Automatic passenger counting (APC) in public transport has been introduced in the 1970s and has been rapidly emerging in recent years. APC systems, like all other measurement devices, are susceptible to error, which is treated as random noise and is required to not exceed certain bounds. The demand for very low errors is especially fueld by applications like revenue sharing, which is in the billions, annually. As a result, both the requirements as well as the costs heavily increased. In this work, we address the latter problem and present a solution to increase the efficiency of initial or recurrent (e.g. yearly or more frequent) APC validation. Our new approach, the partitioned equivalence test, is an extension to this widely used statistic hypothesis test and guarantees the same bounded, low user risk while reducing effort. This can be used to either cut costs or to extend validation without cost increase. While the pre-classification itsself can be arbitrary, we evaluated multiple use cases: entirely manual and algorithmic resp. artificial intelligence assisted workflows. For former, by restructuring the evaluation of manual counts, our new statistical test can be used as a drop-in replacement for existing test procedures. The largest savings, however, result from latter algorithmic use cases: Due to the user risk being as bounded as in the original equivalence test, no additional requirements are introduced. Algorithms are allowed to be failable and thus, our test does not require the availability of general artificial intelligence. All in all, automatic passenger counting as well as the equivalence test itself can both benefit from our new extension.
翻译:在公共运输中自动计票(APC)是1970年代引入的,近年来也迅速出现。APC系统与所有其他测量装置一样,容易出错,被作为随机噪音处理,要求不超过一定限度。对非常低误差的需求特别受到收入分享等应用的推动,每年有数十亿个,因此,要求和费用都大大增加。在这项工作中,我们处理后一个问题,并提出提高APC初始或经常性(例如每年或更频繁)验证效率的解决方案。我们的新办法,即分解等值测试,是这一广泛使用的统计假设测试的延伸,保证相同的受约束的低用户风险,同时减少努力。这可以用来降低成本或扩大验证范围,而不会增加成本。虽然分类前本身可能具有任意性,但我们评估了多种使用案例:完全手工和算法的重整人工情报协助工作流程。前,通过调整手动计数,我们新的统计测试可以用来作为现有测试程序的递减替换替代。因此,最大的预估值本身不能作为原定等值测试的自动测试结果。因此,最大的自动测试需要从原定值本身的自动测试进行额外的计算。