Package 'binomCI'

Confidence Intervals for a Binomial Proportion.

Description

Functions to compute 12 confidence intervals for a binomial proportion.

Details

Package:	binomCI
Type:	Package
Version:	1.3
Date:	2026-02-01
License:	GPL-2

Maintainers

Michail Tsagris [email protected].

Note

I would like to express my acknowledgements to Marc Girondot for spotting an error in the "Wilson" method in two extreme cases, when $x=1$ and when $n-x=1$. He also proposed a modification that exists in the package "Hmisc" and the relevant paper to cite is Agresti & Coull (1998).

Herman Callaert pointed out to me a limitation of the Agresti & Coull (1998) CI, and I fixed it. Their formula is valid only for $95\%$ confidence.

Author(s)

Michail Tsagris [email protected].

References

Agresti, A. & Caffo, B. (2000). Simple and effective confidence intervals for proportions and differences of proportions result from adding two successes and two failures. The American Statistician, 54(4), 280–288.

Agresti, A. & Coull, B. A. (1998). Approximate is better than "exact" for interval estimation of binomial proportions. The American Statistician, 52(2): 119–126.

Brown, L. D., Cai, T. T. & DasGupta, A. (2001). Interval estimation for a binomial proportion. Statistical Science, 16(2): 101-133.

Brown, L. D., Cai, T. T. & DasGupta, A. (2002). Confidence intervals for a binomial proportion and asymptotic expansions. The Annals of Statistics, 30(1): 160-201.

Cameron, E. (2011). On the estimation of confidence intervals for binomial population proportions in astronomy: the simplicity and superiority of the Bayesian approach. Publications of the Astronomical Society of Australia, 28(2): 128–139.

Newcombe, R. G. (1998). Two-sided confidence intervals for the single proportion: comparison of seven methods. Statistics in Medicine, 17(8): 857–872.

Pan, W. (2002). Approximate confidence intervals for one proportion and difference of two proportions. Computational statistics & Data Analysis, 40(1): 143-157.

Pires, A. M. & Amado, C. (2008). Interval estimators for a binomial proportion: Comparison of twenty methods. REVSTAT-Statistical Journal, 6(2): 165-197.

Ranucci, G. (2009). Binomial and ratio-of-Poisson-means frequentist confidence intervals applied to the error evaluation of cut efficiencies. arXiv preprint arXiv:0901.4845.

Sauro, J. & Lewis, J. R. (2005, September). Estimating completion rates from small samples using binomial confidence intervals: comparisons and recommendations. In Proceedings of the Human Factors and Ergonomics Society Annual Meeting (Vol. 49, No. 24, pp. 2100-2103). Sage CA: Los Angeles, CA: SAGE Publications.

Somerville, M. C. & Brown, R. S. (2013). Exact likelihood ratio and score confidence intervals for the binomial proportion. Pharmaceutical Statistics, 12(3): 120-128.

Thulin, Mans. The cost of using exact confidence intervals for a binomial proportion. (2014): 817-840. Electronic Journal of Statistics 8(1): 817-840.

Vollset, S. E. (1993). Confidence intervals for a binomial proportion. Statistics in Medicine, 12(9): 809-824.

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Confidence Intervals for a Binomial Proportion.

Description

Confidence Intervals for a Binomial Proportion.

Usage

binomCI(x, n, a = 0.05)

Arguments

x	The number of successes.
n	The number of trials.
a	The significance level to compute the $(1-\alpha)\%$ confidence intervals.

Details

The confidence intervals are:

Jeffreys:

$\left[ F(\alpha/2; x+0.5, n-x+0.5), F(1-\alpha/2; x+0.5, n-x+0.5) \right],$

where $F(\alpha, a, b)$ denotes the $\alpha$ quantile of the Beta distribution with parameters $a$ and $b$, $Be(a, b)$.

Wald:

$\left[ \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right],$

where $\hat{p}=\frac{x}{n}$ and $Z_{1-\alpha/2}$ denotes the $1-\alpha/2$ quantile of the standard normal distribution. If $\hat{p}=0$ the interval becomes $(0 , 1 - e^{\frac{1}{n}\log({\alpha}{2})})$ and if $\hat{p}=1$ the interval becomes $(e^{\frac{1}{n}\log({\alpha}{2})}, 1)$.

Wald corrected:

$\left[ \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} - \frac{0.5}{n}, \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} + \frac{0.5}{n} \right],$

and if $\hat{p}=0$ or $\hat{p}=1$ the previous (Wald) adjustment applies.

Wald BS:

$\left[ \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n-Z_{1-\alpha/2}-2Z_{1-\alpha/2}/n-1/n}} - \frac{0.5}{n}, \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n-Z_{1-\alpha/2}-2Z_{1-\alpha/2}/n-1/n}} + \frac{0.5}{n} \right],$

and if $\hat{p}=0$ or $\hat{p}=1$ the previous (Wald) adjustment applies.

