1. STATISTICS ELEMENTARY MARIO F. TRIOLA
Section Contingency Tables: Independence and Homogeneity
MARIO F. TRIOLA
EIGHTH
EDITION

2. Definition Contingency Table (or two-way frequency table)
a table in which frequencies correspond to two variables.
(One variable is used to categorize rows, and a second variable is used to categorize columns.)
page 589 of text

3. Definition Contingency Table (or two-way frequency table)
a table in which frequencies    correspond to two variables.
(One variable is used to categorize rows, and a second variable is used to categorize columns.)
Contingency tables have at least two rows and at least two columns.
Emphasize the difference between a multinomial table and a contingency table is the number of rows.

4. Definition Test of Independence
tests the null hypothesis that the    row variable and column variable in    a contingency table are not related. (The null hypothesis is the statement that the row and column    variables are independent.)
page 590 of text
One way to explain why the null is always the variables are independent is to say ‘independence means no (= 0) relationship’. This will relate to previous problems where the null has equality.

5. Assumptions 1. The sample data are randomly selected.
2. The null hypothesis H0 is the statement that the row and column variables are         independent; the alternative hypothesis H1 is     the statement that the row and column     variables are dependent.
3. For every cell in the contingency table, the expected frequency E is at least 5. (There is no requirement that every observed frequency must be at least 5.)
bottom of page 590

6. Test of Independence Test Statistic
X2 = 
(O - E)2 E
Critical Values
1. Found in Table A-4 using
degrees of freedom = (r - 1)(c - 1)
r is the number of rows and c is the number of columns
2. Tests of Independence are always right-tailed.
page 591 of text
Same chi-square formula as for multinomial tables.

7. E = Total number of all observed frequencies in the table
(row total) (column total)
(grand total)
Total number of all observed frequencies in the table
page 592 of text
Example of finding an E value on this page.

8. Tests of Independence
H0: The row variable is independent of the column variable
H1: The row variable is dependent (related to) the column variable
This procedure cannot be used to establish a direct cause-and-effect link between variables in question.
Dependence means only there is a relationship between the two variables.

9. Expected Frequency for Contingency Tables
page 593 of text

10. Expected Frequency for Contingency Tables
row total column total
grand total
grand total
grand total

11. Expected Frequency for Contingency Tables
row total column total
grand total
grand total
grand total
n • p
(probability of a cell)
The probability of one cell is computed by an application of the multiplication rule for independent events: P(A and B) = P(A)*P(B). See Chapter 3, Section 3-4, page
The Expected Value (Chapter 4, Section 4-2, page 189) of one cell is found by n times p.

12. Expected Frequency for Contingency Tables
row total column total
grand total
grand total
grand total
n • p
(probability of a cell)

13. Expected Frequency for Contingency Tables
row total column total
grand total
grand total
grand total
n • p
(probability of a cell)
E =
(row total) (column total)
(grand total)

14. Is the type of crime independent of whether the criminal is a stranger?12 39
379
106
727
642
Homicide
Robbery
Assault
Stranger
Acquaintance
or Relative
Exercise #11 on page 601.

15. Is the type of crime independent of whether the criminal is a stranger?
Assault
Row Total
Homicide
Robbery
1118
787
1905 12 39 51
379
106
485
727
642
1369
Stranger
Acquaintance
or Relative
Column Total
Row, column, and the grand total must be calculated first.

16. Is the type of crime independent of whether the criminal is a stranger?
Robbery
Assault
Row Total
Homicide
1118
787
1905 12 39 51
379
106
485
727
642
1369
Stranger
Acquaintance
or Relative
Column Total
E =
(row total) (column total)
(grand total)

17. Is the type of crime independent of whether the criminal is a stranger?
Robbery
Assault
Row Total
Homicide
1118
787
1905 12 39 51
379
106
485
727
642
1369
Stranger
Acquaintance
or Relative
(29.93)
Column Total
E =
(row total) (column total)
(grand total)
The Expected Value (E) must be computed for each cell using the correct row total and column total for that individual cell.
E =
(1118)(51)
= 29.93
1905

18. Is the type of crime independent of whether the criminal is a stranger?
Homicide
Robbery
Assault
Row Total
1118
787
1905 12 39 51
379
106
485
727
642
1369
Stranger
Acquaintance
or Relative
(29.93)
(284.64)
(803.43)
(21.07)
(200.36)
(565.57)
Column Total
E =
(row total) (column total)
(grand total)
E =
(1118)(51)
E =
(1118)(485)
= 29.93 =
1905
1905
etc.

