Department of Epidemiology and Biostatistics, Memorial Sloan-Kettering Cancer Center, New York, NY 10021, USA.
Stat Med. 2013 May 20;32(11):1865-76. doi: 10.1002/sim.5601. Epub 2012 Sep 13.
One of the most basic biostatistical problems is the comparison of two binary diagnostic tests. Commonly, one test will have greater sensitivity, and the other greater specificity. In this case, the choice of the optimal test generally requires a qualitative judgment as to whether gains in sensitivity are offset by losses in specificity. Here, we propose a simple decision analytic solution in which sensitivity and specificity are weighted by an intuitive parameter, the threshold probability of disease at which a patient will opt for treatment. This gives a net benefit that can be used to determine which of two diagnostic tests will give better clinical results at a given threshold probability and whether either is superior to the strategy of assuming that all or no patients have disease. We derive a simple formula for the relative diagnostic value, which is the difference in sensitivities of two tests divided by the difference in the specificities. We show that multiplying relative diagnostic value by the odds at the prevalence gives the odds of the threshold probability below which the more sensitive test is preferable and above which the more specific test should be chosen. The methodology is easily extended to incorporate combinations of tests and the risk or side effects of a test.
最基本的生物统计学问题之一是比较两种二进制诊断测试。通常，一种测试的灵敏度更高，另一种测试的特异性更高。在这种情况下，一般需要根据疾病的阈值概率（即患者选择治疗的概率）进行定性判断，以权衡灵敏度提高是否会导致特异性降低。在这里，我们提出了一种简单的决策分析解决方案，其中灵敏度和特异性由一个直观的参数加权，即患者选择治疗的疾病阈值概率。这给出了一个净收益，可以用来确定在给定的阈值概率下，两种诊断测试中的哪一种会产生更好的临床结果，以及是否有任何一种测试优于假设所有或没有患者患有疾病的策略。我们推导出了一个简单的相对诊断价值公式，它是两种测试的灵敏度差异除以特异性差异。我们表明，将相对诊断价值乘以患病率的比值，可以得到阈值概率以下更敏感的测试更优的概率，以及阈值概率以上更特异的测试应被选择的概率。该方法很容易扩展到包含测试组合和测试的风险或副作用。
