Obtains the power given sample size or obtains the sample size given power for the Wilcoxon test for two-sample ordinal response.
getDesignTwoOrdinal(
beta = NA_real_,
n = NA_real_,
ncats = NA_integer_,
pi1 = NA_real_,
pi2 = NA_real_,
allocationRatioPlanned = 1,
rounding = TRUE,
alpha = 0.05
)An S3 class designTwoOrdinal object with the following
components:
power: The power to reject the null hypothesis.
alpha: The two-sided significance level.
n: The maximum number of subjects.
ncats: The number of categories of the ordinal response.
pi1: The prevalence of each category for the treatment group.
pi2: The prevalence of each category for the control group.
meanscore1: The mean midrank score for the treatment group.
meanscore2: The mean midrank score for the control group.
allocationRatioPlanned: Allocation ratio for the active treatment
versus control.
rounding: Whether to round up sample size.
The type II error.
The total sample size.
The number of categories of the ordinal response.
The prevalence of each category for the treatment group.
Only need to specify the valued for the first ncats-1 categories.
The prevalence of each category for the control group.
Only need to specify the valued for the first ncats-1 categories.
Allocation ratio for the active treatment versus control. Defaults to 1 for equal randomization.
Whether to round up sample size. Defaults to 1 for sample size rounding.
The significance level. Defaults to 0.025.
Kaifeng Lu, kaifenglu@gmail.com
A two-sample ordinal response design is used to test whether the ordinal response distributions differ between two treatment arms. Let \(\pi_{gi}\) denote the prevalence of category \(i\) in group \(g\), where \(g=1\) represents the treatment group and \(g=2\) represents the control group.
The parameter of interest is $$\theta = \sum_{i=1}^{C} w_i (\pi_{1i} - \pi_{2i})$$ where \(w_i\) is the midrank score for category \(i\). The Z-test statistic is given by $$Z = \hat{\theta}/\sqrt{Var(\hat{\theta})}$$ where \(\hat{\theta}\) is the estimate of \(\theta\).
The midrank score \(w_i\) for category \(i\) is calculated as: $$w_i = \sum_{j=1}^{i} \pi_j - 0.5\pi_i$$ where \(\pi_i = r\pi_{1i} + (1-r)\pi_{2i}\) denotes the average prevalence of category \(i\) across both groups, and \(r\) is the randomization probability for the active treatment.
To understand the midrank score, consider \(n\pi_i\) subjects in category \(i\). The midrank score is the average rank of these subjects: $$s_i = \frac{1}{n\pi_i} \sum_{j=1}^{n\pi_i} ( n\pi_1 + \cdots + n\pi_{i-1} + j)$$ This simplifies to $$s_i = n\left(\sum_{j=1}^{i} \pi_j - 0.5\pi_i\right) + \frac{1}{2}$$ By dividing by \(n\) and ignoring \(\frac{1}{2n}\), we obtain the midrank score \(w_i\).
The variance of \(\hat{\theta}\) can be derived from the multinomial distributions and is given by $$Var(\hat{\theta}) = \frac{1}{n}\sum_{g=1}^{2} \frac{1}{r_g} \left\{\sum_{i=1}^{C} w_i^2\pi_{gi} - \left(\sum_{i=1}^{C} w_i\pi_{gi} \right)^2\right\}$$ where \(r_g\) is the randomization probability for group \(g\).
The sample size is chosen such that the power to reject the null hypothesis is at least \(1-\beta\) for a given significance level \(\alpha\).
(design1 <- getDesignTwoOrdinal(
beta = 0.1, ncats = 4, pi1 = c(0.55, 0.3, 0.1),
pi2 = c(0.214, 0.344, 0.251), alpha = 0.025))
Run the code above in your browser using DataLab