Chapter 11 Hypothesis Testing IV (Chi Square). Chapter Outline Introduction Bivariate Tables The Logic of Chi Square The Computation of Chi Square.

Chapter 11

Hypothesis Testing IV (Chi Square)

Chapter Outline

Introduction Bivariate Tables The Logic of Chi Square The Computation of Chi Square The Chi Square Test for Independence The Chi Square Test: An Example

Chapter Outline

An Additional Application of the Chi Square Test: The Goodness-of-Fit Test

The Limitations of the Chi Square Test Interpreting Statistics: Family Values

and Social Class

In This Presentation

The basic logic of Chi Square. The terminology used with bivariate

tables. The computation of Chi Square with

an example problem. The Five Step Model

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).

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.

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.

Tables

There will be as many cells as there are scores on the two variables combined.

Each cell reports the number of times each combination of scores occurred.

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 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.

Low High

High 8 5 13

Low 4 8 12

12 13 25

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

Example of Computation Expected frequencies:

Low High

High 6.24 6.76 13

Low 5.76 6.24 12

12 13 25

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

8 6.24

25 25

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 -1.76

8 6.24 1.76

25 25 0

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 3.10

4 5.76 -1.76 3.10

8 6.24 1.76 3.10

25 25 0

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

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 sampling distribution. The chi square test is non-parametric.

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

Step 2 State the Null Hypothesis

H1: The variables are dependent Another way to state the H1:

H1: fo ≠ fe

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

Calculate the Test Statistic

χ2 (obtained) = 2.00

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.

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 high in gun sales

and cities high in homicide rate were low in gun sales.

As homicide rates increase, gun sales decrease. This relationship is not significant but does have a clear pattern.

Low High

High 8 (66.7%) 5 (38.5%) 13

Low 4 (33.3%) 8 (61.5%) 12

12 (100%) 13 (100%) 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.