Minimum number of specimens for a medical experiment Could someone validate the following - for the minimum number of specimens required. 
A pathogen has 8 different forms. 
Laboratory specimens (small organisms) are infected with the pathogen. 
For each different form of the pathogen: 2 organisms are to be dissected on day 3, another 2 organisms are to be dissected on day 7, and another 2 on day 14. The dissection determines whether "Y - organism still infected" or "N - no longer infected".
What is the minimum sample size of the organisms required so that there is adequate power in the signal (80% confidence level)? Is it 8 * 8 *3? 
 A: The sample size depends on whether you want to analyze each of the eight pathogens separately, or if the form of the pathogen doesn't matter. The former case are actually eight experiments, the latter just one.
Cochran (1953:54) provides a formula for calculating of the sample size $n_{\scriptsize0}$ of a dichotomous outcome such as "still infected" or "no longer infected",
$$n_{\scriptsize0}=\frac{Z^2pq}{e^2},$$
where $Z$ is the value that cuts off an area of size $\alpha$ at the tails of the standard normal distribution (hence assuming normality of $p$), $p$ the probability of the population value, $q$ it's counter-probability, and $e$ the tolerated error margin.
Now, since the $\alpha$ of your desired confidence level of $80\% $ is $1-.8=.2$, for a two-sided test you get a $Z_{\frac\alpha2}=Z_{0.1}=1.282$. Since $p$ is unknown, we assume maximum variability and use $p=0.5$. For $e$ we use the common tolerated probability of 0.05 to falsely reject the null hypothesis (p-value).
$$n_{\scriptsize0}=\frac{1.28^2\times 0.5 \times (1-0.5)}{0.05^2}$$
and get $n_{\scriptsize0}=164$ (rounded up) as the required sample size for one experiment.
