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...
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