# Absolute values -- parametric or non-parametric test?

I am looking at the absolute difference in values between pairs of numbers. I am then comparing these absolute values between two groups (e.g. Males and Females). Since these are absolute values I am not sure whether it is appropriate to use parametric statistics (e.g. Independent t-test) or whether it is better to be using non-parametric tests (i.e. Mann-Whitney U)?

Specifically I am worried that being absolute values (created by ignoring the signs) they cannot fit the assumption of normality needed?

The largest sample size I'm looking at is 19 in Group 1 and 17 in Group 2 (however there are also some smaller groups).

Any help would be appreciated (and explanations why would be great, as I'm trying to wrap my head around all this). Thanks!

• To clarify - in effect, you're testing for a difference in the spread of the two numbers between two groups? Sep 13, 2013 at 18:40
• @jbowman I'm not sure if I entirely understand your question. I am ultimately trying to test if there is a difference between the two groups (e.g. does one group tend to have larger absolute values than the other group?)
– Lexi
Sep 13, 2013 at 21:04
• The way I read the first paragraph, it looks like you have two numbers per observation, and you're calculating the absolute value of the difference between them. This absolute value is what you're trying to test for. If the numbers measure the same thing, e.g., observations of "something" before and after an event, the absolute value of the difference can be thought of as a measure of spread (variability). Sep 14, 2013 at 16:17

Well, one "Lexy" to another (even if you spell yours with an "I" ;):

You won't go wrong by using a nonparametric test. Without seeing the distributions of the absolute values of your actual data, one cannot really say whether normality is a good assumption or not. However, a paired test, like the sign-rank test, which is a test for stochastic dominance, with the null hypothesis H$_{0}\text{: P}(X_{\text{A}} > X_{\text{B}}) = 0.5$ with H$_{\text{A}}\text{: P}(X_{\text{A}} > X_{\text{B}}) \ne 0.5$, should work just fine. What is your sample size?

The first thing to understand is that the assumption of normality when using the t-test can never be proven. One can only say that the data look approximately normal, and if this is the case, the t-test may be appropriate.

So, even though the exclusion of negative values is technically a violation of the normality assumption, it might not be an important violation if the distribution of differences appears to be bell shaped. Of course, a bell shape for absolute values can happen only if the distribution of differences is centered pretty far enough away from zero so that the truncation point at zero is not obvious.

• A difference of absolute values can be centered on zero. Jul 20, 2014 at 22:26
• @Alexis, where are you going with that? Your comment does not negate anything that I stated. Jul 20, 2014 at 23:02
• It seems like the first sentence of your second paragraph is false, I wonder if you are conflating |A| - |B| (gives real values, could have a normal distribution), with |A - B| (has non-negative distribution). Since the paired t test is a one-sample test of d, the issue of non-negative d is a misstep. Jul 21, 2014 at 1:13