Stats week 4 exercise 36

Week 4
Exercise 36
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 treatment for OA. The treatment may have side effects, which can lead to some women dropping out of the study.