How to determine experiment sample size for a target CI width? I'm working on an experiment to measure property $X$ of some pieces of equipment. Assuming there is an excess of equipment available for testing, and barring any funding restrictions, what is the best procedure for choosing how many measurements to make, so that I have a set confidence interval width?
Valid values for property $X$ are strictly positive, but their range is quite wide; typical values can reasonably be as little as $0.01$ and as large as $3000$. For this particular test, I expect the measurements to fall in the $3-20$ range. 
 A: I am hardly a good statistician, but I'll try to contribute. Can't comment for now (not enough Reputation) so directly as an answer. Assuming you want a CI at 95% and normal distribution of your measurements, CI definition is
$CI= 1.96* (sd /\sqrt{N})$
with sd the standard deviation and N the number of samples.
Your measurements will be between $3-20$, so you might assume worst-case-scenario with large sd.  Based on thin air I assumed a sd of $15$ but very simple (and perhaps unnecessary) R simulations showed sd to be between $4$ and $6.5$ for an expected range of $3-20$. A sd of $6.5$ may then help choose a "cautious" (i.e, quite large) sample size.
EDIT: simulations for sd estimation
# Ranges for outcomes
widest_range=seq(from=0,to=3000,by=1)
likely_range=seq(from=3,to=20,by=1)

# matrix to store the results
nb_sims=30
Mat_Results=matrix(0,ncol=2,nrow=nb_sims)
colnames(Mat_Results)=c("Widest_range","Likely_range")

# sims
for (i in (1:nb_sims)) {
  # sampling assuming 10 measures (EDIT: uniform distribution)
  measures_widest=sample(widest_range, 10, replace = FALSE, prob = NULL)
  measures_likely=sample(likely_range, 10, replace = FALSE, prob = NULL)
  # sd
  Mat_Results[i,1]=sd(measures_widest)
  Mat_Results[i,2]=sd(measures_likely)
}

x11()
par(mfrow=c(1,2))
boxplot(Mat_Results[,1],main="SD with widest range")
boxplot(Mat_Results[,2],main="SD with likely range")

A: In my experience, taking a small sample (less than 10, or 5% of the overall number of equipments) doing several measurements gives an indication of how many measurements need to be done to have a good CI.
For instance, 
if I have 50 pieces, I am going to take 5 pieces, do several measurements until I find a balance for each of the 5 pieces, ranging from 3 measurements to 10 first. If I don't find a balance, I'll redo the measurements, but this time 11 to 20 times (and so on). Once I have a smoothed average of measurements, I'll deploy the method to the overall pieces.
This will give you an estimated standard deviation.
Based on that and your level of confidence, you can, then, build your CI with the formula I assume you already know.
