I have two populations with n=18 and I'm trying to find out if it makes sense to compare them with a t-test. I ran a Shapiro-Wilk test in SigmaPlot 12.5 for both populations seperately and these are the results:

population1:    W-Statistic = 0.900       P  = 0.057    Passed
population2:    W-Statistic = 0.912       P  = 0.094    Passed

However, if I'm trying to run a t-test, it says:

Normality Test (Shapiro-Wilk)   Failed  (P = 0.003)

Here it seems that there is only one P-value for both populations, which is a bit confusing to me. Does anyone have an idea how the P-value might have been calculated here and why it can be that low, even if it is much higher for both populations tested seperately?

This is the underlying data...

pop1:       pop2:
6.0696      6.4659
6.8833      6.2842
5.9243      5.9193
6.5391      7.526
7.2505      6.71
6.4299      7.3117
4.9903      13.5116
4.8506      9.1565
4.7737      7.7016
6.9384      8.5998
6.6842      9.2543
6.614       10.2234
6.3128      9.7079
6.3533      7.8677
6.2728      8.7079
7.4372      9.405
7.2657      8.5998
7.1165      8.3411
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  • $\begingroup$ Please explain what you were doing when you were "trying to run a t-test." For instance, were you pairing these data or not? Which of the many possible t-tests were you carrying out? $\endgroup$ – whuber Dec 29 '16 at 21:17
  • $\begingroup$ I was running the t-test in SigmaPlot 12.5, where I used the option “compare two groups” -> “t-test”. The t-test is not further specified, I was just trying to test if the two populations have significant different means. As the program does the normality test by default before running the t-test, I thought this would be a kind of standard procedure to do. My understanding was that I should test both populations seperately for normality. However, it seems that the program paired the data for the normality test and I was wondering if this might be reasonable? $\endgroup$ – Matthias Dec 30 '16 at 11:19
  • $\begingroup$ I expect the test was applied to residuals (in this case, data minus each samples own mean). $\endgroup$ – Glen_b -Reinstate Monica Dec 31 '16 at 1:18

Imagine samples from two normally-distributed populations with vastly different means. Pooling the samples thus ignoring which group the samples were from would give you a bimodal and decidedly non-normal distribution.

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  • $\begingroup$ The OP after editing did the Shapiro-Wilk tests are done separately so there is no issue of pooling the samples. The only assumption in the t test is that the two populations have the same variance. Under that assumption the pooled estimate of variance can be justified because it provides an estimate of the common variance under the null hypothesis. $\endgroup$ – Michael R. Chernick Dec 29 '16 at 22:44
  • $\begingroup$ Now what I have said so far is what would be done for two independent normal samples. However since the OP has not specified that and the sample sizes are equal the observations can be correlated in pairs and the paired t test could be applied. The assumption of independence can't be ignored because the paired data can be viewed as n independent normals (the paired differences are normally distribution with a common variance). The t statistic has a t distribution with n-1 df when the null hypothesis holds. In the independent sample case it has 2n-2 degrees of freedom. $\endgroup$ – Michael R. Chernick Dec 29 '16 at 22:53
  • $\begingroup$ In this example pairing leads to a t with 17 df while the unpaired independent case gives a t with 34 df. That was Bill Huber's point. $\endgroup$ – Michael R. Chernick Dec 29 '16 at 22:56

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