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Example 1b. Regular Two-Sided CI with Tailored One-Sided Test

Traditional 95% CI, [LCL(0.025), UCL(0.025)]. Test whether the AUC exceeds the "essential" threshold of 0.80, i.e., H0: WMWprob ≤ 0.80 vs. H1: WMWprob > 0.80.

As long as subject-matter researchers pay homage to p-values, statisticians should at least give them something useful. Here, testing whether AUC in only better than chance, H0: WMWprob ≤ 0.50 vs. H1: WMWprob > 0.50, is not useful. Testing H0: WMWprob ≤ 0.80 vs. H1: WMWprob > 0.80 is somewhat arbitrary, but such thresholds are set routinely in research. e.g., "the vaccine needs to be 70% effective." To obtain a p-value for H0: WMWprob ≤ 0.80, simply add "WMWprob0 = 0.80" to the WMW() call used in Example 1a.

    Ex1b <- WMW(Y=Rating, Group=TrueDiseaseStatus,

                GroupLevels=c("Abnormal", "Normal"),

                Alpha=c(0.025, 0.025), WMWprob0=0.80)

It is critical to know how the one-sided (directional) hypothesis structure is polarized. With WMW(), lower p-values in the table indicate less chance that the AUC (WMWprob) is less than 0.80, and thus greater chance that it exceeds.0.80. Here, p = 0.0096, which a devoted frequentist would say is strong evidence that AUC exceeds 0.80.

Hypotheses Tested
**************************************************
    H0: WMWprob <= 0.80        H0: WMWodds <= 4.00
    H1: WMWprob >  0.80        H1: WMWodds >  4.00
**************************************************

********************************************************************
Parameter  Estimate      0.95 CI*      One-Sided Hypothesis    P**    
....................................................................
 WMWprob     0.893    [0.818, 0.940]   H0: WMWprob <= 0.80  0.00960  
 WMWodds     8.361    [4.493, 15.559]  H0: WMWodds <= 4.00  0.00960  
********************************************************************
*CI error rates (alphaL, alphaU): (0.025, 0.025)
 CI Method: coupling Sen (1967) & Mee (1990)
**Normal(0, 1) test statistic, Z = 2.34
  P-value for H0: WMWprob >= 0.80:  1 - 0.00960 = 0.99040
  Two-sided p-value (H0: WMWprob = 0.80): 0.0192

Note that a footnote provides the two-sided p-value, 2*min(p, 1–p); here, 0.0192. This tests H0: WMWprob = 0.80, but it is illogical to think that any true WMWprob could be exactly 0.800000.... On the other hand, it makes perfect sense to ask whether this AUC is below 0.80 or above 0.80. Thus, equivalently, we can perform the usual "two-sided" test by considering whether p is "significantly" near 0.00 or 1.00. Ever smaller p-values give greater support for inferring that WMWprob exceeds WMWpro0; ever larger p-values give greater support for inferring that WMWprob is less than WMWprob0. To conduct these two conjoined one-sided tests at the overall 0.05 Type I error rate, just consider whether p < 0.025 or p > 0.975. If either is true, then 2*min(p, 1–p) < 0.05. There is never a true need to report a traditional two-sided p-value.

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