Percentiles of a distribution I'm measuring error on the median of my data using bootstrapping. I would like to obtain I sigma error bars on my data, so I'm measuring the 16th and 84th percentiles of my data. Should I divide these percentile values by the square root of the number of data or not?
Thanks in advance.
 A: Since you are bootstrapping, why not take all of your observed medians and calculate the sample standard deviation $s$ and use that as your estimator of $\sigma$? 
A: If I am understanding your intent, then the answer is "no", don't divide.  The sample size is taken into account as part of the bootstrapping process.  Of course the values that are using will mostly be meaningful if everything is normally distributed, the fact that you are bootstrapping makes that seem an unlikely assumption.  Skewness could really throw this off.  Are you dividing the distance from 16th to 84th by 2?  If not that could be why you think you need to dived by something.
In general (and good to do here to check if I understood you and even if this method will work for your case) you can answer questions like this through a little simulation.  Simulate a data set and do the above analysis, calculate the value both ways and see which is closest to the "true" value from the simulation.  Repeat the process with a different sample size, also simulate from different distributions.  It should become fairly clear after a few simulations whether to divide or not.
A: Following along the lines developed by Greg and soakley, here's a function that calculates the (bootstrapped) standard errors of an estimate of the median:
median.w.se = function(vec,B){
  # Inputs:  vector of data (vec)
  #          number of bootstrap replicates (B)
  # Outputs: list with estimates of median and standard error
  empty.vec = rep(NA,B)
  for(i in 1:B){
  curr.sample = sample(vec,length(vec),replace = TRUE)
  curr.med    = median(curr.sample)
  empty.vec[i] = curr.med
  }
  lst = list(median = mean(empty.vec), se = sd(empty.vec))
  return(lst)
}

data(iris)
median.w.se(iris$Sepal.Length,1000)

