哪种是测试非数字数据之间显着差异的正确方法？ 哪个是正确的事后？

Question

I'm working with non numeric data that looks something like this:

Origin	ESBL
Hospital	ESBL
Hospital	Non-ESBL
Hospital	ESBL
City	ESBL
Hospital	Non-ESBL
City	ESBL
Country	ESBL
Hospital	ESBL

And I want to compare if there is a statistical association between the origin and the variable ESBL.

So far I have tried generating a contingency table in R using:

cont_tab<-table(data$Origin, data$ESBL)

and the running a chi squared test for independence:

chi_test<-chisq.test(cont_tab)

After this, I get that there is indeed independency:

X-squared = 17.306, df = 2, p-value = 0.0001746

But now I want to know which are the combinations that are responsible for this values (ESBL-Hospital, Non-ESBL-Hospital, ESBL-City and so on).

I have tried running multiple Fisher tests:

Library(RVAideMemoire)
multifish<-fisher.multcomp(cont_tab)

But I don't really get what I want:

            ESBL Non-ESBL
  Hospital   46      122
  City       27       21
  Country    56       69

Am I doing anything wrong? Is there a better approach for this?

Thanks!!!

Answer 1

I think the "final result" you are showing is actually cont_tab . When I run your code, cont_tab looks like the result you are showing as being the output from fisher.multicomp :

cont_tab <- table(data$Origin, data$ESBL)

cont_tab
#>           
#>            ESBL Non-ESBL
#>   Hospital   46      122
#>   City       27       21
#>   Country    56       69

Whereas, if I run fisher.multcomp on cont_tab , I get:

library(RVAideMemoire)

fisher.multcomp(cont_tab)
#> 
#>         Pairwise comparisons using Fisher's exact test for count data
#> 
#> data:  cont_tab
#> 
#>         Hospital  City
#> City    0.001313     -
#> Country 0.004249 0.234
#> 
#> P value adjustment method: fdr

We can see in it (as expected) that Hospital is significantly different from both City and Country , but there is no significant difference between City and Country .

^{Created on 2022-12-13 with reprex v2.0.2}

---

Data inferred from question

data <- data.frame(
  ESBL = factor(c(rep(c("ESBL", "Non-ESBL"), times = c(46, 122)),
                  rep(c("ESBL", "Non-ESBL"), times = c(27, 21)),
                  rep(c("ESBL", 'Non-ESBL'), times = c(56, 69)))),
  Origin = factor(rep(c('Hospital', 'City', 'Country'), 
                      times =  c(168, 48, 125)), 
                  c('Hospital', 'City', 'Country')))

Answer 2

Take a look at https://pbil.univ-lyon1.fr/ADE-4/ade4-html/discrimin.coa.html for a completely different, academically pristine approach implemented in the R package ade4. Your variable ESBL would play the role of the class variable in a Discriminant Correspondence Analysis (DCA).