Charles Ragin’s Qualitative Comparative Analysis

1. Charles Ragin’s Qualitative Comparative Analysis • Boolean Algebra • Either A or B = A+B Example: To purchase this car (P) you need either cash (A) or credit (B) P=A+B A and B =A*B Example: To drive this car (D) you have to have both a license (L) and an insurance (I) D=L*I If you also have to buy the car then D=P*L*I D=(A+B)*L*I=A*L*I+B*L*I in words: You either have to have cash and a license and insurance or you have to have credit and a license and insurance Notation: not P is p (not A is a etc.) Example: p=a*b d=??? d=l+i

2. Charles Ragin’s Qualitative Comparative Analysis (cont.) • TRUTH TABLE • Example: Causes of regime Downfall • A=conflict between older and younger military officers • B= Death of a powerful dictator • C=CIA dissatisfaction with the regime •  0=No 1=Yes  • A B C D=Downfall # of cases (total=32 cases) • 0 0 0 0 9 • 1 0 0 1 2 • 0 1 0 1 4 • 0 0 1 1 3 • 1 1 0 1 2 • 0 1 1 1 7 • 1 0 1 1 2 • 1 1 1 1 3 • D=Abc+aBc+abC+ABc+aBC+AbC+ABC • D=A+B+C

3. Simplifying Boolean Expressions • RULE: Look for two terms that are the same except for the fact that one element is present in one and absent in the other. • E.g. ABC and ABc are such pair of terms. AB works with or without C. So C is irrelevant, and both can be simplified into AB. • Similarly, ABcd and ABcD are pairs and both can be simplified into ABc. • BUT ABC and aBc are not pairs, and nor are ABc and ABd. To find the pairs, • create a table of all the terms and find the pairs, indicate their simplified forms put 0 otherwise. (The table will have content only above the diagonal.) • See if any term is left without a pair. Write that down. That term will enter the final equation. • Take the simplified forms, create a table and look for pair again. • Repeat 2. • Repeat 3 and 4 until you have only terms with a single element. (With 3 causes you will have two tables. With 4 you will have three. In general, you will have k-1 tables where k is the number of causes.)

4. Simplifying D=Abc+aBc+abC+ABc+aBC+AbC+ABC

5. Checking Results • D=Abc+aBc+abC+ABc+aBC+AbC+ABC • Simplifies to • D=A+B+C • Checking Results • Create a table of the original terms by the terms in the simplified form. • Write a 1 if the simplified term is part of an original term (write 0 otherwise). • Does each original term have at least a 1?

6. Checking ResultsD=Abc+aBc+abC+ABc+aBC+AbC+ABCD=A+B+C

7. Charles Ragin’s Qualitative Comparative Analysis (cont.) • Example: Causes of successful strikes • A= booming product market • B= threat of sympathy strikes • C= Large Strike Fund • A B C S=Success Number of cases • 1 0 1 1 5 • 0 1 0 1 4 • 1 1 0 1 3 • 1 1 1 1 2 • 1 0 0 0 2 • 0 0 1 0 2 • 0 1 1 0 1 • 0 0 0 0 3 • S=AbC+aBc+ABc+ABC if both aBc and ABc work we can ignore A/a if B is present but C is absent ( c )Bc • if both AbC and ABC work we can ignore B/b if AC are present  AC • if both ABc  and ABC work we can ignore C/c if AB are present  AB • S=Bc+AC+AB • This further simplifies to S=Bc+AC (see later)

8. Simplifying S=AbC+aBc+ABc+ABC

9. Checking ResultsS=AbC+aBc+ABc+ABC  S=Bc+AC+AB

10. Simplifying even further • The previous table revealed that we can drop AB and still each original term will have at least one 1. So the best and final model is: • S=Bc+AC

11. Example: Causes of successfully passing the class • A= Having taken other sociology classes B= Having studied a lot (more than 3 hours a week) • C= Having attended the lectures regularly D= Having a roommate/friend who took this class before • A B C D S=Success Number (Total: 70) • 0 0 0 0 0 3 • 1 0 0 0 0 5 • 0 1 0 0 1 2 • 0 0 1 0 0 4 • 0 0 0 1 0 7 • 1 1 0 0 1 4 • 1 0 1 0 1 5 • 1 0 0 1 1 8 • 0 1 1 0 1 3 • 0 0 1 1 0 4 • 0 1 0 1 1 4 • 1 1 1 0 1 9 • 1 1 0 1 1 3 • 1 0 1 1 1 5 • 0 1 1 1 1 2 • 1 1 1 1 1 2 • S= aBcd+ABcd+ AbCd+AbcD+aBCd+ aBcD+ABCd+ABcD+AbCD+aBCD+ABCD  S=B+AC+AD

12. SimplifyingS= aBcd+ABcd+ AbCd+AbcD+aBCd+ aBcD+ABCd+ABcD+AbCD+aBCD+ABCD

15. CheckingS= aBcd+ABcd+ AbCd+AbcD+aBCd+ aBcD+ABCd+ABcD+AbCD+aBCD+ABCD B+AC+AD