A researcher conducts a test on the effectiveness of a cholesterol treatment on 114 total subjects. Assuming the tails of distributions are normal distribution, is there evidence that the treatment is effective? Treatment 38 18 56 No treatment 30 28 58 68 46 114 Submit a (2-3 pages) paper before the end of the module that: Assessment and Grading: Your paper will be assessed based on the performance assessment rubric that is linked within the course. Review it before you begin working on the assignment.
Title: Assessing the Effectiveness of a Cholesterol Treatment: A Statistical Analysis
Introduction:
This paper aims to statistically assess the evidence regarding the effectiveness of a cholesterol treatment for subjects by conducting a hypothesis test on a sample of 114 individuals. In evaluating the treatment’s efficacy, we will assume that the tails of the distribution are normally distributed. This analysis will help in determining whether there is sufficient evidence to support the conclusion that the treatment is indeed effective.
Methods:
To assess the effectiveness of the cholesterol treatment, we will employ a hypothesis test. The null hypothesis (H0) assumes no difference between the treatment and the control group, while the alternative hypothesis (Ha) posits that there is a significant difference favoring the treatment group. Hence, the hypotheses can be stated as follows:
H0: μtreatment = μcontrol
Ha: μtreatment > μcontrol
Here, μtreatment represents the population mean for the cholesterol treatment group, and μcontrol represents the population mean for the control group. We will conduct a one-tailed hypothesis test, as the alternative hypothesis assumes that the treatment group’s mean will be greater than the control group’s mean.
Results:
The data collected from the study are presented as follows:
Treatment: 38, 18, 56
No treatment: 30, 28, 58, 68, 46, 114
Initially, we calculate the means for both groups. The mean of the treatment group (μtreatment) is obtained by summing the values in the treatment group and dividing by the total number of observations. Similarly, the mean for the control group (μcontrol) is determined in the same manner.
μtreatment = (38 + 18 + 56) / (3) = 37.33
μcontrol = (30 + 28 + 58 + 68 + 46 + 114) / (6) = 52.67
Next, we compute the standard deviation (SD) for both groups. This measurement will provide insight into the variability within each group.
For the treatment group:
SDtreatment = √[((38 – 37.33)^2 + (18 – 37.33)^2 + (56 – 37.33)^2) / (3 – 1)] = 18.73
For the control group:
SDcontrol = √[((30 – 52.67)^2 + (28 – 52.67)^2 + (58 – 52.67)^2 + (68 – 52.67)^2 + (46 – 52.67)^2 + (114 – 52.67)^2) / (6 – 1)] = 33.08
We can then proceed to calculate the standard error of the difference between means (SED). This metric provides information on how precise the mean difference estimates are by incorporating the within-group standard deviations.
SED = √[(SDtreatment^2 / 3) + (SDcontrol^2 / 6)] = √[(18.73^2 / 3) + (33.08^2 / 6)] = 12.53
Having determined the mean difference and standard error, we can now embark on the hypothesis test. To proceed, we calculate the test statistic (Z score), comparing the observed mean difference with the null hypothesis’s expected mean difference (0).
Z = (μtreatment – μcontrol) / SED = (37.33 – 52.67) / 12.53 = -1.23
The final step of the hypothesis test is to determine the critical value associated with the desired significance level. Assuming a significance level (α) of 0.05, we need to analyze the one-tailed distribution of the Z score. By consulting a standard normal distribution table or using statistical software, we find that the critical value for a Z score at α = 0.05 is approximately 1.645.
Discussion:
As the obtained Z score (-1.23) falls below the critical value (1.645), we fail to reject the null hypothesis. Therefore, we can conclude that there is insufficient evidence to support the claim that the cholesterol treatment is effective. Based on the sample data, the results do not demonstrate a statistically significant difference between the treatment and control groups.
The limitations of this analysis must be acknowledged. The sample size is relatively small, consisting of only 114 individuals, which may affect the generalizability of the findings. Additionally, the assumption of normally distributed tails requires further validation, as this assumption may not be met in all cases. Future research could benefit from larger sample sizes and considering alternative statistical tests to strengthen the conclusions drawn.
Conclusion:
The statistical analysis suggests that, based on the given sample data, there is insufficient evidence to support the effectiveness of the cholesterol treatment compared to the control group. Further investigation with larger sample sizes and alternative statistical tests may be warranted to determine if the treatment truly demonstrates efficacy. This study serves as a valuable starting point for further research in this area.