SPSS Statistics added the Dunn or Dunn-Bonferroni post hoc method following a significant Kruskal-Wallis test In Version 19 and later

Analyze>Nonparametric Tests>Independent Samples (not Legacy Dialogs)

Specify (click the Fields Tab) one or more appropriate dependent variables and a grouping factor with more than two levels, pairwise comparisons (click the Setting tab) using the Dunn-Bonferroni approach are automatically produced for any dependent variables for which the Kruskal-Wallis test is significant.

In the Output viewer, double click the Hypothesis Test Summary to activate the Model Viewer output. Look in the lower right of the screen for the View drop-down menu and select Pairwise Comparisons to see the post-hoc test results.

Note: SPSS performs Bonferroni adjustment by multiplying the Dunn’s P value by the number of comparisons.

Rather than attempt to answer your question, I'll pose one of my own:  Why did you use the Krusal-Wallis test rather than a one-way ANOVA?  If it was to deal with heterogeneity of variance, you'd be far better off (IMO) using a form of ANOVA (Welch or Brown-Forsythe) that allows for heterogeneity of variance.  Rank-based tests, when used as tests of location, are very sensitive to heterogeneity of variance and small differences in skewness.  See the links below for simulation studies that demonstrate this.  HTH.

http://www.tandfonline.com/doi/abs/10.1207/S15328031US0204_03#.VHziiMny34s

Article The Wilcoxon-Mann-Whitney test under scrutiny

Behind Bruce's point, which needs to be stressed, is another one: why did you do a one-way analysis of variance? When you make a general comparison of more than two groups, you test a very vague hypothesis: there is (some kinda) difference between…

The fact that you are then following it with post-hoc tests suggests that it's the wrong hypothesis test. Can you rewrite your model as binary comparisons?

For example, you might have measured smoking as

• - never smoked
• - smoker
• - ex-smoker

You can think of this variable as measuring two things:

• Did the person ever smoke?
• Do they still smoke?

And a regression model will give you the effects of a) ever having smoked (lasting damage) and b) smoking now

The other point is that Wilcoxon Mann-Whitney and related tests are not testing a hypothesis equivalent to OLS methods. ANOVA and regression compare means, while WMW methods calculate the probability that a member of one group will score higher than a member of another group. Being sensitive to differences in the shape of the distribution, which Bruce points out, is not an adverse side-effect. It is the actual purpose of the test, because differences in distribution shape will lead to members of one group out-scoring members of another. 

You should not be surprised to get differences in statistical significance between a t-test and a WMW test, then. The hypotheses they test are simply different, and not comparable. I did a little paper on this a while ago, if it's of any interest. 

http://www.stata-journal.com/article.html?article=st0253

Not that I feel comfortable adding to the already high-quality answers by experts in the field, I'd also like to point out the possibility of transforming your data to address potential skewedness/non-normality (such as a log2-transformation) and then performing a two-way ANOVA.

May I ask the group, what is wrong with performing an ANOVA or Kruskal-Wallis as a global comparison 3+ groups and then comparing the difference in means / ranks in a pair-wise comparison (of course taking the EWER etc. into account and p-adjusting when necessary etc.)...? Which I think is sort of what Morgane was asking...?

The regression approach I personally would feel is more advanced and sometimes goes beyond a simple group-wise comparisons...no?

mann whitney u

As you already must realise there are many choices for a post hoc test. Not to disagree with comments made so far I shall assume you are testing with equal variances assumed? This provides the largest number of post hoc tests available for Oneway ANOVA. Exoerience shows Oneway ANOVAs do not always resolve with post hoc tests. Over many years and possibly having observed thousands of tests two emerge as most popular (Tukey's HSD and Duncan's test). If I had to pick between the two, I have observed the largest number of successful resolutions I have observed to be obtained by using the Duncan test (also known as Multiple Range Test). This provides different critical difference values for particular comparisons of means depending on how adjacent the means are. A note of caution: the Duncan test has been criticized for not providing sufficient protection against alpha slippage, however Tukey's HSD is, in theory, is the post hoc procedure of choice as it provides true correction of alpha slippage for the number of comparisons made but does not sacrifice statistical power (however the latter should not be seen as Hobson's choice?)

