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assignments0001 - ,/ / Brittany Poyer 3/23/07 ANOV A...

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,/ / Brittany Poyer 3/23/07 ANOV A assignment #9 1. The null hypothesis is exercise duration has no effect on strength test results among the 30 subjects. 2. The research hypothesis for the ANOV A is the longer the subjects exercises, the stronger they will be. 3. The mean of all 30 scores was 84.23. The standard deviation of all 30 scores was 10.45. 4. The SST was 3169.37. 5. The explained sum of squares-between-as a result of the treatment, was 1414.47. 6. The error sum of squares was 1754.90. 7. The degree of freedom for exercise duration is equal to 2 because there are 3 different groups in the class, to get df you need to subtract one. (3-1=2) 8. The degrees of freedom for residual is calculated by subtracting one from 30 to get the degrees of freedom, then subtract the explained degree of freedom from that, resulting in the value for residual degree of freedom. N - ~ 9. If the null is true, then both mean square values should be equal to 64.996 10. The F value of 10.881 was obtained by dividing the mean square between from the mean square within. (707.233/64.996=10.881) 11. You should reject the null hypothesis because if the null was true, the f ratio should be one. In this case, the f ratio was 10.881, which means we should reject it. 12. The longer a person spends exercising, the stronger the person will become 13. the mean for 30 min: 75.60,45 min: 84.70 and 60 min: 92.40. 14. the null hypothesis is that there is no difference of the mean test scores between each of the classes. 15. the mean for all 60 scores was 26.25. The standard deviation for all 60 scores was 12.18867. 16. The SS for between: 8375.25, the SS for within: 390.0 and SS total: 8765.25. 17. the degrees of freedom are 3 and 56 because there are 4 groups in the class (4- 1=3) there is a total of 60 (df=n-k) which is the total subjects and n is 4 which is the number of groups (60-4=56) 18. the value of the F ratio is 400.867. 19. the significance value for this f ratio was .000 which means it is extremely significant 20. the MS for error (within) is so small (6.964) because there is little variation when looking at the difference within a class. 21. I would reject the null because when looking at results between the classes, there would be large variation among the classes. When looking at the pattern to measure Integrative Physiology knowledge, this pattern does make sense. I would expect to see a greater variation between the classes (comparing f vs h vs j vs s. .. ) because a senior would be expected to know more than a freshman per say, ultimately scoring higher percentages. When comparing within the classes however, I would expect little variation because you are comparing, for example, sophomores versus sophomores knowledge-they have all been equally prepared,

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This note was uploaded on 04/03/2009 for the course IPHY 2800 taught by Professor Decker,mik during the Spring '08 term at Colorado.

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assignments0001 - ,/ / Brittany Poyer 3/23/07 ANOV A...

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