# Which test to use when comparing multiple sets of proportions (with unknown sample size)?

I have two conditions, about 50 samples for each condition, and for each sample - a set of 9 proportions. The proportions are estimates of the proportion of 9 cell types of this patient.

Clarification : The estimates are of the proportion of the cell types in the tissue of the specific sample. It was arrived at using deconvolution software. The software does not estimate the total number of cells, hence I have the estimate of the proportion but not the sample size (not the denominator\numerator of the proportion).

I want to test statistically the proportions of which cell type change between the two conditions.

Clarification : There are 9 estimates of proportions for each sample, which correspond to 9 distinct categories (cell types). Every sample has all 9 categories represented. I want to test across which categories (cell-types), the change is significant between the two conditions. About 50 samples in each condition).

Cell A and cells C proportions probably change between the two conditions (G1&G2) while cell B proportions definitely do not.

What statistical test would be suitable?

I thought of averaging all proportions for each of the groups and applying something like a t-test, but the proportions are bounded between zero and one, hence I don't think it would be suitable. Also, (not sure how important this is) I have no theoretical reasons to suppose normal distribution.

The second problem I see is that the proportions of the different cells are not independent. If cell A proportions are extremely high in a specific patient, the proportions of other cells are more likely to be low. Hence, I am more likely to get false positives compared to a situation where the proportions of each cell type would be independent from the others.

• First, a proportion has a numerator and a denominator. For a statistical comparison one needs to know both. For example, the difference between 3 and 7 successes in 10 in not significant, while the difference between 31 and 68 in 100 is very highly significant. // Second, please list exactly the comparisons do you want to make? Jun 16, 2021 at 15:41
• @BruceET Thank you. I have added clarifications to the OP.
– Sam
Jun 19, 2021 at 12:51
• Would more clarifications help?
– Sam
Jun 23, 2021 at 8:57
• Even an imperfect answer would be of help.
– Sam
Jun 24, 2021 at 6:45

In this context the estimates of cell proportions are based on deconvolution of things like RNA sequencing (RNA-seq) gene-expression data based on signatures specific to cell types. Although concern about knowing both the numerator and denominator of a proportion is valid in general, it might not be such a problem in this particular context if the total number of RNA-seq reads (often in the millions) is comparable among samples. This answer proceeds under that assumption.

If the cell proportions all add up to 1 for each sample, you are absolutely correct that "the proportions of the different cells are not independent." Then you have what's called compositional data with only 8 independent cell proportions for 9 cell types. Depending on how you choose to proceed, you should use your understanding of the subject matter to choose one to omit from the analysis or to use as a reference category. You can still show the results for the omitted cell type but you can't do statistical inference on it, as its value is completely determined by the values of the other 8 cell types. In terms of correlations among cell types leading to a further lack of "independence," that can be dealt with as discussed below.

With about 50 samples for each condition, you are probably in a regime where you can take advantage of the central limit theorem. Even though the individual observations aren't normally distributed, the sampling distribution of their mean values will tend toward a normal distribution so that standard statistical tests can be used. In that case you can treat this as a simple multivariate (multiple-outcome) analysis of variance (MANOVA). You have a single binary predictor (G1/G2) and multiple continuous outcomes (each cell-type proportion, except for the cell type that you have chosen to omit if they add up to 1). The point estimates of G1/G2 differences will be the same as for separate analyses, but MANOVA takes correlations among outcomes into account to set confidence limits etc. The manova() command in R is a starting point (best would be if you have approximately the same number of cases in each of G1 and G2). You provide a matrix of outcomes to the function, instead of a vector as you would in a univariate ANOVA or regression.

It seems like it should also be possible to handle this similarly to a multinomial logistic regression, in which you have multiple cells and each individual cell can belong to one and only one of multiple classes. In that situation one class is chosen as a reference and the log-odds of being in each of the other classes with respect to that reference class is modeled. If you had real data on individual cells that would work well, but that's not the case here. There might be a way to set up artificial data on "virtual cells," in which the number of "cells" in each class for each individual is chosen to represent the underlying uncertainty in class membership arising from the deconvolution. That is, you choose a total number of "cells" for each individual based on the corresponding uncertainty in class proportions and the corresponding multinomial variance (higher uncertainty means fewer "cells"; that's related to the point that BruceET was getting at in a comment, your confidence in a proportion depends on the number of total observations), and then distribute the "cells" into classes according to the estimated proportions. Someone might have worked that out type of approach in detail for cell-type RAN-seq deconvolution data (check in Bioconductor), but I'm not aware of it.

My guess is that with 50 cases per group, the MANOVA (with one cell type omitted if necessary) will work OK and deal with the issue of correlations among cell types.

• Even if I choose only 8 out of 9 cell types, I don't think that the rest will be independent. Suppose there are three cell types, each with the average proportion of 30% (the rest have small proportions). Then if any cell type of these 3 goes strongly up (say to 80%), both of the other two will have to go down. There is no one single cell type I can remove that would make the others independent.
– Sam
Jun 27, 2021 at 14:22
• When you say "The point estimates of G1/G2 differences will be the same as for separate analyses", by separate analysis you mean t-tests?
– Sam
Jun 27, 2021 at 14:24
• @Sam there are two types of non-independence here. First is the mathematically constrained non-independence if all cell-type probabilities equal 1. That must be accounted for by removing one cell type (MANOVA) or using it as the reference (multinomial model). The second non-independence, beyond that, is the set of correlations in terms of which cell types go up when others go down. Those correlations must be taken into account when estimating the significance of a difference in cell-type proportions between G1 and G2.
– EdM
Jun 27, 2021 at 14:53
• @Sam a "point estimate" is a single estimated value, for example the estimated difference in cell type K proportion between G1 and G2. In MANOVA, the estimated proportion differences between G1 and G2 for each cell type will be the same as if you did separate calculations for each. Statistical tests evaluate whether a point estimate is "significantly" different from some specified value, often 0. That requires an estimate of the uncertainty in the estimate. That's where a set of t-tests and MANOVA differ. MANOVA takes the correlations among cell types into account; separate t-tests don't.
– EdM
Jun 27, 2021 at 14:59
• @Sam MANOVA is a standard procedure, though I’m not aware of uses in this particular application. I suspect that there will be implementations of some type on Bioconductor that are oriented toward this type of study.
– EdM
Jun 29, 2021 at 11:42