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Statistics Hlt362V

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HLT 362V
Exercise 36
Questions to be graded 1-10

1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F ₍₁‚ ₂₂₎ = 9.619, p=0.005. Discuss each aspect of these results.

Answer: The F value at 5% level of significance shows a significant difference between the control and the treatment groups. Therefore, the null hypothesis should be rejected because the p value 0.005 is less than 0.05 which means that the control and the treatment groups are different.

2. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI PMR treatment on the patients’ mobility level. Should the null hypotheses be rejected for the difference the two groups in change in mobility scores over 12 weeks? Provide a rationale for your answer.

Answer: The null hypothesis would be that the means between the two groups, the control group and the treatment group will be equal. Since the p value is 0.005 0.01, the result is not statistically significant. Therefore, I would accept the null hypothesis because there is no difference between the control group and the treatment group.

6. Can ANOVA be used to test proposed relationship or predicted correlations between variables in a single group? Provide a rationale for your answer.

Answer: No. ANOVA cannot be used to test proposed relationship or predicted correlations between variables in a single group because ANOVA is a technique that is used to compare the means or averages of more than two groups. In other words you are comparing and contrasting your interpretation with your actual result or finding.

7. If a study had a result of F ₍₂‚ ₁₄₇₎ = 4.56, p=0.003, how many groups were in the study, and what was the sample size?

Answer: F ₍₂‚ ₁₄₇₎;

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