Confidence interval for the mean
A confidence interval gives a range that covers the true mean with a chosen level of confidence.
Formulas
- 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
Your turn
Three problems on this topic. Comma or point, fractions like 1/6 work too.
Task
How large must the sample be so that the 99% confidence interval for μ is at most ±1.0 wide? σ = 15, z = 2.576.
Minimum n
Comma or point; fractions like 1/6 work too.
Worked example
A sample of n = 64 has mean x̄ = 77.1; σ = 12 is known. Find the 99% confidence interval for μ (z = 2.576).
Worked solution
- 1.σ / √n = 12 / 8 = 1.5000
- 2.Margin: z · σ / √n = 2.576 · 1.5000 = 3.8640
- 3.[77.1 − 3.8640; 77.1 + 3.8640] = [73.24; 80.96]
Typical mistake: Dividing by n instead of √n.
Sample: n = 36, x̄ = 44.8, s = 9; σ unknown. Find the 95% confidence interval (t quantile with 35 degrees of freedom: 2.030).
Worked solution
- 1.σ unknown → t distribution with n − 1 = 35 degrees of freedom
- 2.t · s / √n = 2.030 · 9 / 6 = 3.0450
- 3.[41.75; 47.85]
Typical mistake: Using z = 1.96 instead of the t quantile when σ is unknown.
Frequently asked questions
When do I use t instead of z?
When σ is unknown and estimated by s from the sample, especially for small samples.