized tests
like the CompTIA exams?
To answer this, we’ll evaluate a sample dataset and compute:
- Case-wise differences
- Mean of these differences
- Standard deviation and standard error
- Perform a hypothesis test
Sample Data
| Subject | A | B | C | D | E | F | G | H | I | J |
|---|---|---|---|---|---|---|---|---|---|---|
| Before | 700 | 840 | 830 | 860 | 840 | 690 | 830 | 1180 | 930 | 1070 |
| After | 720 | 840 | 820 | 900 | 870 | 700 | 800 | 1200 | 950 | 1080 |
Step 1: Calculate the Differences
For each subject:
\[ d_i = \text{After} - \text{Before} \]
| Subject | Difference \(d_i\) |
|---|---|
| A | 20 |
| B | 0 |
| C | -10 |
| D | 40 |
| E | 30 |
| F | 10 |
| G | -30 |
| H | 20 |
| I | 20 |
| J | 10 |
Step 2: Compute Summary Statistics
Mean of the Differences
\[ \bar{d} = \frac{\sum d_i}{n} = \frac{110}{10} = 11 \]
Variance and Standard Deviation
Sample variance:
\[ s_d^2 = \frac{\sum (d_i - \bar{d})^2}{n - 1} \]
Alternatively, use:
\[ s_d = \sqrt{\frac{\sum d_i^2 - n\bar{d}^2}{n - 1}} \]
From the data:
\[ \sum d_i^2 = 5900,\quad \bar{d} = 11,\quad n = 10 \]
\[ s_d = \sqrt{\frac{5900 - 10(11^2)}{9}} = \sqrt{\frac{4690}{9}} \approx 22.83 \]
Step 3: Hypothesis Test
Null Hypothesis:
\(H_0: \mu_d = 0\) (no difference)
Alternative Hypothesis:
\(H_1: \mu_d \ne 0\) (a significant
difference exists)
Use the test statistic:
\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{11}{22.83 / \sqrt{10}} \approx 1.52 \]
Compare the computed t-statistic to the critical value from a t-distribution with \(n - 1 = 9\) degrees of freedom at your chosen significance level (e.g., 0.05).
Conclusion
- The mean improvement is 11 points.
- The standard deviation shows a spread of ±22.83.
- The t-test assesses whether this improvement is statistically significant.