Statistics in dental materials research: 2 things to plan ahead

A couple of statistical issues should be considered when designing a study in dental materials science.

(1) What statistical test(s) will be used to test the null hypothesis/hypotheses?

It is recommended to design the experiment in such a way that it is possible to test the data using one “global” test such as analysis of variance (ANOVA). The effect of one independent factor on one response variable is tested using one-way ANOVA in many papers on dental materials. Such study design includes e.g. the comparison of the degree of conversion (DC) of several materials cured under the same curing conditions (light, intensity, time, distance). The null hypothesis would be that there is no difference between the means for different materials. So, the independent factor is ‘material’ and the response variable is ‘DC’.

Two-way ANOVA is used to test the effect of two independent factors, e.g. material and light-curing unit (e.g. 3 materials are cured with either a halogen or an LED light-curing unit). Testing for interaction between the two factors shows whether or not the differences caused by one factor are consistent on different levels of the other factor. If so, the interaction is not significant (e.g. the DC may be higher in each material when cured with a halogen unit than an LED unit). Alternatively, if these differences are not consistent, then the interaction is significant (the DC may be higher in some materials when cured with a halogen and in others when cured with an LED unit). In this case, a series of one-way ANOVA must be used to examine this interaction more closely. This will, however, result in multiple testing which by default increases the chance of making the Type I Error (rejecting the null hypothesis when it is true) and some sort of correction is necessary to keep the overall significance level at the usual alpha=0.05. This correction most often means a decrease in the individual alpha value which also reduces the power of the statistical test.

Three-way ANOVA is sometimes used in dental materials science to study the effect of three independent factors on a particular response variable (e.g. the effect of material, light-curing unit and curing time on the DC of resin-based composites). Researchers are often tempted to test more and more factors in order to make their experiments robust. However, one has to keep in mind that the interpretation of three-way ANOVA is more difficult that that of two-way ANOVA and the post-test corrections may significantly reduce the power of the test. These are by no means the only tests used and are only an indication of the type of studies carried out in dental materials science.

(2) Power and sample size – power is the probability of not making Type II Error (failing to reject the null hypothesis when is false) i.e. power is the probability of correctly rejecting the null hypothesis when the difference between the groups truely exists. In sample size calculation prior to an experiment, the power of 80% is generally used as the cut-off point. So, when calculating the number of samples for each group, we need to know the following: number of levels (groups) that we will be comparing; significance level; power; estimated standard deviation (determined in a pilot study or taken from the literature) and the difference between the groups that we consider clinically relevant and don’t want to miss in our statistical testing. This last one may be tricky, because we often don’t know what difference between the groups is clinically relevant in a way that might affect the clinical performance of the tested materials. In this case, we may base our decision on the literature data or we can do a pilot study to find out the likely difference in the response variable between our groups which we would then use in the sample size calculations. Alternatively, if we already have a pre-determined number of samples in each group, we may be able to determine the power of our statistical test (i.e. how certain we are that our conclusion is correct).

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