The test is appropriate if the assumption of independent observation is met and the outcome is ordinal, which should be stated in the report. If SPSS is used then check the test statistics table result, to find the p-value associated with Asymptotic Significance row which indicate the p-value to be interpreted. If less than 0.05 there is statistically significant difference between the outcome of the 3 independent groups (nurses, physicians and pharmacists). On the other hand If the p-value is greater than 0.05, then the finding is not statistically significant. If the p-value is less than 0.05, then a Mann-Whitney U test or a Dunn for pairwise comparisons in a post hoc manner is indicated.

The following is an example for reporting Kruskal-Wallis test:

Kruskal-Wallis Test was conducted to examine the differences on renal dysfunction according to the types of medication taken. No significant differences (Chi square = 3.71, p = .39, df = 6) were found among the five categories of participants (none, ACE inhibitors, ARB, ACE inhibitor and ARB, NSAID, and ACE inhibitor or ARB and NSAID).

Dear Olaide Bamidele Edet,,

thank you for your response. actually, I have used bonferroni correction when P values were less than 0.05.

Dear Muayyad Ahmad ,,

thank you for your response . Its very helpful

Dear Dr.Abdulsalam

Krukal Wallis test used for not normal continuous data to compare among more than two groups for you ( physicians, pharmacists and nurses), Now if the p>0.05, it means the three groups have no significant difference in "score" but if less than 0.05 we have significant difference among groups but to know this significant difference where you have to compare between:

Physician and pharmacists

Physician and nurses

pharmacists and nurses

Maybe you will find significant in more than one comparisons

Generally Mann Whitney U test used for pairwise comparison, in SPSS you can do both tests(Kruskal and Mann Whitney U) at once... I advice you to read more about bonferroni correction it is not always helpful..

Good luck

Dear Dr. Louai Saloumi,,

Thank you very much ..

I agree with Muayyad Ahmed answer on this.........

Thank you for this post, l categorically benefited from the points raised which l subscribe to as well.

I found this example elsewhere:

A Kruskal-Wallis H test showed that there was a statistically significant difference in pain score between the different drug treatments, χ2(2) = 8.520, p = 0.014, with a mean rank pain score of 35.33 for Drug A, 34.83 for Drug B and 21.35 for Drug C.

After that post hoc test is required

Tashi Dendup I think this example can be found on that website here: https://statistics.laerd.com/spss-tutorials/kruskal-wallis-h-test-using-spss-statistics.php

They explain a lot about the Kruskal-Wallis Test in general, so I thought it might be useful to post a link.

@ Tashi Dendup , please how do i obtain the mean rank? from which table ?

I have done a Kruskal-wallis test but my output doesn't seem to have chi-square, have I done something wrong along the way?

Jamie McNulty , the chi-square value is the Kruskal-Wallis H Value in the Test statistics table. I don't think you have done anything wrong, if you can identify the Kruskal-Wallis H Value.

The SPSS version 25 doesn't display a chi-square value

I should also that if significant differences were detected using a global test, which in this case is the Kruskal-Wallis test, you would want to conduct a posthoc test. You should apply the Conover test as posthoc test for pairwise multiple comparisons of the ranked data (i.e. use Conover test as posthoc if Kruskal-Wallis p-value < 0.05). I don't use SPSS, but in RStudio, the command line is very straightforward.

I got feedback from a journal peer reviewer recently stating the following - "Please use the Kruskal-Wallis H test statistic when reporting results of that test, i.e.,

H(3)=3.514, p=.319." My understanding was the test statistic was χ2 (chi quared lovingly edited by researchgate). Who is correct?

Hi, Karl E. Bridges , May I know that the best way to report the Kruskal-Wallis H test is like H(3)=3.514, p=.319 ?

I never got an answer - this thread is very dated but I still persist with chi squared. My searches on the internet appear to recommend chi squared more so than H.

How to visualize the Kruskal-Wallis H test? ANy ideas for graph - seven groups were pairwise tested?

Olga Gyarfasova perhaps as a heatmap similar to a correlation matrix? If you are working in R there are a few suitable plotting techniques in various packages, e.g. ggplot2 and plonly.

@olga Gyarfasova try check this link may help https://www.datanovia.com/en/lessons/kruskal-wallis-test-in-r/

No

I think Chi square values are reserved for Chi square test. APA style requires that you Report Kruskal Wallis statistic, the H statistic.

