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Chapter 11 Hypothesis Testing IV (Chi Square)

Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic Chi Square is a test of significance based on bivariate tables. We are looking for significant

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Page 1: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

Chapter 11

Hypothesis Testing IV (Chi Square)

Page 2: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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).

Page 3: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

The relationship of homicide rate and gun sales

Low homicide

High homicide

Totals

Low gun sales

8 5 13

High gun sales

4 8 12

Totals 12 13 25

Page 4: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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 columns3. Subtotals are called marginals.4. N is reported at the intersection of row and

column marginals.

Page 5: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

TablesTitle

Rows Column 1 Column 2

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

Page 6: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

Example of Computation

Problem 11.2 Are the homicide rate and volume of gun

sales related for a sample of 25 cities?

Page 7: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 8: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 9: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

Example of Computation Expected frequencies:

Low High

High 6.24 6.76 13

Low 5.76 6.24 12

12 13 25

Page 10: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 11: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 12: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 13: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 14: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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.

Page 15: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 16: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

Step 2 State the Null Hypothesis

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

H1: fo ≠ fe

Page 17: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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

Page 18: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

Calculate the Test Statistic

χ2 (obtained) = 2.00

Page 19: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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.

Page 20: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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.

Page 21: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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%) 5 (38.5%) 13

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

12 (100%) 13 (100%) 25

Page 22: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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.

Page 23: Chapter 11 Hypothesis Testing IV (Chi Square). Basic Logic  Chi Square is a test of significance based on bivariate tables.  We are looking for significant

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