Confidence interval calculator for a mean (t or z), a proportion (Wilson or Wald) or a difference of two means, with the margin of error and both bounds.
Actualizado Ejemplos verificados: 7
We can be 95% confident the true mean lies between 69.6594 and 75.1406 (72.4 ± 2.7406). The confidence describes the method — about 95% of intervals built this way contain the true value — not a probability for this one interval.
A confidence interval is a point estimate plus or minus a margin of error, and the margin is a critical value times the standard error. For a mean the standard error is s/√n and the critical value comes from Student's t with n − 1 degrees of freedom, or from the normal distribution when σ is known. For a proportion the Wilson score interval is the default, with the Wald formula p̂ ± z√(p̂(1 − p̂)/n) offered for comparison.
Survey results, lab measurements and A/B tests are reported this way. The default sample of 36 with mean 72.4 and standard deviation 8.1 gives SE = 1.35, t = 2.030 on 35 df and a 95% interval of 69.66 to 75.14, or 72.4 ± 2.74.
The 95% describes the procedure: across many samples, about 95% of intervals built this way contain the true value. Any single interval either contains it or does not.
Fuente de comprobación: t₀.₉₇₅,₃₅ = 2.030 (t table); Python bisection on the A&S 26.7.3 closed form gives 2.0301079283; margin = t·8.1/6
Fuente de comprobación: z₀.₉₇₅ = 1.959964 (z table; Python NormalDist().inv_cdf(0.975)); margin = z·8/6
Fuente de comprobación: Wilson score formula evaluated in Python with z = NormalDist().inv_cdf(0.975)
Fuente de comprobación: p̂ ± z√(p̂(1−p̂)/n) in Python
The method captures the true value in 95% of repeated samples. For the default data the interval is 69.66 to 75.14, but there is not a 95% probability that the true mean lies in that particular range: the true mean is fixed, and this interval either contains it or not. Nor does the interval hold 95% of individual values, which spread far wider (SD 8.1); that needs a prediction or tolerance interval.
Multiply the critical value by the standard error. For a mean, E = t × s/√n: with s = 8.1 and n = 36, E = 2.030 × 1.35 = 2.74. For a proportion, E ≈ z√(p̂(1 − p̂)/n): 54% of 1,000 respondents gives 1.96 × 0.0158 ≈ 0.031, or ±3.1 percentage points. Because n sits under a square root, quadrupling the sample size halves the margin.
Only when the population standard deviation σ is known, which is rare outside textbook problems and long-running process data. With σ estimated from the sample, use t. Its 95% critical value is larger for small samples, 2.262 for n = 10 against 1.960 for z, and approaches z as n grows: 2.030 at n = 36 and 1.984 at n = 101.
The Wald interval p̂ ± z·SE covers the true proportion less often than stated when n is small or p̂ is near 0 or 1; Brown, Cai and DasGupta (2001) found its coverage can fall far below 95% even with hundreds of observations. Wilson stays close to the stated level and never leaves the range 0 to 1. With 0 successes in 20 trials Wald gives the zero-width interval [0, 0], while Wilson gives 0 to 0.161.
Yes, for the matching test: a 95% interval for a difference in means excludes 0 exactly when a two-sided test at α = 0.05 using the same method (Welch or pooled) rejects no difference. The default two-group summaries give 3.5 ± 4.0, from −0.5 to 7.5, which includes 0, so the difference is not significant at 5%. The interval also shows how large the effect could plausibly be, which a p-value does not.
La precisión depende de tus datos y de los supuestos del método. El cálculo decimal usa 50 cifras significativas, pero las estimaciones, los métodos numéricos y los datos de origen pueden ser menos precisos; el redondeo mostrado no elimina esos límites. Ejemplos resueltos comprobados con fuentes independientes: 7. Por ejemplo, «Mean with sample SD, n = 36 (defaults)» se comprueba con t₀.₉₇₅,₃₅ = 2.030 (t table); Python bisection on the A&S 26.7.3 closed form gives 2.0301079283; margin = t·8.1/6.
NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.1 Confidence limits for the mean; NIST/SEMATECH e-Handbook, §7.2.4.1 Confidence intervals for a proportion (Wilson and normal approximation); Brown, Cai & DasGupta (2001). Interval estimation for a binomial proportion. Statistical Science 16(2), 101–133.
Esta calculadora incluye 7 ejemplos resueltos con respuestas de fuentes independientes. Forman parte del conjunto de pruebas y también puedes ejecutarlos aquí.
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