0
$\begingroup$

Considering the following hypothetical dataset of lesion occurrence in kidney for ~200 patients with disease X:

location of kidney left right
location 1 112 101
location 2 54 39
location 3 37 33
location 4 112 95
location 5 24 18

** patients can have the lesions in multiple locations within the kidney

** Location is where the surgeon found the lesion within the kidney (like renal cortex,renal medulla...).

What test should I use to show that lesion occurrence has a right/left preference with disease X? At this stage, we don't consider the difference among the locations within the kidney. I was told to use the multinomial logistic regression analysis, but I can't figure out why and how to interpret the results.

$\endgroup$
4
  • $\begingroup$ Are these different anatomical locations within a kidney? How distinct are the locations? Also, please edit the question to explain how the location matters here, as if all you want is "to show that disease X [is] more likely occur on the left kidney than right kidney" the location within the kidney doesn't matter: all you need to know is whether left or right kidney had the disease. Or do you want to know whether the (within-kidney?) locations have a right/left preference for having disease? Please provide that information by editing the question; comments are easy to overlook. $\endgroup$
    – EdM
    Jan 10, 2023 at 15:27
  • $\begingroup$ Thanks for comment. Edits are made. I put some context to my question. Basically. the table is just a surgeon observation of lesion found in different parts of kidney from patients with disease X. I want to know whether the locations have a right/left preference for having disease. $\endgroup$
    – PacMan
    Jan 10, 2023 at 21:12
  • $\begingroup$ Do you have data in the following (or similar) structure? SubjectID, Location, Laterality, Presence of disease? Would be a candidate for GLMM since subjects with advanced disease are simply more likely to have disease everywhere. $\endgroup$
    – AdamO
    Jan 10, 2023 at 21:33
  • $\begingroup$ Yes, I do. Can you elaborate more on the use of GLMM? $\endgroup$
    – PacMan
    Jan 11, 2023 at 2:12

1 Answer 1

0
$\begingroup$

Multinomial regression only works when categories are mutually exclusive. For example, that would be appropriate if you cared about disease locations within the kidney and only one location per kidney was possible. That's not your situation.

If all you care about is Left versus Right Laterality and all patients only have disease in 1 kidney, then all you need to do is a simple binomial proportion test (e.g., binomial.test() in R). You just note the Left versus Right location for each patient, regardless of how many lesions there were in the kidney or their corresponding locations. The test is whether the fraction of patients with Left-side lesions is significantly different from 1/2.

A generalized linear mixed model (GLMM) would allow more detailed analysis of Laterality and Location. The form of generalized linear model could be a binomial regression for the probability of lesionPresence as a function of both Laterality and Location, with a random effect (leading to a mixed model) to account for differences among individuals in overall probabilities of having lesions. The simplest data setup would be 10 rows for each individual, each row annotated with the Laterality, one of the 5 Location values, SubjectID, and diseasePresence coded as 1/0 for presence/absence at that combination of values.

The default in R for binomial modeling is logistic regression. The structure of the model in R with the lme4 package could be:

lesionModel <- glmer(diseasePresence ~ Laterality * Location + (1| SubjectID),
                       data = yourData, family = binomial)

The "*" interaction operator allows for differences in Location depending on Laterality. The (1|SubjectID) term specifies a random intercept to allow for overall differences in diseasePresence among patients.

Be careful in interpreting the individual coefficients returned by the model. In logistic regression the Intercept will be the estimated log-odds at the reference values of both Laterality and Location. The individual coefficient for Laterality will be for the difference from that in log-odds between the non-reference and reference Laterality at the reference Location. The 4 individual coefficients for Location will be for the differences in log-odds from the reference Location at the reference Laterality, and the 4 interaction coefficients will be the extra differences at those values of Location at the non-reference Laterality.

It will probably be simplest and safest to use the default "Type II" analysis of variance provides by the Anova() function in the R car package to summarize the results. That will provide estimates of the significance of overall differences associated with Laterality and Location, and (via the estimate for the interaction terms) of whether Laterality is significantly different depending on Location. You then can use post-modeling tools like those in the emmeans package to evaluate predictions for particular combinations of Laterality and Location, transformed if you wish back to the probability scale that can be easier to think about than log-odds.

$\endgroup$
2
  • $\begingroup$ Thank you for the well-explained answer. binomial proportion test is the first we came up with, but patients can have disease in both kidneys. Is that any way to modify the test? I am asking because my boss is quite keen on using that. $\endgroup$
    – PacMan
    Jan 16, 2023 at 20:18
  • $\begingroup$ @PacMan simplest is to remove patients with disease in both kidneys from the calculations, as they have no Laterality preference. Then just do the binomial test on the patients with unilateral disease; those patients include all the information about potential Lateralitypreference, if you don't want to further examine Location within the kidney. That's similar to what McNemar's test does with contingency tables on paired data. The binomial mixed model would be best for extending to Location information. $\endgroup$
    – EdM
    Jan 16, 2023 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.