power.equivalence.md {MBESS} | R Documentation |
A function to calculate the power of the two one-sided tests prodedure (TOST). This is
the probability that a confidence interval lies within a specified equivalence
interval. (See package equivalence
, function tost
.)
power.equivalence.md(alpha, logscale, ltheta1, ltheta2, ldiff, sigma, n, nu)
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 |
ltheta1 |
lower limit of equivalence interval |
ltheta2 |
upper limit of equivalence interval |
ldiff |
true difference (ratio on log scale) in treatment means |
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 |
power |
Power of TOST (probability that (1- 2*alpha ) confidence interval will lie
within (theta1 , theta2 ) given sigma , n , and nu ) |
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.
power.equivalence.md.plot
, power.density.equivalence.md
, tost
# 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, the # difference in the mean bioavailability is 0.05 (5 percent), and we choose # n=24, corresponding to 22 degrees of freedom. We need to test # bioequivalence at the 5 percent significance level, which corresponds to # having a 90 percent confidence interval lying within (-0.20, 0.20). Then # the power will be 0.8029678. This corresponds to Phillips (1990), # Table 1, 5th row, 5th column, and Figure 3. Use power.equivalence.md(.05, FALSE, -.2, .2, .05, .20, 24, 22) # If the formulations are compared on the logarithmic scale with # theta1 = 0.80, theta2 = 1.25, n=18 (16 degrees of freedom), and # a ratio of test to reference of 1.05. Then the power will be 0.7922796. # This corresponds to Diletti, Table 1, power=.80, CV=.20, ratio=1.05, and Figure 1c. Use power.equivalence.md(.05, TRUE, .8, 1.25, 1.05, .20, 18, 16)