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Sample size justification is a very crucial part in the design of clinical trials. In this paper, the authors derive a new formula to calculate the sample size for a binary outcome given one of the three popular indices of risk difference. The sample size based on the absolute difference is the fundamental one, which can be easily used to derive sample size given the risk ratio or OR.
Keywords: odds ratio, relative risk, risk differenceSample size calculation is an essential part in the design of clinical trials. In many cases, a primary outcome of interest is compared between two groups (namely control and treatments groups) in a trial. Usually, the sample size calculation is related to the test statistic used in such comparisons. However, as discussed in section 2, more assumptions are needed to uniquely determine the sample size.
Suppose we want to design a clinical trial to determine if the treatment effect of a new drug is better than that of the current one. The most popular design is to assign patients randomly to the treatment group (the new drug) and the control group (current drug). If the outcome is a success or a failure, it is a binary variable. Generally, for binary outcomes, there are three popular measures of treatment difference: risk difference, relative risk and OR. 1 See Feng and colleagues 2 for relationships among these three measures of effect size.
Formulas for sample size estimation for the binary outcome has been well developed and incorporated in many statistical software packages. See, for example, the formula of Chow and colleagues 3 (4.2.2). In this paper, the authors derive another formula using more information in the hypotheses. Once the sample size formula based on risk difference is obtained, the sample size formulas for the other two indices can be acquired easily.
In this paper, the authors consider the sample size calculation in the parallel design. This paper is organised as follows. In section 2, the authors derive a new formula based on the null hypothesis of the success rate in the control group and the proposed difference of success rates of two groups under the alternative hypothesis. The authors compare it with the formula in Chow and colleagues. 3 . Sections 3 and 4 derive a formula for OR and relative risk. The conclusion and discussion are in section 5.
Suppose the true success rates of two groups are p 1 and p 2, respectively. Since p 1, p 2∈(0,1), let Θ=(0,1) × (0,1) be the parameter space and Θ0=θ ∈ Θ:p1=p2>. Usually, given the significance level α and power 1−β, the sample size calculation depends on the null and alternative hypotheses about the parameters. In the current context, we want to see if the success rates of two groups are the same. Here we consider several scenarios of the hypotheses.
The null and alternative hypotheses are specified as:
H 0 : p 1 = p 2 a n d H 1 : p 1 ≠ p 2The specification in 1 is the same as:
H 0 : θ ∈ Θ 0 a n d H 1 : θ ∈ Θ ∖ Θ 0 .
Although we can test H 0 with data in data analysis, the specification in equation 1 does not offer us enough information to calculate the sample size, as the alternative hypothesis lacks specific details about the treatment effect. Both the null and alternative hypotheses in equation 1 are composite. In fact, any p 1 and p 2 are potential candidates for H 1 as long as they are not equal to each other. However, the power to reject the null hypothesis depends on the true success rates in two groups if they are different. For example, p 1 = 0.1 , p 2 = 0.2 and p 1 = 0.2 , p 2 = 0.3 both satisfy the alternative hypothesis. We have different powers to reject the hypothesis that they have the same success rates in these two cases.
The hypotheses are specified as:
H 0 : p 1 = p 2 a n d H 1 : p 2 = p 1 + Δ .where Δ is a prespecified known constant. Although both null and alternative hypotheses are still composite, the alternative hypothesis in equation 2 is much simpler than that in equation 1. It turns out that we still do not have sufficient information to determine the sample size. For example, consider the following two special cases:
Case 1. The hypotheses are:
H 0 : p 1 = p 2 = 0.1 a n d H 1 : p 1 = 0.1 , p 2 = 0.3.Case 2. The hypotheses are:
H 0 : p 1 = p 2 = 0.4 a n d H 1 : p 1 = 0.4 , p 2 = 0.6.In both cases, p 2 = p 1 + 0.2 under the alternative hypothesis. However, in the following sections, we will show that the sample sizes in these two cases are different. Given the difference of success rates, usually it is much easier to reject the null hypothesis in case 1 than in case 2.
The null and alternative hypotheses are:
H 0 : p 1 = p 2 = p 0 a n d H 1 : p 1 = p 0 , p 2 = p 0 + Δ ,where p 0 and Δ are prespecified constants. Without the loss of generality, we assume that Δ>0 in the following discussion. It turns out that we can uniquely determine the sample size in this case.
We derive a sample size formula based on the hypotheses specified in 3 using the large sample theory. 4 The typical way is to first derive the asymptotic distribution of a test statistic under the null and alternative hypothesis followed by solving an equation to obtain the sample size formula (with the given significance level and power) (see, eg, Tu et al 5 ).
Although the treatment and control groups have the same sample size in many studies, it is unnecessary in practice. Some studies intentionally assign more patients in one group. Suppose the sample size in groups 1 and 2 are n and nκ , respectively, where κ is a prespecified positive constant. Group 2 has more (less) subjects than group 1 depending on if κ > 1 ( 0 ≤ κ ≤ 1 ) . If κ = 1 , the two groups have an equal sample size.
Let p ^ 1 = m 1 n and p ^ 2 = m 2 n κ denote the estimates of p 1 and p 2 , where m 1 ( m 2 ) denote the number of events of success in group 1 (equation 2). According to the central limit theorem, 6
n ( p ^ 1 − p 1 ) → N ( 0 , p 1 ( 1 − p 1 ) ) , n κ ( p ^ 2 − p 2 ) → N ( 0 , p 2 ( 1 − p 2 ) ) ,as n is large enough.
Under the hypothesis of p 1 = p 2 = p 0 , the variances of p ^ 1 and p ^ 2 are p 0 ( 1 − p 0 ) / n and p 0 ( 1 − p 0 ) / ( n κ ) , respectively. To test the null hypothesis that p 1 = p 2 , we consider the following test statistics:
T = p ^ 2 − p ^ 1 p 0 ( 1 − p 0 ) n + p 0 ( 1 − p 0 ) n κ .Then T → N ( 0 , 1 ) as n grows unbounded.
Let Φ be the distribution of standard normal distribution. For each η ∈ ( 0 , 1 ) , let z η be such that Φ ( z η ) = η , that is, z η is the ( 100 × η )th percentile of the standard normal distribution. Given the significance level α , we reject the hypothesis of p 1 = p 2 i f | T | > z 1 − α / 2 . Note that: