# 6a.5 - Comparative Treatment Efficacy Studies

6a.5 - Comparative Treatment Efficacy Studies

Suppose that a comparative treatment efficacy (CTE) trial consists of comparing two independent treatment groups with respect to the means of the primary clinical endpoint. Let $$\mu_1$$ and $$\mu_2$$ denote the unknown population means of the two groups, and let $$\sigma$$ denote the known standard deviation common to both groups. Also, let $$n_1$$ and $$n_2$$ denote the sample sizes of the two groups.

The treatment difference in means is $$\Delta = \mu_1 -\mu_2$$ and the null hypothesis is $$H_0\colon \Delta = 0$$. The test statistic is

$$Z = \left( \bar{Y}_1 - \bar{Y}_2 \right) / \sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

which follows a standard normal distribution when the null hypothesis is true. If the alternative hypothesis is two-sided, i.e., $$H_1 \colon \Delta \ne 0$$, then the null hypothesis is rejected for large values of |Z|.

Under a particular alternative where there might be some difference $$\Delta, \Delta = \mu_1 - \mu_2$$,

$$Z = \left( \bar{Y}_1 - \bar{Y}_2 - \Delta \right)/ \sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

Suppose we let $$AR = \dfrac{n_1}{n_2}$$ denote the allocation ratio $$\left(AR\right)$$, (in most cases we will assign $$AR = 1$$ to get equal sample sizes). If we wish to a have large enough sample size to detect an effect size Δ with a two-sided, α-significance level test with $$100 \left(1 - \beta \right)\%$$ statistical power, then

$$n_2 = \left( \frac{AR+1}{AR}\right) \left( z_{1-\alpha/2}+z_{1-\beta} \right)^2\sigma^2/\Delta^2$$

and $$n_1 = AR \times n_2$$.

Note this formula matches the sample size formula in our FFDRG text on p. 180, assuming equal allocation to the two treatment groups and multiplying the result here by 2 to get 2N, which FFDRG uses to denote the total sample size.

If the alternative hypothesis is one-sided, then $$Z_{1 - α}$$ replaces $$Z_{1 - \frac{\alpha}{2}}$$ in either formula.

Notice that the sample size expression contains $$\left(\dfrac{\sigma}{\Delta}\right)^2$$, the square of the effect size expressed in standard deviation units. Thus, sample size is a quadratic function of the effect size and precision. As the variance gets larger, it has a quadratic effect on the sample size. For example, reducing the effect size by one-half quadruples the required sample size.

Although this sample size formula assumes that the standard deviation is known so that a z test can be applied, it works relatively well when the standard deviation must be estimated and a t-test applied. A preliminary guess of σ must be available, however, either from a small pilot study or a report in the literature. For smaller sample sizes $$\left(n_1 ≤ 30, n_2 ≤ 30 \right)$$ percentiles from a t distribution can be substituted, although this results in both sides of the formula involving $$n_2$$ so that it must be solved iteratively:

$$n_2 = \left( \dfrac{AR+1}{AR}\right) \left( t_{n_1+n_2-2,1-\alpha/2}+t_{n_1+n_2-2,1-\beta} \right)^2\sigma^2/\Delta^2$$

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