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I can't seem to find a general method for deriving standard errors anywhere. I've looked on google, this website and even in text books but all I can find is the formula for standard errors for the mean, variance, proportion, risk ratio, etc... and not how these formulas were arrived at.

If any body could explain it in simple terms or even link me to a good resource which explains it I'd be grateful.

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I provide a general simple model and apply it, with all details worked out, in the post at stats.stackexchange.com/a/18609/919. This and many other posts on standard errors (almost a thousand to date) can be found by searching our site for "standard error" –  whuber Mar 7 at 14:21

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up vote 3 down vote accepted

What you want to find is the standard deviation of the sampling distribution of the mean. I.e., in plain English, the sampling distribution is when you pick $n$ items from your population, add them together, and divide the sum by $n$. We than find the variance of this quantity and get the standard deviation by taking the square root of its variance.

So, let the items that you pick be represented by the random variables $X_i, 1\le i \le n$, each of them identically distributed with variance $\sigma^2$. They are independently sampled, so the variance of the sum is just the sum of the variances. $$ \text{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n\text{Var}\left(X_i\right) = \sum_{i=1}^n\sigma^2 = n\sigma^2 $$

Next we divide by $n$. We know in general that $\text{Var}(kY)=k^2 \text{Var}(Y)$, so putting $k=1/n$ we have

$$ \text{Var}\left(\frac{\sum_{i=1}^n X_i}{n}\right) = \frac{1}{n^2} \text{Var}\left(\sum_{i=1}^n X_i\right) = \frac{1}{n^2} n\sigma^2 = \frac{\sigma^2}{n} $$

Finally take the square root to get the standard deviation $\dfrac{\sigma}{\sqrt{n}}$. When the population standard deviation isn't available the sample standard deviation $s$ is used as an estimate, giving $\dfrac{s}{\sqrt{n}}$.

All of the above is true regardless of the distribution of the $X_i$s, but it begs the question of what do you actually want to do with the standard error? Typically you might want to construct confidence intervals, and it is then important assign a probability to constructing a confidence interval that contains the mean.

If your $X_i$s are normally distributed, this is easy, because then the sampling distribution is also normally distributed. You can say 68% of samples of the mean will lie within 1 standard error of the true mean, 95% will be within 2 standard errors, etc.

If you have a large enough sample (or a smaller sample and the $X_i$s are not too abnormal) then you can invoke the central limit theorem and say that the sampling distribution is approximately normally distributed, and your probability statements are also approximate.

A case in point is estimating a proportion $p$, where you draw $n$ items each from a Bernouilli distribution. The variance of each $X_i$ distribution is $p(1-p)$ and hence the standard error is $\sqrt{p(1-p)/n}$ (the proportion $p$ is estimated using the data). To then jump to saying that approximately some % of samples are within so many standard deviations of the mean, you need to understand when the sampling distribution is approximately normal. Repeatedly sampling from a Bernouilli distribution is the same as sampling from a Binomial distribution, and one common rule of thumb is to approximate only when $np$ and $n(1-p)$ are $\ge5$. (See wikipedia for a more in-depth discussion on approximating binomial with normal. See here for a worked example of standard errors with a proportion.)

If, on the other hand, your sampling distribution can't be approximated by a normal distribution, then the standard error is a lot less useful. For example, with a very skewed, asymmetric distribution you can't say that the same % of samples would be $\pm1$ standard deviation either side of the mean, and you might want to find a different way to associate probabilities with samples.

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Thanks, this approach makes sense and I can see how it applies to the mean but I can't see how to extend it to other statistics. For example, how would I find the standard error of a rate? or a rate ratio? –  Daniel Gardiner Mar 7 at 15:38
    
I've updated my post. The key point is that quantities like mean, variance, etc -- and hence stderr -- can be found for any distribution. But to make probability statements you need to know something about the distribution, be it normal, binomial or whatever. So the stderr can always be found, but how useful it is depends on the situation. –  TooTone Mar 7 at 17:03

The standard error is the standard deviation of the statistic (under the null hypothesis, if you're testing). A general method for finding standard error would be to first find the distribution or moment generating function of your statistic, find the second central moment, and take the square root.

For example, if you're sampling from a normal distribution with mean $\mu$ and variance $\sigma^2$, the sample mean $\bar{X}=\frac{1}{n}\sum_{i=1}^{n} X_i$ is normally distributed with mean $\mu$ and variance $\sigma^2/n$. This can be derived from three properties:

  1. The sum of independent random variables is normal,
  2. $\mathrm{E}\left[\sum_{i=1}^{n} a_i X_i\right] = \sum_{i=1}^{n} a_i \mathrm{E}\left[ X_i \right]$,
  3. If $X_1$ and $X_2$ are independent, $\mathrm{Var}\left(a_1 X_1 + a_2 X_2 \right) = a_1^2 \mathrm{Var}\left(X_1\right) + a_2^2 \mathrm{Var}\left( X_2 \right)$.

Thus the standard error of the sample mean, which is the square root of its variance, is $\sigma/\sqrt{n}$.

There are shortcuts, like you don't necessarily need to find the distribution of the statistic, but I think conceptually it's useful to have the distributions in the back of your mind if you know them.

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