# Difference between two groups of people, each person "is" several characteristics

I have two groups of people, A and B (let's say 15 and 25 people).

Each person in each group is characterized by a bucket of features (bucket = 6-18 features). Each feature, during qualitative phase of analysis, is assigned to a single category.

So let's say person number 1:
— belongs to group A
— has features "f1", "f3", "f5", ... (6 in total)

And person number 2:
— belongs to group B
— has features "f3", "f7", "f9", ... (8 in total)

And so on.

Now, the question is: how to prove statistically that two groups are different.

My idea is to "pour" all the features of all the people of group A into basket A. Then to "pour" all the features of all the people of group B into basket B. And then to compare two baskets using chi-square.

Do I miss anything with such approach? Is there a way to compare groups without ignoring the fact that "features" are naturally "grouped" into "persons"?

The best way to deal with this analyse I see is the set a data table looking like this:

ind Gr f1 f2 f3 f4 f5 ..................

1 A 1 0 1 0 1 ...................

2 B 0 0 1 0 1 ...................

Then each person has a modality for all the feature. After that, you can study the characteristic of all feature in each group.

You can use a PCA on the feature to compute an indicator. After you do a t-test on the indicator and the group.

@Abdoul_Haki has suggested something which is to me reminiscent of Correspondence Analysis and proposed a t-test on an indicator. I think Correspondence Analysis will be quite useful as a descriptive method. Concerning the t-test on an indicator, I believe you are better served by a Hotelling's t-test. Basically you code each feature with 0 (absence) or 1 (presence) and perform a test of equality of the mean vectors.

The test requires normality (and variables which are 0-1 clearly violate it) but is quite robust. On the other hand, you can use a permutation test for the same test statistic if you fear Hotelling's approximation might be inadecuate (which need not be, given your group sizes).