One post-hoc you could do is pairwise Mood's median tests, followed by a p-value adjustment for the multiple tests.  That is, you run the test on Group A vs. B, A vs. C, and B vs C.  Then apply a p-value adjustment like Holm or Benjamini-Hochberg.

An example of this is here

http://rcompanion.org/handbook/F_09.html

I think using confidence intervals fro each median is another good approach.  Please be sure to use confidence intervals appropriate for medians.  These are different than those for means.  I recommend using bootstrap methods for the confidence intervals.

Some examples here:

http://rcompanion.org/handbook/E_04.html

As far determining the practical significance, it really depends on what you are measuring.  One question to answer:  Are the differences between the medians large enough to be practically important?   Are there any other factors like cost that can negate these differences in the real world?

Thank you sir for your reply - I shall explore the examples you listed.

An alternative for the Mood Test is to use the ranks of the data. Then, use Tukey's test on the ranks. Compare the results with what you got with the Mood test.

Thank you sir for your reply - but just to check, are these options mentioned available in Minitab, and where can I find them?

This is because when I do a Mood Median Test, there seem to be no options within the test to conduct a post hoc test, unlike ANOVA where there is the option for Tukey/Games-Howell post hoc tests when I select the option of ANOVA itself.

Many thanks for your advice.

Just convert the data into ranks, from 1 to N. Input the ranks into minitab; and use ANOVA with Tukey's test.

Good luck.

Dear sir,

When you mentioned to convert the data into ranks, are you referring to the following:

http://support.minitab.com/en-us/minitab/17/topic-library/minitab-environment/calculator-and-matrices/column-calculator-functions/rank-function/

Many thanks for your assistance. I appreciate it.

Yes, that would work. 

Thank you sir

Mood's median test and anova by ranks are two different tests. 

It is better to decide what you are actually trying to test rather than just use any statistical test someone mentions.

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I also caution you against adopting the idea that anova on ranks can be performed accurately on any data set.  I'll start you with the Wikipedia article on the topic, e.g.

"The ANOVA on ranks has never been recommended when the underlying assumption of homogeneous variances has been violated, either by itself, or in conjunction with a violation of the assumption of population normality."

( https://en.wikipedia.org/wiki/ANOVA_on_ranks#Violating_homoscedasticity ).

You might look at Conover and Iman, 1981 ( http://people.umass.edu/bioep740/topics/amstat-1985-iman.pdf ), for general background on this technique.

*   *   *

My honest advice:  decide what you are trying to test. 

If you want to test the hypothesis tested by a median test, use that. 

If you want to test the hypothesis tested by Kruskal-Wallis test, use that. 

If you want to test the hypothesis tested by an anova of rank-transformed data, use that.

Noted with thanks sir

Do a test of variance homogeneity first if you use any of the ANOVA tests. While the Mood Test does not require the assumption of variance homogeneity, it has low power. Run the Brown and Forsythe Test (also called Modified Levene Test) first. If this test is non-significant, run a rank based ANOVA next, followed by Tukey's Test on the ranks. If the Brown and Forsythe Test is significant, the Mood Test would still work.

Thank you sir for the advice given

You can use the non-overlap determination, this is advocated in the six-sigma approach i.e. that the 95 percent confidence intervals of the individual group medians can help to find where the significant difference is. The rule is: If there is no overlap between the confidence intervals, a significant difference can be assumed. Or  as is outlined above by Raid the Tukey test may suffice.

There are some youtubes on this see: https://www.youtube.com/watch?v=BenyNtRVHcQ

also

https://www.youtube.com/watch?v=9FZUS5QKGAU

Supporting what Robert has said above, using a schematic box plot ( a box plot with approximtae 95% confidence intervals for the medians shown as notches) is a simple method for "testing" for equal population medians. SAS calls them Schematic Plots.

SAS example: https://support.sas.com/documentation/cdl/en/statug/63347/HTML/default/viewer.htm#statug_boxplot_sect029.htm