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I have a sample of 350 and I have to compare the anxiety score among different variables. I wanted to apply the z test to compare the means but my data is not normally distributed. Is there an alternative test I can use for testing the same but on my data ?

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In short because this has been discussed many times before:

  • if you have a sample then you would usually use a t-test rather than a z-test, unless you know the population standard deviation (unlikely),
  • a t-test (or z-test) does not require the data to be normally distributed, but rather the means of the populations to be normally distributed, which is often not an issue and is satisfied thanks to the CLT.
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Depending on the structure of your data, there are many possible approaches. A good place to start is the form of your response variable anxiety score. If this is categorical, ranked data (for example 1: none, 2: rarely, 3: somewhat, 4: often, 5: always), then it can be described as ordinal.

$t$-tests compare means (as you suggest), but comparing means of ordinal data is problematic - intuitively for example the gaps between consecutive numbers might not considered equally large. More commonly one would consider measures of spread like the median.

Wikipedia has an article on Ordinal data that might be helpful.

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  • $\begingroup$ If my hypothesis is that females have a higher anxiety score than males with score ranging from 0-21 , what would be the appropriate test to compare the scores between these two groups. I'm sorry I am new at statistical analysis. $\endgroup$ Jul 30 '20 at 10:40
  • $\begingroup$ I would start with a box plot of the two groups. To compare medians can use Mann-Whitney test, to compare means can use $t$-test. If you wish to control for other factors, then investigate (ordinal) regression. $\endgroup$
    – deeprich
    Jul 30 '20 at 19:57

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