@Liang Ma, it is possible to get a significant post hoc test even with a non SIG KW test, but it is meaningless. As a rule post hoc tests as the name implies means you only go for it after an omnibus test has shown a significant difference. Otherwise you don't.

@ Wenyan Xu, Kruskal-Wallis test results should be reported with an H statistic, degrees of freedom and the P value; thus H (3) = 8.17, P = .013. Please note that the H and P are capitalized and italicized as required by most Referencing styles.

The abstract to this article attached has an example of reporting Kruskal-Wallis H statistic; you may find it useful:

https://doi.org/10.9734/ijpss/2020/v32i830320

Can i use Kruskal Wallis test for this type of data?

Initially I designed the research to apply ANOVA test however this data is in not normally distributed (bimodal distribution).

Which test best matches with this types of data?

@Madhav Koirala, the data on infestation appears to be measured in ratio scale. For that type of data you better use ANOVA to test for differences between the categories. ANOVA is more robust than KW where the data is measured on a ratio scale.

@Madhav Koirala, sorry, I had not gotten your additional comments. Where data is not normally distributed, Welch's ANOVA is recommended.

My SPSS , doesn't give me chi squared value when I conduct Kruskal Wallis test. It gives me H value, df, and p value. Is it okay If I report these three? Phys Rev paper shows people mostly reporting chi square, df, and p value.

Aesha Bhansali I copy from another RG question:

Kruskal-Wallis test results should be reported with an H statistic, degrees of freedom and the P value; thus H (3) = 8.17, P = . 013. Please note that the H and P are capitalized and italicized as required by most Referencing styles.

My version of SPSS (26.0) gives me the chi squared value; in fact, in my last work (under review) I typed like this: c^2 (2, 122) = 28.69, p < .001.

You can also find different choices here:

- APA:

https://www.slideshare.net/plummer48/reporting-a-kruskal-wallis-test

https://guides.library.lincoln.ac.uk/c.php?g=110730&p=4638045

- Handbook of biological statistics:

http://www.biostathandbook.com/kruskalwallis.html

I guess you can do both, it basically depends on reviewers and journals' editorial policies.

Thanks a lot ! Kindly help me with this too !:

What is difference between Mann-whitney and independent t-statistics..?

I am reporting Kruskal Wallis for three group comparisons, and so subsequently I am reporting Mann Whitney to compare each groups.

But for the gender comparison I used independent t-statistics.

But I am told, we can't report different statistics in a paper ..

@Aesha Bhansali, once you have used the Kruskall-Wallis you need to consistently use it. You do not have to use t-test at the same time as the the Kruskall-Wallis (KW). For gender, KW will still flag differences if they exist.

The post-hoc test of Mann-Whitney will serve to show you the groups that are significantly different from each other once the KW has shown that significant differences exist.

Okay ! thank you !

Aesha Bhansali the main difference lies in the fact that t-tests are parametric tests, while Mann-Whitney and Kruskal-Wallis are non-parametric.

Given your non-normal distribution, I'd stick to non-parametric tests throughout the whole paper. In most cases you won't lose your significance ;)

Hello, thanks again ! Can anyone help me with this too?

Andy field says:

"For the Mann–Whitney test, we need to report only the test statistic (which is denoted by U) and its significance. Of course, we really ought to include the effect size as well. So, we could report something like:

 Depression levels in ecstasy users (Mdn = 17.50) did not differ significantly from alco- hol users (Mdn = 16.00) the day after the drugs were taken, U = 35.50, z = −1.11, ns, r = −.25 "

my U value is 24861.000 , and W value is 39057.000 .. should I report these or just z, r and p values (z, r values are in line of the example above) ?

Aesha Bhansali Hi. I know it's a bit late but perhaps for future reference, when conducting the Kruskal Wallis, in order to get a result for Chi Square ensure that you tick the median box in addition to the option for Kruskal Wallis.

How much is your data far from a normal distribution?

What test did you run?

What is your sample size?

You can transform the U statistic into an AUC, which is bounded between 0 and 1 but beware

KW extends the MWN test on ranks: as such it is a test on stochastic dominance, it does not compare any location, not even the mode

You can get small pvalues even if there is no change in location

If your data is unimodal and symmetric, treat them as normal, at least you can interpret them in a straightforward way,

going nonparametric makes sense only if your data is symmetric, unimodal and with equal variability, which - personally - seems to be a stronger requirement than normality

The welch's test is known to perform well even tho data are non-normal, you just need to check if the sampling distribution is close to normal

The following is an example for reporting Kruskal-Wallis test:

Kruskal-Wallis Test was conducted to examine the differences on renal dysfunction according to the types of medication taken. No significant differences (Chi square = 3.71, p = .39, df = 6) were found among the five categories of participants (none, ACE inhibitors, ARB, ACE inhibitor and ARB, NSAID, and ACE inhibitor or ARB and NSAID).