I do the following:

1. rank all data 

2. apply the ANOVA F test to the ranks

3. use Tukey's test on the ranks.

Background: The KW Test is simply SSTRT/MSE when using the ranks. It is very similar to the F test applied to the ranks. F=MSTRT/MSE, where MSTRT is the mean square treatments, and MSE is the mean square errorr.

No special software is required. Any statistics software will do the job.

Check for equal variances with the Brown & Forsythe test. 

If you have some idea of the distribution(s) underlying your data, why not use maximum likelihood estimation and likelihood ratio tests using this (these) distribution(s), provided that the resulting models are nestable?

Or (even better) dispense with the testing nonsense entierely and estimate the distributions of the differences (with Bayes factors thrown in if your reviewers insist on model comparison...) ? This, of course, entails stating (eliciting) a prior distribution, and might get you questions from reviewers unfamiliar with Bayesian methodology...

While I agree with Bruce et al., there are some simple follow-up tests for Kruskal Wallis.

If you have only three groups then a variant of Fisher's LSD is a possibilty. This would just involve uncorrected pairwise tests such as Mann-Whitney etc. Type I error protection is provided by the omnibus test of the Kruskal Wallis. With more than 3 groups this approach is inadequate because it doesn't protect against type I error for partial null hypotheses (only for the complete null that mean ranks are equal for all groups).

For more than 3 groups I'd suggest the Hochberg procedure as this is more powerful than most of the other options and just requires p values from pairwise tests to compute.

Thom makes an excellent, but often overlooked point about the special case of 3 groups.  (David Howell's chapter on multiple comparison procedures in Statistical Methods for Psychology has a nice discussion of why Fisher's LSD controls the family-wise alpha properly when there are 3 groups.) 

Yes - David Howell explains it clearly (though I think my own book "Serious Stats" does also!).

Using ranks in the ANOVA F test takes into account the relative levels, and it compares the mean ranks. In this sense, it combines the best features of the Kruskal Wallis Test with the ANOVA F test. Post-hoc  procedures are straightforward, and they are based on the ordered data (ranks). 

Perhaps the conover test is a good posthoc test after kruskal wallis

http://www.brightstat.com/index.php?option=com_frontpage&Itemid=1

http://www.ncbi.nlm.nih.gov/pubmed/?term=18653259

Hi,

A non-parametric version of repeated-measure ANOVA, which is called Friedman test. If you are comparing the data against a between-subject factor, you can use a Kruskal-Wallis test. Otherwise, you should need to use a Friedman test.

Example:  

The two tests, Kruskal-Wallis and Friedman tests only support a one-way analysis. This means that you can compare the data only across one factor and unfortunately, you cannot easily extend these tests to two-way or mixed-design as we can do with ANOVA then you have to use a non-parametric test.  The post non-parametric methods equivalent to two-way repeated-measure or mixed-design ANOVA.

The Friedman test is an extension of the sign test - not of one-way ANOVA.

https://seriousstats.wordpress.com/2012/02/14/friedman/

Rank transformation followed by ANOVA is generally better.

More generally, rank transformation and related tests aren't great for two-way designs. Testing an interaction model on ranked data introduces serious difficulties because additivity of ranks might not imply additivity of the raw scores (and vice versa).

Conover's text points out that only normal quantile rank (also called normal scores test) based two way ANOCA F test has an interaction test that is supported with valid theory.  Using the regular ranks works well fir the main effects but not fir interaction tests. The normal quantiles F test is best suited for the two way model. 

Friedman's test is a test for a block design. One factor is the treatment factor while the second factor is the blocking factor. No interactiobs are allowed in this model.

ranks work well in obe factor models. This is like the Kruskal Walis test.

Thom: make sure that normal quantiles sre used and not the nasic ranks.

i often use this method with other models, such as a three factor model with a covariate, say. I explained the aplication of normal quantiles in a recent article on shark human interactions in Animal Cognition. It was featured in a Springer Verlag press release on it. 

