| 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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