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**STATISTICS ELEMENTARY MARIO F. TRIOLA**

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

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**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

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**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.

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**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.

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**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

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**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.

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**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.

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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.

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**Expected Frequency for Contingency Tables**

page 593 of text

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**Expected Frequency for Contingency Tables**

row total column total grand total grand total grand total

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**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.

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**Expected Frequency for Contingency Tables**

row total column total grand total grand total grand total n • p (probability of a cell)

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**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)

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**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.

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**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.

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**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)

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**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

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**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.

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**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

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**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

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**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

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**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.

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**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 =

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**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 =

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**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

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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.

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**Relationships Among Components in X2 Test of Independence**

Figure 10-8 page 595 of text

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**Definition Test of Homogeneity**

test the claim that different populations have the same proportions of some characteristics page 596 of text

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**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.

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Copyright © 2010 Pearson Addison-Wesley. All rights reserved. Chapter 13 One-Factor Experiments: General.

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