power.equivalence.md.plot | R Documentation |
A function to plot the power of the two one-sided tests prodedure (TOST) for various alternatives. (See also package equivalence
, function tost
.)
power.equivalence.md.plot(alpha, logscale, theta1, theta2, sigma, n, nu, title2)
alpha |
alpha level for the 2 t-tests (usually alpha=0.05).
Confidence interval for full test is at level 1- 2* |
logscale |
whether to use logarithmic scale |
theta1 |
lower limit of equivalence interval |
theta2 |
upper limit of equivalence interval |
sigma |
|
n |
number of subjects per treatment (number of total subjects for crossover design) |
nu |
degrees of freedom for |
title2 |
Title appearing at bottom of plot |
power |
Plot of power of TOST (probability that (1-2* |
Kem Phillips; kemphillips@comcast.net
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.
## Not run:
# 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')
## End(Not run)
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