From: Eric N Lundquist
Sent: Friday, October 11, 2002

"Standard error" means the scatter of all the possible sample means around the population mean of a continuous variable (like height, age, test score). NOT the scatter of particular scores; it's the scatter of the MEANS of all the samples (of a given size "n") you could take of those scores. A proportion is just the mean of a discrete variable (yes-or-no, success-or-failure, 0-or-1, etc.) -- usually one with only two categories, but the math can be extended if you want. So standard error of the mean and standard error of a proportion are the same thing but for different kinds of variables, and with different formulas involved.

Let me assume you know what a sample's standard deviation is and how to calculate it. In words instead of symbols (cause I can't type them), the variance is

"(sum of [(x - mean)squared]) / n"

(or if you're estimating the population s.d., you divide by (n - 1) instead, as you probably know -- let's forget that for now). And the standard deviation is the square root of that.

Standard error of the mean says, take a random sample of size n -- say you take a measure on 10 people. Then take another random sample of size n (ten more people). Then another. Keep doing it. For every sample you take you can find the mean of those ten scores. Every sample mean will be a little different, but they'll all be dancing around the true mean in the population. Like, you might measure the average pregancy duration for ten women and find it's 38.5 weeks, and in another group of ten it might be 39.2 weeks, and in another maybe it's 40.1 weeks. If you keep taking those samples forever, you know the mean of ALL the sample means is going to be 39 weeks, because that's the average human gestation period. (Isn't it?) So the mean of all the sample means is equal to the population mean. You could also find the standard deviation of all those sample means, but that's NOT equal to the population standard deviation. Cause think about it, the mean of a sample tends to be closer to the population mean than just one particular point would be, otherwise we'd just ask one person our questions instead of getting an average from a bunch of people to represent the population. So this standard deviation of all the sample means will be smaller than the population standard deviation of individual scores.

You might use the exact same math you use when you find the standard deviation of the scores in one sample. Just, instead of "x" being one of the scores and "the mean" being the sample mean, you would have "x" be one of the means from one of your samples and "the mean" would be the whole population mean. Instead of asking "how are the scores in a sample distributed around the sample mean?", you would be asking "how are all the possible sample means distributed around the population mean?" It's literally the standard deviation of the sample means around the population mean, but it's called the standard error of the mean so as not to confuse the two ideas.

The math is really easy though. Realistically you don't actually take multiple sample means and use the same old basic std.dev. formula, because you would have to take every possible sample of a given size. How many samples of size ten could you take of all the women who have ever given birth? It might as well be infinite. So you just stick to the one sample you do have and through someone's magical formula which you can just trust, it turns out all you have to do to estimate the standard error of the mean is, take your one sample's standard deviation, and divide by the square root of the sample size:

std. err. of mean = "(std. dev.) / (square root of n)"

And since the std. dev. squared is the variance, sometimes that formula is written as

std. err. of mean = "square root of (variance / n)"

which is the same thing algebraically.

Standard error of a proportion is the exact same thing, except the formula for the standard deviation is different. Proportions are for things like coin tosses or yes / no responses (or yes / no / undecided if you want to make more categories, but that gets more complicated). A pregnancy can last 273 days, or 274, or 275, 277, 282, 296 etc. - it's a continuous variable with lots of possible values. But coin tosses aren't - they can only be heads or tails, or numerically, 1 or 0. So the proportion of heads also tells you the proportion of tails. That makes the math a lot simpler -- the mean proportion of heads is the probability of a head (=.5). That, times the number of observations or tosses, is the number of heads you'd expect to see. If you toss the coin 30 times you'd expect 15 heads.

(It's not always .5 though; you might be doing some genetics thing where you expect .75 of a couple's kids to have brown eyes and .25 to have blue eyes, or something like that. Same formula works.)

You probably wouldn't get exactly 15 heads out of 30 because there's some chance in the throwing, which means variability in the observations. It would be kind of weird if it were always EXACTLY half heads. So there's a standard deviation for the proportion of heads, and the formula for it is just

std.dev. = "square root of [probability of heads)x(probability of tails)]"

and the probability of tails is just (1-probability of heads); obviously if you're getting half heads, the other half is tails. So you can write it as

std.dev. = "square root of [(probability of heads)x (1 - probability of heads)]"

and by the way, that number squared is the variance,

variance = [(probability of heads)x (1 - probability of heads)]

I found a guy who did the algebra if you don't believe it, http://www.tufts.edu/~gdallal/psd.htm

The whole story about the standard error of the mean applies again here, so now that you have the formula for the standard deviation and variance of your sample proportion, you again have

std. err. of mean = "(std. dev.) / (square root of n)"
which becomes

"square root of [(probability of heads)x (1 - probability of heads)] / (square root of n)"

Alternately, you also had

std. err. of mean = "square root of (variance / n)"

which would be

"square root of [(probability of heads)x (1 - probability of heads) / n]"