1. Hypothesis Testing IV (Chi Square)
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
Low homicide
High homicide
Totals
Low gun sales 8 5 13
High gun sales 4 12 25

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 Title Rows Column 1 Column 2 Row 1 cell a cell b Row Marginal 1
cell c
cell d
Marginal 2
Column N

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.
Low
High 8 5 13 4 12 25

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:
Low
High
6.24
6.76 13
5.76 12 25

10. Example of Computation
A computational table helps organize the computations. fo fe
fo - fe
(fo - fe)2
(fo - fe)2 /fe 8
6.24 5
6.76 4
5.76 25

11. Example of Computation
Subtract each fe from each fo. The total of this column must be zero. fo fe
fo - fe
(fo - fe)2
(fo - fe)2 /fe 8
6.24
1.76 5
6.76
-1.76 4
5.76 25

12. Example of Computation
Square each of these values fo fe
fo - fe
(fo - fe)2
(fo - fe)2 /fe 8
6.24
1.76
3.10 5
6.76
-1.76 4
5.76 25

13. Example of Computation
Divide each of the squared values by the fe for that cell. The sum of this column is chi square fo fe
fo - fe
(fo - fe)2
(fo - fe)2 /fe 8
6.24
1.76
3.10
.50 5
6.76
-1.76
.46 4
5.76
.54 25
χ2 = 2.00

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 . The apparent pattern may be sampling error.
Low
High
8 (66.7%)
5 (38.5%) 13
4 (33.3%)
8 (61.5%) 12
12 (100%)
13 (100%) 25

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