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I measure the weight of 100 people and pick 20 lots of 5 people from these 100. Therefore n = 5, repeated 20 times. The 100 people represent the population and groups of 5 people represent samples from the population.

The mean weights (kg) of my 20 samples of n = 5 are 76, 71, 81, 75, 77, 78, 91, 80, 70, 78, 72, 78, 75, 75, 79, 70, 80, 74, 75, 75.

In R, I can combine these mean weights into a vector called x using this code:

x <- c(76, 71, 81, 75, 77, 78, 91, 80, 70, 78, 72, 78, 75, 75, 79, 70, 80, 74, 75, 75)

If I wanted to calculate the standard error of the mean weights, would I simply do the code below?

sd(x)

When we repeatedly sample from a population and calculate the standard deviation of the sampling distribution of the mean, should the standard error still be refered to as the standard error or just the standard deviation of means?

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  • $\begingroup$ What you are computing is the standard deviation of those means, so you should call it that way. It's not the standard deviation of the weights in your population. $\endgroup$ – Cristián Antuña Mar 3 '15 at 20:11
  • $\begingroup$ So the standard error should only be refered to as the standard error when we calculate just 1 mean? $\endgroup$ – luciano Mar 3 '15 at 20:15
  • $\begingroup$ What do you mean by 'just 1 mean'? $\endgroup$ – Cristián Antuña Mar 3 '15 at 20:19
  • $\begingroup$ If we take 1 sample, calculate the mean of the sample and calculate sqrt(sd(x)/n), should we use the term 'standard error of the mean'? If we take lots of samples, calculate the means of these samples and calculate the standard deviation of the means of these samples, should we use the term 'standard deviation of the means'? $\endgroup$ – luciano Mar 3 '15 at 20:30
  • $\begingroup$ If you have 1 sample and calculate its standard deviation, then you just quote it like it (because it is defined that way, it is understood that you refer to the sd in your sampled population). If you are calculating the sd of several means then the statistic (i.e. what you have calculated) refers not to the original population but to a new one - the vector of means. So in this last case, you are calculating the standard deviation of computed means. It is not common usage so that's why I'd quote it carefully, trying to be as careful as I can. $\endgroup$ – Cristián Antuña Mar 3 '15 at 20:39
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I think the confusion here is partly due to the looseness in the way we typically talk about these topics (myself included). There isn't really any such thing as "the standard error". There is a standard error of the mean, and a standard error of the variance, etc., but not a standard error per se.

The standard deviation of a set of means drawn from the same population is an estimate of the standard error of the mean. The standard deviation of a sample, divided by $\sqrt N$, is also an estimate of the standard error of the mean. This was Fisher's original insight leading to the ANOVA test (see my answer here: How does the standard error work?).

Two more notes:

  • If we take your example literally, you have a finite population, so you may want to use the finite population correction. (We also have a tag for this topic on CV: .)
  • When you bootstrap, although you are drawing from a finite dataset, the assumption is that the population is infinite. You are simply assuming that the population mirrors the sample, and therefore you can use the sample as your population by sampling with replacement (which makes the population infinite).
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If I wanted to calculate the standard error of the mean weights, would I simply do the code below?

sd(x)

Yes

When we repeatedly sample from a population and calculate the standard deviation of the sampling distribution of the mean, should the standard error still be refered to as the standard error or just the standard deviation of means?

You have to be careful here. This is not the repeated sampling one would use to get the standard deviation of the 5-sample mean. You'd have to sample with replacement for that, i.e. after getting 5-people sample, you'd return them back to the population before sampling randomly again.

What you are doing is slightly different. You got the standard deviation of the partition. You split 100 into 20 partitions, and got the standard deviation of the mean of the parts of this partition.

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  • $\begingroup$ +1, the point about the partition is a good addition to this thread. $\endgroup$ – gung Mar 4 '15 at 21:00
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If your samples are selected by simple random sampling (which I believe is the default way most sample functions in R packages work), you can estimate the standard error of the mean of your population using the standard error of the mean of your sample using the formula ($\sigma_{s}$) :

$$\widehat{\sigma_{p}} = \sqrt{\left(\left(1-\dfrac{n}{N}\right) \cdot \dfrac{1}{n}\right)} \cdot \dfrac{N}{N-1} \cdot \sigma_s$$

with $N = 100$ and $n=5$.

The estimator $\widehat{\sigma_{p}}$ is (almost) unbiased, so you can improve this estimator using all your samples by taking the mean of $\widehat{\sigma_{p}}$ over the 20 samples.

In R, something like that should work :

sdEstimator <- function(sample) {
  return(sqrt((1-n/N)*1/n) * N / N-1 * sd(sample))
}

And assuming you put all sdEstimator values in sdEstVector, your new estimator can be computed with:

mean(sdEstVector)
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