Question

Not sure how to go about answering this problem below, I get stuck in the treatment part and uncertain if I am in putting the information correctly and interperting the results correctly.

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 1=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 (±) 0.40 for the non-inferiority parameter (d).

Parameter	Estimate	Standard Error
Intercept	-2.57	1.11
Treat (b1)	0.52	0.46
age	0.03	0.02

1. Based on the above table, what is the 95% confidence interval for the coefficient for the treatment effect (b₁)?

A. the p-value for testing the superiority of the new treatment greater than or less than 0.05?

B. What can you conclude from the above analyses concerning the superiority of the new treatment?

C. Using the above non-inferiority parameter, what can the investigators conclude about the non-inferiority of the new treatment? Justify your response.

Answer 1

95% confidence interval
(b^ +- 1.96 * se (b^)
= (0.52 +- 1.96 * 0.46)
= ( -0.3816 , 1.4216 )

A)
p-value > 0.05 as 0 is present in 95% confidence interval
other method
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.

B)
since Treat is not significant
we conclude that new treatment is not superior

C)
As coefficient of treatment=0.52>δ=0.40, so the treatment is non-inferior.