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 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 to test proposed relationships or predict correlations between the variables in the single group. ANOVA test is not applicable because an ANOVA test is used to identify differences in relationship within groups not just a single group.
7. If a study had a result of F (2, 147) = 4.56, p= 0.003, there would be 2 groups with a sample size of 149.
8. The researchers state 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...
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