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 suggest that the groups are significantly different and the null hypothesis should be rejected.
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.
* Treatment group mean = control group mean
* With the p-value being < the alpha, the null hypothesis would be rejected indicating the difference in the mean mobility scores.
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?
* The p-value of 0.005 suggests that the results are statistically significant.
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?...

...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 suggest that the groups are significantly different and the null hypothesis should be rejected.
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: Treatment group mean=control group mean
With the p-value being < the alpha, the null hypothesis would be rejected indicating the difference in the mean mobility scores.
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: the p-value of 0.005 suggests that the results are statistically significant.
4. If the researchers had set the level of significance or α = 0.01, would the results of p = 0.001 still be statistically...

...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: The result was statistically significant with a probability score of p < 0.001,
4. If the researchers had set the level of significance or = 0.01, would the results of p 0.001 still be statistically significant? Provide a rationale for your answer.
Answer: Yes, because 0.001 < 0.01 is 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 rationale for your answer.
Answer: OVA 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. 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: The study had 149 subjects and 2 groups.
8. The researchers state that the sample for their study was 28 women with a diagnosis of OA,...

...
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 F(1, 22) 9.619, p 0.005. Discuss each aspect of these results.
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.
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?
If the researchers had set the level of significance or = 0.01, would the results of p 0.001 still be statistically significant? Provide a rationale for your answer.
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?
Can ANOVA be used to test proposed relationships or predicted correlations between...

...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 rejection of the null hypothesis, meaning that the control and treatment groups are found to be 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.
The null hypothesis is that the mean mobility scores for both groups are the same. As stated previously, since the p-value 0.005 < 0.05 this means I reject the null hypothesis. So the mean "difficulty with mobility score" for both groups must be 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?
Yes the result was statistically significant at probability p < 0.001, according to the text....

...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 scores for the control group and the treatment group are equal. The null hypothesis is to be rejected because the p value of 0.005 is less than 0.05. Therefore, the mean difficulty with mobility score for both groups is different.
3. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Based on the information provided by the text, this result was definitely statistically significant at a probability of p < 0.001.
4. If the researchers had set at the level of significance or a= 0.01, the result of p = 0.001 would still be statistically significant because p of 0.001 is less than 0.01. The null hypothesis would be accepted.
5. If F (3, 60) = 4.13, p= 0.04 and a= 0.001, the result not will not be statistically significant because the p of 0.04 is greater than 0.001. Therefore the null hypothesis would be accepted and conclude that there is truly no difference between the groups.
6. An ANOVA test cannot be used...

...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 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...

...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 significant
5. Because 0.04 > 0.01 there is no statistical difference and will accept the null hypothesis.
6. ANOVA cannot be used to test proposed relationships or predicted correlations between a single group. ANOVA is used to test relationships within various groups and among the groups.
7. There were 149 subjects and the sample size was 2 in the study.
8. A weakness for the study is the size of the sample. More subjects would have made the test better. A strength would be that the study had a control group to examine the treatment over the 12 weeks time and the significant difference between the 2 groups.
9. In my opinion the study results indicated that the treatment for pain and mobility with GI and PMR has made a significant difference for the women in the treatment group. F(1,26)=4.406, p=0.046. 0.046<0.001 rejecting the null hypothesis.
10. A possible problem with the study is the amount of women in the study. The study would be better with more women to study the...