Standard error of the mean
Take a sample from a population, calculate the mean of that sample, put everything back, and do it over and over. How much do those sample means tend to vary from the "average" sample mean? This is what the standard error of the mean measures. Its longer name is the standard deviation of the sampling distribution of the sample mean. Created by Sal Khan.
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- Can you help me understand when we use (sigma)/(√n) rather than just using sigma? For example, when we are calculating z-scores, I am so confused as to when we say (x-bar - mu) / ((sigma)/(√n)) rather than (x-bar - mu) / (sigma). Can anyone help?(6 votes)
- The proof is based on this basic property of random variables:
For three random variables X, Y and Z,
(1) if Z=mX + mY = m(X + Y),
(2) then Var[Z] = m^2*Var[X] + m^2*Var[Y] = m^2(Var[X] + Var[Y]).
Similarly, by taking a sample of size n and calculating its mean, we are creating a new random variable x-bar that is the sum of many random variables. Think of x-bar as Z, 1/n as m, and each sample point x(i) as X, Y, etc.:
(1) x-bar = (x(1) + x(2) + ... + x(n))/n = 1/n * (x(1) + x(2) + ... + x(n))
What is the variance of this new random variable x-bar? Apply the property from above:
(2) Var[x-bar] = 1/n^2 * (Var[x(1)] + Var[x(2)] + ... + Var[x(n)])
When we take a sample of n data points, each individual point x(i) 'inherits' the population's variance: Var[x(i)] = σ^2. This means we can simplify:
(2) Var[x-bar] = 1/n^2 * (σ^2 + σ^2 + ... + σ^2) = n/n^2 * σ^2 = σ^2/n
This is just the variance for one sample. The sampling distribution is a combination of all these new random variables:
(1) Distribution = x-bar(1) + x-bar(2) + ... + x-bar(n)
So the sampling distribution has variance:
(2) Var[Distribution] = 1/n^2 * (σ^2 + σ^2 + ... + σ^2) = n/n^2 * σ^2 = σ^2/n
Finally, the sampling distribution's standard error is the square root of the sampling distribution's variance: σ/√n.(5 votes)
- I read some book before that a normal distribution have a kurtosis of 3, how come the java have a kurtosis close to zero if it is approximately normal distributed?(6 votes)
- I think the java application assumes 0 kurtosis for the normal curve. In other words, it subtracts 3 from the kurtosis achieved.(6 votes)
- if the original data set only had 10000 data points, and i selected a sample size of n=10000, calculated x_bar 100 times, and created a frequency distribution, wouldn't that just be a vertical bar? In that case the distribution doesn't look very normal at all.
It would have no tails and no peaks, so how can the distribution look increasingly normal as n->∞ ? Are we assuming an arbitrarily large original data set? Wouldn't it make more sense to say that the distribution looks increasingly normal with n as it initially increases, and then decreasingly normal with n as it approaches the size of the total original data set? Can we take partial derivatives and minimize for skew and kurtosis?
Also, if we can always get to an arbitrarily small variance by increasing n, aren't we losing the meaning of the data? Isn't it like blurring an image to the point where it's all just one color? At some point the image is no longer recognizable.
Do we just keep iterating through variances until we're happy? Is there a heuristic for preserving data integrity (in the non-image case where it's not as easy to identify whether something is representative of the original data)?(6 votes)
- If the population has N=10000, and the sample has n=10000, then there is no need to think about the sampling distribution. The sampling distribution is a way to describe how a statistic behaves from sample to sample, but if we sampled the whole population, then we can calculate the parameters directly.
More generally though, you seem to be getting at the idea of what happens as n->∞. Yes, it's true that the standard error of the mean gets smaller and smaller as n increases, but it won't get to the point of a distribution that's just a single vertical bar (we'd call it a degenerate distribution). That's too far out into n being large, it may be what "will eventually happen", but we can never actually get to that point.
And also, yes, we often assume that the population size is arbitrarily large relative to the sample size (quite often we assume that the population is infinite in size). In cases where the sample is large relative to the population (such as when N=10000 and n=9000) there are corrections that can be made to account for this fact.
> "Can we take partial derivatives and minimize for skew and kurtosis?"
