Dear Makula!
❝ please tell me how to calculate power of pilot study by using statistical equations. 
I'm not sure whether I've got your question correctly...
You have a performed a pilot study, evaluated it (point estimate, intra-subject CV) and now want to calculate some kind of
a-posteriori power
'by hand'?
In the strict statistical sense this is not correct.
1
But if you want to play around with numbers (nothing for M$-Excel or even most of the basic software packages), just to give you an idea:
Quoting (a little modified from) Diletti
et al.2
Bioequivalence is concluded if the following two conditions hold true:
$$\small{t_1=\frac{\bar{X}_\textrm{T}-\bar{X}_\textrm{R}-\log_{e}(\theta_1)}{\hat{\sigma}\times\sqrt{2/n}}\geq t_{1-\alpha,n-2}\,\wedge}\tag{1}$$ $$\small{t_2=\frac{\bar{X}_\textrm{T}-\bar{X}_\textrm{R}-\log_{e}(\theta_2)}{\hat{\sigma}\times\sqrt{2/n}}\leq t_{1-\alpha,n-2},\;\,}\tag{2}$$ where \(\small{t_{1-\alpha,n-2}}\) denotes the \(\small{1-\alpha}\)-quantile of the central \(\small{t}\)-distribution with \(\small{n-2}\) degrees of freedom. Using this procedure for bioequivalence assessment, the sample size determination should be based on the corresponding power function. The power of a statistical test is the probability that the hypothesis \(\small{H_0}\), in this case bioinequivalence, is rejected if the alternative hypothesis \(\small{H_1}\), in this case bioequivalence, is true. In other words, the probability of correctly accepting bioequivalence is called the power of the test procedure:
\(\small{p(t_1\geq t_{1-\alpha,n-2}\;\wedge\;-t_2\leq t_{1-\alpha,n-2}\;|}\) bioequivalence holds).
The probability of concluding bioinequivalence, if bioequivalence is true, is the so-called producer risk, and this is equal to 1-power.
Owen3 has shown that the vector \(\small{(t_1,t_2)}\) has a bivariate noncentral \(\small{t}\)-distribution with \(\small{n-2}\) degrees of freedom and noncentrality parameters
$$\small{\delta_1=\frac{\log_{e}(\mu_\textrm{T}/\mu_\textrm{R})-\log_{e}(\theta_1)}{\sigma\times \sqrt{2/n}}}$$ $$\small{\delta_2=\frac{\log_{e}(\mu_\textrm{T}/\mu_\textrm{R})-\log_{e}(\theta_2)}{\sigma\times \sqrt{2/n}}}$$ and that the above expression for the power can be calculated by the difference of two definite integrals which depend on \(\small{\delta_1}\) and \(\small{\delta_2}\). The exact form of these integrals and an algorithm for their calculation has been provided by Owen (1965, pp 437–446) […]
Since the noncentral
t-distribution is not given in 'standard software' (although available in SAS) you may go
e.g. for
![[image]](img/uploaded/Rlogo_15_12.svg)
. Then start coding
Another source of the noncentral
t-distribution is
StaTable available from 'Cytel'.
If your study was inbalanced within sequences \(\small{n_1\neq n_2}\), substitute \(\small{2n}\) in all equations by \(\small{\frac{1}{2n_1}+\frac{1}{2n_2}}\), and use \(\small{t_{1-\alpha,n_1+n_2-2}}\) instead.
If you want an out-of-the-box solution you may try
StudySize by CreoStat HB.
- Hoenig JM, Heisey DM. The Abuse of Power. The Pervasive Fallacy of Power Calculations for Data Analysis. Am Stat. 2001;55(1):19–24. doi:10.1198/000313001300339897.
free resource.
- Diletti E, Hauschke D, Steinijans VW. Sample size determination for bioequivalence assessment by means of confidence intervals. Int J Clin Pharm Ther Toxicol. 1991;29(1):1–8.
- Owen DB. A special case of a bivariate non-central t-distribution. Biometrika. 1965;52:437–46.