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For the second part of my paper I collected data on whether patients' condition was acute/chronic/unknown and also collected their opinion to a statement Y on a likert scale

I want to compare whether there is a link between acute/chronic/unknown condition and opinion of statement Y , so I decided to do a kruskal wallis test. Here is the data:

enter image description here

For the test I convert the likert scale to a 1-5 scale and did the calulcation. I have two questions I could do with some help in:

(1) Have I communicated the method of analysis appropriately in my "analysis" section and if I have presented results appropriately?

Analysis: "Likert-scale data was treated as ordinal (1-5) and subsequently analysed using the Kruskal-Wallis H test and post-hoc Dunn’s test with p-values adjusted according to the Benjamini-Hochberg FDR method."

Results: "There was no significant difference of opinion of statement Y between self-reported acute (mean Likert score: 2.84 ± 1.180 ), chronic (mean Likert score: 3.15 ± 1.058) unknown disease (mean Likert score: 3.06 ± 1.079) status (H (2, n = 235) = 3.49, p = .175)."

Bear in mind I used an online website to do these and it offered the Benjamini-Hochberg FDR method for correction of p value in post hoc analysis (although that wasn't needed here). do I need to mention the n = 235 in presenting the results or is that not necessary?

Any other comments will be appreciated

Thanks

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I am puzzled about two parts of your question. (a) The Kruskal-Wallis test does not show significant results, so it is completely inappropriate to do any kind of ad hoc tests (looking for non-existent significant differences among groups). Why then are you even mentioning ad hoc tests.

xa = rep(1:5, c(10,12,25,8,7))
xc = rep(1:5, c(10,14,58,22,15))
xu = rep(1:5, c(7,5,24,14,4))
x = c(xa,xc,xu)
g = rep(1:3, c(length(xa),length(xc),length(xu)))
kruskal.test(x ~ g)

        Kruskal-Wallis rank sum test

data:  x by g
Kruskal-Wallis chi-squared = 3.4893, df = 2, 
  p-value = 0.1747

Boxplots show essentially equal medians:

boxplot(x ~ g, col="skyblue2", pch=19)

enter image description here

(b) The Kruskal-Wallis test has nothing to do with group means. So, why are you showing overlapping confidence intervals for population means?

If you going to mention that there were 235 subjects in the three groups combined, the natural way to do that would be to include a 'Total' row and column in your data table, showing 235 as the Grand Total.

BTW, a chi-squared test on a table of your Likert counts does not reject the null hypothesis that responses are homogeneous among the groups.

TAB = rbind(c(10,12,25, 8, 7),
            c(10,14,58,22,15),
            c( 7, 5,24,14, 4))
chisq.test(TAB)

        Pearson's Chi-squared test

data:  TAB
X-squared = 9.0368, df = 8, p-value = 0.3392

If you had the same proportions in a study with twice as many subjects (all counts approximately doubled), then the chi-squared test might have had the power to reject the hypothesis of homogeneity. But if you actually doubled the overall sample size, the proportions might be quite different---perhaps still failing to reject the null hypothesis.

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  • $\begingroup$ Sorry for the confusion the reason I included the posthoc analysis in the "Analysis" is because I have it in my paper and it is relevant for other calculations which require it since the KW test rejects the null hypothesis. Forgot to take it out here. And with regard to the "Total" row/column, forgot to mention but I am not including the table in the paper, this was just my data on excel. So I was wondering whether it was reasonable to include a (n=235) as I have done above, when describing the results. Many thanks $\endgroup$ Jun 19 '20 at 17:16
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    $\begingroup$ I think I will take out the CI as they are quite uselss, I included them almost as a habit with means. I will need to reorganise the way I am describing my results but I want to include the mean/median so that this can be used to compare the direction of difference in distribution if any arise. $\endgroup$ Jun 19 '20 at 17:18

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