1. The researchers found a signifi cant 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.

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. 3. The researchers stated that the participants in the intervention group reported a reduction in mobility diffi culty at week 12. Was this result statistically signifi cant, and if so at what probability? 4. If the researchers had set the level of signifi cance or α = 0.01, would the results of p = 0.001 still be statistically signifi cant? Provide a rationale for your answer. 5. If F(3, 60) = 4.13, p = 0.04, and α = 0.01, is the result statistically signifi cant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?

274 EXERCISE 36 • Analysis of Variance (ANOVA) I
6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a rationale for your answer.
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?
8. The researchers state that the sample for their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the study strengths and/or weaknesses in this statement. 9. In your opinion, have the researchers established that guided imagery (GI) with progressive muscle relaxation (PMR) reduces pain and decreases mobility diffi culties in women with OA? 10. The researchers stated that this was a 12-week longitudinal, randomized...

...Kellci Tessendorf
HLT 362V
15 March 2015
■ EXERCISE 36\
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.
Since the F value is significant, based on the p-value of 0.005 which is less than 0.05, making it sufficient to...

...
HLT 362 Module 4 Exercise 36
IF You Want To Purchase A+ Work Then Click The Link Below , Instant Download
http://hwnerd.com/HLT-362-Module-4-Exercise-36-539555.htm?categoryId=-1
If You Face Any Problem E- Mail Us At Contact.Hwnerd@Gmail.Com
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....

...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: 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 suggest that the...

...Exercise 36
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 rejection of...

...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: 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. State the...

...Marcia Landell
Applied Statistics Week 6: Analysis of Variance (ANOVA)
Exercise 36 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 scores for the...

...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.
* 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 suggest that the...

...Exercise 36 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 probability score...