A modified rule of three for the one-sided binomial confidence interval

Lonnie Turpin; Jeanne-Claire Patin; William Jens; Morgan Turpin

doi:10.1515/ijb-2022-0061

Article

A modified rule of three for the one-sided binomial confidence interval

Lonnie Turpin , Jeanne-Claire Patin , William Jens and Morgan Turpin

Published/Copyright: September 4, 2023

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From the journal The International Journal of Biostatistics Volume 20 Issue 2

Abstract

Consider the one-sided binomial confidence interval L , 1 containing the unknown parameter p when all n trials are successful, and the significance level α to be five or one percent. We develop two functions (one for each level) that represent approximations within α / 3 of the exact lower-bound L = α ^1/n. Both the exponential (referred to as a modified rule of three) and the logarithmic function are shown to outperform the standard rule of three L ≃ 1 − 3/n over each of their respective ranges, that together encompass all sample sizes n ≥ 1. Specifically for the exponential, we find that exp − 3 / n is a better lower bound when α = 0.05 and n < 1054 and that exp − 4.6569 / n is a better bound when α = 0.01 and n < 209.

Keywords: one-sided binomial confidence interval; exponential approximation; logarithmic approximation; rule of three; optimization

MSC Classification: 62F25; 90C30

Corresponding author: Lonnie Turpin, Jr., McNeese State University, Lake Charles, LA 70605, USA, E-mail: lturpin@mcneese.edu

Acknowledgments

We would like to thank the anonymous reviewers for the generous feedback that greatly improved this manuscript. Additionally, we are grateful to Dr. Matiur Rahman for the many helpful discussions, as well as the editor for the constructive comments on the overall readability and structure of the paper.

Research ethics: Not applicable.
Author contributions: The authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Competing interests: The authors state no conflict of interest.
Research funding: None declared.
Data availability: Not applicable.

Appendix: Resolving the open problem from [8]

Fix α = 0.01 and maintain A _0.01 = 32 (for brevity, we will drop the subscript for the remainder of this section as the focus is only on the one-percent level). We wish to find an estimate (denoted by B) over the sample size n for which the inequality L − A > α holds. That is, a B such that L − B ≤ α for the range b ≤ n ≤ A − 1. Consider

B = 1 − − ln α n + n 1 A + 2 + ln α ln A 3 − 2 × − ln α + α A + ln α − 1 Γ n + 1 − α ⁡ ln A 3

where Γ ⋅ denotes the gamma function. Let the absolute value of β ₂ + β ₃ follow

β 2 + β 3 ≤ 1 ln A

such that β 1 , β 2 , β 3 ∈ R . We begin by expressing B in the target form

B = A + − ln α n − 2 + β 1 + β 2 Γ n + 1 − β 3

B = 1 − − ln α n + − ln α n − 2 + 1 A + 2 + ln α ln A 3 + α A + ln α − 1 − n ! α ⁡ ln A 3 n !

by the identity Γ n + 1 = n ! and a simple rearranging of terms.

Lemma 1

The range n < A is invalid as n = 1 implies L − B > α .

Proof

When n = 1, we get 1 − − ln α / n = 1 + ln α , the expression

− ln α n − 2 + β 1 = − ln α ,

and the equivalency

β 2 Γ n + 1 − β 3 = β 2 − n ! β 3 n !

is reduced to β ₂ − β ₃. Thus, A = 1 + ln α and B = − ln α + β 2 − β 3 . Following [8]; we solve L − B in the form L − A − B as

L − B = α − 1 + ln α − − ln α + β 2 − β 3 = α − 1 − ln α + ln α − β 2 + β 3 = α − 1 − β 2 + β 3

where we let α − 1 − β 2 + β 3 be a function of β ₂ + β ₃ as

f β 2 + β 3 = α − 1 − β 2 + β 3 .

