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I'm newbie and bit lost... Can someone help me read this result from SPSS 1-sample Kolmogorov-Smirnov test (poisson distribution).

                                        SUBKEY1   SUBKEY2  SUBKEY3  SUBKEY4    SUBKEY5

N                                128       128       128      128    128

Poisson Parametera,,b     Mean          .4609     .4609    .4922     .5156  .4922

Most Extreme Differences  Absolute  .092      .092     .103      .113   .103

                      Positive  .079      .079     .088      .095   .088

                      Negative     -.092     -.092    -.103     -.113      -.103

Kolmogorov-Smirnov Z                1.037    1.037    1.171     1.276   1.171

Asymp. Sig. (2-tailed)              .233      .233     .129      .077   .129

From the result above, I define (in simple words):

Ho = satisfied
Ha = not satisfied

but I don't know how to specify the level significance for this test and I don't know whether have to read Asymp. Sig. (2-tailed) (p-value) or Kolmogorov-Smirnov Z (D value) to be compare with critical value...

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If $\alpha = .05$ is conventional in your field, I would simply state, "A one-sample Kolmogorov-Smirnov test failed to reject the null hypothesis that the data followed the Poisson distribution (D = .092, .092, .103, .113, and .103 respectively for variables 1-5, N = 128 each, and p > .05 each)." You can also report that you set $\alpha = .05$ in your methods section. Note that $Z \approx \sqrt{n} D$ in the SPSS output and that the "Asymp. Sig." (p-values) in the output should be based on critical values of the Kolmogorov distribution to which $Z$ converges in distribution (see here for a description). However, you don't need to find or report these critical values -- I would just report the test statistics and p-value in the manner above.

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  • $\begingroup$ Hi, lockedoff... tq for the answer... $\endgroup$
    – blossom emerald
    Commented Sep 9, 2011 at 6:28
  • $\begingroup$ thanks for reply...i really appreciate that. But, i just want to make sure with what i understand that you state above...is it you compare the Dmax value (D = .092, .092, .103, .113, and .103) with the critical value with α=.05 (n>35 = 1.36/sqrt(n))? which is, if the D < critical value(0.120) than you fail to reject the null hypothesis? or you just compare the p-value>0.05? I'm still newbie in this field so, there a lot of question to be ask... =) $\endgroup$
    – blossom emerald
    Commented Sep 9, 2011 at 6:46
  • $\begingroup$ I just want to make sure with what i understand that you state above...is it you compare the Dmax value (D = .092, .092, .103, .113, and .103) with the critical value with α=.05 (n>35 = 1.36/sqrt(n))? which is, if the D < critical value(0.120) than you fail to reject the null hypothesis? or you just compare the p-value>0.05? I'm still newbie in this field so, there a lot of question to be ask... =) $\endgroup$
    – blossom emerald
    Commented Sep 11, 2011 at 15:04
  • $\begingroup$ Yes, based on a quick web search, it appears that the critical value for $D$ can be approximated by $1.36/\sqrt{n}$ for $\alpha = .05$. I would rather rely on the p-value reported by the program than an imprecise approximation, but either works. $\endgroup$
    – lockedoff
    Commented Sep 13, 2011 at 17:16
  • $\begingroup$ It is correct if I've got higher value of p-value from 1-sample K-S its means i got better result than the other sample that i compared? eg: Variable_1 (p-value)= 0.09 and Variable_2 (p-value) = 0.06 means that variable_1 is better distribution than variable_2 or its just the same because that accept value is >0.05 so, there is no which result better than other? $\endgroup$
    – blossom emerald
    Commented Oct 18, 2011 at 8:04

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