Learning Objectives
By the end of this chapter, you will be able to:
- Understand when to use paired t-test
- Calculate the difference scores and their statistics
- Perform paired t-test step by step
- Interpret results in context
- Distinguish between independent and paired samples
When to Use Paired t-Test
Use paired t-test when:
- Same subjects measured twice (before-after)
- Matched pairs (twins, siblings, matched characteristics)
- Related samples (husband-wife, left-right measurements)
flowchart TD
A[Two groups to compare]
B{Are observations<br/>paired/matched?}
C[Independent t-test]
D[Paired t-test]
A --> B
B -->|No| C
B -->|Yes| D
Examples of Paired Data
| Type | Example |
|---|---|
| Before-After | Weight before and after diet |
| Pre-Post | Test scores before and after training |
| Matched Pairs | Twin study comparing treatments |
| Same Subject | Left eye vs right eye measurements |
| Related Samples | Husband and wife income comparison |
The Paired t-Test Procedure
Step 1: Calculate Differences
For each pair: $d_i = x_{1i} - x_{2i}$
Step 2: Calculate Statistics of Differences
Mean of differences: \(\bar{d} = \frac{\sum d_i}{n}\)
Standard deviation of differences: \(s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}\)
Step 3: Calculate Test Statistic
\[t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}\]Degrees of freedom: df = n - 1 (where n = number of pairs)
Step-by-Step Example 1: Before-After Training
Problem: A training program was conducted for 8 employees. Their performance scores (out of 100) are:
| Employee | Before | After |
|---|---|---|
| 1 | 65 | 72 |
| 2 | 70 | 78 |
| 3 | 55 | 60 |
| 4 | 80 | 85 |
| 5 | 68 | 75 |
| 6 | 72 | 80 |
| 7 | 58 | 65 |
| 8 | 75 | 82 |
Test at α = 0.05 if training improved performance.
Solution:
Step 1: State hypotheses
- $H_0: \mu_d = 0$ (no improvement)
- $H_1: \mu_d > 0$ (improvement - after is higher)
Let d = After - Before (so positive d means improvement)
Step 2: Calculate differences
| Employee | Before | After | d = After - Before | d² |
|---|---|---|---|---|
| 1 | 65 | 72 | 7 | 49 |
| 2 | 70 | 78 | 8 | 64 |
| 3 | 55 | 60 | 5 | 25 |
| 4 | 80 | 85 | 5 | 25 |
| 5 | 68 | 75 | 7 | 49 |
| 6 | 72 | 80 | 8 | 64 |
| 7 | 58 | 65 | 7 | 49 |
| 8 | 75 | 82 | 7 | 49 |
| Sum | 54 | 374 |
Step 3: Calculate mean and SD of differences
\[\bar{d} = \frac{54}{8} = 6.75\] \[s_d = \sqrt{\frac{374 - \frac{(54)^2}{8}}{8-1}} = \sqrt{\frac{374 - 364.5}{7}} = \sqrt{\frac{9.5}{7}} = \sqrt{1.357} = 1.165\]Step 4: Calculate test statistic
\[t = \frac{6.75 - 0}{1.165/\sqrt{8}} = \frac{6.75}{0.412} = 16.38\]Step 5: Find critical value
- df = 8 - 1 = 7
- Right-tailed, α = 0.05
- From t-table: t* = 1.895
Step 6: Decision
- t = 16.38 > 1.895
- Reject H₀
Step 7: Conclusion At the 0.05 level of significance, there is strong evidence that the training program significantly improved employee performance scores.
Step-by-Step Example 2: Policy Impact Assessment
Problem: Blood pressure readings for 10 patients before and after a new medication:
| Patient | Before | After | d = Before - After |
|---|---|---|---|
| 1 | 142 | 138 | 4 |
| 2 | 155 | 145 | 10 |
| 3 | 140 | 142 | -2 |
| 4 | 148 | 140 | 8 |
| 5 | 165 | 155 | 10 |
| 6 | 150 | 148 | 2 |
| 7 | 138 | 130 | 8 |
| 8 | 160 | 152 | 8 |
| 9 | 145 | 140 | 5 |
| 10 | 158 | 150 | 8 |
Test at α = 0.01 if medication reduces blood pressure.
Solution:
Step 1: State hypotheses
- $H_0: \mu_d = 0$ (no reduction)
- $H_1: \mu_d > 0$ (reduction - before is higher)
Let d = Before - After (positive d means reduction)
Step 2: Calculate statistics
- $\sum d = 61$
- $\sum d^2 = 521$
- n = 10
Step 3: Calculate test statistic
\[t = \frac{6.1 - 0}{4.07/\sqrt{10}} = \frac{6.1}{1.287} = 4.74\]Step 4: Find critical value
- df = 10 - 1 = 9
- Right-tailed, α = 0.01
- From t-table: t* = 2.821
Step 5: Decision
- t = 4.74 > 2.821
- Reject H₀
Step 6: Conclusion At the 0.01 level of significance, there is strong evidence that the medication significantly reduces blood pressure.
Step-by-Step Example 3: Two-Tailed Paired Test
Problem: Reaction times (milliseconds) of 6 subjects using right and left hands:
| Subject | Right | Left |
|---|---|---|
| 1 | 350 | 380 |
| 2 | 320 | 340 |
| 3 | 400 | 390 |
| 4 | 280 | 310 |
| 5 | 360 | 350 |
| 6 | 310 | 325 |
Test at α = 0.05 if there is a difference.
Solution:
Step 1: State hypotheses
- $H_0: \mu_d = 0$ (no difference)
- $H_1: \mu_d \neq 0$ (two-tailed)
Step 2: Calculate differences (d = Right - Left)
| Subject | d | d² |
|---|---|---|
| 1 | -30 | 900 |
| 2 | -20 | 400 |
| 3 | 10 | 100 |
| 4 | -30 | 900 |
| 5 | 10 | 100 |
| 6 | -15 | 225 |
| Sum | -75 | 2625 |
Step 3: Calculate statistics \(\bar{d} = \frac{-75}{6} = -12.5\)
\[s_d = \sqrt{\frac{2625 - \frac{(-75)^2}{6}}{5}} = \sqrt{\frac{2625 - 937.5}{5}} = \sqrt{337.5} = 18.37\]Step 4: Calculate test statistic \(t = \frac{-12.5 - 0}{18.37/\sqrt{6}} = \frac{-12.5}{7.50} = -1.67\)
Step 5: Find critical value
- df = 6 - 1 = 5
- Two-tailed, α = 0.05
- From t-table: t* = ±2.571
Step 6: Decision
- |t| = 1.67 < 2.571
- Fail to Reject H₀
Step 7: Conclusion At the 0.05 level of significance, there is insufficient evidence to conclude that reaction times differ between right and left hands.
Confidence Interval for Mean Difference
\[\bar{d} \pm t^* \times \frac{s_d}{\sqrt{n}}\]Example 4: 95% CI for Mean Improvement
Using Example 1 data:
- $\bar{d}$ = 6.75
- $s_d$ = 1.165
- n = 8
- df = 7, t* = 2.365 (two-tailed, α = 0.05)
\(95\% \text{ CI} = 6.75 \pm 2.365 \times \frac{1.165}{\sqrt{8}}\) \(= 6.75 \pm 2.365 \times 0.412 = 6.75 \pm 0.97\) \(= (5.78, 7.72)\)
Interpretation: We are 95% confident that the mean improvement in scores after training is between 5.78 and 7.72 points.
Decision Flow: Independent vs Paired
flowchart TD
A[Comparing two means]
B{Same subjects<br/>measured twice?}
C{Subjects<br/>matched?}
D[Paired t-test]
E[Independent t-test]
A --> B
B -->|Yes| D
B -->|No| C
C -->|Yes| D
C -->|No| E
Why Not Use Independent t-Test for Paired Data?
| Aspect | Paired t-test | Independent t-test |
|---|---|---|
| Controls for individual differences | Yes | No |
| df | n - 1 | n₁ + n₂ - 2 |
| Power | Higher (removes variation) | Lower |
| When to use | Related samples | Unrelated samples |
Using independent t-test on paired data:
- Ignores the pairing
- Loses statistical power
- May miss real differences
Summary Table
| Component | Formula |
|---|---|
| Difference | $d_i = x_{1i} - x_{2i}$ |
| Mean difference | $\bar{d} = \frac{\sum d_i}{n}$ |
| SD of differences | $s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}$ |
| Test statistic | $t = \frac{\bar{d}}{s_d/\sqrt{n}}$ |
| Degrees of freedom | df = n - 1 |
| Confidence interval | $\bar{d} \pm t^* \times \frac{s_d}{\sqrt{n}}$ |
Practice Problems
Problem 1
A weight loss program measured 10 participants:
| Person | Before | After |
|---|---|---|
| 1 | 85 | 80 |
| 2 | 92 | 88 |
| 3 | 78 | 76 |
| 4 | 95 | 90 |
| 5 | 88 | 82 |
| 6 | 76 | 74 |
| 7 | 82 | 78 |
| 8 | 90 | 85 |
| 9 | 84 | 80 |
| 10 | 88 | 84 |
Test at α = 0.05 if the program reduced weight.
Problem 2
Calculate a 95% confidence interval for the mean weight loss in Problem 1.
Problem 3
Job satisfaction scores (1-10) before and after policy change:
| Employee | Before | After |
|---|---|---|
| 1 | 6 | 8 |
| 2 | 5 | 6 |
| 3 | 7 | 7 |
| 4 | 4 | 6 |
| 5 | 6 | 7 |
| 6 | 5 | 8 |
Test at α = 0.05 if satisfaction improved.
Problem 4
Test scores of students taught by two different methods (same students, both methods):
| Student | Method A | Method B |
|---|---|---|
| 1 | 75 | 78 |
| 2 | 82 | 85 |
| 3 | 68 | 70 |
| 4 | 90 | 88 |
| 5 | 85 | 86 |
Test at α = 0.05 if there is a significant difference.
Summary
- Paired t-test is used for dependent/matched samples
- Calculate differences first, then test if mean difference ≠ 0
- df = n - 1 (number of pairs minus 1)
- More powerful than independent t-test when pairing is relevant
- Common applications: before-after studies, matched designs
Next Topic
In the next chapter, we will study the Chi-Square Test for testing categorical data - association between variables and goodness of fit.
