Internal Assessment - Analysis and Statistics

[@geoprofi]

The normality tests are supposed to be used as a battery - some are more sensitive towards certain departures than others. If one test results in a p-value smaller than the threshold, you should reject the normality assumption and in your case the Shapiro-Francia test shows a p-value of 0 (meaning p < 0.0000), so you should do that. While the Shapiro-Wilk is the most popular, the Shapiro-Francia is generally better, as stated in the supporting text beneath the calculator: "in fact performs better than the Shapiro-Wilk test as it is more sensitive against most distributions even for sample sizes smaller than 50 [1]." With such small sample sizes (15 data points) most tests of normality would have very low power so non-rejection of normality is far from proof that the normality assumption is adequate. Given that, the fact that the Shapiro-Francia shows such an extreme p-value should be a great indicator that your data is likely not normally distributed. And in fact it is not. A simple histogram plot of your data points reveals it is closer to being uniformly distributed than it is to being normally distributed: imgur.com/a/4znB54v P.S. I'm the author of the normality calculator.

[@ibpsychsurvival-mistykarma6902]

Georgi Georgiev Hey Georgi, thank you for the feedback! I appreciate it :) I’ve always used Shapiro-Will, and was aware that with such small samples, normality tests are affected. So from what I’ve understood, the Shapiro-Francia would be less sensitive and hence more powerful than the Shapiro-wilk in the case of small samples?

[@geoprofi]

@@ibpsychsurvival-mistykarma6902 I think it goes 'more sensitive and hence more powerful', but yes, as this one example also demonstrates, it has greater power towards a greater number of departures from normality.

[@ibpsychsurvival-mistykarma6902]

Georgi Georgiev okay, got it! Thanks again 😀