Kruskal-Wallis and Dunn test will work.

What assumption of the K-W test are you concerned about ?

Actually, anova will mathematically work, though the obvious difference in variance among treatments will violate the assumptions.

Sal Mangiafico Thank you so much for your help! I may be incorrect in remembering this but I was advised as an undergrad that although data don't need to be normal for KW, the datasets being compared should have approximately similar distributions.

Hello Jemima Becker. I have no idea what your raw data look like. Your graph suggests there are 3 independent groups of observations. You also said you "quantified the number of cells expressing different markers". That sounds to me as if a given cell expresses or does not express--Yes or No. I speculate, therefore, that in each group, you have some number of cells, and some proportion of those cells express a given marker. But then you said n = 4. If that means n = 4 cells per group, then the proportions expressing would be limited to 0, .25, .5, .75, and 1. But your graph shows that cannot be so. Hence my confusion.

Can you clarify what an (independent) observation is, what n = 4 means, and how the percentages you show were computed? Apologies if this is completely clear to everyone else. ;-)

Thanks, Bruce Weaver . I had the same question.

Bruce Weaver ah my apologies - I don't think I explained this clearly! n=4 corresponds to the four individual animals in each group. I counted the total number of GFP+ cells (which was approx 200 in each animal) and then the number of those that were GFP+Ki67+, with the value plotted on the graph being %Ki67+ ie ((GFP+Ki67+)/(Total GFP+))*100, with one value plotted per animal.

I have four animals expressing each genetic construct (Ev, 58i, and 59i)

Thank you for clarifying, Jemima Becker. So the raw data for each animal is a proportion. Therefore, you should use a model intended for analyzing proportions--e.g., beta regression or some flavour of logistic regression. If you say what software you use, someone may be able to give more specific advice.

Bruce Weaver I usually use GraphPad Prism or R! Thank you so much!

Aren’t you losing information by coding the outcome as binary? I wonder if a permutation or bootstrap test might be superior?

Bruce Weaver Just saw the later post - if the underlying data are proportions then that’s definitely the way to go.

When the standard deviation is zero, this means zero variability and that all values of the variable are constant. No inference could be made on such data.

When someone contradicts your answer, does not mean "Gross nonesence". We are open to all path of thinking. I believe my answer is correct, there is something wrong on the data starting from coding. Thanks for the reply, I truly respect your point of view, however I assure you something is wrong with the data.

@ Jochen Wilhelm

When someone contradicts your answer, does not mean "Gross nonesence". We are open to all path of thinking. I believe my answer is correct, there is something wrong on the data starting from coding. Thanks for the reply, I truly respect y​o​u​r​

… Read mor

@Jochen Wilhelm

No comments. Wishing you the best in your research work

Jemima Becker . I did just want to follow up on your question about Kruskal-Wallis.

It's not true that the Kruskal-Wallis test assumes that groups have distributions of equal spread or shape.

The assumptions of the test are sometimes presented this way, in order to force the test to be a test of simple location shift. That is, if you start with groups with distributions of similar shape and spread, then a significant result corresponds to a difference in medians. (And actually a difference in means, and any other measure of location).

Good text books will either not bother with this assumption, or present two different sets of assumptions, one for the location shift test, and one for the test of stochastic equality.

Sal Mangiafico thank you so much for your advice and clear explanation - KW seems the way to go!

One sample t test, since constant is also true value.

Just because your SD is zero does not mean something is wrong with your data. Although it doesn't happen that often, we all need to realize this outcome is POSSIBLE (albeit not all that often). Pragmatically, look at your data range. If that is the same number (both high and low), it is not possible for the SD to be anything but zero (all the data has the same data point/result). The comment by Matej Jovan is another approach, as most statistical software packages actually DEFAULT the SD to zero for testing (test value). Once (and if) you find statistical significance (p = .05 or below for most social research or p = .01 for medical research???) with your data, the KW test has been backed up. When done, write up your methodology to include everything you did to validate your SD being zero. Further, Jochen Wilhelm

Afterthought: when we get into the weeds of inferential statistics, we often forget to refer back to our descriptive statistics to help us explain results. As I recall, if the SD is zero, the mean, mode and median should all be the same constant value. That may be an even better write-up for your methodology, findings and conclusions.

This is a fascinating and illuminating thread, and I'm glad I read through it.

With my limited and rudimentary stats understanding, though, I felt it might be useful to chip in some slightly broader, less mathematically bounded, but more biologically-directed advice as to how to handle "yes/no" results like this.

Biology is messy, and not necessarily always statistically tractable. Biology never claimed to be, and nor does statistical signifcance necessarily equate to biological relevance. Sometimes a change means a lot for your biological model but has no statistical value at all. Sometimes a change is incredibly statistically significant but of no biological consequence at all.

