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
).

The relationship of homicide rate and gun sales
Totals Low homicide
8
High homicide
5 13 Low gun sales
High gun sales
Totals
4
12
8
13
12
25

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.

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

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

Example of Computation
 The bivariate table showing the relationship between homicide rate (columns) and gun sales
(rows). This 2x2 table has 4 cells.
High
Low
Low
8
4
12
High
5
8
13
13
12
25

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

Example of Computation
 Expected frequencies:
High
Low
Low
6.24
5.76
12
High
6.76
6.24
13
13
12
25

5
4 f o
8
8
25
Example of Computation
 A computational table helps organize the computations.
f o
- f e
(f o
- f e
) 2 (f o
- f e
) 2 /f e f e
6.24
6.76
5.76
6.24
25

5
4 f o
8
8
25
Example of Computation
 Subtract each f e from each f o
. The total of this column must be zero.
(f o
- f e
) 2 (f o
- f e
) 2 /f e f e
6.24
6.76
5.76
6.24
25 f o
- f e
1.76
-1.76
-1.76
1.76
0

Example of Computation
 Square each of these values
5
4 f o
8
8
25 f e
6.24
6.76
5.76
6.24
25 f o
- f e
1.76
-1.76
-1.76
1.76
0
(f o
- f e
) 2
3.10
3.10
3.10
3.10
(f o
- f e
) 2 /f e

5
4 f o
8
8
25
Example of Computation
 Divide each of the squared values by the f cell. The sum of this column is chi square e for that f e
6.24
6.76
5.76
6.24
25 f o
- f e
1.76
-1.76
-1.76
1.76
0
(f o
- f e
) 2
3.10
3.10
3.10
3.10
(f o
- f e
) 2 /f e
.50
.46
.54
.50
χ 2 = 2.00

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.

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

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
:

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

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

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.

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.

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
Low
8 (66.7%)
4 (33.3%)
12 (100%)
High
5 (38.5%)
8 (61.5%)
13 (100%)
13
12
25

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.

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