This is what the data looks like in a graph

Hello Daniel,

What I understood is that you don't have any repetition (ie the value in the table come from a only one shot count) is that right ?

In that case, you can not compute basic statistic test (t-test) with you data. So you can not test hypothesis which mean that you can not get statistical difference.

Repetition, mean and variance are the base of statistic tests. Moreover, prior statistic analysis, especially with t-test or ANOVA, you have to check some assumptions (eg verify if data satisfy necessary hypothesis : independence, variance homogeneity and normality).

Please consider this : it is really important to design your study (statistics tests included) prior doing measurements. Otherwise, you may be like wanted to cook a cake without the right ingredients.

What you can do (which is even more interesting with more than 2 groups) is having look at PCA or FA. You may see correspondence (or not) between category.

They are not really difficult to compute.

https://www.theanalysisfactor.com/the-fundamental-difference-between-principal-component-analysis-and-factor-analysis/

Best regards.

Maybe you could run the Cochran-Mantel-Haenszel test, followed by chi-square tests for each Category. In attachment an R script to do these tests.

The CMH test is sign., which means that the Groups (Control vs. Misidentified) interacts with the Categrories:

Cochran-Mantel-Haenszel test data: Data.xtabs Cochran-Mantel-Haenszel M^2 = 658.15, df = 23, p-value < 2.2e-16

For "Cell Rich" Category e.g. , the chi-square test is not sign.:

Response yes no Group misident 34 22 control 23 33 > (x (x

As I understand your problem is that you have two frequency distributions and you want to know if the data is sufficient to show that these two samples are from different populations (that is, you want to reject the null hypothesis that both samples are taken from a the same population).

This is done by a Chi-squared test. You can use an on-line tool like this:

https://www.icalcu.com/stat/chisqtest.html

Your data gives a Chi-squared value of 70.5 on 23 degrees of freedom, resulting in a p-value of 0.00000102 (it's enough to say that p < 0.001).

Interpretation: if both samples were taken from the same population, differences in the frequency distributions of two samples at least as big as the one observed were extremely unexpected. You can therefore reject the null hypothesis. Your data is sufficient to demonstrate that the two samples were not taken from the same population. The difference in the frequency distributions is statistically significant.

Note that this is not a "prove" (as you said) that there are differences. The observed differences are extremely unlikely under the hypothesis that both samples were taken from the same population. Extremely unlikely does not mean logically false or impossible. The conclusion is only that it is reasonable, given your data, to believe that the two samples are from two different populations, rather than from one population but so unlucky that such big differences occured.

Cyril Iaconelli

Yes I don't have any repetition. I admit I have tried to do stats after recording the data which is why I'm probably struggling so much.

Jochen Wilhelm

Yes thats correct I am trying to prove that these are not from the same population. Thank you for your help

Thank you both for your help

@Daniel Smith

I didn't want to be rude and sorry if I looked like it. Actually, I'm happy that Jochen Wilhelm answered well your question.

Sure the two groups looked like independent and test tell us there is good probability that they are, but unfortunately chi-square won't tell more...

Maybe I did not quite understand what you are trying to find out. I thought you wanted to know whether the values observed are different in the misidentified set compared to those in the control set. With the Cochran-Mantel-Haenszel test you can test these differences over all categories, i.e. controling for categories. This test is not sign. here (I tested the wrong hypothresis in my former answer, sorry!), but for some categories there are differences.

E.g. for the category Severe:

Frequency Table Group --------Response yes -- no -- Total misident ---------------------46 ---10 ---- 56 control ------------------------28 ---28 ---- 56

The chi-square is very sign.:

X-squared = 11.511, df = 1, p-value = 0.000692

There are clearly much more 'yes' values observed when 'misidentified'. The row percentages for yes are: 82% for 'misidentified' vs. 50% for control.