Hlt 362 Exercise 36

Topics: Statistical hypothesis testing, Statistical significance, Statistics Pages: 3 (846 words) Published: March 16, 2013
Exercise 36

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.

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.

Yes this still implies statistical significance because 0.001 < 0.01

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?

In this case the result would not be considered statistically significant because 0.04...
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