EXERCISE 36 Questions to be graded
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
F-value shows significance between the two groups. The null hypothesis should be rejected because the P-Value is 0.005 which would mean that the groups are different.

2. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI with PMR treatment on patients’ mobility level. Should the null hypothesis be rejected for the difference between 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 for each groups the control and the treatment will be equal. The hypothesis will be rejected as the probability of happening is 0.005. Mobility scores are different.

3. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Was this result statistically significant, and if so at what probability?

Answer
Results are statically significant P=0.005
results are significant at exactly the 99.5% level (very high)

4. If the researchers had set the level of significance or α = 0.01, would the results of p = 0.001 still be statistically significant? Provide a rationale for your answer.

Answer
Yes p of 0.001 is less than 0.01 and therefore significant.

5. If F(3, 60) = 4.13, p = 0.04, and α = 0.01, is the result statistically significant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?

Answer
It would not be significant at the P=0.04>0.01 the null hypothesis would be accepted

6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a...

...EXERCISE36 Questions to be graded
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.
* The F-value suggests that there is a significant difference between the results of the control and treatment groups. The P-value of 0.005 is < the alpha of 0.05. This...

...EXERCISE36 Questions to be Graded
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 suggests there is a significant difference between the results of the control and treatment groups. The P-value of 0.005 is < the alpha of 0.05.This...

...significant.
5. If F (3, 60) 4.13, p 0.04, and = 0.01, is the result statistically significant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?
Answer: The 0.04 > 0.01 would indicate that there is no statistical significance and except the null and conclude that there is no difference between the groups.
6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a...

...
HLT 362 Module 4 Exercise36
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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...

...Exercise36
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.
The F-value is high enough at the 5% level of significance to suggest a significant difference between the control and treatment groups. The p-value 0.005 < 0.05 hence this suggests a...

...Marcia Landell
Applied Statistics Week 6: Analysis of Variance (ANOVA)
Exercise36 Analysis of Variance (ANOVA) I
1. A major significance is identifiable between the control group and the treatment group with the F value at 5% level of significance. The p value of 0.005 is less than 0.05 indicating that the control group and the treatment group are indeed different. Based on this fact, the null hypothesis is to be rejected.
2. Null hypothesis: The mean mobility...

...Exercise36 Answers
1. Since the F value is significant, based on the p-value of 0.005 which is less than 0.05 which is sufficient to reject the null hypothesis. This suggests that there is a difference in the control and treatment groups.
2. Since the p- value is less than 0.05 and therefor the null hypothesis can be rejected. This presents that the mean, difficulty and mobility scores, must be different
3. The result was statistically significant with a...

...Week 4
Exercise36
1. F is the statistic for ANOVA and the group df=1 and the error for df is 22. The f ratio is 9.619, which is significant for 0.005.
2. I would reject the null hypothesis because p=0.005 which is less than 0.05. This suggests there is a difference between the 2 groups.
3. There was a significant reported change of the p<0.0001 at week 12 compared with the control group, which there was no change.
4. Yes because 0.001<0.01 which is...