# 14.3 - A Special Case with Drug Combinations

14.3 - A Special Case with Drug Combinations

A special case of a partial factorial design that occasionally is used in clinical research is the incomplete 2 × 2 factorial design with three treatment groups consisting of drug A, drug B, and drug A in combination with drug B:

Placebo A + Active B

Active A + Placebo B

Active A + Active B

Notice that the Placebo A + Placebo B group is not included in the design, hence the incompleteness. The incomplete factorial design has become popular.

Why?

Combination therapies can be marketable and profitable. If a company can combine the active ingredients for treatment A and treatment B into one pill/tablet/capsule, more symptoms are relieved with one dose of medicine. For example, combining an antihistamine with a decongestant for cold symptoms produces a new cold remedy that will alleviate two major symptoms with one capsule. Additionally once the company has created the new combination product, the company applies for a new patent, extending the years of profitable returns from the research dollars expended to develop the intial products. Approval of a combination therapy however, requires evidence demonstrating the superiority of the AB combination therapy to the A monotherapy and the B monotherapy. A logical experimental design to demonstrate these results would be the incomplete factorial.

Suppose that the response is continuous and that we want to compare the means $$\mu_{A}, \mu_{B}$$, and $$\mu_{AB}$$, which represent the population means for the A monotherapy, the B monotherapy, and the AB combination therapy, respectively. The research objective is to show the superiority of the combination therapy over the individual therapies.

Assuming that the higher response is more beneficial, a one-sided hypothesis testing format can be constructed as $$H_{0}: {\mu_{A} ≥ \mu_{AB} \text{ or } \mu_{B} ≥ \mu_{AB}}$$ versus $$H_{1}: {\mu_{A} < \mu_{AB} \text{ and }\mu_{B} < \mu_{AB}}$$

Notice that the null hypothesis indicates that the AB combination therapy is not better than at least one of the monotherapies, whereas the alternative indicates that the AB combination is better than the A monotherapy and the B monotherapy.

How do we do this?

The appropriate test statistic to use for this situation is called the “min” test. If the data are normally distributed, construct two two-sample t statistics, one comparing the AB combination therapy to the A monotherapy (call it $$t_{A}$$) and the other comparing the AB combination therapy to the B monotherapy (call it $$t_{B}$$).

$$t_A=(\bar{Y}_{AB}-\bar{Y}_A)/s \sqrt{\dfrac{1}{n_{AB}}+\dfrac{1}{n_{A}}}$$ , $$t_B=(\bar{Y}_{AB}-\bar{Y}_B)/s \sqrt{\dfrac{1}{n_{AB}}+\dfrac{1}{n_{B}}}$$

where

$$\bar{Y}_{A}=\dfrac{1}{n_A}\sum_{i=1}^{A}Y_{A, i} , \bar{Y}_{B}=\dfrac{1}{n_B}\sum_{i=1}^{B}Y_{B, i} , \bar{Y}_{AB}=\dfrac{1}{n_AB}\sum_{i=1}^{AB}Y_{AB, i}$$

and

$$s^2= \dfrac{1}{n_{A}+n_{B}+n_{AB}-3} \left( \sum_{i=1}^{n_A}\left( Y_{A,i}-\bar{Y}_A \right)^2 + \sum_{i=1}^{n_B}\left( Y_{B,i}-\bar{Y}_B \right)^2 + \sum_{i=1}^{n_{AB}}\left( Y_{AB,i}-\bar{Y}_{AB} \right)^2 \right)$$

The null hypothesis is rejected at the $$\alpha$$ significance level in favor if the alternative hypothesis when each of $$t_{A}$$ and $$t_{B}$$ is statistically significant at the $$\alpha$$ significance level.

It is called the min test because in this situation it is comparable to rejecting the null hypothesis if

$$\text{minimum}(t_A, t_b) > t_{n_A+n_B+n_{AB}-3, 1-\alpha}$$

As a simple example, suppose that:

$$\bar{Y}_A=20, \bar{Y}_B=21, \bar{Y}_{AB}=24, n_A=n_B=n_{AB}=50, s=10$$

Then $$t_{A} = 2, t_{B} = 1.5$$, and $$\text{minimum}\left(t_{A}, t_{B}\right) = 1.5$$, which is not greater than $$t_{147, 0.95} = 1.66$$. Thus, the null hypothesis cannot be rejected at the 0.05 significance level, i.e., the AB combination is not significantly better than the A monotherapy and the B monotherapy. It is close, but there clearly is not enough statistical evidence to show significant difference.

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