Calculate sample size for clinical trial without raw data Usually, we calculate the sample size needed for a clinical trial based on a former clinical trial which shows there is a significant effect. What if there is no such a trial? What if there is no information? How do we decide the sample size? Any idea?

It would be unusual (if not irresponsible) for a clinical trial to be conducted on a new drug in a complete vacuum of
information. You need to try to find a suitable sample
size for such a trial, given the information you do have
along with customary practice for clinical trials.
Suppose the proposed clinical trial is for a new drug that has
determined to be safe to use on human subjects. Now it is
time to assess its effectiveness.
An existing drug for the same condition gives favorable
results with 70% of subjects, and you consider this "cure rate" to be accurate. The new drug will be considered
worth additional attention, if it raises the cure rate to 80%.
Suppose you want to work at the 5% level of significance
and you want power (probability of detection of increased
cure rate as above) to be about 95%. Then you have
enough information, even if partially speculative, to determine a suitable sample size.
Here is a 'power and sample size' procedure from a recent
release of Minitab. Other statistical software programs
and some online 'calculators' work similarly. You will
need about 800 suitable subjects randomized into two
groups of 400 each, one receiving the old drug and one receiving the new.
Power and Sample Size 

Test for Two Proportions

Testing comparison p = baseline p (versus >)
Calculating power for baseline p = 0.7
α = 0.05

              Sample  Target
Comparison p    Size   Power  Actual Power
         0.8     404    0.95      0.950370

The sample size is for each group.


There are several versions of tests comparing two
proportions. Here is an example of results from prop.test
in R for a trial with 320 cures for the new drug and 280 for
the old drug out of 400 in each group. (Of course you can't
expect sample proportions will always come so close to
theoretical population proportions).
prop.test(c(320, 280), c(400,400), alt="g", cor=F)

        2-sample test for equality of proportions 
        without continuity correction

data:  c(320, 280) out of c(400, 400)
X-squared = 10.667, df = 1, p-value = 0.0005454
alternative hypothesis: greater
95 percent confidence interval:
 0.04997373 1.00000000
sample estimates:
prop 1 prop 2 
   0.8    0.7 

A simulation in R of $100\,000$ trials with binomial data can be used to determine whether a trial
with 400 subjects in each group will perform as expected.
The result is a simulated power very nearly 95%.
set.seed(2022)
pv = replicate( 10^5,
 prop.test(c(rbinom(1,400,.8),rbinom(1,400,.7)),
           c(400,400), alt="g", cor=F)$p.val )
mean(pv <= 0.05)
[1] 0.94826      # aprx power

