power.equivalence.md {MBESS}

Power of Two One-Sided Tests Procedure (TOST) for Equivalence

Description

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 also package equivalence, function tost.)

Usage

power.equivalence.md(alpha, logscale, ltheta1, ltheta2, ldiff, sigma, n, nu)

Arguments

alpha

\alpha level (Type I error rate) for the two 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 (square root of the mean square error from ANOVA, or coefficient of variation)

n

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

nu

degrees of freedom for sigma

Value

power

Power of TOST; the probability that the confidence interval will lie within ['theta1', 'theta2'] given sigma, n, and nu

Author(s)

Kem Phillips; [e-mail]

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, 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)