# Another t-statistic normality assumption

I'm using a one sample t-test to compare individual case specimens to a know sample. The sample is skewed data which contains the absolute value of differences between left/right continuous data. The t-statistic being calculated here is t=case1 - mean (sample) /sd (sample). Given that the sample size is very large (more than 4000) does the central limit come into play? As far as I understand it, the sample means approach normality, which I've tested by randomly sampling from the sample and plotting the means of those samples. Also, if the mean is approximately normal, is it still valid to use a 2-tail test? A friend of mine is arguing that we need to use a one tail since the sample is skewed in one direction.

Edited: I've run the same test using bootstrapping and transformation for normality both of which achieved very similar results to using the data as is.

• This is not clear to me. Are you trying to compare each individual observation to the known sample? Jan 26, 2017 at 15:02
• Yeah, where the individual comparison consists of a single absolute value of differences.
– JJL
Jan 26, 2017 at 15:23
• Do I understand correctly that you are comparing individual observations (specimens) to the mean of your sample using one sample t test? If so, then your approach is not valid in my opinion. What is your actual research question? Jan 26, 2017 at 15:34
• It's not so much a question as it is a method. The concept is to determine if the absolute value of differences is significantly different than what you would expect from the known sample. The actual comparison may be left data from an individual different than the right data, in which case it would come back as significantly different.
– JJL
Jan 26, 2017 at 15:41