Agresti and Coull:

$\left[ \hat{\theta} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{n+4}}, \hat{p} - Z_{1-\alpha/2} \times \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{n+4}} \right],$

where $\hat{\theta}=\frac{x+2}{n+4}$. Herman Callaert pointed out to me that in their 1998 the authors of this interval stated that this method is valid for $95\%$ confidence intervals, but not for other confidence levels such as $90\%$ or $99\%$. Hence, this interval is of $95\%$ confidence always.

Wilson:

$\left[ \frac{x_b}{n_b} - \frac{Z_{1-\alpha/2}\sqrt{n}}{n_b} \times \sqrt{\hat{p}(1-\hat{p})+Z_{1-\alpha/2}/4}, \frac{x_b}{n_b} + \frac{Z_{1-\alpha/2}\sqrt{n}}{n_b} \times \sqrt{\hat{p}(1-\hat{p})+Z_{1-\alpha/2}/4} \right],$

where $x_b=x+Z_{1-\alpha/2}^2/2$ and $n_b=n+Z_{1-\alpha/2}^2$.

Score:

$\left[ \frac{x+Z_{1-\alpha/2}^2-c}{n+Z_{1-\alpha/2}^2} , \frac{x+Z_{1-\alpha/2}^2+c}{n+Z_{1-\alpha/2}^2} \right],$

where $c=Z_{1-\alpha/2}\sqrt{x-x^2/n+Z_{1-\alpha/2}^2/4}$.

Score corrected:

$\left[ \frac{\ell_1}{n+Z_{1-\alpha/2}} , \frac{\ell_2}{n+Z_{1-\alpha/2}} \right],$

where $\ell_1=b_1+0.5Z_{1-\alpha/2}^2-Z_{1-\alpha/2}\sqrt{b_1-b_1^2/n+0.25Z_{1-\alpha/2}^2}$, $\ell_2=b_2+0.5Z_{1-\alpha/2}^2+Z_{1-\alpha/2}\sqrt{b_2-b_2^2/n+0.25Z_{1-\alpha/2}^2}$ and $b_1=x-0.5$, $b_2=x+0.5$.

Wald-logit:

$\left[ 1-(1+e^{b-c})^{-1}, 1-(1+e^{b+c})^{-1} \right],$

where $b=\log(\frac{x}{n-x})$ and $c=\frac{Z_{1-\alpha/2}}{\sqrt{n\hat{p}(1-\hat{p})}}$. If $\hat{p}=0$ or $\hat{p}=1$ the previous (Wald) adjustment applies.

Wald-logit corrected:

$\left[ 1-(1+e^{b-c})^{-1}, 1-(1+e^{b+c})^{-1} \right],$

where $b=\log(\frac{\hat{p}_b}{\hat{q}_b})$, $\hat{p}_b=x+0.5$, $\hat{q}_b=n-x+0.5$ and $c=\frac{Z_{1-\alpha/2}}{\sqrt{(n+1)\frac{\hat{p}_b}{n+1}(1-\frac{\hat{p}_b}{n+1})}}$.

Arcsine:

$\left\lbrace \sin^2\left[sin^{-1}(\sqrt{\hat{p}})-0.5\frac{Z_{1-\alpha/2}}{\sqrt{n}}\right], \sin^2\left[sin^{-1}(\sqrt{\hat{p}})+0.5\frac{Z_{1-\alpha/2}}{\sqrt{n}}\right] \right\rbrace.$

If $\hat{p}=0$ or $\hat{p}=1$ the previous (Wald) adjustment applies.

Exact binomial:

$\left[ (1+\frac{a_1}{d_1})^{-1}, (1+\frac{a_2}{d_2})^{-1} \right],$

where $a_1=n-x+1$, $a_2=a_1-1$, $d_1=x-F(\alpha/2,2x,2a_1)$, $d_2=(x+1)F(1-\alpha/2,2(x+1),2a_2)$ and $F(\alpha,a,b)$ denotes the $\alpha$ quantile of the F distribution with degrees of freedom $a$ and $b$, $F(a, b)$.

Value

A list including:

prop	The proportion.
ci	A matrix with 12 rows containing the 12 different $(1-\alpha)\%$ confidence intervals.

Author(s)

Michail Tsagris.

R implementation and documentation: Michail Tsagris [email protected].

See Also

binomCIs

Examples

binomCI(45, 100)

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Confidence Intervals for many Binomial Proportions.

Description

Confidence Intervals for many Binomial Proportions.

Usage

binomCIs(x, n, a = 0.05)

Arguments

x	A vector with the number of successes.
n	A vector with the number of trials.
a	The significance level to compute the $(1-\alpha)\%$ confidence intervals.

Value

A list with the the first element being the vector with the proportions and the rest 12 items contain the $(1-\alpha)\%$ confidence intervals.

Author(s)

Michail Tsagris.

R implementation and documentation: Michail Tsagris [email protected].

See Also

binomCI

Examples

x <- sample(40, 10)
n <- rep(40, 10)
binomCIs(x, n)

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