MBESS (version 4.1.0)

power.equivalence.md.plot: Plot power of Two One-Sided Tests Procedure (TOST) for Equivalence

Description

A function to plot the power of the two one-sided tests prodedure (TOST) for various alternatives. (See also package equivalence, function tost.)

Usage

power.equivalence.md.plot(alpha, logscale, theta1, theta2, sigma, n, nu, title2)

Arguments

alpha

alpha level for the 2 t-tests (usually alpha=0.05). Confidence interval for full test is at level 1- 2*alpha

logscale

whether to use logarithmic scale TRUE or not FALSE

theta1

lower limit of equivalence interval

theta2

upper limit of equivalence interval

sigma

sqrt(error variance) as fraction (root MSE from ANOVA, or coefficient of variation)

n

number of subjects per treatment (number of total subjects for crossover design)

nu

degrees of freedom for sigma

title2

Title appearing at bottom of plot

Value

References

Diletti, E., Hauschke D. & Steinijans, V.W. (1991) Sample size determination of bioequivalence assessment by means of confidence intervals, International Journal of Clinical Pharmacology, Therapy and Toxicology, 29, No. 1, 1-8.

Phillips, K.F. (1990) Power of the Two One-Sided Tests Procedure in Bioquivalence, Journal of Pharmacokinetics and Biopharmaceutics, 18, No. 2, 139-144.

Schuirmann, D.J. (1987) A comparison of the two one-sided tests procedure and the power approach for assessing the equivalence of average bioavailability, Journal of Pharmacokinetics and Biopharmaceutics, 15. 657-680.

Examples

# Suppose that two formulations of a drug are to be compared 
# on the regular scale using a two-period crossover design, 
# with theta1 = -0.20, theta2 = 0.20, rm(CV) = 0.20, and 
# we choose 
n<-c(9,12,18,24,30,40,60)

# corresponding to 
nu<-c(7,10,16,22,28,38,58)

# degrees of freedom.  We need to test bioequivalence at the 
# .05 significance level, which corresponds to having a .90 confidence
# interval lying within (-0.20, 0.20). This corresponds to 
# Phillips (1990),  Figure 3.  Use

power.equivalence.md.plot(.05, FALSE, -.2, .2, .20, n, nu, 'Phillips Figure 3')

# If the formulations are compared on the logarithmic scale with 
# theta1 = 0.80, theta2 = 1.25, and 

n<-c(8,12,18,24,30,40,60)

# corresponding  to 
nu<-c(6,10,16,22,28,38,58)

# degrees of freedom. This corresponds to Diletti, Figure 1c. Use

power.equivalence.md.plot(.05, TRUE, .8, 1.25, .20, n, nu, 'Diletti, Figure 1c')