Diagnosis Congestive Heart Failure Essay
1. Think about a real world: Diagnosis Congestive heart failure (CHF), non compliance with daily weight or diet/situation.
Address using an independent or related samples t test.
2. Identify the independent (grouping) and dependant (response) variables important to study
3. Explain whether an independent sample or related sample t test is appropriate and why
4. Generate a hypothesis, including null and alternative hypothesis
5. Describe what information the effect size will tell you and what information the effect size will tell you and what information the p value or critical value approach will not
6. Using realistic numbers for the degrees of freedom, sample size and t statistic, report hypothetical results in 2-3 sentences
Solution: …show more content…
(3) This analysis corresponds to an independent-samples design, because the treatments (diet/no diet) are applied to different subjects.
(4) We are interested in the following research question:
Is there a difference in the incidence of CHF for the diet and no-diet group?
The following hypotheses are used:
[pic]
where [pic]represents the mean CHF incidence …show more content…
(6) The following hypothetical results could be obtained:
“A sample of n = 25 subjects was used for each treatment. For the diet group, the mean CHF incidence is M = 2.3% (SD = 0.9%), and for the no-diet group, the mean CHF incidence is M = 3.1% (SD = 1.1%). There is enough evidence to support the claim that there is a significant difference in the mean incidence for the diet and no-diet group, t(48) = -2.81, p = 0.0071 < 0.05. The effect size is d = 0.796”
Appendix: Calculations of the hypothetical analysis
We are interested in testing
[pic]
which corresponds to a two-tailed independent samples t-test. Before performing a t-test, we need to test whether the variances can be assumed to be equal or not. We need to test
[pic]
The F-statistics is computed as
[pic]
The lower and upper critical values for[pic] and df1 = 24 and df2 = 24 are
[pic]
which means that we fail to reject the null hypothesis of equal variances. Observe that we are assuming that the variances are equal, so the t-statistics is computed
Address using an independent or related samples t test.
2. Identify the independent (grouping) and dependant (response) variables important to study
3. Explain whether an independent sample or related sample t test is appropriate and why
4. Generate a hypothesis, including null and alternative hypothesis
5. Describe what information the effect size will tell you and what information the effect size will tell you and what information the p value or critical value approach will not
6. Using realistic numbers for the degrees of freedom, sample size and t statistic, report hypothetical results in 2-3 sentences
Solution: …show more content…
(3) This analysis corresponds to an independent-samples design, because the treatments (diet/no diet) are applied to different subjects.
(4) We are interested in the following research question:
Is there a difference in the incidence of CHF for the diet and no-diet group?
The following hypotheses are used:
[pic]
where [pic]represents the mean CHF incidence …show more content…
(6) The following hypothetical results could be obtained:
“A sample of n = 25 subjects was used for each treatment. For the diet group, the mean CHF incidence is M = 2.3% (SD = 0.9%), and for the no-diet group, the mean CHF incidence is M = 3.1% (SD = 1.1%). There is enough evidence to support the claim that there is a significant difference in the mean incidence for the diet and no-diet group, t(48) = -2.81, p = 0.0071 < 0.05. The effect size is d = 0.796”
Appendix: Calculations of the hypothetical analysis
We are interested in testing
[pic]
which corresponds to a two-tailed independent samples t-test. Before performing a t-test, we need to test whether the variances can be assumed to be equal or not. We need to test
[pic]
The F-statistics is computed as
[pic]
The lower and upper critical values for[pic] and df1 = 24 and df2 = 24 are
[pic]
which means that we fail to reject the null hypothesis of equal variances. Observe that we are assuming that the variances are equal, so the t-statistics is computed