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As part of a survey I had to conduct Kruskal-Wallis H test. The aim of my study is to understand whether the current waste management practices in households have any effect on people's interests in selecting a centralized waste treatment programme.

Current management practices (independent variable) are divided into 3 groups: Door 2 Door, Self and Other Methods The interests (dependent variable) are on ordinal scale from 1 to 4.

So I know that, if the dependent variable has similarly shaped distribution for all groups of independent variable, then we can use the Kruskal-Wallis H test to make inferences about the difference in medians between my groups of independent variable (i.e., whether there is a current management effect on median interest score).

enter image description here

This is the SPSS result I obtained; do the distributions look similar?

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One answer is that the box plots you have chosen (or that SPSS chooses -- see below) are very poor for comparing data of this kind.

Ordinal scores -- in your case from 1 to 4 -- are such that the median and quartiles must be either integers (e.g. 1, 2, 3, 4) or half-integers (e.g. 1.5, 2.5).

Detailed study of your box plots shows that (for example) for door-to-door, the upper quartile is reported at 3.5. Therefore 25% of values are equal to the maximum 4. Since no lower whisker is shown, more than 25% of values are equal to the minimum 1. These are very important details, but it requires thorough understanding of box plots and detective work to read them off the plots. Box plots were not designed for, and do not work well with, this kind of data.

If you work through the other categories, similar partial inferences can be made, e.g. that for other methods more than 25% of values are equal to the maximum 4.

The major point, therefore, is that the box plots are an indirect way of showing the distributions, which would be much more clearly shown by a bar chart, a mosaic plot or spine plot, or even a table of relative frequencies.

Beyond that the question seems muddled. If the distributions are similar, meaning identical, then a Kruskal-Wallis test is unnecessary any way! So, what do you understand by similar here? Similar or dissimilar distribution shape is not quantifiable using ranks alone. A chi-square test would be an answer to whether the distributions are similar, but it would be imperfect insofar as the ordinal character of the scores is ignored. Conversely, I think that most careful accounts of the Kruskal-Wallis test would advise caution with a large number of tied ranks, such as you inevitably have.

I don't use SPSS and I can't read all your screen image, but the most important detail I can't see is your sample size. Test assumptions are typically most crucial for small sample sizes. But you should be able to give us the data as a 3 x 4 table of frequencies.

EDIT:

The chart posted allows a chi-square test. I can't edit it in here because the graphics format isn't compatible.

Here's the result I get (with Stata, but that's immaterial here).

          observed frequency
          expected frequency

------------------------------------------
          |              col              
      row |      1       2       3       4
----------+-------------------------------
        1 |     22       8      27      10
          | 20.770   6.700  26.130  13.400
          | 
        2 |      3       2       4       5
          |  4.340   1.400   5.460   2.800
          | 
        3 |      6       0       8       5
          |  5.890   1.900   7.410   3.800
------------------------------------------

5 cells with expected frequency < 5

         Pearson chi2(6) =   6.3346   Pr = 0.387
likelihood-ratio chi2(6) =   7.9963   Pr = 0.238

You're in a bind. Your distributions do look different, but your sample size is not large enough to confirm that with this test.

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  • $\begingroup$ The sample size is 100. Population is 890,000. The frequency chart: ibb.co/bOi1y5 @nickcox $\endgroup$ – Rahul Ramesh Apr 9 '17 at 10:58

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