Michael Brenndoerfer | Data & AI, Private Equity, Technology

Michael Brenndoerfer

The T-Test › The One-Sample T-Test

Michael BrenndoerferJanuary 3, 20265 min read

Machine Learning from ScratchAll chapters

Part 2 of 9Hypothesis Testing

Chapter 15The T-Test › The One-Sample T-Test

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The one-sample t-test compares a sample mean to a hypothesized population value when the population variance is unknown. This is the simplest and most direct application of the t-distribution.

The Complete ProcedureLink Copied

Hypotheses:

$H_0: \mu = \mu_0$ (population mean equals the hypothesized value)
$H_a: \mu \neq \mu_0$ (two-sided), or $\mu > \mu_0$ / $\mu < \mu_0$ (one-sided)

Test statistic:

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

where:

$\bar{x}$ = sample mean
$\mu_0$ = hypothesized population mean
$s$ = sample standard deviation (with Bessel's correction: divide by $n-1$ )
$n$ = sample size
$s / \sqrt{n}$ = standard error of the mean

Distribution under $H_0$ : $t \sim t_{n-1}$ (t-distribution with $n-1$ degrees of freedom)

P-value calculation:

Two-sided: $p = 2 \times P(T > |t|)$ where $T \sim t_{n-1}$
One-sided (greater): $p = P(T > t)$
One-sided (less): $p = P(T < t)$

Decision rule: Reject $H_0$ if $p < \alpha$ , or equivalently, if $|t| > t_{\alpha/2, n-1}$ for two-sided tests.

Worked Example: Quality ControlLink Copied

A coffee shop claims their large drinks contain 16 ounces. A quality inspector measures 15 randomly selected drinks to verify this claim.

In[4]:

Code

import numpy as np
from scipy import stats

# Measurements from 15 randomly selected large drinks (in ounces)
drinks = [
    15.2,
    15.8,
    16.1,
    15.4,
    15.9,
    15.3,
    15.7,
    16.0,
    15.5,
    15.2,
    15.8,
    15.6,
    15.9,
    15.4,
    16.2,
]

# Hypothesized mean (the claimed amount)
mu_0 = 16.0

# Step 1: Calculate sample statistics
n = len(drinks)
x_bar = np.mean(drinks)
s = np.std(drinks, ddof=1)  # ddof=1 for Bessel's correction

# Step 2: Calculate standard error
se = s / np.sqrt(n)

# Step 3: Calculate t-statistic
t_stat = (x_bar - mu_0) / se

# Step 4: Determine degrees of freedom
df = n - 1

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

# Step 6: Get critical value for comparison
t_crit = stats.t.ppf(0.975, df=df)

# Step 7: Calculate 95% confidence interval
ci_lower = x_bar - t_crit * se
ci_upper = x_bar + t_crit * se

Let's walk through the calculation step by step:

Step 1: Sample Statistics

\bar{x} = \frac{1}{15}\sum_{i=1}^{15} x_i = \frac{15.2 + 15.8 + \cdots + 16.2}{15} = 15.667 \text{ oz}

s = \sqrt{\frac{\sum_{i=1}^{15}(x_i - \bar{x})^2}{15-1}} = 0.309 \text{ oz}

Step 2: Standard Error

SE = \frac{s}{\sqrt{n}} = \frac{0.309}{\sqrt{15}} = 0.080 \text{ oz}

Step 3: T-Statistic

t = \frac{\bar{x} - \mu_0}{SE} = \frac{15.667 - 16.0}{0.080} = \frac{-0.333}{0.080} = -4.18

Step 4: P-Value

With df = 14, we find the probability of observing $|t| \geq 4.18$ under $H_0$ :

p = 2 \times P(T_{14} > 4.18) = 0.0009

Out[5]:

Console

One-Sample T-Test: Coffee Shop Drink Size
=======================================================

Data: 15 drink measurements (ounces)
Hypothesized mean: mu_0 = 16.0 oz

Step-by-step calculation:
-------------------------------------------------------
1. Sample mean:       x_bar = 15.667 oz
2. Sample std dev:    s = 0.324 oz
3. Standard error:    SE = s/sqrt(n) = 0.324/sqrt(15) = 0.084 oz
4. T-statistic:       t = (x_bar - mu_0)/SE = (15.667 - 16.0)/0.084 = -3.980
5. Degrees of freedom: df = n - 1 = 15 - 1 = 14
6. P-value (two-sided): p = 0.0014

Decision (alpha = 0.05):
-------------------------------------------------------
Critical value: t_crit = +/-2.145
|t| = 3.980 > 2.145 = t_crit

95% Confidence Interval: [15.487, 15.846] oz
Note: CI does not contain 16.0, consistent with rejecting H_0

Conclusion:
-------------------------------------------------------
Reject H_0. There is significant evidence (p = 0.0009) that
the true mean drink size differs from the claimed 16 oz.
The drinks appear to be underfilled by about 0.33 oz on average.

Out[6]:

Visualization

T-distribution with observed test statistic and rejection regions highlighted. — Visualization of the one-sample t-test for the coffee shop example. The t-distribution shows the sampling distribution under the null hypothesis. The observed t-statistic of -4.18 falls far into the left tail, well beyond the critical value, leading to rejection of H_0.