power.ancova: Power Calculation for ANCOVA

In stamats/MKpower: Power Analysis and Sample Size Calculation

Power Calculation for ANCOVA

Description

Compute sample size for ANCOVA.

Usage

power.ancova(n = NULL, mu = NULL, var = 1, nr.covs = 1L, group.ratio = NULL, 
             contr.mat = NULL, sig.level = 0.05, power = NULL, n.max = 1000L,
             rel.tol = .Machine$double.eps^0.25)

Arguments

n	vector of sample sizes per groups.
mu	vector of mean values of the groups.
var	error variance.
nr.covs	number of covariates (larger or equal than 1).
group.ratio	vector of group sizes relative to group 1; i.e., first entry should always be one. If NULL, a balanced design is used.
contr.mat	matrix of contrasts (number of columns must be idential to number of groups). If NULL, standard ANCOVA contrasts are used; see examples below.
sig.level	significance level (type I error probability)
power	power of test (1 minus type II error probability)
n.max	maximum sample size considered in the computations.
rel.tol	relative tolerance passed to function integrate.

Details

Exactly one of the parameters n and power must be passed as NULL, and that parameter is determined from the other.

The function includes an implementation of the exact approach of Shieh (2020). It is based on the code provided in the supplement of Shieh (2020), but uses integrate instead of the trapezoid rule and uniroot for finding the required sample size.

Value

Object of class "power.htest", a list of the arguments (including the computed one) augmented with a note element.

Author(s)

Matthias Kohl [e-mail]

References

G. Shieh (2020). Power Analysis and Sample Size Planning in ANCOVA Designs. Psychometrika 85:101-120. \Sexpr[results=rd]{tools:::Rd_expr_doi("10.1007/s11336-019-09692-3")}.

S.E. Maxwell and H.D. Delaney (2004). Designing experiments and analyzing data: A model comparison perspective (2nd ed.). Mahwah, NJ: Lawrence Erlbaum Associates.

See Also

power.anova.test, power.t.test

Examples

## Default matrix of contrasts
## 3 groups
cbind(rep(1,2), -diag(2))
## 4 groups
cbind(rep(1,3), -diag(3))

## Table 1 in Shieh (2020)
power.ancova(mu=c(400, 450, 500), var = 9900, power = 0.8)
power.ancova(n = rep(63/3, 3), mu=c(400, 450, 500), var = 9900)
power.ancova(mu=c(400, 450, 500), var = 9900, power = 0.8, nr.covs = 10)
power.ancova(n = rep(72/3, 3), mu=c(400, 450, 500), var = 9900, nr.covs = 10)

## Table 2 in Shieh (2020)
power.ancova(mu=c(400, 450, 500), var = 7500, power = 0.8)
power.ancova(n = rep(48/3, 3), mu=c(400, 450, 500), var = 7500)
power.ancova(mu=c(400, 450, 500), var = 7500, power = 0.8, nr.covs = 10)
power.ancova(n = rep(60/3, 3), mu=c(400, 450, 500), var = 7500, nr.covs = 10)

## Table 3 in Shieh (2020)
power.ancova(mu=c(400, 450, 500), var = 1900, power = 0.8)
power.ancova(n = rep(18/3, 3), mu=c(400, 450, 500), var = 1900)
power.ancova(mu=c(400, 450, 500), var = 1900, power = 0.8, nr.covs = 10)
power.ancova(n = rep(27/3, 3), mu=c(400, 450, 500), var = 1900, nr.covs = 10)

## ANOVA approach for Table 1-3
power.anova.test(groups = 3, between.var = var(c(400, 450, 500)), 
                 within.var = 10000, power = 0.8)
power.anova.test(n = 63/3, groups = 3, between.var = var(c(400, 450, 500)), 
                 within.var = 10000)

## Table 4 in Shieh (2020)
power.ancova(mu=c(410, 450, 490), var = 9900, power = 0.8)
power.ancova(n = rep(96/3, 3), mu=c(410, 450, 490), var = 9900)
power.ancova(mu=c(410, 450, 490), var = 9900, power = 0.8, nr.covs = 10)
power.ancova(n = rep(105/3, 3), mu=c(410, 450, 490), var = 9900, nr.covs = 10)

## Table 5 in Shieh (2020)
power.ancova(mu=c(410, 450, 490), var = 7500, power = 0.8)
power.ancova(n = rep(72/3, 3), mu=c(410, 450, 490), var = 7500)
power.ancova(mu=c(410, 450, 490), var = 7500, power = 0.8, nr.covs = 10)
power.ancova(n = rep(84/3, 3), mu=c(410, 450, 490), var = 7500, nr.covs = 10)

## Table 6 in Shieh (2020)
power.ancova(mu=c(410, 450, 490), var = 1900, power = 0.8)
power.ancova(n = rep(24/3, 3), mu=c(410, 450, 490), var = 1900)
power.ancova(mu=c(410, 450, 490), var = 1900, power = 0.8, nr.covs = 10)
power.ancova(n = rep(33/3, 3), mu=c(410, 450, 490), var = 1900, nr.covs = 10)

## ANOVA approach for Table 4-6
power.anova.test(groups = 3, between.var = var(c(410, 450, 490)), 
                 within.var = 10000, power = 0.8)
power.anova.test(n = 96/3, groups = 3, between.var = var(c(410, 450, 490)), 
                 within.var = 10000)

###############################################################################
## Example from Maxwell and Delaney (2004) according to Shieh (2020)
###############################################################################
## ANCOVA (balanced design)
power.ancova(n = rep(30/3, 3), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898)
power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8)
power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.9)

## ANOVA
power.anova.test(n = 30/3, groups = 3, between.var = var(c(7.5366, 11.9849, 13.9785)), 
                 within.var = 29.0898)
power.anova.test(groups = 3, between.var = var(c(7.5366, 11.9849, 13.9785)), 
                 within.var = 29.0898, power = 0.8)
power.anova.test(groups = 3, between.var = var(c(7.5366, 11.9849, 13.9785)), 
                 within.var = 29.0898, power = 0.9)
                 
## ANCOVA - imbalanced design
power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, 
             group.ratio = c(1, 1.25, 1.5))
power.ancova(n = c(13, 16, 19), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898,  
             group.ratio = c(1, 1.25, 1.5))
power.ancova(mu=c(7.5366, 11.9849, 13.9785), var = 29.0898, power = 0.8, 
             group.ratio = c(1, 0.8, 2/3))
power.ancova(n = c(17, 14, 12), mu=c(7.5366, 11.9849, 13.9785), var = 29.0898,  
             group.ratio = c(1, 0.8, 2/3))

stamats/MKpower documentation built on April 10, 2024, 3:34 p.m.