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Chapter 11

Chapter 11. Hypothesis Testing IV (Chi Square). 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 ).

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Chapter 11

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  1. Chapter 11 Hypothesis Testing IV (Chi Square)

  2. 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 (fo) and those that would be expected by random chance (fe).

  3. The relationship of homicide rate and gun sales

  4. Tables • Notice the following about these tables 1. Table must have a title 2. Independent vrble must go into columns and if percentaged, must percentage within columns 3. Subtotals are called marginals. 4. N is reported at the intersection of row and column marginals.

  5. Tables

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

  7. Example of Computation • The bivariate table showing the relationship between homicide rate (columns) and gun sales (rows). This 2x2 table has 4 cells.

  8. Example of Computation • Use Formula 11.2 to find fe. • 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

  9. Example of Computation • Expected frequencies:

  10. Example of Computation • A computational table helps organize the computations.

  11. Example of Computation • Subtract each fe from each fo. The total of this column must be zero.

  12. Example of Computation • Square each of these values

  13. Example of Computation • Divide each of the squared values by the fe for that cell. The sum of this column is chi square

  14. 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 distribution of the parameters. The chi square test is non-parametric.

  15. Step 2 State the Null Hypothesis • H0: The variables are independent • Another way to state the H0, more consistent with previous tests: • H0: fo = fe

  16. Step 2 State the Null Hypothesis • H1: The variables are dependent • Another way to state the H1: • H1: fo ≠ fe

  17. Step 3 Select the S. D. and Establish the C. R. • Sampling Distribution = χ2 • Alpha = .05 • df = (r-1)(c-1) = 1 • χ2 (critical) = 3.841

  18. Calculate the Test Statistic • χ2 (obtained) = 2.00

  19. Step 5 Make a Decision and Interpret the Results of the Test • χ2 (critical) = 3.841 • χ2 (obtained) = 2.00 • The test statistic is not in the Critical Region. Fail to reject the H0. • There is no significant relationship between homicide rate and gun sales.

  20. 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.

  21. Interpreting Chi Square • Cities low on homicide rate were low in gun sales and cities high in homicide rate were high in gun sales. • As homicide rates increase, gun sales increase. This relationship is not significant but does have a clear pattern.But the pattern may be just sampling error.

  22. 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.

  23. Additional limits • If there are more than four categories in either variable, the use of chi square is questionable. • If one of the cells has a frequency less than 5 (as in our example), the use of chi square is questionable

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