Chapter 11, 12, 13, 14 and 16 Association at Nominal and Ordinal Level The Procedure in Steps.
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Transcript of Chapter 11, 12, 13, 14 and 16 Association at Nominal and Ordinal Level The Procedure in Steps.
Chapter 11, 12, 13, 14 and 16
Association at Nominal and Ordinal Level
The Procedure in Steps
The Procedure in Steps for Nominal Variables
Step 1: Make Tables
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.
Step 1: Make 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.
Step 1: Make TablesTitle
Rows Columns
Row 1 cell a cell b Row Marginal 1
Row 2 cell c cell d Row Marginal 2
Column Marginal 1
Column Marginal 2
N
Example of Table The bivariate table showing the relationship
between gender (columns) and party preference (rows).
Female Male
Labour 8 5 13
Conservatives 4 8 12
12 13 25
Example of Table The bivariate table showing the relationship
between gender (columns) and party preference (rows).
Female Male
Labour 66.7% 38.5%
Conservatives 33.3% 61.5%
100% (N=12)
100% (N=13)
Step 2: What is the pattern/direction of the association?
See percentages in Table. Female voters tend to have party
preference for Labour and male voters have party preference for Conservatives
This relationship does have a clear pattern
But is it also significant?
Step 3: Is the Association between the Variables Significant?
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).
Example of Computation
Use Formula 11.2 to find fe.
Multiply column and row marginals for each cell and divide by N. For Problem above
(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:
Female Male
Labour 6.24 6.76 13
Conservatives 5.76 6.24 12
12 13 25
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 3.10 .46
4 5.76 -1.76 3.10 .54
8 6.24 1.76 3.10 .50
25 25 0 χ2 = 2.00
Example of Computation See Chapter 11 of Healey (pp. 286-289) for
five-step model for Chi Square Test to find out whether variables are independent/ whether the association between the variables is significant or not
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
gender and party preference
Interpreting Chi Square The chi square test tells us only if the
variables are independent or not. 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
Step 4: If an Association does Exist, how Strong is it?
It is always useful to compute column percentages for bivariate tables.
But, it is also useful to have a summary measure – a single number – to indicate the strength of the relationship.
For nominal level variables, there are the following commonly used measures of association: Phi Cramer’s V Lambda
Nominal Measures: Phi See Healey, formula 13.1, p. 342 Phi is used for 2x2 tables. The formula for Phi:
Nominal Measures: Cramer’s V See Healey, formula 13.2, p. 343 Cramer’s V is used for tables larger than 2x2. Formula for Cramer’s V:
Nominal Measures: Lambda See Healey, formula 13.3, p. 348 When dependent and independent variables are clear Formula for Lambda:
Step 5: Is there still an Association, if Control Variables are Added?
See Chapter 16 in Healey See week 10 of this course
The Procedure in Steps for Ordinal Variables
Steps 1 ,2, 3, 5 are similar to those for nominal variables
Only step 4 is different, because you need other measures of association
Step 4: If an Association does Exist, how Strong is it?
For ordinal level variables, there are the following commonly used measures of association: Spearman’s Rho (if there is ranking of
scores, see Healey pp. 376-382) Gamma (Formula 14.1, see Healey, p.
366). Is the strength of the relationship significant? Test whether gamma is significant. See Healey, pp. 380-381
An Ordinal Measure: Gamma
interpret the strength of gamma. e.g. gamma is 0.61. This is a strong association. In
addition to strength, gamma also identifies the direction of the relationship.
This is a positive relationship: e.g. as education increases, income increases.
In a negative relationship, the variables would change in the different direction.
Test whether the strength of Gamma is significant
Value Strength
Between 0.0 and
0.30Weak
Between 0.30 and
0.60Moderate
Greater than 0.60
Strong