Hypothesis Testing IV (Chi Square)

1
Chapter 11

• Hypothesis Testing IV (Chi Square)

2
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

3
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

4
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

5
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).

6
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.

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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.

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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.

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Tables
Title Title Title
Rows Columns ? Columns ? Columns ? Columns ?
Row 1 cell a cell b Row Marginal 1 Row Marginal 1
Row 2 cell c cell d Row Marginal 2 Row Marginal 2
Column Marginal 1 Column Marginal 2 N N
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Example of Computation

• Problem 11.2
• Are the homicide rate and volume of gun sales
  related for a sample of 25 cities?

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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
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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
• (1312)/25 156/25 6.24
• (1313)/25 169/25 6.76
• (1212)/25 144/25 5.76
• (1213)/25 156/25 6.24

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Example of Computation

• Expected frequencies

Low High
High 6.24 6.76 13
Low 5.76 6.24 12
12 13 25
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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
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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
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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
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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 ?2 2.00
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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.

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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

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Step 2 State the Null Hypothesis

• H1 The variables are dependent
• Another way to state the H1
• H1 fo ? fe

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Step 3 Select the S. D. and Establish the C. R.

• Sampling Distribution ?2
• Alpha .05
• df (r-1)(c-1) 1
• ?2 (critical) 3.841

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Calculate the Test Statistic

• ?2 (obtained) 2.00

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Step 5 Make a Decision and Interpret the Results
of the Test

• ?2 (critical) 3.841
• ?2 (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.

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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.

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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
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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.