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I am working with some bird mortality data across 9 years of surveys and have run up against a problem estimating the standard deviation.

Across the 9 years, survival estimates were .47,.60,.36,.58,.57,.50,.57,.46, and .40. Within each of the annual survival estimates, let's say there were 100 data points every year.

what is the standard deviation and standard error, assuming all values are drawn from the same population (not a legitimate assumption, I know)?

Is the standard deviation simply sqrt(p * (1-p))? that just seems too large? I am not sure how this would account for dispersion differences. For example, you could imagine another scenario where the survival estimates had a greater spread but the same mean (e.g. .1,.2,.3,.8,.7,.8,.6,.4,.6) that would have the same overall mean and thus would arrive at the same standard deviation.

Here is some R code that I have used to try and figure it out. dat is the simulated survival (1) and mortality(0) data, while prop represents the proportions.

set.seed(250)
Randprop = c(.47,.60,.36,.58,.57,.50,.57,.46,.40)
func = function(x){return(rbinom(100,1,x))}
dat = lapply(prop,func)
prop = unlist(lapply(dat,mean))

Maybe what I want is simply the standard deviation ( e.g. sd(prop) in R), but I can't really think of a justification, other than the fact that it would include information on the variation between the sampled proportion vlaues.

As for the standard error, is N = 9 (the number of years) or is N = total number of data points (9 * 100).

Any help would be appreciated.

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  • $\begingroup$ There seems to be some confusion expressed here. In particular, sd simply computes an (adjusted) standard deviation. It has nothing to do with any assumptions of normality. You might therefore want to rephrase your question in a way that distinguishes basic statistics, like the SD, from your objectives. In particular, what are you actually trying to find out about these data? $\endgroup$
    – whuber ♦
    Commented Jun 8, 2017 at 17:01
  • $\begingroup$ Thanks. I edited my question to reflect that sd() simply returns the standard deviation of a series of numbers and is distribution agnostic. Is that correct? $\endgroup$
    – user44796
    Commented Jun 8, 2017 at 19:00
  • $\begingroup$ First $n=100$ the sample size. On the other hand, 9ie the number of times the exercise was replicated. Ie number of replications. $\endgroup$
    – Onyambu
    Commented Nov 20, 2020 at 1:25
  • $\begingroup$ In R you could simulate the proportions: ie replicate(5,rowMeans(matrix(rbinom(9*100, 1, Randprop), 9))) each column is a replicate of the Randprop vector $\endgroup$
    – Onyambu
    Commented Nov 20, 2020 at 4:58

1 Answer 1

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If all your values are from the same sample size then you can compute their average to obtain the overall rate, 0.501. You can make a better estimate, particularly if the sample sizes vary by multiplying each value by the sample size to get the annual number, total these as the numerator (say s) and total the sample sizes as the denominator (say n) and compute the rate, p = s/n. The variance and standard error use the standard formulae. That is se = sqrt(p(1-p)/n).

If your sample sizes are all 100, then you will get an overall rate of 0.501 and a se of 0.017.

Doubtless your sample sizes are not all equal or equal to 100.

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    $\begingroup$ In light of the remark about the possibility that "the survival estimates had a greater spread," wouldn't it behoove one to--at the very least--perform a check for overdispersion? And if that's the case, just what would your "se" actually mean? $\endgroup$
    – whuber ♦
    Commented Jun 8, 2017 at 19:12
  • $\begingroup$ The se is only appropriate if the rates are homogeneous. $\endgroup$
    – David Smith
    Commented Jun 8, 2017 at 19:20
  • $\begingroup$ My question is this. I have two set of proportional sample values. both have the same mean proportion. One set is 47,.60,.36,.58,.57,.50,.57,.46, and .40. The second set is .1,.2,.3,.8,.7,.8,.6,.4,.6. Intuitively, I would think the first set would have a smaller standard deviation relative to the second set. Or do you simply go by the standard deviation formula for the binomial distribution? That doesn't seem intuitive to me. As for the overdispersion, not sure how to test it for binomial data like this. Any suggestions would be appreciated. $\endgroup$
    – user44796
    Commented Jun 8, 2017 at 21:02

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