1
$\begingroup$

I want to compare the proportions of 2 groups statistically. My data look like this (the actual dataset contains 700,000 rows):

GROUP A GROUP B GROUP C  
0.7     0.2     0.1  
0.3     0.4     0.3  
0.2     0.23    0.57  
…       …       …  

So every row sums to 1. Moreover, the data are paired in the sense that every row is a proportion for the same subject.

Now I want to know whether the average proportion of GROUP A is significantly higher than the average proportion of GROUP B. Does anybody have an idea about the appropriate test for this? I could not find a specific test for this problem so far on the internet. I assume that a regular t-test is not appropriate in this case because I'm working with proportions.

$\endgroup$

1 Answer 1

1
$\begingroup$

You are comparing means of two quantitative variables, not two proportions, although the items in your population are proportions. Therefore, your problem is in the scope of t-test.

Anyway, your two populations (group A and group B) are not independent, at least because some combinations are impossible, and they aren't normally distributed because range is bounded to a short interval - short compared to the standard deviation I would expect from your 3 rows example. Here might arise the concern about whether t-test assumptions are met - strictly, they aren't - and if that will affect t-test outcomes. Since your population is huge and your distribution don't seem to be highly skewed I would say t-test will work fine but I'd do the following:

  • Check for big deviations from normality: an histogram and a mesure of kurtosis and skewness might be enough if the histogram looks fine and kurtosis and skewness aren't far from those of a normal (let's say not farther than 2 times).
  • Perform a t-test.
  • Perform a non parametric test with the same data, as the Wilcoxon rank-sum test.

In fact, if t-test and Wilcoxon rank-sum test agree on similar p-values you can even skip the first step. If they don't agree, probably a more deep check should be needed to see what's happening with your data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.