It is clear that f β 2 + β 3 has its only real root r at r = −0.99. Considering

β 2 + β 3 ≤ 1 ln A = β 2 + β 3 ∈ − 1 ln A , 1 ln A ,

we get α − 1 − β 2 + β 3 > α since r ∉ − 1 / ln A , 1 / ln A .□

Theorem 2

Since b > 1 for the range b ≤ n ≤ A − 1, we claim that

B = 1 − − ln α n + − ln α n − 2 + 1 A + 2 + ln α ln A 3 + α A + ln α − 1 − n ! α ⁡ ln A 3 n !

for 2 ≤ n ≤ A − 1.

Proof

Solving the argument

b : = arg min n ≥ 2 α − 1 − − ln α n + − ln α n − 2 + 1 A + 2 + ln α ln A 3 + α A + ln α − 1 − n ! α ⁡ ln A 3 n ! ≥ 0

yields b = 2. Let us define the objective function as follows

f β 1 , β 2 , β 3 = ∑ n = b A − 1 α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! .

We now use the optimization model below

min ∑ n = b A − 1 α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! s . t . α − α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! ≥ 0 , ∀ n = b , b + 1 , … , A − 1 β 2 + β 3 ≤ 1 ln A β 1 , β 2 , β 3 ∈ R

resulting in β 1 * = 0.3796 , β 2 * = − 0.1478 , β 3 * = 0.0113 for b = 2. The estimates

β 1 = 1 A + 2 + ln α ln A 3 = 0.3813 ,

β 2 = α A + ln α − 1 = − 0.1547 , and β 3 = α ⁡ ln A 3 = 0.0116 are approximately equal to β 1 * , β 2 * , and β 3 * , respectively. Denoting ϵ = f β 1 , β 2 , β 3 − f β 1 * , β 2 * , β 3 * , we get ϵ = 0.0025 revealing the estimates β 1 , β 2 , β 3 to be an ϵ-optimal solution.□

Lemma 2

Suppose we relax the assumption β 2 + β 3 ≤ 1 / ln A . A solution for n ≥ 1 remains infeasible, and has no effect on the solution for n ≥ 2.

Proof

We begin by simply plugging in the current values at n = 1 giving

L − B = α − 1 + ln α + − ln α + β 2 − β 3 = α − 1 + ln α − ln α + β 2 − β 3 = α − 1 + α A + ln α − 1 − α ⁡ ln A 3 = 0.01 − 1 − 0.1547 − 0.0116 = − 0.8237

which violates the criterion L − B ≤ α . To check this logic, let us revisit the argument for b by modifying the boundary from n ≥ 2 to n ≥ 1 as

c : = arg min n ≥ 1 α − 1 − − ln α n + − ln α n − 2 + 1 A + 2 + ln α ln A 3 + α A + ln α − 1 − n ! α ⁡ ln A 3 n ! ≥ 0 ,

which still results in c = 2. This shows there exists no solution for c = 1, and thereby no feasible solution for n ≥ 1, under the current estimates. To fully rule out n ≥ 1 being feasible, we eliminate the constraint β 2 + β 3 ≤ 1 / ln A , and obtain the following system

min ∑ n = c A − 1 α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! s . t . α − α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! ≥ 0 , ∀ n = c , c + 1 , … , A − 1 β 1 , β 2 , β 3 ∈ R

The result is no solution for c = 1, and thereby no feasible solution for n ≥ 1. Solving for c = 2, we get the identical solution β 1 * = 0.3796 , β 2 * = − 0.1478 , β 3 * = 0.0113 as before for b = 2. Therefore, the initial assumption β 2 + β 3 ≤ 1 / ln A has no effect on the solution for n ≥ 2.□

Remark 4

Fix L = α ^1/n and A = 1 − − ln α / n such that α = 0.01, A = 32, and n is finite. Empirically, we find L − B ≤ α in the form

B = 1 − − ln α n + − ln α n − 2 + 1 A + 2 + ln α ln A 3 + α A + ln α − 1 − n ! α ⁡ ln A 3 n !

for 2 ≤ n ≤ A − 1, and L − A ≤ α if n ≥ A. This guarantees the existence of logarithmic approximations within α of L for n ≥ 2. The optimization of

α − α 1 n − 1 − − ln α n + − ln α n − 2 + β 1 + β 2 − n ! β 3 n ! ≥ 0

for n = 2, 3, …, A − 1 by changing β 1 , β 2 , β 3 ∈ R validates our results for n ≥ 2 and effectively rules out any n < 2. This resolves the open problem posed in the closing statement of [8].

