School of Psychological Science, La Trobe University, Australia.
Risk Anal. 2010 Mar;30(3):512-23. doi: 10.1111/j.1539-6924.2009.01337.x. Epub 2009 Dec 17.
Elicitation of expert opinion is important for risk analysis when only limited data are available. Expert opinion is often elicited in the form of subjective confidence intervals; however, these are prone to substantial overconfidence. We investigated the influence of elicitation question format, in particular the number of steps in the elicitation procedure. In a 3-point elicitation procedure, an expert is asked for a lower limit, upper limit, and best guess, the two limits creating an interval of some assigned confidence level (e.g., 80%). In our 4-step interval elicitation procedure, experts were also asked for a realistic lower limit, upper limit, and best guess, but no confidence level was assigned; the fourth step was to rate their anticipated confidence in the interval produced. In our three studies, experts made interval predictions of rates of infectious diseases (Study 1, n = 21 and Study 2, n = 24: epidemiologists and public health experts), or marine invertebrate populations (Study 3, n = 34: ecologists and biologists). We combined the results from our studies using meta-analysis, which found average overconfidence of 11.9%, 95% CI [3.5, 20.3] (a hit rate of 68.1% for 80% intervals)-a substantial decrease in overconfidence compared with previous studies. Studies 2 and 3 suggest that the 4-step procedure is more likely to reduce overconfidence than the 3-point procedure (Cohen's d = 0.61, [0.04, 1.18]).
当可用数据有限时,征求专家意见对于风险分析很重要。专家意见通常以主观置信区间的形式征求;然而,这些往往存在很大的过度自信。我们研究了征求意见的问题格式的影响,特别是征求意见过程中的步骤数量。在 3 点征求意见过程中,要求专家提供下限、上限和最佳猜测,这两个限制形成了一定置信水平(例如 80%)的区间。在我们的 4 步区间征求意见过程中,专家还被要求提供现实的下限、上限和最佳猜测,但不指定置信水平;第四步是评估他们对产生的区间的预期置信度。在我们的三项研究中,专家对传染病的发病率进行了区间预测(研究 1,n = 21 和研究 2,n = 24:流行病学家和公共卫生专家),或海洋无脊椎动物种群(研究 3,n = 34:生态学家和生物学家)。我们使用荟萃分析综合了我们的研究结果,发现平均过度自信为 11.9%,95%CI [3.5, 20.3](80%区间的命中率为 68.1%)-与之前的研究相比,过度自信有了实质性的降低。研究 2 和 3 表明,4 步程序比 3 点程序更有可能降低过度自信(Cohen's d = 0.61,[0.04, 1.18])。