How to use
- Choose Mean or Proportion.
- For a mean, enter mean, SD and n or paste raw data; for a proportion, enter successes and sample size.
- Choose the confidence level and read the interval.
Options explained
- Mean
- Mean ± t × SD ÷ √n; with a known population σ, z is used instead of t.
- Proportion
- The Wilson interval is far more accurate than Wald (p ± z√(p(1−p)/n)) for small samples or proportions near 0 or 100%, and is recommended.
Worked example
A sample of n = 25 has mean 50 and SD 10. The standard error is 10 ÷ √25 = 2, and the t value with 24 degrees of freedom is 2.0639 at 95%:
margin = 2.0639 × 2 = 4.128
95% CI = 50 ± 4.128 = 45.87 to 54.13
99% CI: t = 2.7969, margin 5.594, so 44.41 to 55.59A higher confidence level gives a wider interval.
Why Wilson for proportions
For 8 successes out of 20 (p = 0.40), the Wald interval is 18.5% to 61.5% and the Wilson interval is 21.9% to 61.3%. The difference is larger at the extremes: 0 out of 20 gives a Wald interval of exactly 0% to 0%, which says there is no uncertainty at all, while Wilson gives 0% to 16.1%.
Wilson centre = (p + z²/(2n)) ÷ (1 + z²/n)
half-width = z × √(p(1−p)/n + z²/(4n²)) ÷ (1 + z²/n)Frequently asked questions
What does a 95% confidence interval mean?
If you repeated the sampling many times, about 95% of such intervals would contain the true value. It does not mean a 95% chance for this particular interval.
When should I use z instead of t?
Use t unless you truly know the population standard deviation.
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