In a previous article, we outlined the required steps for sample size calculations when comparing 2 proportions. In this article, we will perform a sample calculation for comparisons of 2 rates using information on lingual retainer bond failures from the study by Lie Sam Foek et al. Before we proceed, we need to define rate and clarify its difference from a proportion. A proportion is the number of failures divided by the total number of participants. In the example used in the previous article, there were 105 failures out of 277 participants over the duration of the study; this gave a failure proportion of 105/277 = 0.38, or 38%. Risk also can be used instead of proportion, and we can say that the risk of lingual retainer failure over the entire followup duration is 38%. This proportion or risk does not include a time element and is just telling us that, over the entire observation period, 38% of the retainers failed.
However, what if we would like to know whether the failures occurred early or late in the followup period? Additionally, in those types of studies, participants are entering (bonding of fixed retainer) and exiting (broken retainer or lost to followup) the study at different times. It is evident that not all patients enter and exit the study at the same time, and failures are counted when they occur. The participants are contributing various amounts of time (also termed survival time ) to the study; survival time is better used with rates. But what is the definition of rate ? The rate is given by the following formula: number of failures/total followup times of all participants. The total followup period is calculated by adding up the times that each participant contributed to the study. For example, if we have 5 participants who were followed for 3, 4, 6, 7, and 8 months, respectively, 28 months (3 + 4 + 6 + 7 + 8) indicates the total followup time. If there were 2 failures during the followup, then the proportion of failures would be 2/5 = 0.4, or 40%; the rate of failures would be 2/28 months = 0.071 failures per personmonths. By digging further into the article, we see that the median followup period was about 20 months. So, if we assume that all patients were followed for an average of about 20 months, then the total followup time is 277 × 20 = 5540 months. We can then calculate the rate of failures in this study as 105/5540 = 0.019 failures per personmonths. The rate can be converted by rescaling the months into years (5540/12 = 461.67), and it would become 105/461.67 = 0.23 failures per personyears. It could be further rescaled per 100 personyears by multiplying the rate by 100 to become 23 failures per 100 personyears. The rate of 23 failures per 100 personyears indicates that, for every 100 persons followed on average for 1 year, we expect 23 lingual retainers to fail. The term rate is often used erroneously in orthodontics to indicate the proportion of failures and not rates as they are described above.
Let us briefly remind ourselves of the information we need before we proceed with the example: (1) the research question; (2) the principal outcome measure of the trial; (3) λ1, the expected rate of the standard or control treatment; (4) λ2, the expected rate of the alternative or new treatment, and hence the minimum clinically important difference in rates between treatment arms that we would like to detect is λ2 − λ1; and (5) the degree of certainty that we want to detect in the expected treatment difference (power or 1 – type II error: 1 – β) and the level of significance (type I error, α).
We are interested in conducting a trial in which we will compare lingual retainer failure rates between retainers bonded with conventional acid etching vs bonding with selfetching primers. We would like to determine whether there are differences at the 5% statistical level and with 90% power. The sequence for conducting the sample calculation for this study will be as follows.
 1.
We must make assumptions regarding the expected rate of failures in the control arm, which will be the conventional acidetching group in this example. Two possible sources that will help us to determine the expected rate of failure in the control arm could be previously published studies or piloting the trial we are designing. Lie Sam Foek et al found a failure rate of 0.23 per personyears, as we calculated previously, with conventional etching.
 2.
We must decide what is an acceptable difference to be observed that has clinical importance. If we want to detect a 25% reduction in the failure rate in the selfetching group, we will assume a failure rate of 0.23 × 0.75 = 0.1725, or approximately 0.17 failures per 100 personyears.
 3.
We must determine the test size and power, and take α = 0.05, or 5%, and power, 1 – β = 0.90, or 90%.
So far, we have the failure rate in the control arm, the minimum difference to be detected, and the desired significance and power levels. To carry out this calculation, we will use the formula described by Smith and Morrow. It assumes independently distributed outcomes, equal numbers of participants per arm, and no losses to followup:
This is the formula for sample calculation for rates, where PY is the required size per trial arm expressed in personyears, λ1 is the expected failure rate per year in subjects having standard treatment, λ2 is the the expected failure rate per year in subjects having the new treatment, and f (α, β) is a function of α and β, values for which are given in the Table .
α 2sided 
β  

0.05 95% power 
0.1 90% power 
0.2 80% power 
0.5 50% power 

0.05  13.00  10.51  7.85  3.84 
0.01  17.81  14.88  11.68  6.64 
By substituting the selected values in this formula: