Michael Brenndoerfer | Data & AI, Private Equity, Technology

Michael Brenndoerfer

The T-Test › The Paired T-Test

Michael BrenndoerferJanuary 3, 20265 min read

Machine Learning from ScratchAll chapters

Part 2 of 9Hypothesis Testing

Chapter 15The T-Test › The Paired T-Test

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The paired t-test is used when observations come in matched pairs, where each pair shares some characteristic that makes them non-independent. Common scenarios include:

Before-after measurements: Same subjects measured twice
Matched case-control studies: Cases matched to controls on key characteristics
Twin studies: Comparing outcomes between twins
Left-right comparisons: Same subject, different sides

Why Pairing MattersLink Copied

Consider testing whether a training program improves performance. You could:

Independent design: Randomly assign people to training or control groups, compare group means
Paired design: Measure each person before and after training, analyze the changes

The paired design is often more powerful because it eliminates between-subject variability. People differ enormously in baseline ability, but by looking at changes within each person, we focus only on the training effect.

The MathematicsLink Copied

The paired t-test is simply a one-sample t-test on the differences:

d_i = x_{i,\text{after}} - x_{i,\text{before}}

Test statistic:

t = \frac{\bar{d}}{s_d / \sqrt{n}}

where:

$\bar{d}$ = mean of the paired differences
$s_d$ = standard deviation of the differences
$n$ = number of pairs
$df = n - 1$

Null hypothesis: $H_0: \mu_d = 0$ (no average change)

Worked Example: Blood Pressure MedicationLink Copied

A study tests whether a new medication reduces blood pressure. Ten patients have their systolic blood pressure measured before starting medication and again after 4 weeks of treatment.

In[11]:

Code

import numpy as np
from scipy import stats

# Blood pressure measurements (mmHg)
before = np.array([145, 152, 148, 155, 160, 142, 158, 165, 150, 155])
after = np.array([138, 145, 140, 150, 152, 140, 148, 158, 145, 148])

# Step 1: Calculate differences (after - before)
differences = after - before
n = len(differences)

# Step 2: Calculate mean and std of differences
d_bar = np.mean(differences)
s_d = np.std(differences, ddof=1)

# Step 3: Calculate standard error
se_d = s_d / np.sqrt(n)

# Step 4: Calculate t-statistic
t_stat = d_bar / se_d
df = n - 1

# Step 5: Calculate p-value (two-sided)
p_value = 2 * stats.t.sf(abs(t_stat), df=df)

# Step 6: Confidence interval for mean change
t_crit = stats.t.ppf(0.975, df=df)
ci = (d_bar - t_crit * se_d, d_bar + t_crit * se_d)

# Verify with scipy
t_scipy, p_scipy = stats.ttest_rel(after, before)

Out[12]:

Console

Paired T-Test: Blood Pressure Medication Study
============================================================

Data (n = 10 patients):
------------------------------------------------------------
Before: [np.int64(145), np.int64(152), np.int64(148), np.int64(155), np.int64(160), np.int64(142), np.int64(158), np.int64(165), np.int64(150), np.int64(155)]
After:  [np.int64(138), np.int64(145), np.int64(140), np.int64(150), np.int64(152), np.int64(140), np.int64(148), np.int64(158), np.int64(145), np.int64(148)]
Change: [np.int64(-7), np.int64(-7), np.int64(-8), np.int64(-5), np.int64(-8), np.int64(-2), np.int64(-10), np.int64(-7), np.int64(-5), np.int64(-7)]

Step-by-step calculation:
------------------------------------------------------------
1. Mean change:     d_bar = -6.60 mmHg
2. Std of changes:  s_d = 2.17 mmHg
3. Standard error:  SE = s_d/sqrt(n) = 2.17/sqrt(10) = 0.69
4. T-statistic:     t = d_bar/SE = -6.60/0.69 = -9.616
5. Degrees of freedom: df = n - 1 = 9
6. P-value (two-sided): 0.000005

95% CI for mean change: [-8.15, -5.05] mmHg

Interpretation:
------------------------------------------------------------
The average blood pressure decreased by 6.6 mmHg (p < 0.001).
We are 95% confident the true mean reduction is between
5.0 and 8.2 mmHg.

Out[13]:

Visualization

Paired data plot showing before and after blood pressure measurements connected by lines. — Individual patient changes in blood pressure before and after medication. Each line connects measurements from the same patient. The consistent downward slope indicates most patients experienced a reduction.

Histogram showing distribution of blood pressure changes. — Distribution of changes in blood pressure. The paired t-test detects this reduction with high significance (p < 0.001).

Why Not Use an Independent T-Test?Link Copied

You might wonder: what if we ignored the pairing and used an independent two-sample t-test? Let's compare:

In[14]:

Code

# Paired t-test (correct)
t_paired, p_paired = stats.ttest_rel(after, before)

# Independent t-test (ignoring pairing)
t_indep, p_indep = stats.ttest_ind(after, before)

print("Comparison of paired vs. independent analysis:")
print("-" * 50)
print(f"Paired t-test:      t = {t_paired:.3f}, p = {p_paired:.6f}")
print(f"Independent t-test: t = {t_indep:.3f}, p = {p_indep:.4f}")
print()
print("The paired test is much more powerful because it eliminates")
print("between-subject variability, focusing only on within-subject changes.")

Out[14]:

Console

Comparison of paired vs. independent analysis:
--------------------------------------------------
Paired t-test:      t = -9.616, p = 0.000005
Independent t-test: t = -2.233, p = 0.0385

The paired test is much more powerful because it eliminates
between-subject variability, focusing only on within-subject changes.

The paired test detects the effect with much higher confidence because it accounts for the correlation between before and after measurements within each subject.