Confidence interval with the t distribution
When σ is estimated from the sample, use the t quantile with n − 1 degrees of freedom instead of z.
Worked example
Sample: n = 16, x̄ = 59.3, s = 5; σ unknown. Find the 95% confidence interval (t quantile with 15 degrees of freedom: 2.131).
Worked solution
- 1.σ unknown → t distribution with n − 1 = 15 degrees of freedom
- 2.t · s / √n = 2.131 · 5 / 4 = 2.6638
- 3.[56.64; 61.96]
Typical mistake: Using z = 1.96 instead of the t quantile when σ is unknown.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task
Sample: n = 36, x̄ = 34.3, s = 6; σ unknown. Find the 95% confidence interval (t quantile with 35 degrees of freedom: 2.030).
Lower limit (2 decimal places)
Upper limit (2 decimal places)
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- σ known: x̄ ± z · σ / √n
- σ unknown: x̄ ± t(n − 1) · s / √n
- Minimum n for margin E: n ≥ (z · σ / E)², round up
Typical mistake: Using z = 1.96 instead of the t quantile when σ is unknown.
Frequently asked questions
What is the typical mistake in "Confidence interval with the t distribution"?
Using z = 1.96 instead of the t quantile when σ is unknown.
When do I use t instead of z?
When σ is unknown and estimated by s from the sample, especially for small samples.