n.wilcox.ord: Sample size for Wilcoxon-Mann-Whitney for ordinal data In samplesize: Sample Size Calculation for Various t-Tests and Wilcoxon-Test

Description

Function computes sample size for the two-sided Wilcoxon test when applied to two independent samples with ordered categorical responses.

Usage

 `1` ```n.wilcox.ord(power = 0.8, alpha = 0.05, t, p, q) ```

Arguments

 `power` required Power `alpha` required two-sided Type-I-error level `t` sample size fraction n/N, where n is sample size of group B and N is the total sample size `p` vector of expected proportions of the categories in group A, should sum to 1 `q` vector of expected proportions of the categories in group B, should be of equal length as p and should sum to 1

Details

This function approximates the total sample size, N, needed for the two-sided Wilcoxon test when comparing two independent samples, A and B, when data are ordered categorical according to Equation 12 in Zhao et al.(2008). Assuming that the response consists of D ordered categories C_1 ,..., C_D. The expected proportions of these categories in two treatments A and B must be specified as numeric vectors p_1,...,p_D and q_1,...,q_D, respectively. The argument t allows to compute power for an unbalanced design, where t=n_B/N is the proportion of sample size in treatment B.

Value

 `total sample size` Total sample size `m` Sample size group 1 `n` Sample size group 2

Ralph Scherer

References

Zhao YD, Rahardja D, Qu Yongming. Sample size calculation for the Wilcoxon-Mann-Whitney test adjsuting for ties. Statistics in Medicine 2008; 27:462-468

Examples

 ```1 2 3 4 5``` ```## example out of: ## Zhao YD, Rahardja D, Qu Yongming. ## Sample size calculation for the Wilcoxon-Mann-Whitney test adjsuting for ties. ## Statistics in Medicine 2008; 27:462-468 n.wilcox.ord(power = 0.8, alpha = 0.05, t = 0.53, p = c(0.66, 0.15, 0.19), q = c(0.61, 0.23, 0.16)) ```

Example output

```\$`total sample size`
[1] 8390

\$m
[1] 3943

\$n
[1] 4447
```

samplesize documentation built on May 29, 2017, 6:59 p.m.