Rindskopf confidence interval for Binomial proportion

Arguments

x

A number or a vector with the number of successes.

n

A number or a vector with the number of trials.

conf.level

Confidence level for the returned confidence interval. By default, it is 0.95.

Value

A vector with the lower and upper limits.

Details

The expression to calculate the confidence interval according to the Rindskopf approach is given by:

$\phi=\text{logit}(\pi)=\log\left(\frac{\pi}{1 - \pi}\right)$,

where the maximum likelihood estimator for $\phi$ is:

$\hat{\phi}_{ML}=\log\left(\frac{x + 0.5}{n - x + 0.5}\right)$,

and its standard error is:

$\text{se}(\hat{\phi}_{ML})=\sqrt{\frac{1}{x + 0.5} + \frac{1}{n - x + 0.5}}$.

The adjustment of adding 0.5 successes and non-successes ensures that intervals can also be computed for the cases where $x=0$ or $x=n$ (where otherwise the maximum likelihood estimator and standard error would be infinite).

Since the scale of $\phi$ is $(- \infty, \infty)$, this interval respects the boundary constraints. Back-transformation to the scale of $\pi$ is performed using the inverse logit function:

$\pi=\text{expit}(\phi)=\frac{\exp(\phi)}{1 + \exp(\phi)}$.

Thus, the confidence interval for $\pi$ in the original scale is the Rindskopf confidence interval, as proposed by Rindskopf.

References

Rindskopf, D. (2000). Commentary: Approximate is better than “exact” for interval estimation of binomial proportions. The American Statistician, 54, 88.

See also

Author

David Esteban Cartagena Mejía, dcartagena@unal.edu.co

Examples

ci_p_rindskopf(x=15, n=50, conf.level=0.95)
#>           [,1]
#> [1,] 0.1938182
#> [2,] 0.4422596