19. Is the type of crime independent of whether the criminal is a stranger?
X2 = 
(O - E )2 E
Homicide
Robbery
Forgery
(E) 12 39
379
106
727
642
(284.64)
(803.43)
Stranger
Acquaintance
or Relative
(29.93)
(O - E )2
[10.741] E
(21.07)
(200.36)
(565.57
The relative magnitude - (O-E)2/E - of each cell must next be computed.
(O -E )2
( )2
Upper left cell: = = E
29.93

20. Is the type of crime independent of whether the criminal is a stranger?
X2 = 
(O - E )2 E
Homicide
Robbery
Forgery
(E) 12 39
379
106
727
642
(29.93)
(284.64)
[31.281]
(803.43)
[7.271]
Stranger
Acquaintance
or Relative
(O - E )2
[10.741] E
(21.07)
[15.258]
(200.36)
[44.439]
(565.57)
[10.329]
(O -E )2
( )2
Upper left cell: = = E
29.93

21. Is the type of crime independent of whether the criminal is a stranger?
X2 = 
(O - E )2 E
Homicide
Robbery
Forgery
(E) 12 39
379
106
727
642
(29.93)
(284.64)
[31.281]
(803.43)
[7.271]
Stranger
Acquaintance
or Relative
(O - E )2
[10.741] E
(21.07)
[15.258]
(200.36)
[44.439]
(565.57)
[10.329]
The sum of each cell’s relative magnitude gives the test statistic chi-square value.
X2 = =
Test Statistic

22. X2 = 119.319 X2 = 5.991 (from Table A-4) Test Statistic Critical Value
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
X2 = (from Table A-4)
Critical Value
The test statistic chi-square values need to be compared with the chi-square critical value found in Table A-4.

23. X2 = 119.319 X2 = 5.991 (from Table A-4) Test Statistic Critical Value
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
X2 = (from Table A-4)
Critical Value
Fail to Reject
Independence
Reject
Independence
 = 0.05
X2 = 5.991
Sample data: X2 =

24. X2 = 119.319 X2 = 5.991 (from Table A-4) Test Statistic Critical Value
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
X2 = (from Table A-4)
Critical Value
Fail to Reject
Independence
Reject
Independence
 = 0.05
Reject independence
X2 = 5.991
Sample data: X2 =

25. X2 = 119.319 X2 = 5.991 (from Table A-4) Test Statistic Critical Value
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
X2 = (from Table A-4)
Critical Value
Fail to Reject
Independence
Reject
Independence
 = 0.05
Reject independence
X2 = 5.991
Sample data: X2 =
Claim: The type of crime and knowledge of criminal are independent
Ho : The type of crime and knowledge of criminal are independent
H1 : The type of crime and knowledge of criminal are dependent

26. X2 =
Test Statistic
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of freedom
X2 = (from Table A-4)
Critical Value
Fail to Reject
Independence
Reject
Independence
 = 0.05
Reject independence
X2 = 5.991
Sample data: X2 =
It appears that the type of crime and knowledge of the criminal are related.

27. Relationships Among Components in X2 Test of Independence
Figure 10-8
page 595 of text

28. Definition Test of Homogeneity
test the claim that different populations have the same proportions of some characteristics
page 596 of text

29. How to distinguish between a test of homogeneity and a test for independence:
Were predetermined sample sizes used for different populations (test of homogeneity), or was one big sample drawn so both row and column totals were determined randomly (test of independence)?
The key to identifying it is a test of homogeneity is the predetermined sample sizes.
See the discussion in the middle of page 596 and the example at the bottom of the page
Problem #5 and 6 are problems involving tests of homogeneity.