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Module 18: One Way ANOVA PowerPoint PPT Presentation

Module 18: One Way ANOVA. This module begins the process of using variances to address questions about means. Strategies for more complex study designs appear in a subsequent module. Reviewed 11 May 05 /MODULE 18. Independent Random Samples from Two Populations of Serum Uric Acid values.

Module 18: One Way ANOVA

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Module 18: One Way ANOVA

This module begins the process of using variances to address questions about means. Strategies for more complex study designs appear in a subsequent module.

Reviewed 11 May 05 /MODULE 18

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Independent Random Samples from Two Populations of Serum Uric Acid values

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Serum Acid SS (Total) Worksheet

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SS (Within) and SS (Among) worksheet

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SS (Within) = SS (sample 1) + SS (sample 2)

= 0.824 + 1.669

= 2.490

SS (Within) = 2.49

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1. The hypothesis: H0: 1 = 2 vs H1: 1  2

2. The assumptions: Independent random samples , normal distributions,

3. The -level : = 0.05

4. The test statistic: ANOVA

5. The rejection region: Reject H0: 1 = 2 if

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6. The result:

7. The conclusion: Reject H0:

Since F = 10.86 > F0.95(1,18) = 4.41

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Testing the Hypothesis that the Two Serum Uric Acid Populations have the Same Mean

1. The hypothesis: H0: 1 = 2vs H1: 1  2

2. The -level: = 0.05

3. The assumptions: Independent Random Samples Normal Distribution

4. The test statistic:

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5. The reject region:Reject H0 if t is not between ± 2.1009

6. The result:

7. The conclusion:Reject H0 : 1 = 2since t is not between  2.1009

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Example 2

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SS(Among) = 11.236 + 25.921 + 22.201 - 57.685

= 1.673

SS(Within) = 0.824 + 1.669 + 1.169

= 3.662

SS(Total) = 1.673 + 3.662 = 5.335

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• The hypothesis: H0: µ1=µ2=µ3 ,vs. H1: µ1≠ µ2≠ µ3

• 2.The assumptions: Independent random samples normal distributions

• 3. The -level : = 0.05

• 4. The test statistic: ANOVA

Testing the Hypothesis that the Three populations have the same Average Serum Uric Acid Levels

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Example 3

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SS(Among) = 111,936 + 94,478 + 93,123 - 299,001

= 536.47

SS(Within) = 708 + 202 + 1,134

= 2,043.70

SS(Total) = 536 + 2,043 = 2,580.17

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• The hypothesis:

• 2.The assumptions: Independent random samples normal distributions

• 3. The -level : = 0.05

• 4. The test statistic: ANOVA

Testing the Hypothesis That the Three Populations Have the Same Average Blood Pressure Levels

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For Group 1, first child,

Individual effect = = 100 - 105.8 = - 5.8

100 = 99.83 + 5.97 + ( -5.80)

Xij = + i + ij

Yij = + i + ij

Individual Value

=

Overall Mean

+

Group Effect

+

Individual Effect

Group

Effect

Random

Effect

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Pulmonary Function Equipment Comparison

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All

Four

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Example: AJPH, Sept. 1997; 87 : 1437

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ANOVA for Anxiety at Baseline

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ANOVA for Psychosocial Adjustment to illness at 3 months

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Example: AJPH, August 2001; 91:1258

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