Why shapiro wilk don't change when standardized (Z and T)score?
the data is 13 cases and variables are cognitive test. The test have automatic transformation in software but i want calculate t score because parameters ara based in other population. However, i have tested shapiro-wilk with original T score and it´s normal distribution.
Why do you expect that a z-transformation changes the outcome of the Shapiro-Wilk or Kolmogorov-Smirnov test? With such a transformation you do not change the shape of the distribution.
Any linear transformation of a probability distribution function changes the mean and standard deviation of the distribution but not its shape. Hence getting the Z score which is a linear transform will not change the shape.
Non-linear transforms do change the shape.
as they told you, unfortunally you can't change the shape of your distribution by transforming your raw scores into Z or T-scores.
Andy Field's chapter 5 of his book ("discovering statistics using SPSS") is very helpful because he explains what to do in this cases when data don't follow normal distributions (modify/delate outliers, transform data to correct skewness, using non-parametrical test, etc.)
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