This is fine: as the investigator, it's your job to determine which applies in which situation, and there is little value getting hung up on specifics, when the conclusion is...honestly pretty obvious.

The situation here is "no cells express ki67 in the control group. None. Not a single one, despite counting hundreds of the bastards in four separate biological replicates".

Now, _yes_ this could be that expression is just really, really low, and you just happened to miss those one or two rare positive cells that were totally there. This is, in fact, probably the closest approximation to biological reality (and actually detecting one or two cells would be ideal, because it would ironically make the stats much easier), but the fact remains: you detected zero cells.

This is fine, and "we detected zero positive cells" is absolutely a legitimate statement to make in a manuscript. It potentially makes stats more difficult, because zero values are always more challenging for statistical comparisons, but it doesn't change the fact that you detected zero positive cells.

So...my advice would be: don't do the stats for "zero" vs "some", because that isn't really helpful nor informative. Even a stats-naive reader can interpret that "zero" vs "some" is indicative of an increase. Just do the stats for the actual values that actually have expression. It's obviously not significant between 58i and 59i just by _eye_, and likely no statistical test will change this, but you do the analysis just so you can say it isn't significant with the backing of MATHS. This isn't, and shouldn't be, necessary when the conclusions are as obvious as they are here, but you do this anyway because there will always be a reviewer #2 who picks up on it.

Your scenario here is thus "in the absence of knockdown, zero Ki67 positive cells were detected (200-400 cells per replicate, N=4), indicating a very low proliferation rate. Knockdown, however, resulted in modest but detectable Ki67 expression (~0.5-1.5% of total cell count), though this did not vary significantly between treatment groups."

That's...that's basically a conclusion which is supported by the data, does not require esoteric statistical tests to handle zero values, and is also demonstrably obvious just by looking at the data.

Great reply, Jochen Wilhelm

I'm curious, though: "completely negative if zero or just a single positive cell is observed" seems to be an actively unhelpful stipulation here, given that a single positive cell would turn that "zero" into something, which would make the comparison here less statistically challenging (ostensibly).

If we take your hypothetical and change it to "completely negative if zero positive cells are observed", how does that change the interpretation?

I.e. "in one group we NEVER observed a single positive cell (1000 cells per group, N=4), while in the other two groups (same sample size) we typically observed 5-20 positive cells, i.e. 0.5-2% of total cell counts"

I might naively assume "never seeing a phenomenon" vs "seeing it routinely, albeit at low but comparable frequencies" is reasonably clear cut, if considering a population of some 1000 cells per assessment, but perhaps that's a...qualified interpretation? How many times would I need to repeat this to be convinced to statistically justified levels of confidence, for example?

To add in some additional complexity, I'd point out that ki67 is pretty transiently expressed, so even in a population of actively dividing cells, you'll be looking at a 'snapshot', where some will be negative purely because they're nowhere near G2/M: how would this adjust the interpretation? How many random fields of cells (or individual cells) might be needed to conclude "these are not an actively dividing population" from "these are a minimally mitotically active population"?

I want to stress: these are genuine questions, and I always appreciate your answers. Statistics is not something I am in any way good at, but I'm always eager to improve.

John Hildyard on the topic of Ki67 transience - I am planning to repeat this using BrdU labelling (administered over a 7 day period) because of the low number of ki67+ cells - so i can (hopefully) ID cells that have gone through S-phase during this period, rather than the Ki67 snapshot that I've quantified here.

I suppose the unclear results from the Ki67 tell me a few things. I find a larger number of GFP+ labelled cells (they are labelled in the V-SVZ) ending up in the olfactory bulb (OB) upon treatment with the 58i and 59i - and this is up to a four-fold increase. This could be due to changes in migration rate, proliferation, and cell death.

I have ruled out cell death and I doubt that the modest increase (0 cells to 2-5 GFP+Ki67+ cells per animal) can account for the scale of increase in GFP+ cells in the OB. I suspect some combination of increased rate of migration + small increase in rate of proliferation gives this effect.

Jemima Becker Have you considered other (corroborating) assays, like qPCR for gene expression? I've found that I can detect bulk changes in Ki67 readily via grind-and-find qPCR analysis, whereas detecting it via ISH or IHC is more stochastic (very much "there/not there").

You could also measure expression of GFP, both in brain subregions and globally: if GFP expression is comparable after labelling, but then increases globally over time in 58i/59i treated groups only, this strongly implies proliferation regardless of where the cells are going.

But yeah: BrdU is a smart idea. Good luck!