power.equivalence.md  R Documentation 
A function to calculate the power of the two onesided tests prodedure (TOST). This is
the probability that a confidence interval lies within a specified equivalence
interval. (See also package equivalence
, function tost
.)
power.equivalence.md(alpha, logscale, ltheta1, ltheta2, ldiff, sigma, n, nu)
alpha 
alpha level for the 2 emphttests (usually alpha=0.05).
Confidence interval for full test is at level 1 2* 
logscale 
whether to use logarithmic scale ( 
ltheta1 
lower limit of equivalence interval 
ltheta2 
upper limit of equivalence interval 
ldiff 
true difference (ratio on log scale) in treatment means 
sigma 

n 
number of subjects per treatment (number of total subjects for crossover design) 
nu 
degrees of freedom for 
power 
Power of TOST; the probability that the confidence interval will lie within ['theta1', 'theta2'] given 
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 OneSided Tests Procedure in Bioquivalence, Journal of Pharmacokinetics and Biopharmaceutics, 18, No. 2, 139–144.
Schuirmann, D.J. (1987). A comparison of the two onesided tests procedure and the power approach for assessing the equivalence of average bioavailability, Journal of Pharmacokinetics and Biopharmaceutics, 15. 657–680.
# Suppose that two formulations of a drug are to be compared on # the regular scale using a twoperiod 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)
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