1. Chapter 11 Hypothesis Testing IV (Chi Square)

2. Chapter Outline  Introduction  Bivariate Tables  The Logic of Chi Square  The Computation of Chi Square  The Chi Square Test for Independence  The Chi Square Test: An Example

3. Chapter Outline  An Additional Application of the Chi Square Test: The Goodness-of-Fit Test  The Limitations of the Chi Square Test  Interpreting Statistics: Family Values and Social Class

4. In This Presentation  The basic logic of Chi Square.  The terminology used with bivariate tables.  The computation of Chi Square with an example problem.  The Five Step Model

5. Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant differences between the actual cell frequencies in a table (f o ) and those that would be expected by random chance (f e ).

6. Tables  Must have a title.  Cells are intersections of columns and rows.  Subtotals are called marginals.  N is reported at the intersection of row and column marginals.

7. Tables  Columns are scores of the independent variable. There will be as many columns as there are scores on the independent variable.  Rows are scores of the dependent variable. There will be as many rows as there are scores on the dependent variable.

8. Tables  There will be as many cells as there are scores on the two variables combined.  Each cell reports the number of times each combination of scores occurred.

9. Tables Title RowsColumns  Row 1cell acell bRow Marginal 1 Row 2cell ccell dRow Marginal 2 Column Marginal 1 Column Marginal 2 N

10. Example of Computation  Problem 11.2 Are the homicide rate and volume of gun sales related for a sample of 25 cities?

11. Example of Computation  The bivariate table showing the relationship between homicide rate (columns) and gun sales (rows). This 2x2 table has 4 cells. LowHigh 8513 Low4812 1325

12. Example of Computation  Use Formula 11.2 to find f e.  Multiply column and row marginals for each cell and divide by N. For Problem 11.2  (13*12)/25 = 156/25 = 6.24  (13*13)/25 = 169/25 = 6.76  (12*12)/25 = 144/25 = 5.76  (12*13)/25 = 156/25 = 6.24

13. Example of Computation  Expected frequencies: LowHigh 6.246.7613 Low5.766.2412 1325

14. Example of Computation  A computational table helps organize the computations. fofo fefe f o - f e (f o - f e ) 2 (f o - f e ) 2 /f e 86.24 56.76 45.76 86.24 25

15. Example of Computation  Subtract each f e from each f o. The total of this column must be zero. fofo fefe f o - f e (f o - f e ) 2 (f o - f e ) 2 /f e 86.241.76 56.76-1.76 45.76-1.76 86.241.76 25 0

16. Example of Computation  Square each of these values fofo fefe f o - f e (f o - f e ) 2 (f o - f e ) 2 /f e 86.241.763.10 56.76-1.763.10 45.76-1.763.10 86.241.763.10 25 0

17. Example of Computation  Divide each of the squared values by the f e for that cell. The sum of this column is chi square fofo fefe f o - f e (f o - f e ) 2 (f o - f e ) 2 /f e 86.241.763.10.50 56.76-1.763.10.46 45.76-1.763.10.54 86.241.763.10.50 25 0 χ 2 = 2.00

18. Step 1 Make Assumptions and Meet Test Requirements  Independent random samples  LOM is nominal Note the minimal assumptions. In particular, note that no assumption is made about the shape of the sampling distribution. The chi square test is non- parametric.

19. Step 2 State the Null Hypothesis  H 0 : The variables are independent  Another way to state the H 0, more consistent with previous tests: H 0 : f o = f e

20. Step 2 State the Null Hypothesis  H 1 : The variables are dependent  Another way to state the H 1 : H 1 : f o ≠ f e

21. Step 3 Select the S. D. and Establish the C. R.  Sampling Distribution = χ 2  Alpha =.05  df = (r-1)(c-1) = 1  χ 2 (critical) = 3.841

22. Calculate the Test Statistic  χ 2 (obtained) = 2.00

23. Step 5 Make a Decision and Interpret the Results of the Test  χ 2 (critical) = 3.841  χ 2 (obtained) = 2.00  The test statistic is not in the Critical Region. Fail to reject the H 0.  There is no significant relationship between homicide rate and gun sales.

24. Interpreting Chi Square  The chi square test tells us only if the variables are independent or not.  It does not tell us the pattern or nature of the relationship.  To investigate the pattern, compute %s within each column and compare across the columns.

25. Interpreting Chi Square  Cities low on homicide rate were high in gun sales and cities high in homicide rate were low in gun sales.  As homicide rates increase, gun sales decrease. This relationship is not significant but does have a clear pattern. LowHigh 8 (66.7%)5 (38.5%)13 Low4 (33.3%)8 (61.5%)12 12 (100%)13 (100%)25

26. The Limits of Chi Square  Like all tests of hypothesis, chi square is sensitive to sample size. As N increases, obtained chi square increases. With large samples, trivial relationships may be significant.  Remember: significance is not the same thing as importance.