Hello Wi Aziza. A Chi2 test of association may be suitable, but remember that it is an approximate test that depends on expected frequencies (given a true H0) not being too small. It would be easier for readers to advise you if you showed us the 2x2 table. You don't need to upload an image--just report the 4 cell counts. Thanks.

Hi Bruce Weaver

I attached a file with my 2x2 table in it. Much appreciated

Hi Wi Aziza . I do not see an attachment. But note that you could just report the 4 cell counts like this:

a = Ctrl group, Data

b = Ctrl group, No data

c = Trt, Data

d = Trt, No data

Thanks.

Bruce Weaver okay, here's my expected values:

Ctrl group, data= 161

Ctrl group, no data= 39

Trt, data= 80

Trt, no data= 20

Thank you

The observed values in the attachment are 155, 86, 45, 14.

Thanks. If you plug those counts into an online calculator like this one...

http://vassarstats.net/tab2x2.html

you'll see that the minimum expected frequency (given a true H0) is 19.67. A commonly used rule of thumb for 2x2 tables is that the Chi2 test approximation is pretty good so long as all expected frequencies are equal to 5 or more. Some more statistically conservative folks say 10 or more. So even the more conservative folks would agree that you can safely use a Chi2 test for your data.

EDIT (22-Nov-2022)

PS- Here are my notes on the common rules of thumb regarding expected frequencies for Chi2 tests.

• https://sites.google.com/a/lakeheadu.ca/bweaver/Home/statistics/notes/chisqr_assumptions

Below I have the R code for this analysis, if you don't want to use the online calculator. You can run it at https://rdrr.io/snippets/ without installing software.

Note the table with the proportions in each row. That is probably important information to consider as well.

Not to be confusing, I programmed three ways to calculate a chi-square test of independence: a) With Yate's correction, b) Without correction, and c) by Monte Carlo simulation. The result for c) should be similar to the result for Fisher's Exact Test from the online calculator.

Another option is to use the free gui-based software Jamovi. I've attached three screenshots of the input and results from Jamovi.

The plot may be more useful if it were done with the proportions rather than counts.

Matrix = matrix(c(155, 86, 45, 14), nrow=2,

dimnames = list(c("Control", "Intervention"),

c("Improved","NotImproved")))

Matrix

prop.table(Matrix, margin=1)

chisq.test(Matrix)

chisq.test(Matrix, correct=FALSE)

chisq.test(Matrix, simulate.p.value=TRUE, B=10000)