Back when I was a student, one of my profs came up with an interesting idea for a two-way analysis of ranks that was based on the additivity of independent chi-squares.  See the link below.  (I apologize for the pop-ups that are likely to occur--I really should move that note from Angelfire to a Google Sites page sometime.)

http://www.angelfire.com/wv/bwhomedir/notes/alternative_to_anova.txt

From what Bruce has posted for us:

> Could anyone tell me what kind of test I should be use to test a mixed model? I've

> got 1 between-subject and 1 within-subject effects. And although my dependent

> variable is continuous, it is highly non-normal. I know how to test each single

> effect. But don't know about their interaction effect.

>

> Any help would be much appreciate.

>

> Helen

--------------- >8 ----------------

I have a copy of an unpublished manuscript by G. Rolfe Morrison from the

Pscyhology Department at McMaster University (in Hamilton, Ontario) that

adddresses this issue. The paper is called "Rank order alternatives to

the two way analysis of variance". I'm sure he would be happy to send you

a copy. (Maybe if enough people ask him to send copies, I'll finally be

able to convince him to submit it for publication!) His e-mail address is

. The snail-mail address is:

Psychology Department

McMaster University

Hamilton, Ontario

CANADA L8S 4K1

For a problem like yours, he uses the following example: Two independent

groups of subjects are given the same picture sorting task. Each subject

in Group A is shown 4 photos of "physicians" and asked to rank them from

1 to 4 in terms of trustworthiness. Subjects in Group B are shown the

same photos, but are told they are psychiatrists.

If you treat each group separately, you can use Friedman ANOVA. The test

statistic is distributed (at least approximately) as chi-square with k-1

degrees of freedom (i.e., df = 3 in this case). Because the groups are

independent, you have 2 independent chi-squares, each with df = 3. The

sum of two independent chi-squares is itself a chi-square with df = to

the sum of the df for the components. In other words, if you add

together the two Friedman test statistics, the sum is approximately

distributed as chi-square with df = 2(k-1) = 6 for this example.

It might help if I give some numbers to make the example more concrete.

For Morrison's example:

Friedman statistic for Group A = 12.0 (df = 3)

Frideman statistic for Group B = 3.9 (df = 3)

Total = 15.9 (df = 6)

The next step is to calculate the Friedman statistic on ALL THE DATA,

as if there was only one group of subjects. For Morrison example:

Friedman statistic for ALL data = 2.55 (df = 3)

This statistic provides a test of the MAIN EFFECT of the variable with

repeated measures. For Morrison's example, it is not significant.

The final step is a test of the AxB "order" interaction. The statistic

for this test is obtained by subtraction as follows:

AB statistic = Total - Main Effect

= 15.9 - 2.55 = 13.35

df = 6 - 3 = 3

This statistic is distributed (approximately) as chi-square with df = 3.

For Morrison's example, the AB order interaction is significant. This

means that the two groups are not ranking the photos the same way.

One important thing to note is that this procedure allows you to detect

"order" interactions only (i.e., where the B treatments are ordered

differently at the levels of A). It cannot detect an interaction where

the effects are going in the same direction, but the size of the effect

is larger in one case than the other.

Morrison also discusses a test for the case where you have two variables

that are manipulated between groups. His methods don't work if both

variables are manipulated within-subjects though, because the chi-square

values are not mutually independent in that case.

Hope this helps.

Dear Dr Comby,

I'll try not to reformulate your question. I belive that the Nemenyi-Damico-Wolfe-Dunn procedure (post-hoc on joint ranks) is the more powerful procedure and it is easily done in R, the post-hoc p-values are calculated based on a multivariate T-distribution; a procedure developed by, among others, Frank Bretz who is a well-known expert on the subject of post-hoc tests.

