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### Course: AP®︎/College Statistics > Unit 9

Lesson 6: Sampling distributions for sample means# 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.

## Want to join the conversation?

- 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?(7 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.(6 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?(7 votes)
- I think the java application assumes 0 kurtosis for the normal curve. In other words, it subtracts 3 from the kurtosis achieved.(7 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)?(7 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.(5 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)

- does this formula work for any statistic? or only for the sample distribution of the sample mean?(3 votes)
- The formula given specifically applies to the sampling distribution of the sample mean. It doesn't apply to other statistics directly, but similar principles about the distribution of sample statistics can apply more broadly.(1 vote)

- 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)

## Video transcript

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.