Skewed, non-normal data and central tendency tests

#1
Hi everyone, I have about 10,000 observations for a variable with a skewed, non-normal distribution (min=-2000, max=3000, mean=-0.30, median=0.003). I would like to perform a t-test against the null that the mean = 0. Although the sample is large, the non-normality gives me pause, which leads to a Wilcoxon Signed Rank test. Problem here is the signs of the t-test and the Signed Rank Test are opposite. Under consideration now are: 1) performing these two tests on normalized data (i.e., all values between 0 and 1); and 2) a randomization test. I'm not sayiing that these two considerations are necessarily appropriate. So, before exhausting the necessary resources for these two considerations, I would like to hear the opinions of the members of this forum for how they would proceed. I am grateful for any insight. Thank you, Rick
 

Miner

TS Contributor
#2
With that large a sample size (n=10,000), you are perfectly safe using a t-test. Normality is only a concern with small sample sizes (e.g., n ~< 10).
 

hlsmith

Less is more. Stay pure. Stay poor.
#3
Agreed. Make a histogram for your data as well - since given the closeness of mean and median, they don't see terribly skewed base on those markers.
 
#4
A histogram reveals that the distribution is very leptokurtic. Any insight for the conflicting signs between the t-test and the Signed Rank test?
 

Karabiner

TS Contributor
#5
Are you absolutely sure that one test says A > B and the other says A < B?
Sometimes there are coding errors or interpretation errors.

With kind regards

Karabiner
 

Miner

TS Contributor
#6
The two tests are checking different things. The t-test is a means test. The Wilcoxon signed rank test is not. It is median test of the differences.
 

hlsmith

Less is more. Stay pure. Stay poor.
#7
Well your mean is negative and median is positive, does that answer your question? Do you need a test for this comparison to zero? Couldnt you just report a histogram with the mean and median markered. What does a formal test really solve here. Also, are you saying they are opposite via something like two pvalue that are pretty much the same, but one marginally passes a cutoff?