The One-Sample t-Test

Question: Is there a significant difference between the mean of your sample and a known benchmark?

Use it when: You have one group and want to compare its mean to a fixed value.

Hypotheses:

Test statistic:

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

where $\bar{x}$ is the sample mean, $mu$ is the benchmark, $s$ is the sample standard deviation, and $n$ is the sample size.

In Python: scipy.stats.ttest_1samp(sample, popmean=benchmark)

One-Sample t-Test

Choose a scenario

Adjust inputs

Sample mean (x̄)4.7

Std dev (s)0.3

Sample size (n)100

Benchmark (μ₀)4.5

t-Distribution (df = 99)

Shaded tails = p-value region. Orange marker = t-statistic.

Results

Standard error

0.0300

t-statistic

6.667

Degrees of freedom

p-value (two-tailed)

< 0.001

Interpretation (α = 0.05)

p = < 0.001 < 0.05 — reject H₀. The sample mean (4.7) differs significantly from the benchmark (4.5).

Step-by-step

1. SE = s / √n = 0.3 / √100 = 0.0300

2. t = (x̄ − μ₀) / SE = (4.7 − 4.5) / 0.0300 = 6.667

3. df = n − 1 = 100 − 1 = 99

4. p = < 0.001 (significant at α = 0.05)

p-value region (shaded tails)t = observed t-statistic−t = mirror (two-tailed)

Enter sample statistics (mean, SD, n). Compute the t-statistic and p-value, and see where the statistic falls on the t-distribution.