Best regards,

Fredrik

PS. How to do all pairwise-comparisons (write to me if any problems):

Install R from www.r-project.org, start and type what's after >

Rconsole> install.packages(c("coin", "multcomp"))

Rconsole>library(coin)

Rconsole>library(multcomp)

Rconsole>?kruskal_test

## You will have to fill in your own data instead of Gizzard-data

## substitute your dependent varaible for 46, 28, 46,... below

## substitute the the Group they belong to in site-variable; if you have only

## 5 in Group I (the five first entries of length (46,28,46,37,32 below)) then write

## rep("I",5). If you have five Groups write  rep("V", number of observations in this 

## Group) etc. Suppose that the last 5 entries belong (in length) to Group V, then it ## would be

##  ..., rep("IV", 10), rep("V",5))))

### Length of YOY Gizzard Shad from Kokosing Lake, Ohio,

### sampled in Summer 1984, Hollander & Wolfe (1999), Table 6.3, page 200

> YOY kw kw

> pvalue(kw)

### Nemenyi-Damico-Wolfe-Dunn test (joint ranking)

### Hollander & Wolfe (1999), page 244

### (where Steel-Dwass results are given)

> if (require("multcomp")) {

NDWD

I draw up in horror when I see a test for all possible pairwise comparisons. 

The first thing it suggests is a researcher who is letting the data generate their hypotheses. This breaks the logic of statistical testing : that the model is specified a priori and the parameters are calculated from the data. 

The second thing is that there is no point in controlling the experiment-wise error rate when you have no experimental hypothesis.

Essentially, if your hypothesis entails pairwise comparisons, then run an appropriate model using binary variables. Post-hoc tests are not a substitute for a properly-formulated research question. 

First, a correction: It is Peter H Westfall that has written Multiple Comparisons and Multiple Tests Using SAS, Second Edition together with Randall D. Tobias, Russell D. Wolfinger that I thought about. Westfall is also behind the multcomp package. Sorry for that. I hope that Frank Bretz can be called expert on the subject anyway, I think very highly of him and the other authors of multcomp.

I agree with Dr Conroy, that if the hypothesis is stated such that one aims to find out only if treatments are better than a control/standard treatment, then it is of course suitable to do the analysis accordingly. But I maintain that there are situations were one would like to know which of all pairs differ.

A simple example, suppose that there are three treatments for depression. Standard, Group Therapy and Cognitive Therapy. Suppose that we have an adequate measure of depression and ideally one random person from each group in Group therapy.  Suppose that we believe that Standard therapy is quite inefficient. Then it would be interesting to know if the 2 treatments are better than Standard treament, but also if the two new treatments differ. I see no problem in that, nor do many, if not most, authors of books in experimental design. I think one has to separate the logical null hypothesis (what one believes a priori) and the statistical null (the reverse of the logical null); i.e. if I belive that there are differences between groups, then the statistical null is that there are none. Note that if I believe that there are differences between some of the pairwise comparisons, then a way of testing that is to assume that there are no differences between groups. It may lead to inconclusive results, such that we can not say if treatment A differs from treatment B, but that's a power issue that may occur in any experiment. Differences has to be clinically important.

Dr Conway sugest that if all pairwise comparisons are done post-hoc, then no experiment was done since the hypothesis was not stated prior to the experiment. I disagree, by the argument above. However, I'd agree that sometimes there isn't a clear hypothesis, for instance in genome-wide association studies, where many single nucleotide polymorphisms are tested, in fact it's not even an experiment! Many fine statisticians obviously try to find differences in such situations without apparent inspirational distress, among them Dr Conroy (Clin Transplant 2013: 27: 379–387). Of course, here pairwise-comparisons would be preposterous [no pun intended] when one has almost a million SNPs and a couple of hundred patients!

A final note, an attractive alternative if you have a prior rank of your hypotheses is to divide the critical p-value into portions and spend these on the hypotheses at hand, for instance 0.02 on both comparisons against the Standard treatment in the example above and 0.01 on the difference between Group and Cognitive Therapy.

Fredrik, for your example with 3 groups, you could use Fisher's LSD.  When there are 3 groups, it controls the family-wise alpha at the per-contrast alpha.  See the chapter on multiple comparison procedures in Dave Howell's Statistical Methods for Psychology for an example.  See also the paper linked below.  From the abstract of that article:  "Fisher's LSD procedure is known to preserve the experimentwise type I error rate at the nominal level of significance, if (and only if) the number of treatment groups is three."

