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Can anyone give me some insight into where this calculation for the KS test statistic is going wrong (see figure 1)? I ran the test in SPSS and SAS as a check. I have used the same process for other data and gotten the correct KS statistic (see figure 2).

I suspect that the presence of duplicate values in the figure 1 data is operative. If this is the case, does anyone know how SPSS and SAS adjusts the KS test stat for dups?

Any assistance is appreciated.

Figure 1 enter image description here

Figure 2 enter image description here

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  • $\begingroup$ I failed to mention that I suspect that the presence of duplicate values in the figure 1 data is operative. If this is the case, does anyone know how SPSS and SAS adjusts the KS test stat for dups? $\endgroup$
    – Dan
    Aug 18 '18 at 23:18
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    $\begingroup$ Please merge your accounts so you can edit your original post. $\endgroup$
    – Glen_b
    Aug 18 '18 at 23:31
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    $\begingroup$ (1) It's not about making an adjustment, but calculating the empirical distribution function correctly. What proportion of observations in the Figure 1 data are less than or equal to 76? (2) What does the presence of duplicates in itself tell you about the null hypothesis of normality? $\endgroup$ Aug 19 '18 at 8:04
  • $\begingroup$ Not the question but this test is a poor test for normality: necessarily it's most sensitive to differences n the middle of the distribution, the opposite of what is usually helpful. Indeed all tests are poor tests of normality: using a normal quantile plot (normal probability plot, probit plot, yet other names) is typically a better idea. $\endgroup$
    – Nick Cox
    Oct 4 '21 at 11:24
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I was originally using this formula for the column entitled "Abs Diff" in my spreadsheet above:

enter image description here

However, after further research I found this formula which indeed matches the KS test statistic values produced by SPSS and SAS in all cases.

enter image description here

Thanks to everyone who viewed this post.

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It appears that the SPSS K-S Algorithm uses the sample mean and sample standard deviation in its determination of its F(x) value. I am looking at a simple example of a K-S test in Canavos (1984) - Applied probability and statistics, pp. 343-344, they show a sample of n = 16 observations where they calculate the KS test statistic.

They use a hypothetical population mean and sigma of mu = 985 and sigma = 50. The sample mean = 980.50 and a sample standard deviation of 61.29.

In the Canavos example, when they calculate their KS statistic using the m = 985 and sigma = 50 values, their hand calculated KS value is 0.1207. When I ran it in both SPSS and MINITAB, I got a value of 0.1750. So, I did an EXCEL hand calculation using the sample mean and sample standard deviation values of 980.50 and 61.29, respectively. Then I got the same value of 0.17488 ~ 0.175. So that is what is going on with software packages like SPSS and MINITAB.

v/r Kenneth Lewis Virginia State University Department of Psychology

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  • $\begingroup$ Oh, SPSS and MINITAB also use S(x) = (i - 1)/N instead of i/N for their S(x) values. I hope what I wrote makes sense. $\endgroup$ Oct 4 '21 at 4:16

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