Factorial trial

In certain situations, it is possible to evaluate 2 or more interventions or parameters simultaneously in a single trial. This can be accomplished by using a factorial design, which takes the form of a 2 × 2 in the simplest scenario. The advantages of the factorial design are related to the fact that 2 or more parameters can be assessed at the same time, thus creating a more efficient trial in terms of resources, including sample size, compared with 2 separate trials for the assessment of the 2 parameters. However, certain assumptions must not be violated for the factorial design to be appropriate. The assumptions are that the 2 parameters can be combined and that there is no interaction (when the effect of 1 variable on an outcome is not modified according to the level of the second variable). We will add a small parenthesis here and use an example from drug therapy to briefly explain what we mean by interaction . We have an interaction (or effect modification) if, for example, the effect of a systolic blood pressure lowering drug differs between men and women. In this scenario, we say that there is an interaction between drug and sex, and therefore we cannot give an overall estimate of the blood pressure lowering effect of the drug that will apply to both men and women. When there is an interaction, we must consider the effect of the drug on lowering blood pressure separately for men and women, and we should give a separate effect estimate for each.

When the main reason for the trial is to compare the impact of 2 different interventions in the same trial, the approach to sample size calculations is relatively straightforward. The most common method is to consider the trial as 2 separate 2-arm trials. The sample size for each separate comparison is calculated, and whichever of these calculations results in the largest number of patients provides the basis for the overall sample size. These separate calculations are likely to be similar if the same outcome is used for both.

Let’s use an example. We are interested in evaluating radiographically the change in inclination of the maxillary incisors during retraction in Class II Division 1 patients treated with maxillary first premolar extractions. We would like to compare the change in maxillary incisor inclination for 2 wire types (0.019 × 0.025-in stainless steel and 0.19 × 0.25-in beta-titanium alloy) and 2 bracket types (self-ligating and conventional) in the same trial. The details for this 2 × 2 factorial design are shown in Table I .

Table I
Factorial design assessing simultaneously the effects of wire type and bracket type on torque loss during maxillary anterior tooth retraction in premolar extraction patients
Parameter: wire
0.19 × 0.25-in stainless steel 0.19 × 0.25-in beta-titanium alloy
Parameter:
bracket
Self-ligating A B
Conventional C D

If there is no interaction between wire type and bracket type, then we can conduct the following comparisons of the main effects.

  • 1.

    A + B vs C + D. This compares the change in incisor inclination between the 2 bracket types and assumes that the change in inclination for each bracket does not differ as the type of wire varies (no interaction assumption). In other words, the change in inclination between brackets is the same regardless of using the stainless steel wire or the beta-titanium alloy wire.

  • 2.

    A + C vs B + D. This compares the change in inclination between the wire types and assumes that the change in incisor inclination for each wire type does not differ as the type of bracket varies (no interaction assumption). In other words, the change in inclination between wires is the same regardless of using self-ligating or conventional brackets.

In a typical factorial trial, we can decide on the required sample size by calculating the sample size for both parameters and then taking the larger sample size (if different). So let’s say that the expected change in inclination in the stainless steel wire is 10° (μ1) and that we would like to be able to observe a 3° difference (μ 1 – μ 2 ), if such a difference exists, between the 2 wire types with alpha of 0.05 and power of 90%; we assume that σ 1 = σ 2 = σ = 5°.

We will use the same formula that we used during sample calculations for 2 independent means:

n=f(α,β)x2σ2(μ1μ2)2=10.525232=58.33,
n = f ( α , β ) x 2 σ 2 ( μ 1 − μ 2 ) 2 = 10.5 ∗ 2 ∗ 5 2 3 2 = 58.33 ,
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Apr 8, 2017 | Posted by in Orthodontics | Comments Off on Factorial trial
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