Abstract

We construct a point and interval estimation using a Bayesian approach for the difference of two population proportion parameters based on two independent samples of binomial data subject to one type of misclassification. Specifically, we derive an easy-to-implement closed-form algorithm for drawing from the posterior distributions. For illustration, we applied our algorithm to a real data example. Finally, we conduct simulation studies to demonstrate the efficiency of our algorithm for Bayesian inference.

Highlights

  • Misclassifications can occur in binomial data due to human errors or imprecise diagnostic procedures

  • In the traffic accident case, over-reporting occurs when a driver has lied to a police officer about not wearing a seatbelt, while the hospital record examination showed otherwise, for an example see the U.S National Highway Traffic Safety Administration, NHTSA [3]

  • Since there is no personally identifiable information (PII) of any human subject involved, and only summary statistics are displayed, an International Review Board (IRB) approval or an exemption to collect such data and publish results based on Table 3 is not required

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Summary

Introduction

Misclassifications can occur in binomial data due to human errors or imprecise diagnostic procedures. For one-sample data with both types of misclassification errors, Raats and Moors [7] derived both an exact confidence interval and a Bayesian credible interval for the true proportion parameter, and Lee [9] reported a Bayesian interval estimation for the true binomial proportion parameter. For two-sample data with two types of misclassification errors, Prescott and Garthwaite [10] proposed a Bayesian credible interval, and Morrissey and Spiegelman [11] derived likelihood-based confidence intervals for the odds ratio Their methods are computationally burdensome, not guaranteed to converge to the true parameter value, and hard to reproduce by other practitioners. Misclassification error occurs when Fij differs from Tij. In the validation sub-study, for i = 1, 2, j = 0, 1, and k = 0, 1, we use nijk to denote the number of items in Sample i classified as j and k by the inerrant device and the errant device, respectively. The statistical hypothesis testing in (2) can be rejected if this CI does not contain the number zero

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