HTH.

http://www.ncbi.nlm.nih.gov/pubmed/17128424

Dear Bruce,

You're quite right of course. But still, it also a situation where all pairwise comparisons are of interest, which I think they'd be if I included a fourth treatment, say Garden Therapy, too.

Thank you for the reference! I've wanted for some time to know the exact argument that the LSD has this nice property. Now I have an excellent starting point.

,Yours Sincerely,

Fredrik

Fredrik, I don't have a copy of it (yet), but I am told (by Thom Baguley, the author) that the book Serious Stats is another reference that supports the use of Fisher's LSD when there are 3 groups.  Cheers!

http://www.palgrave.com/page/detail/?k=9780230577183

Bruce, how would Fisher's LSD handle the non-normal data issue? I thought that the main reason for using KW Test was the fact that Morgane  has non-normal data.

See my earlier comment. Fisher's LSD doesn't make any correction on the individual follow-up tests, instead the Type I error protection comes from the omnibus F test. By analogy one can make the same argument for KW followed by appropriate pairwise tests.

The problem with Fisher's LSD is that it only protects against the complete null hypothesis that all population means are equal (or mean ranks for KW). With more than 3 means if the complete null hypothesis is false it no longer implies at most one Type I error.

Thom, how is the LSD test better than using, say, Tukey's  Test?  I fail to see a true advantage for using the LSD Test. It has an inflated Type I Error Rate other than depending on depending on an umbrella coverage from the F test. 

Key references are probably the following:

Shaffer, J. P. (1986). Modified sequentially rejective multiple test procedures. Journal of the American Statistical Association, 81, 826-831.

Shaffer, J. P. (1995). Multiple hypothesis testing. Annual Review of Psychology, 46, 561-576.

The Fisher LSD test has no inflated type I error rate if you have only 3 means.

The point is that that if you reject mu1=mu2=m3 (via the omnibus test) and it is not true, you have made at least 1 type I error. the null hypotheses for the remaining pairwise tests are:

mu1 = mu2

mu2 = mu3

mu1 = mu3

If one of these is false then at least one other must be false (given that we have rejected the null that they are all equal with the omnibus test). Thus at most one partial null can be true an thus only one Type I error errors is possible at this stage of the analysis. It follows that one does not need to correct the individual tests to protect multiple partial null hypotheses. Tukey's HSD is very conservative but corrects for all partial null hypotheses - the Shaffer approach is more efficient.

Shaffer's paper explains it in detail. Hope this is clear (I'm writing this fairly late …)

Hi,

You can consult the following paper:

Joaquín Derrac, Salvador García, Daniel Molina, Francisco Herrera, A practical tutorial on the use of nonparametric statistical tests as a methodology for comparing evolutionary and swarm intelligence algorithms, Swarm and Evolutionary Computation, Volume 1, Issue 1, March 2011, Pages 3-18, ISSN 2210-6502, http://dx.doi.org/10.1016/j.swevo.2011.02.002.

http://www.sciencedirect.com/science/article/pii/S2210650211000034

There is an exact post-hoc test for Kruskal Wallis after Schaich and Hamerle or post-hoc test after Conover (already suggested by Vincent Bloks)

Both tests are illustrated on brightstat.com and brightstat's webapp lets you compute the critical differences and perform the posth-hoc analysis right away.

https://secure.brightstat.com/index.php?p=c&d=1&c=2&i=7

Resolving the problem

SPSS Statistics added the Dunn or Dunn-Bonferroni post hoc method following a significant Kruskal-Wallis test in the NPTESTS procedure (Analyze>Nonparametric Tests>Independent Samples in the menus) in Version 19. It is not available in earlier versions.

In Version 19 and later, if you specify Analyze>Nonparametric Tests>Independent Samples, specify one or more appropriate dependent variables and a grouping factor with more than two levels, pairwise comparisons using the Dunn-Bonferroni approach are automatically produced for any dependent variables for which the Kruskal-Wallis test is significant. If you are looking at Model Viewer output, you need to double click on the Hypothesis Test Summary to activate the Model Viewer. When the Model Viewer opens, look in the lower right of the screen for the View drop-down menu and select Pairwise Comparisons to see the post-hoc test results.

