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Module 18: One Way ANOVAPowerPoint Presentation

Module 18: One Way ANOVA

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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

18 - 1

Independent Random Samples from Two Populations of Serum Uric Acid values

18 - 2

Serum Acid SS (Total) Worksheet Uric Acid values

18 - 3

SS (Within) and SS (Among) worksheet Uric Acid values

18 - 4

SS (Within) = SS (sample 1) + SS (sample 2) Uric Acid values

= 0.824 + 1.669

= 2.490

SS (Within) = 2.49

18 - 1

1. The hypothesis: Uric Acid values 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: Uric Acid values

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: Populations have the Same MeanReject 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 Populations have the Same Mean

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18 - 12 Populations have the Same Mean

SS(Among) = 11.236 + 25.921 + 22.201 - 57.685 Populations have the Same Mean

= 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: Populations have the Same Mean 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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18 - 15 Populations have the Same Mean

Example 3 Populations have the Same Mean

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18 - 17 Populations have the Same Mean

SS(Among) = 111,936 + 94,478 + 93,123 - 299,001 Populations have the Same Mean

= 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: Populations have the Same Mean
- 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

18 - 19

18 - 20 Populations have the Same Mean

18 - 21 Populations have the Same Mean

For Group 1, first child, Populations have the Same Mean

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 Populations have the Same Mean

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18 - 24 Populations have the Same Mean

Example: Populations have the Same MeanAJPH, Sept. 1997; 87 : 1437

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ANOVA for Anxiety at Baseline Populations have the Same Mean

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ANOVA for Psychosocial Adjustment to illness at 3 months Populations have the Same Mean

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Example: Populations have the Same MeanAJPH, August 2001; 91:1258

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18 - 30 Populations have the Same Mean

18 - 31 Populations have the Same Mean

18 - 32 Populations have the Same Mean

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