BBR Session 11: Nonparametric Tests Using the Proportional Odds Model

[f2harrell]

@jlwoodard

would I just start with testing H0 with the LRT p value, then proceed to the contrasts if significant? I also wasn’t sure how to compare A and B directly to get the contrast, confidence interval, z and p value.

Start with developing a statistical analysis plan that is not informed by any examination of associations between X and Y. Pre-specify individual and chunk tests (e.g., LR tests) and contrasts, and don’t make contrasts of interest conditional on something else being arbitrarily “significant”. Comparisons with reference groups, although they are the less accurate Wald tests when using this function, are provided automatically with summary(). But take control using contrast(f, list(group='B'), list(group='A')) to get the B-A contrast on the log odds scale. Antilog to get the B:A odds ratio. Best is to get LR tests and profile likelihood confidence intervals by specifying LR as the type of test to be done by contrast().

@jlwoodard

In the output for the PO ordinal regression model, what do the beta’s reflect? Are they directly interpretable?

You get log odds ratios (differences in log odds) against a reference group when printing the fit object. Whether anti-logging the \beta's or the contrast()'s these are odds ratios.

@jlwoodard

I understand that the contrast shown for the comparison between B and C is the difference between the betas, but how would we interpret the odds ratio for that comparison? Would it be something like “The odds that a score from group C is larger than a score from group B are 3.46?”

No that’s not what ORs mean. Instead a C:B OR is the ratio of odds that Y \geq y if in C to the odds that Y \geq y if in group B, for any y other than the lowest y. This is the identical interpretation to a binary logistic model except you replace y with 1.

[f2harrell]

@jlwoodard

If I selected a particular intercept, such as y>=86 (beta=0.1122) and I converted the logit beta to a probability, could I say then that the probability is 52.8% that a randomly selected score in the no/mild reference group has a score greater than or equal to 86 and a 94.6% probability that a randomly selected score from the moderate/severe group would have a score greater than or equal to 86 (I just added 2.7586 to the 0.1122 beta and converted that logit to a probability)?

Let’s call the intercepts \alpha's to avoid confusion with the slopes (\beta's). Drop the phrase “randomly selected” as that has nothing to do with the simple non-pairwise probabilities being computed here. To get P(Y \geq 86 | B) take the expit of the sum of the intercept for 86 and the slope for B. This is the estimated probability of a response equally or exceeding 86 for an observation from group B.