I suppose it may be possible, but not really meaningful. Neither of those numbers are strictly positive, so minimizing with respect to them wouldn't help regulate us to a Normal distribution.
> "Also, if we can always get to an arbitrarily small variance by increasing n, aren't we losing the meaning of the data? Isn't it like blurring an image to the point where it's all just one color? At some point the image is no longer recognizable."
We can do this (within reason, sometimes it's just too expensive to collect a lot of observations). However, we aren't losing the meaning of the data. The sampling distribution isn't meant to reflect the original data in the least bit, it's meant to give us information on the population mean (because the sample mean will tend to be around the population mean). When the standard error gets very small, we can estimate the population mean with much more precision.(4 votes)
- In addition to varying the sample sie (n) shouldn't variation in the number of trials (say, 10 x n versus 10,000 x n) impact the degree to which the sampling distribution fits the normal curve?(6 votes)
- Yes, you are absolutely right. The central limit theorem states that in large samples (n), the sampling distribution of the sample mean (xbar) is approximately normal no matter how the population is distributed. But it ALSO dictates that as the number of samples increase, the distribution approaches normality :)(1 vote)
- So, what are the assumptions for the CLT to be true? Of course, if the distribution is Cauchy, the CLT doesn't apply. Is all you need, a finite standard deviation? Don't the samples have to be independent as well? I suppose that may be the most difficult condition to meet in the real world? Do these same, rather outlandish, assumptions apply to the law of large numbers?(3 votes)
- This can get a bit tricky. For the "typical" CLT, we assume that the samples are all independent draws from a population with a constant mean, and a constant, finite variance.
There are generalizations to the CLT which relax these assumptions. I think the least restrictive one says something like - all samples must be from populations with finite mean and variance. They don't necessarily need to have the same mean or variance, and don't necessarily need to be independent (though I believe those thing affect the tate of convergence, so the "n>30" rule wouldn't work).
As to the law of large numbers, I believe that's more thinking in terms of estimating a parapeter. I believe that there is an assumption that the observations come from the same population (constant parameter values) and are independent. I don't think there is any need for the mean or variance to be finite, unless that's what the LLN is being applied to. It's just about probability of convergence, so as long as the parameter you're interested in is finite, other parameters shouldn't really matter.
This is all recollection off the top of my head, but I'm pretty confident.(4 votes)
- Given that the size of a sample is 30 ( n=30 ).
I know that the population mean ( "mu" ) is equal to the mean of the repeated sample means ( it means that we have collected so many samples and each sample has a sample size of 30).
For the population s.d. ( "sigma" ), it could be found if we divide the standard deviation of the repeated sample means by the square root of the sample size ( n=30 ), we therefore can estimate the population mean by using the confidence interval analysis.
My question is:
We often estimate the sigma ( the population s.d. ) by simply using the s ( the sample s.d.), which is the s.d. of just one sample ( with a size of 30), in the above formula.
However, this s is not the s.d. of the repeated sample mean.
What is the reasoning behind or is there something I got wrong?
Thank you so much : ](2 votes)
- I think you've misunderstood something along the way. An interval estimate for the population mean, mu, is:
xbar +/- T * s / sqrt(n)
where s is the standard deviation of the original data, it is NOT the standard deviation of the repeated sample means (the standard error of the sample mean, or just the standard error, SE). The SE is the entire value of s / sqrt(n). Of these, s is the estimate of the population standard deviation, the SE is not an estimate of sigma (it's an estimate of sigma / sqrt(n) ).(5 votes)
- Doesn't the standard error depend on three factors - standard deviation of the original distribution, size of the sample and also the number of repetitions? Why is the number of repetitions not present in the formula?
For example, If the number of repetitions approaches infinity, then wouldn't the standard error approach 0 irrespective of n?(2 votes)
- What do you mean by the "number of repititions" ? The formula for the SE is SE = sigma / sqrt (n).
You might be thinking of when Sal plots the histogram of the sample mean for many replications. If so, then the SD (not SE) of this will be roughly equal to sigma/sqrt (n). However, this is just to illustrate the effect, the number of replications(3 votes)
- is there any one who can explain this problems why answer is 0.0166 ? A normal distributed population with 200 elements has a means of 60 and a standard deviation of 10. the probability that the mean of a sample of 25 elements taken from this population will be smaller than 56 is ?