If you find significant difference in your data after  after KW, you can apply further the pairwise KW to identify the outlying mean or means among your data.  This is easy to do in R.

As several researchers said, you can use the ANOVA or t tests with your data transformed into ranks. Here one paper that help to you with your question: Article [Rank Transformations as a Bridge Between Parametric and Non...

Another method you can use is through the contrast of your treatments by means of a GLM model and contruction of a dummy data matrix with the coefficients of your model.

Best,

Criss

Shingala, M. C., & Rajyaguru, A. (2015). Comparison of post hoc tests for unequal variance. International Journal of New Technologies in Science and Engineering, 2(5), 22-33.

Hope this help.

How to perform Nemenyi test in Matlab?

I think that there is a valuable option in R. It is very simple and can deal with the most popular corrections for multiplicity. Let's take a look at pairwise.wilcox.test(). Basically I agree that performing pairwise Mann-Whitney test accounting for multiplicity is simple and correct.

SPSS Statistics added the Dunn or Dunn-Bonferroni post hoc method following a significant Kruskal-Wallis test In Version 19 and later

Analyze>Nonparametric Tests>Independent Samples (not Legacy Dialogs)

Specify (click the Fields Tab) one or more appropriate dependent variables and a grouping factor with more than two levels, pairwise comparisons (click the Setting tab) using the Dunn-Bonferroni approach are automatically produced for any dependent variables for which the Kruskal-Wallis test is significant.

In the Output viewer, double click the Hypothesis Test Summary to activate the Model Viewer output. Look in the lower right of the screen for the View drop-down menu and select Pairwise Comparisons to see the post-hoc test results.

Note: SPSS performs Bonferroni adjustment by multiplying the Dunn’s P value by the number of comparisons.

in spss

comparison as post hoc with significant kruskal -wallis test is used dunn's test in path analyze -nonparametric - independent samples -field -customize paired comparison

And how does one determine the effect size?

I think that performing pairwise Mann-Whitney test accounting for multiplicity is simple and correct.

look this calculator: http://astatsa.com/KruskalWallisTest

Using R, and Bonferroni to adjust p-values for multiple comparisons:

Dunn's test:

install.packages("dunn.test")

library(dunn.test)

install.packages("FSA")

library(FSA)

dunnTest(y~x, method="bonferroni")

*other options for method: "none", "sidak", "holm", "hs", "hochberg", "bh", "by"

more details (in R):

help(dunnTest)

help(dunn.test)

The main diffference is there are a control group or not. So if not, is better performing pairwise Mann-Whitney test.

Jose Manuel Azevedo : I just came across this discussion, and saw this site(https://www.statisticshowto.datasciencecentral.com/dunns-test/) that apparently contradicts your information: "Use Dunn’s when you choose to test a specific number of comparisons before you run the ANOVA and when you are not comparing to controls. If you are comparing to a control group, use the Dunnett test instead.". Could you please comment?

https://scholar.google.com/citations?user=PmvtNFUAAAAJ&hl=en

Dunn test is suitable

Do pair wise comparisons using Dunn’s test. Use the Bonferroni correction to account for multiple comparisons thus reducing type 1 error.

The effect size can be calculated as eta squared (η2) or Epsilon squared

E2R.

You can try in IBM spps, stepwise step down, and pairwise with Dunn test

Hi all, I have a question. Is Dunn test a multiple comparison of means?

In case of non-parametric data, I understand that it is better to work with medians. Is that correect?

So that suggests to me that I should use Mann Whitney as post hoc after Kruskal Wallis.

Conover's test with Holm correction, for example.