Although the Kruskal Wallis test is estimated by softwares using the Chi square values, the statistic calculated is the H-statistic . Thus Kruskall Wallis should report H statistic and not Chi square. Please refer to earlier discussions.

Kruskal-Wallis H Test & Conover-Iman

The main reasons that led to selecting the Kruskal-Wallis 𝐻 test were its non-parametric (distribution-free) nature and the option it offers to perform multiple comparisons between many groups, especially when dealing with ranked data and ordinal-level dependent variables (Agresti, 2007; Mangiafico, 2016). The null hypothesis (𝐻0) assumes that the 𝑘 samples belong to the same population and the probability 𝑎 of a type-I error—occurring when a true 𝐻0 is rejected—has been set to .05 (i.e. 95% confidence level).

In terms of statistical significance, the Conover-Iman post-hoc test was chosen to further reject the 𝐻0 strictly because of its more powerful nature compared to the available alternatives (e.g. Dunn’s test) (Conover & Iman, 1979; Conover, 1999).

Since Kruskal-Wallis test of rank is a non-parametric test, the test does require to fulfill the assumptions of normality. We can use test statistic and post hoc (Dunn) to find out the significant pairs. Using the Post hoc one can distinguish the significant pairs and the non-significant pairs. I had asked a question on systematic method of segregating and tabulating pairs using SPSS. Further, more materials are recommended by @ Sajjad Hemati Nourani, Thank you

Thanks to all. Very useful comments for reporting Kruskal-Wallis tests

Kruskal-Wallis test results should be reported with an H statistic, degrees of freedom and the P value. @ Sajjad as given an appropriate example. Reported as for example: H (3) = 8.17, P = .013. Conventionally the H and P are capitalized and italicized as required by most Referencing styles.

Here is an example of my Kruskal-Wallis test results. How should it be expressed in the article?

@Salaahoddin Zarei , this should be reported as: H (6) = 26.96, P < .01

With H and P italized by convention.

Joseph Kipkorir Cheruiyot truly appreciate your attention

reporting the results of Kruskal wellis test differed from researcher to another. some preferred to used mean rank with chi-square value and p-value.

others preferred to use median of each group. moreover some researchers may use arithmetic mean and standard error or standard deviation.

but I think the most reliable method is using median of each group with expression of chi-square value and p-value

you can do the test on spss version 21 or later versions please see this video in the following link

https://youtu.be/a_Wr4-a2k3s

Joseph Kipkorir Cheruiyot sorry in your answer to Salaahoddin you give the degrees of freedom as 7, and I was wondering where you got that/ how you calculated it. I'm using GraphPad also, so get the same output and was unsure where to find the degrees of freedom. Thanks!

@Bridget Ashford, the degrees of freedom for Zarei's case is 6 not 7. The associated df is k-1, where k is the number of treatments/groups being tested for mean or median differences.

Thanks :)

Thanks for this.

Hello to you all. Do you think the following statement is true and in accordance with standards?

Given that the distributions were not normal for the three groups, a Kruskal-Wallis test (H (2) = 3.62, p = 0.14) shows no difference in...

To report the results of Kruskal-Wallis test, you can report the test statistic value, degrees of freedom, and p-value. If the p-value is less than the significance level (usually 0.05), you can reject the null hypothesis and conclude that there is a significant difference between at least two groups.

To report the results of the Kruskal-Wallis test, you should report the H statistic, degrees of freedom and the P value. The H and P should be capitalized and italicized as required by most referencing styles.

To report the results of a Kruskal-Wallis test, you have to state the following:

- The test statistic H and its degrees of freedom (df), which is the number of groups minus one.

- The p-value of the test, which indicates the probability of observing a test statistic as extreme or more extreme by chance, assuming the null hypothesis is true.

- The conclusion of the test, which is whether you reject or fail to reject the null hypothesis. The null hypothesis is that the medians of all groups are equal.

Let us take cognizance of the fact that the Kruskal-Wallis test is an extension of the Mann-Whitney U-test which uses mean ranks to test whether there are differences in datasets from groups. The appropriate Null hypothesis therefore is that the mean ranks are equal. The test statistic (H) is based on mean ranks, not means, not medians.