# Shapiro-Wilk test and t-test

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

• 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?
– whuber
Dec 29, 2016 at 21:17
• 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? Dec 30, 2016 at 11:19
• I expect the test was applied to residuals (in this case, data minus each samples own mean). Dec 31, 2016 at 1:18