binomCI {binomCI}

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}.

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 [e-mail].

See Also

binomCIs

Examples

binomCI(45, 100)