# 6a.8 - Comparing Treatment Groups Using Hazard Ratios

6a.8 - Comparing Treatment Groups Using Hazard Ratios

For many clinical trials, the response is time to an event. The methods of analysis for this type of variable are generally referred to as survival analysis methods. The basic approach is to compare survival curves.

With an event time endpoint, it is mathematically convenient to compare treatment groups (and curves) with respect to the hazard ratio. The survival function for a treatment group is characterized by $$\lambda$$, the hazard rate. At time t, $$\lambda(t)$$ for a treatment group, is defined as the instantaneous risk of the event (or failure) occurring at time t. In other words, given that a subject has survived the event up to time t, the hazard at time t is the probability of the event occurring within the next instant. You can think of the hazard as the slope of the survival curve.

The hazard ratio is defined as the ratio of two hazard functions, $$\lambda_1(t)$$ and $$\lambda_2(t)$$, corresponding to two treatment groups. Typically, we assume proportional hazards, i.e., $$\Lambda= \dfrac{\lambda_1(t)}{\lambda_2(t)}$$ is a constant function independent of time. The graphs on the next two slides illustrate the concept of proportional hazards.

A hazard function may be constant, increasing, or decreasing over time, or even be a more complex function of time. In trials in which survival time is the outcome, an increasing hazard function indicates that the instantaneous risk of death increases throughout the trial.

An example where the hazard function might be decreasing involves the disease ARDS (adult respiratory distress syndrome), whereby the risk of death is highest during the early stage of the disease.

A sample size formula for comparing the hazards of two groups via the logrank test (discussed later in the course) is expressed in terms of the total number of events, E, that need to occur. For a two-sided, α-level significance test with $$100 \left(1 - \beta \right)\%$$ statistical power, hazard ratio Λ, and allocation ratio AR,

$$E=\left( \dfrac{(AR+1)^2}{AR}\right) \dfrac{\left( z_{1-\alpha/2} + z_{1-\beta} \right)^2}{ \left(log_e(\Lambda) \right)^2}$$

NOTE! this formula above matches FFDRG text p. 185 simple formula, if it is assumed that all participants will have an event. However, we most often have censored data, which is a number of participants who do not experience the event before the trial ends.

Since we do not expect all persons in the trial to experience an event, the sample size must be larger than the required number of events.

Suppose that $$p_1$$ and $$p_2$$ represent the anticipated event rates in the two treatment groups. Then the sample sizes can be determined from $$n_2 = \dfrac{E}{(AR \times p_1 + p_2)}$$ and $$n_1 = AR \times n_2$$

If a hazard function is assumed to be constant during the follow-up period [0, T], then it can be expressed as $$\lambda(t) = \lambda = \dfrac{-log_e(1 - p)}{T}$$. In such a situation, the hazard ratio for comparing two groups is $$\Lambda = \dfrac{log_e(1 - p_1)}{log_e(1 - p_2)}$$.

A constant hazard rate, $$\lambda(t) = \lambda$$ for all time points t, corresponds to an exponential survival curve, i.e., survival at time $$t = exp(-\lambda t)$$.

Survival curves plot the probability of the event occurring to a subject over time.

## SAS® Example

### Using PROC POWER to calculate sample size when comparing two hazard functions

An investigator wants to compare an experimental therapy to placebo when the response is time to infection via a two-sided, 0.05-significance level test with 90% statistical power and equal allocation. He plans to follow each patient for one year and he expects that 40% of the placebo group will experience infection and he considers a 20% rate in the therapy group as clinically relevant.

If he assumes constant hazard functions, then

$$\Lambda= \dfrac{log_e(0.6)}{log_e(0.8)} = 2.29$$

Then the number of required events is

$$E = \dfrac{(4)(1.96 + 1.28)^2}{{log_e(2.29)}^2} = 62$$

and the sample sizes are

$$n_2 = \dfrac{E}{(AR \times p_1 + p_2)} = \dfrac{62}{(0.4 + 0.2)} = 104$$ and $$n_1 = 104$$

SAS Example - This is a program that illustrates the use of PROC POWER to calculate sample size when comparing two hazard functions.

***********************************************************************
* This is a program that illustrates the use of PROC POWER to         *
* calculate sample size when comparing two hazard functions.          *
***********************************************************************;

proc power;
twosamplesurvival groupweights=(1 1) alpha=0.05 power=0.9 sides=2
test=logrank curve("Placebo")=(1.01):(0.6) curve("Therapy")=(1.01):(0.8)
groupsurvival="Placebo"|"Therapy" accrualtime=0.01 followuptime=1 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Hazard Functions (1:1 Allocation)";
run;

proc power;
twosamplesurvival groupweights=(1 3) alpha=0.05 power=0.9 sides=2
test=logrank curve("Placebo")=(1.01):(0.6) curve("Therapy")=(1.01):(0.8)
groupsurvival="Placebo"|"Therapy" accrualtime=0.01 followuptime=1 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Hazard Functions (3:1 Allocation)";
run;


Additional comments on this program: Note the curve statements indicate points on the survival curves. In this example, at the end of study, at time 1.01 (followup plus accrual in SAS), the proportion in the placebo group without an event is 0.6 and the proportion remaining the therapy group is 0.8.

SAS PROC POWER for the logrank test requires information on the accrual time and the follow-up time. It assumes that if the accrual (recruitment) period is of duration $$T_1$$ and the follow-up time is of duration $$T_2$$, then the total study time is of duration $$T_1 + T_2$$. It assumes, however, if a patient is recruited at time $$\dfrac{T_1}{2}$$, then the follow-up period for that patient is $$\dfrac{T_1}{2 + T_2}$$ instead of $$T_2$$. This assumption may be reasonable for observational studies, but not for clinical trials in which follow-up on each patient is terminated when the patient reaches time $$T_2$$. Therefore, for a clinical trial situation, set accrual time in SAS PROC POWER equal to a very small positive number. For the given example, SAS PROC POWER yields $$n_1 = 109$$ and $$n_2 = 109$$.

SAS notes for PROC POWER for survival

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