6 Qualitative comparative analysis
An intuitive way to identify causality is to search for patterns in the data of which we believe can be interpreted in a causal fashion. The qualitative comparative analysis (QCA) offers a framework to identify, based on set theory, relationships in terms of necessary conditions and sufficient conditions for a specific outcome.
Suppose I ask my son who he wants to invite to his birthday party, and he replies, “Everybody.” Since this isn’t helpful, I give him a list of eight children I’d consider inviting and ask him to pick which ones he seriously wants to invite. From this list, he selects five children.
Later, I ask my wife why he chose those five. She explains that he invited everyone from his kindergarten group and older girls. As a scientist, I want to logically verify her explanation, so I collected the data shown in Table 6.1. The table shows a so called truthtable which lists all names, with binary values (“0” or “1”) indicating if a child is female, older, or in kindergarten (where “1” means true and “0” means false).
| Name (number) | Female (F) | Age (A) | Kindergarten (K) | Invited (Y) | QCA notation |
|---|---|---|---|---|---|
| Jürgen (1) | 0 | 0 | 0 | 0 | fak |
| Ben (2) | 0 | 1 | 0 | 0 | fAk |
| Ida (3) | 1 | 0 | 0 | 0 | Fak |
| Carl (4 | 0 | 0 | 1 | 1 | faK |
| Paul (5) | 0 | 1 | 1 | 1 | fAK |
| Marie (6) | 1 | 0 | 1 | 1 | FaK |
| Anna (7) | 1 | 1 | 0 | 1 | FAk |
| Pia (8) | 1 | 1 | 1 | 1 | FAK |
Applying QCA to this truth table is straightforward. The basic idea is:
- Look for cases with the outcome \(Y = 1\) (invited) that differ in only one condition.
- When two such cases differ only in one condition, that condition is irrelevant for producing the outcome (in this data) and can be dropped.
- Continue simplifying until you cannot reduce the conditions any further without losing the outcome \(Y = 1\).
In our example:
- Consider the invited children (those with \(Y = 1\)). Cases (4) and (5) have configurations faK and fAK. They differ only in \(A\) (age), so \(A\) does not appear to be important here. We keep \(fK\) as the relevant conjunction.
- Cases (6) and (8) have configurations FaK and FAK. Again, they differ only in \(A\), so \(A\) is not important in this pair either. We keep \(FK\) as the relevant conjunction.
- Looking at cases (7) and (8), with configurations FAk and FAK, we see that they differ only in \(K\) (kindergarten). Hence, \(K\) is not important in this pair, and we keep \(FA\) as the relevant conjunction.
At this point, we have the reduced configurations \(fK\), \(FK\), and \(FA\) for the invited children. Comparing \(fK\) and \(FK\), we see that they differ only in \(F\) (female). Therefore, \(F\) is not relevant for this group either, and we can simplify \(fK\) and \(FK\) to just \(K\).
We are left with two combinations that can explain all invitations: \(K\) and \(FA\). In QCA notation, we write:
\[ Y = K + FA \]
This means that there are two sufficient conditions for being invited:
- being in the same kindergarten (\(K = 1\)), or
- being both a girl (\(F = 1\)) and older (\(A = 1\)).
Now that we have explained the invitations, we should check whether these conditions contradict the cases where the outcome is \(Y = 0\) (not invited). Looking at cases (1) to (3), we see that none of them satisfies either \(K = 1\) or the combination \(FA = 1\). Therefore, there is no contradiction: all non-invited children lack both sufficient conditions we identified for being invited.