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 of p < 0.001. 4. Yes, because 0.001 < 0.01 and would still be significant. 5. 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. NOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group. This is because ANOVA is tests relationships within various groups and among the groups.

7. The study had 149 subjects and 2 groups
8. The strength of the study where that they include a control group to test the dependent variable to examine the differences over time. The weakness of the study comes from the low number of subjects in the study. More subjects would have made the study more creditable. 9. The study results indicated a significant improvement in the pain scores of women with OA who received the treatment of guided imagery (F(1, 26) =4.406, p = 0.046). Thus, the null hypothesis was rejected. But in my opinion I would have liked to have seen a larger number of subjects. Also, including the standard deviations for the treatment and control groups also are needed to calculate the effect size in the study. The effect size is needed to conduct a power analysis to predict the sample size needed for future studies.

10. Possible problems and limitation with the study is that the pain that leads to limited mobility and may lead to disability which can hinder them form taking the treatments. Also, with it being over such 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
6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide rationale for your answer.
ANOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group because it is designed to test for correlations and interactions amongst groups, i.e. in the test group of patients with OA you are testing the correlations between those who...

...PEARSON’S PRODUCT-MOMENT CORRELATION COEFFICIENT
ANSWERS TO EXERCISE 23
Question 1
The r value for the relationship between Hamstring strength index 60o and the Shuttle run test is -0.149. This r value shows a weak correlation between the two variables, as it is less than the 0.3 threshold for significance. Therefore, the r value is not significant.
Question 2
Between r=1.00 and r=-1.00, there is no difference in terms of strength. Both values are on 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....

...Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary Statistics
Mean...

...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...

...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...

...
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...