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18 - 1 Module 18: One Way ANOVA Reviewed 11 May 05 /MODULE 18 This module begins the process of using variances to address questions about means. Strategies.

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Presentation on theme: "18 - 1 Module 18: One Way ANOVA Reviewed 11 May 05 /MODULE 18 This module begins the process of using variances to address questions about means. Strategies."— Presentation transcript:

1 18 - 1 Module 18: One Way ANOVA Reviewed 11 May 05 /MODULE 18 This module begins the process of using variances to address questions about means. Strategies for more complex study designs appear in a subsequent module.

2 18 - 2 Independent Random Samples from Two Populations of Serum Uric Acid values

3 18 - 3 Serum Acid SS (Total) Worksheet

4 18 - 4 SS (Within) and SS (Among) worksheet

5 18 - 1 SS (Within) = SS (sample 1) + SS (sample 2) = 0.824 + 1.669 = 2.490 SS (Within) = 2.49

6 18 - 7 1. The hypothesis: H 0 :  1 =  2 vs H 1 :  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

7 18 - 8 6. The result: 7. The conclusion: Reject H 0 : Since F = 10.86 > F 0.95 (1,18) = 4.41

8 18 - 9 1. The hypothesis: H 0 :  1 =  2 vs H 1 :  1   2 2. The  -level:  = 0.05 3. The assumptions: Independent Random Samples Normal Distribution 4. The test statistic: Testing the Hypothesis that the Two Serum Uric Acid Populations have the Same Mean

9 18 - 10 5. The reject region: Reject H 0 if t is not between ± 2.1009 6. The result: 7. The conclusion: Reject H 0 :  1 =  2 since t is not between  2.1009

10 18 - 11 Example 2

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12 18 - 13 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

13 18 - 14 Testing the Hypothesis that the Three populations have the same Average Serum Uric Acid Levels 1.The hypothesis: H 0 : µ 1 =µ 2 =µ 3, vs. H 1 : µ 1 ≠ µ 2 ≠ µ 3 2.The assumptions: Independent random samples normal distributions 3. The  -level :  = 0.05 4. The test statistic: ANOVA

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15 18 - 16 Example 3

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17 18 - 18 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

18 18 - 19 Testing the Hypothesis That the Three Populations Have the Same Average Blood Pressure Levels 1.The hypothesis: 2.The assumptions: Independent random samples normal distributions 3. The  -level :  = 0.05 4. The test statistic: ANOVA

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21 18 - 22 Individual Effect + Group Effect + Overall Mean = Individual Value Group Effect Random Effect

22 18 - 23 Pulmonary Function Equipment Comparison

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24 18 - 25 All Four

25 18 - 26 Example: AJPH, Sept. 1997; 87 : 1437

26 18 - 27 ANOVA for Anxiety at Baseline

27 18 - 28 ANOVA for Psychosocial Adjustment to illness at 3 months

28 18 - 29 Example: AJPH, August 2001; 91:1258

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