I thought the answer is 0.0228...(2 votes)
- Technically the calculation is slightly different because you're dealing with a finite population and not an infinite population (since N = 200). To calculate the standard error of the mean for a finite population, you multiply the regular standard error of mean by the square root of "(N-n)/(N-1)", where "N" is the size of the population and "n" is the sample size. Then, you just proceed at you would normally when calculating the Z-score.
Try this out and tell me if it works. . . Hope this helps!
Note: With this method you should get a new standard error of the mean of about "1.8755" instead of "2".(3 votes)
- can a distribution have a skew(negative or positive) but still mean be equal to mode?(2 votes)
- The answer is yes.
Let's say we have a population with n elements, with each element different from each other and with a random skew.
We calculate the mean.
Then we imagine a population with n+2 elements, which contains the previous n elements plus 2 elements with values equal to the mean that we calculated previously. This population will have the mean equal to the mode. As for the skew it can be zero, only if the initial skew was zero.(2 votes)
- Is the standard error of the mean; σ/sqrt(n) the same as (Σ(x-xbar)^2/n-1)^1/2 ? if not can you please explain the difference!(1 vote)
- They are not the same. The latter is the sample standard deviation, which is an estimate of σ. We can estimate the standard error of the mean by dividing that quantity by the square root of n.
Population standard deviation: σ
Estimated by: s = (Σ(x-xbar)^2/n-1)^1/2
Population standard error of the mean: σ / sqrt(n)
Estimated by: Std Err = s / sqrt(n)(4 votes)
We've seen in the last several videos, you start off with any crazy distribution. It doesn't have to be crazy. It could be a nice, normal distribution. But to really make the point that you don't have to have a normal distribution, I like to use crazy ones. So let's say you have some kind of crazy distribution that looks something like that. It could look like anything. So we've seen multiple times, you take samples from this crazy distribution. So let's say you were to take samples of n is equal to 10. So we take 10 instances of this random variable, average them out, and then plot our average. We get one instance there. We keep doing that. We do that again. We take 10 samples from this random variable, average them, plot them again. Eventually, you do this a gazillion times-- in theory, infinite number of times-- and you're going to approach the sampling distribution of the sample mean. And n equals 10, it's not going to be a perfect normal distribution, but it's going to be close. It would be perfect only if n was infinity. But let's say we eventually-- all of our samples, we get a lot of averages that are there. That stacks up there. That stacks up there. And eventually, we'll approach something that looks something like that. And we've seen from the last video that, one, if-- let's say we were to do it again. And this time, let's say that n is equal to 20. One, the distribution that we get is going to be more normal. And maybe in future videos, we'll delve even deeper into things like kurtosis and skew. But it's going to be more normal. But even more important here, or I guess even more obviously to us than we saw, then, in the experiment, it's going to have a lower standard deviation. So they're all going to have the same mean. Let's say the mean here is 5. Then the mean here is also going to be 5. The mean of our sampling distribution of the sample mean is going to be 5. It doesn't matter what our n is. If our n is 20, it's still going to be 5. But our standard deviation is going to be less in either of these scenarios. And we saw that just by experimenting. It might look like this. It's going to be more normal, but it's going to have a tighter standard deviation. So maybe it'll look like that. And if we did it with an even larger sample size-- let me do that in a different color. If we do that with an even larger sample size, n is equal to 100, what we're going to get is something that fits the normal distribution even better. We take 100 instances of this random variable, average them, plot it. 100 instances of this random variable, average them, plot it. We just keep doing that. If we keep doing that, what we're going to have is something that's even more normal than either of these. So it's going to be a much closer fit to a true normal distribution, but even more obvious to the human eye, it's going to be even tighter. So it's going to be a very low standard deviation. It's going to look something like that. I'll show you that on the simulation app probably later in this video. So two things happen. As you increase your sample size for every time you do the average, two things are happening. You're becoming more normal, and your standard deviation is getting smaller. So the question might arise, well, is there a formula? So if I know the standard deviation-- so this is my standard deviation of just my original probability density function. This is the mean of my original probability density function. So if I know the standard deviation, and I know n is going to change depending on how many samples I'm taking every time I do a sample mean. If I know my standard deviation, or maybe if I know my variance. The variance is just the standard deviation squared. If you don't remember that, you might want to review