The below is for a hypothetical study demonstrating the application of the conce

Question

The below is for a hypothetical study demonstrating the application of the concepts for superiority and non-inferiority testing. A study investigated a new agent for the treatment of arthritis. Participants were randomized to receive either the current widely used (and effective) treatment or to the new experimental treatment. After 3 weeks they recorded if, based on a clinical assessment, there were improvements in symptoms (recorded as yes/no). Because of the potential large impact of age the investigators used a multiple regression model to test for differences adjusting for age. (Technically since the outcome is binary they used what is called a logistic regression model, but the analytic approach is the same.) Investigators want to test if the new treatment is non-inferior to the current treatment. Treatment is coded as 0=usual treatment and l=new treatment so that a positive coefficient represents greater improvement with the new treatment. The study was done on a large sample and you may use the normal distribution for testing coefficients. Prior to testing the investigators selected (plusminus) 0.40 for the non-inferiority parameter (delta). 1. [1] Based on the above table, what is the 95% confidence interval for the coefficient for the treatment effect (beta_1)? 2. [1] Is the p-value for testing the superiority of the new treatment greater than or less than 0.05? 3. [1] What can you conclude from the above analyses concerning the superiority of the new treatment? 4. [2] Using the above non-inferiority parameter, what can the investigators conclude about the non-inferiority of the new treatment? Justify your response.

Explanation / Answer

Part-1: 95% CI for treatment coefficient = 0.52±1.96*0.46 = (-0.3816        1.4216)

Part-2: As we can use normal distribution so t-statistic =z=0.52/0.46 =1.13

As this is <1.96, so we do not reject the zero coefficient null hypothesis and hence p-value>0.05.

Part-3: As coefficient of treatment is not significant from part-2, so the new treatment is not superior.

Part-4: As coefficient of treatment=0.52>=0.40, so the treatment is non-inferior.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.