References

1. Bol’shev, L. On the construction of confidence limits. Theor Probab Appl 1965;10:173–7. https://doi.org/10.1137/1110022.Search in Google Scholar

2. Crow, E. Confidence intervals for a proportion. Biometrika 1956;43:423–35. https://doi.org/10.2307/2332920.Search in Google Scholar

3. Blyth, C, Still, H. Binomial confidence intervals. J Am Stat Assoc 1983;78:108–16. https://doi.org/10.1080/01621459.1983.10477938.Search in Google Scholar

4. Wang, W. Smallest confidence intervals for one binomial proportion. J Stat Plann Inference 2006;136:4293–306. https://doi.org/10.1016/j.jspi.2005.08.044.Search in Google Scholar

5. Clopper, C, Pearson, E. The use of confidence or fiducial limits illustrated in the case of the binomial. Biometrika 1934;26:404–13. https://doi.org/10.1093/biomet/26.4.404.Search in Google Scholar

6. Agresti, A, Coull, B. Approximate is better than “exact” for interval estimation of binomial proportions. Am Stat 1998;52:119–26. https://doi.org/10.1080/00031305.1998.10480550.Search in Google Scholar

7. Thulin, M. The cost of using exact confidence intervals for a binomial proportion. Electron J Stat 2014;8:817–40. https://doi.org/10.1214/14-ejs909.Search in Google Scholar

8. Turpin, L, Jens, W. On logarithmic approximations for the one-sided binomial confidence interval. J Optim Theor Appl 2018;177:254–60. https://doi.org/10.1007/s10957-018-1257-x.Search in Google Scholar

9. Van Belle, G. Statistical rules of thumb. New York: John Wiley & Sons; 2002.Search in Google Scholar

10. Jovanovic, B, Levy, P. A look at the rule of three. Am Stat 1997;51:137–9. https://doi.org/10.2307/2685405.Search in Google Scholar

11. Cai, T. One-sided confidence intervals in discrete distributions. J Stat Plann Inference 2005;131:63–88. https://doi.org/10.1016/j.jspi.2004.01.005.Search in Google Scholar

12. Hall, P. Improving the normal approximation when constructing one-sided confidence intervals for binomial or Poisson parameters. Biometrika 1982;69:647–52. https://doi.org/10.1093/biomet/69.3.647.Search in Google Scholar

13. Louis, T. Confidence intervals for a binomial parameter after observing no successes. Am Stat 1981;35:154. https://doi.org/10.2307/2683985.Search in Google Scholar

14. Bickel, P, Doksum, K. Mathematical statistics: basic ideas and selected topics. Boca Raton: CRC Press; 2015.10.1201/b19822Search in Google Scholar

15. Hanley, J, Lippman-Hand, A. If nothing goes wrong, is everything all right? Interpreting zero numerators. J Am Med Assoc 1983;249:1743–5. https://doi.org/10.1001/jama.249.13.1743.Search in Google Scholar

16. Loridan, P. Necessary conditions for ϵ-optimality. Math Progr Stud 1982;19:140–52.10.1007/BFb0120986Search in Google Scholar

Received: 2022-05-26

Accepted: 2023-07-12

Published Online: 2023-09-04

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Articles in the same Issue

https://doi.org/10.1515/ijb-2022-0061

Keywords for this article

one-sided binomial confidence interval; exponential approximation; logarithmic approximation; rule of three; optimization