those videos. But if I know the variance of my original distribution, and if I know what my n is, how many samples I'm going to take every time before I average them in order to plot one thing in my sampling distribution of my sample mean, is there a way to predict what the mean of these distributions are? The standard deviation of these distributions. And to make it so you don't get confused between that and that, let me say the variance. If you know the variance, you can figure out the standard deviation because one is just the square root of the other. So this is the variance of our original distribution. Now, to show that this is the variance of our sampling distribution of our sample mean, we'll write it right here. This is the variance of our sample mean. Remember, our true mean is this, that the Greek letter mu is our true mean. This is equal to the mean. While an x with a line over it means sample mean. So here, what we're saying is this is the variance of our sample means. Now, this is going to be a true distribution. This isn't an estimate. If we magically knew the distribution, there's some true variance here. And of course, the mean-- so this has a mean. This, right here-- if we can just get our notation right-- this is the mean of the sampling distribution of the sampling mean. So this is the mean of our means. It just happens to be the same thing. This is the mean of our sample means. It's going to be the same thing as that, especially if we do the trial over and over again. But anyway, the point of this video, is there any way to figure out this variance given the variance of the original distribution and your n? And it turns out, there is. And I'm not going to do a proof here. I really want to give you the intuition of it. And I think you already do have the sense that every trial you take, if you take 100, you're much more likely, when you average those out, to get close to the true mean than if you took an n of 2 or an n of 5. You're just very unlikely to be far away if you took 100 trials as opposed to taking five. So I think you know that, in some way, it should be inversely proportional to n. The larger your n, the smaller a standard deviation. And it actually turns out it's about as simple as possible. It's one of those magical things about mathematics. And I'll prove it to you one day. I want to give you a working knowledge first. With statistics, I'm always struggling whether I should be formal in giving you rigorous proofs, but I've come to the conclusion that it's more important to get the working knowledge first in statistics, and then, later, once you've gotten all of that down, we can get into the real deep math of it and prove it to you. But I think experimental proofs are all you need for right now, using those simulations to show that they're really true. So it turns out that the variance of your sampling distribution of your sample mean is equal to the variance of your original distribution-- that guy right there-- divided by n. That's all it is. So if this up here has a variance of-- let's say this up here has a variance of 20. I'm just making that number up. And then let's say your n is 20. Then the variance of your sampling distribution of your sample mean for an n of 20-- well, you're just going to take the variance up here-- your variance is 20-- divided by your n, 20. So here, your variance is going to be 20 divided by 20, which is equal to 1. This is the variance of your original probability distribution. And this is your n. What's your standard deviation going to be? What's going to be the square root of that? Standard deviation is going to be the square root of 1. Well, that's also going to be 1. So we could also write this. We could take the square root of both sides of this and say, the standard deviation of the sampling distribution of the sample mean is often called the standard deviation of the mean, and it's also called-- I'm going to write this down-- the standard error of the mean. All of these things I just mentioned, these all just mean the standard deviation of the sampling distribution of the sample mean. That's why this is confusing. Because you use the word "mean" and "sample" over and over again. And if it confuses you, let me know. I'll do another video or pause and repeat or whatever. But if we just take the square root of both sides, the standard error of the mean, or the standard deviation of the sampling distribution of the sample mean, is equal to the standard deviation of your original function, of your original probability density function, which could be very non-normal, divided by the square root of n. I just took the square root of both sides of this equation. Personally, I like to remember this, that the variance is just inversely proportional to n, and then I like to go back to this, because this is very simple in my head. You just take the variance divided by n. Oh, and if I want the standard deviation, I just take the square roots of both sides, and I get this formula. So here, when n is 20, the standard deviation of the sampling distribution of the sample mean is going to be 1. Here, when n is 100, our variance-- so our variance of the sampling mean of the sample distribution or our variance of the mean, of the sample mean, we could say, is going to be equal to 20, this guy's variance, divided by n. So it equals-- n is 100-- so it equals one fifth. Now, this guy's standard deviation or the standard deviation of the sampling distribution