See R packages PMCMR or PMCMRplus

Hi , I analyzed my data by kruskal Wallis test for 3 groups and I have a directional hypothesis . I got adjusted p- value by Bonferroni correction for multiple test p=0.060 at 2-sided tests. can I divided p-value by 2 to get p- value at one tail tes.t

Rather than attempt to answer your question, I'll like to proposed an alternative, that has been shown to be effective: ARTool -align-and-rank- transform for a non-parametric ANOVA is a free software that can align your data, so that you can perform ANOVA analysis on non- parametric data. Interaction effects using non-parametric methods has it limitations.

http://depts.washington.edu/acelab/proj/art/index.html

thank you Gem

Hi every body , I compared 2 independent group of students for each exam to measure outcomes. for example exam 1, exam2, exam3, and final exam. is this analysis increase type I error.

Alanazi Amal Hi, what did you do at the end regarding your p value, did you divide them by 2 to address directional hypothesis or not? Gem's recommendation is not an option for me now so it would help if you had an alternative. Thanks!

Hi christodouli, I did not find my answer until now. SPSS give me adjusted p-value at 2-sided and I have a directional hypothesis based on the theory which group 1 is significantly higher than group 3. What I understood from the statistics book, if I have a directional hypothesis, the p-value must be divided by 2.

Hi all

I have 3 groups comparison. my hypothesis group1 is higher than group 2 and group 1 is higher than group 3. I found group 1 is higher than group 3, in this case I will reported p- value one tailed test. I am not care about group2 and 3. what about adjusted p-value

Thank you

As Bruce Weaver mentioned, it is worth considering exactly why you are using a Kruskal-Wallis test in the first place. You mentioned that you want to identify "which groups differ from the rest." What metric are you comparing: means, medians? Ranks tests are interpretable and intuitive for ordinal data that can be ranked, however their use as second-best options to parametric 'equivalents' has been criticized for some time. Importantly, the Kruskal-Wallis test, when used as an alternative to parametric ANOVA, requires data to be homoscedastic.

Hi all.

why that many survey responses would give us less than 5 responses for a single category but more than 5 when the categories are combined?

Thank you

Miky Timothy please can you point me to a source where I can read more about this statement of yours " Importantly, the Kruskal-Wallis test, when used as an alternative to parametric ANOVA, requires data to be homoscedastic."?

Precious Okoroafor, please see the explanation, with example data, in the link below:

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

As a side note; Alanazi Amal, unless I am missing something about your posts, please start a new thread if you have a new question.

Miky Timothy Thanks. Already found it and was ready it when you responded. Please check you message.

Hi all

I f I have a directional hypothesis group 1 is higher than group 2 and group 3. when I did my I analysis I got group 1 is significantly higher than group 3. which p- value I will reported one tailed or two-tailed.

I have 4-points likert scale survey and the total sample size is about 50 students . I want to investigate any relationship between gender and responses( 1= strong disagree,2=disagree, 3=agree, 4=strongly agree). I will use chi square for independence because I have categories variables. My question why a sample of 50 students might give cell frequencies less than 5?

Alanazi Amal If you have expected counts below 5, Fisher's Exact Test is an alternative to a chi-square test.

Thank you Konstantinos Mastrothanasis. yes Fisher's Exact Test is an alternative.

to a chi-square test. my question is why a sample of 50 students might give cell frequencies less than 5.

why that many survey responses would give us less than 5 responses for a single category but more than 5 when the categories are combined?

Hi Konstantinos Mastrothanasis, if I analyze my data by 2x2 table and still I found cell less than 5. in this way I will reported p- value for Fisher's Exact Test.

A Kruskal-Wallis test provided very strong evidence of a difference (p=0.02

A Kruskal-Wallis test provided very strong evidence of a difference (p=0.02

If you have few data, there is a problem whit the distribution of the data and you can´t use some analysis. Is necessary use a chi-square test. If you have more than 30 observations per group, then you can up in the model´s analysis. According whit your type of distribution it is recommended use a Kruskal-Wallis like a post-hoc.

Morgane A. L. Comby The Wilcoxon analysis is a non-parametric test to evaluate the differences between two independent groups of data is like a non-parametric version of the ANOVA. And the Nemenyi test is to prove differences between more than two groups, that you know previously that they are different, and now you want to know ho is different.