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Module 21:  2 For Contingency Tables. This module presents the  2 test for contingency tables, which can be used for tests of association and for differences between proportions. Reviewed 06 June 05 /MODULE 21. Contingency Tables.

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Module 21:  2 For Contingency Tables


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module 21 2 for contingency tables
Module 21: 2 For Contingency Tables

This module presents the 2 test for contingency tables, which can be used for tests of association and for differences between proportions

Reviewed 06 June 05 /MODULE 21

21- 1

slide2

Contingency Tables

Contingency tables are very common types of tables used to summarize. The cells of these tables typically include counts of something and/or percentages. For our purposes, we can use only those tables that have integers that are counts. If you are interested in analyzing data in a contingency table that includes only percentages in the cells, then you must convert these percentages into counts in order to proceed.

21- 2

slide6

Association Hypothesis

H0: There is no association between the classification of the numbers of patients who kept their appointments and the classification of the sex of the patients,

vs.

H1: There is association between the classification of the numbers of patients who kept their appointments and the classification of the sex of the patients.

21- 6

slide7

Proportions Hypothesis

The association hypothesis can also be tested using a test of proportions hypothesis:

H0: PM = PF vs. H1: PMPF

where

PM = Proportion of males in population who “kept their appointments,”

PF = Proportion of females in population who “kept their appointments.”

21- 7

slide8

Contingency Tables

Contingency tables have:

r = number of rows, not counting any totals

c = number of columns, not counting any totals

r  c = number of cells

Approach is to compare observed number in each cell with the number expected under the assumption that the null hypothesis of no association is true.

21- 8

slide9

Calculating Expected Numbers

Calculating the number expected in each contingency table cell under the assumption that the null hypothesis is true is, in practice, quite simple. This number is equal to the total for the row the cell is in times the total for the column the cell is in divided by the overall total for the table. You may be interested in working out why this is true; however, it really isn’t worth the effort.

The algebraic expression is:

21- 9

slide10

Calculating χ2

The value of the test statistic χ2 calculated by comparing the observed number (O) to the expected number (E) in each cell according to the following formula:

The result of this calculation is then compared to the appropriate value in the tables for the χ2 distribution. These are in the Table Module 3: The χ2 Distribution.

Note that you need to look up χ20.95(df) for a test with the -level at  = 0.05.

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slide11

So, for a 22 table,

df = (2-1)(2-1) = 11=1

Reject H0: if 2 20.95(1) = 3.84

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slide12

2 Test for Appointment Keeping Example 1

  • H0: There is no association between the classification of the numbers of patients who kept their appointments and the classification of the sex of the patients,
  • vs.
  • H1: There is association between the classification of the numbers of patients who kept their appointments and the classification of the sex of the patients.
  • The hypothesis: H0: PM = PF vs. H1: PM PF
  • The assumption: Contingency table
  • 3. The  – level: = 0.05

21- 12

slide15

4. The test statistic:

5. The critical region: Reject H0: if 220.95(1) = 3.84

6. The result:2 = 0.86

7. The conclusion: Accept H0; Since 2 = 0.86 < 3.84

21- 15

slide16

The published table as shown on slide 13 indicates a result of

Our result is

If with df = 1; then p cannot be > 0.05

21- 16

slide18

2 Test for Appointment Keeping Example 2

  • H0: There is no association between the classification of the numbers of patients who kept their appointments and the classification of the age of the patients,
  • vs.
  • H1: There is association between the classification of the numbers of patients who kept their appointments and the classification of the age of the patients.
  • The hypothesis: H0: PY = PO vs. H1: PY PO
  • The assumption: Contingency table
  • 3. The  – level:  = 0.05

21- 18

slide20

4. The test statistic:

5. The critical region: Reject H0: if 220.95(1) = 3.84

6. The result:2 = 16.90

7. The conclusion: Reject H0; Since 2 = 16.90 > 3.84

21- 20

slide22

2 Test for Appointment Keeping Example 3

  • H0: There is no association between the classification of the numbers of patients who kept their appointments and the classification of the ethnicity of the patients,
  • vs.
  • H1: There is association between the classification of the numbers of patients who kept their appointments and the classification of the ethnicity of the patients.
  • The hypothesis: H0: PP = PB= Pw vs. H1: PP PB ≠ Pw
  • The assumption: Contingency table
  • 3. The  – level: = 0.05

21- 22

slide24

4. The test statistic:

5. The critical region: Reject H0: if 220.95(2) = 5.99

6. The result:2 = 54.43

7. The conclusion: Reject H0; Since 2 = 54.43 > 5.99

21- 24

slide28

2 Test for Heavy Drinkers Intervention Example 1

  • The Hypothesis
  • H0: There is no association between the classification of the number of male patients according to their change in alcohol consumption and the treatment they received
  • vs.
  • H1: There is an association between the classification of the numbers of male patients according to their change in alcohol consumption and the treatment they received.
  • 2. The assumption: Contingency table
  • 3. The  – level: = 0.05

21- 28

slide30

4. The test statistic:

5. The critical region: for a 3 x 3 table; df = (3-1)(3-1) = 4 Reject H0: if 220.95(4) = 9.49

6. The result:2 = 32.03

7. The conclusion: Reject H0; Since 2 = 32.03 > 9.49

21- 30

slide32

2 Test for Heavy Drinkers Intervention Example 2

  • The Hypothesis
  • H0: There is no association between the classification of the number of male patients according to their posttest hazardous alcohol consumption and the treatment they received.
  • vs.
  • H1: There is an association between the classification of the numbers of male patients to according to their posttest hazardous alcohol consumption and the treatment they received.
  • 2. The assumption: Contingency table
  • 3. The  – level: = 0.05

21- 32

slide34

4. The test statistic:

5. The critical region: for a 2 x 3 table; df = (2-1)(3-1) = 2 Reject H0: if 220.95(2) = 5.99

6. The result:2 = 11.74

7. The conclusion: Reject H0 , since 2 = 11.74 > 5.99

21- 34

slide40

2 Test for the MATISS Project

  • The Hypothesis
  • H0: There is no association between the classification of the number of male patients according to their heart rate category and this classification of their diabetes status.
  • vs.
  • H1: There is an association between the classification of the numbers of male patients to according to their heart rate category and this classification of their diabetes status.
  • 2. The assumption: Contingency table
  • 3. The  – level: = 0.05

21- 40

slide43

4. The test statistic:

5. The critical region: Reject H0: if 220.95(4) = 9.49

6. The result:2 = 15.45

7. The conclusion: Reject H0; Since 2 = 15.45 > 9.49

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