of the sample mean, or the standard error of the mean, is going to the square root of that. So 1 over the square root of 5. And so this guy will have to be a little bit under one half the standard deviation, while this guy had a standard deviation of 1. So you see it's definitely thinner. Now, I know what you're saying. Well, Sal, you just gave a formula. I don't necessarily believe you. Well, let's see if we can prove it to ourselves using the simulation. So just for fun, I'll just mess with this distribution a little bit. So that's my new distribution. And let me take an n-- let me take two things it's easy to take the square root of, because we're looking at standard deviations. So let's say we take an n of 16 and n of 25. And let's do 10,000 trials. So in this case, every one of the trials, we're going to take 16 samples from here, average them, plot it here, and then do a frequency plot. Here, we're going to do a 25 at a time and then average them. I'll do it once animated just to remember. So I'm taking 16 samples, plot it there. I take 16 samples, as described by this probability density function, or 25 now. Plot it down here. Now, if I do that 10,000 times, what do I get? What do I get? All right. So here, just visually, you can tell just when n was larger, the standard deviation here is smaller. This is more squeezed together. But actually, let's write this stuff down. Let's see if I can remember it here. Here, n is 6. So in this random distribution I made, my standard deviation was 9.3. I'm going to remember these. Our standard deviation for the original thing was 9.3. And so standard deviation here was 2.3, and the standard deviation here is 1.87. Let's see if it conforms to our formula. So I'm going to take this off screen for a second, and I'm going to go back and do some mathematics. So I have this on my other screen so I can remember those numbers. So, in the trial we just did, my wacky distribution had a standard deviation of 9.3. When n was equal to 16-- just doing the experiment, doing a bunch of trials and averaging and doing all the thing-- we got the standard deviation of the sampling distribution of the sample mean, or the standard error of the mean. We experimentally determined it to be 2.33. And then when n is equal to 25, we got the standard error of the mean being equal to 1.87. Let's see if it conforms to our formulas. So we know that the variance-- or we could almost say the variance of the mean or the standard error-- the variance of the sampling distribution of the sample mean is equal to the variance of our original distribution divided by n. Take the square roots of both sides. Then you get standard error of the mean is equal to standard deviation of your original distribution, divided by the square root of n. So let's see if this works out for these two things. So if I were to take 9.3-- so let me do this case. So 9.3 divided by the square root of 16-- n is 16-- so divided by the square root of 16, which is 4. What do I get? So 9.3 divided by 4. Let me get a little calculator out here. Let's see. We want to divide 9.3 divided by 4. 9.3 divided by our square root of n-- n was 16, so divided by 4-- is equal to 2.32. So this is equal to 2.32, which is pretty darn close to 2.33. This was after 10,000 trials. Maybe right after this I'll see what happens if we did 20,000 or 30,000 trials where we take samples of 16 and average them. Now let's look at this. Here, we would take 9.3. So let me draw a little line here. Maybe scroll over. That might be better. So we take our standard deviation of our original distribution-- so just that formula that we've derived right here would tell us that our standard error should be equal to the standard deviation of our original distribution, 9.3, divided by the square root of n, divided by square root of 25. 4 was just the square root of 16. So this is equal to 9.3 divided by 5. And let's see if it's 1.87. So let me get my calculator back. So if I take 9.3 divided by 5, what do I get? 1.86, which is very close to 1.87. So we got in this case 1.86. So as you can see, what we got experimentally was almost exactly-- and this is after 10,000 trials-- of what you would expect. Let's do another 10,000. So you got another 10,000 trials. Well, we're still in the ballpark. We're not going to-- maybe I can't hope to get the exact number rounded or whatever. But, as you can see, hopefully that'll be pretty satisfying to you, that the variance of the sampling distribution of the sample mean is just going to be equal to the variance of your original distribution, no matter how wacky that distribution might be, divided by your sample size, by the number of samples you take for every basket that you average, I guess is the best way to think about it. And sometimes this can get confusing, because you are taking samples of averages based on samples. So when someone says sample size, you're like, is sample size the number of times I took averages or the number of things I'm taking averages of each time? And it doesn't hurt to clarify that. Normally when they talk about sample size, they're talking about n. And, at least in my head, when I think of the trials as you take a sample of size of 16, you average it, that's one trial. And you plot it. Then you do it again, and you do another trial. And you do it over and over again. But anyway, hopefully this makes everything clear. And then you now also understand how to get to the standard error of the mean.