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Help with Shapiro-Wilk Test interpretation.

  1. May 30, 2009 #1
    Hi everyone,

    I need to make sure that I'm interpreting the Shapiro WIlk test correctly. This is how I'm doing the interpretations:

    Set 1
    CI = 95%
    n = 15
    Shapiro W = .92
    p = .171

    I think this set is distributed normally because p is the probability that it is not normal, so the probability that it isn't normal is 17.1% right?

    Set 2
    CI = 95%
    n = 15
    Shapiro W = .95
    p = .502

    This set is slightly more probable to be not distributed normally because p is 50.2 %

    Any help appreciated,

  2. jcsd
  3. May 30, 2009 #2
    Since CI = 95% implies a critical "alpha" value of 5%, the null hypothesis of normality cannot be rejected for either set (at the 5% level of statistical significance).
  4. May 30, 2009 #3
    But I can reject Set 1, if I chose an alpha like 20% right?
  5. May 31, 2009 #4
  6. Jun 1, 2009 #5
    Looking at this test more carefully. This test is more for testing whether a sample comes from a population that is not normally distributed.

    I mean if the p > alpha then you can't reject the probability that it might be Normal (but it is just a probability, it doesn't tell you how probable is it that it is normal?). What is a good test to determine whether a distribution is Normal or not?
  7. Jun 2, 2009 #6


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    If p > alpha then you can't reject the NULL HYPOTHESIS that THE DISTRIBUTION IS Normal.

    When testing a hypothesis you cannot ever accept the null hypothesis, you can either reject, or fail to reject. There is no statistical test that will tell you the distribution is normal; they can only tell whether you can or cannot reject normality. See http://www.keithbower.com/Miscellaneous/Don't 'Accept' H0.htm.

    I suggest using tests based on skewness and/or kurtosis; two examples are the Jarque–Bera test and D'Agostino's K-squared test. If you don't need a formal test result, you can also make a Q-Q plot and decide visually.
    Last edited by a moderator: Apr 24, 2017
  8. Jun 3, 2009 #7
    Thank EnumaElish for clarifying that for me :).
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