getDesignTwoOrdinal: Power and Sample Size for the Wilcoxon Test for Two-Sample...

In lrstat: Power and Sample Size Calculation for Non-Proportional Hazards and Beyond

Power and Sample Size for the Wilcoxon Test for Two-Sample Ordinal Response

Description

Obtains the power given sample size or obtains the sample size given power for the Wilcoxon test for two-sample ordinal response.

Usage

getDesignTwoOrdinal(
  beta = NA_real_,
  n = NA_real_,
  ncats = NA_integer_,
  pi1 = NA_real_,
  pi2 = NA_real_,
  allocationRatioPlanned = 1,
  rounding = TRUE,
  alpha = 0.05
)

Arguments

beta	The type II error.
n	The total sample size.
ncats	The number of categories of the ordinal response.
pi1	The prevalence of each category for the treatment group. Only need to specify the valued for the first ncats-1 categories.
pi2	The prevalence of each category for the control group. Only need to specify the valued for the first ncats-1 categories.
allocationRatioPlanned	Allocation ratio for the active treatment versus control. Defaults to 1 for equal randomization.
rounding	Whether to round up sample size. Defaults to 1 for sample size rounding.
alpha	The significance level. Defaults to 0.025.

Details

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.

Value

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.

Author(s)

Kaifeng Lu, [e-mail]

Examples

(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))

lrstat documentation built on Aug. 8, 2025, 7:36 p.m.