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

Lesson 4: Sampling distributions for sample proportions- Sampling distribution of sample proportion part 1
- Sampling distribution of sample proportion part 2
- Normal conditions for sampling distributions of sample proportions
- The normal condition for sample proportions
- Mean and standard deviation of sample proportions
- Probability of sample proportions example
- Finding probabilities with sample proportions
- Sampling distribution of a sample proportion example

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# Sampling distribution of sample proportion part 1

Formulas for the mean and standard deviation of a sampling distribution of sample proportions.

## Want to join the conversation?

- This might be a silly doubt but at8:26its taken the standard deviation for our sample proportion as the standard deviation of binomial random variable X divided by n.

Why?(36 votes)`P-hat = X/n`

Var(P-hat) = Var (X/n)

= (1/n)^2 ⋅ Var(X)

=Var(X)/n^2

=(n⋅p⋅(1-p))/n^2

= p(1-p)/n(23 votes)

- 0:43, what is the Bernoulli random variable? Was it introduced somewhere else?(25 votes)
- IDK if it was introduced before, but Bernoulli random variables can only take values 0 or 1. Failure or success. Yes or No. This variable is binary, because there are 2 possible results, like flipping a coin.(25 votes)

- could anyone explain the mean of sample proportion? (at7:56)(10 votes)
- Let's say you do 99999 times of sampling tests, of sample size 10 each. You find the p̂ of each sample test, and get p̂1, p̂2 all the way to p̂99999.

p̂1 can be written as (X1)/10, p̂2 can be written as (X2)/10....p̂99999 can be written as (X99999)/10, where X1 is Total # of Yellow Gumbals in Trail No.1 & X2 is Total # of Yellow Gumballs in Trail No.2....X99999 is Total # of Yellow Gumballs in Trail No.99999.

Now, μp̂ = (p̂1 + p̂2 +...+ p̂99999) /99999. (The mean of all 99999 p̂)

We can rewrite this as [(X1)/10+(X2)/10+...(X99999)/10]/99999.

We slimplify it to [(X1+x2+...X99999)/10]/99999

= [(X1+x2+...X99999)]/[10*99999]

Since, (X1+x2+...X99999)/99999 = μX (i.e. Sum of all # of Yellow Gumballs in all 99999 trails divided by 99999, which is total number of trails).

We can finally simplify [(X1+x2+...X99999)]/[10*99999] to μX/10,

hence μp̂ = μX/n, and our example, μp̂ = μX/99999.

Sorry for the formatting, write it out on your own and it will all make sense.(27 votes)

- Why do we divide all the statistics (mean, SD) calculated for the sampling distribution by 'n'? I'm really confused. I mean, why do we divide the mean of the random variable X by 'n' to get the mean of the sampling distribution?(10 votes)
- Hi Sal...the video is very informative and well explained.

I have one question:

In the previous videos(https://www.khanacademy.org/math/statistics-probability/summarizing-quantitative-data/variance-standard-deviation-sample/v/sample-variance), you explained about the calculating the sample variance and the unbiased estimator by dividing by n-1. How that is related to this topic of sample proportion. Here also, you calculated the variance of sampling distribution of sample proportion.

Could you please explain the relation?(6 votes)- I've been having a play with this same question in my head. I believe Sal takes the above standard deviation of X approach, because he explicitly states he knows the Population probability (P), 0.6. In the examples where Sal divides by n-1, we're only given sample estimates of P,
*P-hat*, therefore we have to take into account the potential bias introduced, by using estimators instead of actual values. I think the way to mitigate this bias is by dividing by n-1 rather than n, but I might be wrong..

This video:(https://www.khanacademy.org/math/statistics-probability/sampling-distributions-library/sample-proportions/v/sampling-distribution-of-sample-proportion-part-1), uses n-1 in that context and the above is the only way I've been able to reason why!

It would be interesting to see a video where Sal had to infer the probability P, given only a sample size n and the proportion of yellow balls found*x-hat*to see how the approach would differ.(2 votes)

- Why is the square root of p*p-1 equal to the standard deviation? What makes the formula work?(3 votes)
- Good question!

Actually the standard deviation of Y is square root of the quantity p*(1-p).

In the lesson, Y is a random variable that is 1 with probability p, and 0 with probability (1-p).

The mean of Y, mu_Y, is E(Y) = 0*P(Y=0)+1*P(Y=1) = 0(1-p)+1*p = p.

The variance of Y, sigma^2_Y, is by definition the expected value of the squared difference of Y from its own mean.

So the variance of Y is

sigma^2_Y = E[(Y - mu_Y)^2] = E[(Y-p)^2]

= (0-p)^2 P(Y=0) + (1-p)^2 P(Y=1)

= p^2 * (1-p) + (1-p)^2 * p

= p(1-p)[p + (1-p)]

= p(1-p)(1)

= p(1-p).

The standard deviation of Y, sigma_Y, is by definition the square root of its variance, so the standard deviation of Y is sigma_Y = sqrt[p(1-p)].(3 votes)

- Is proportion the same as probability? The proportion is 0.6 at0:36. So the probability of picking a yellow ball out of the machine is 60 percent. So does that mean the probability is the same as proportion? Why use a different word?(2 votes)
- The sample proportion is the experimental probability (based on data), but might not be the same as the theoretical (true) probability.(4 votes)

- At4:40Sal says, "Well, you could say..." and the statement follows on which the rest of video relies: p-hat = X / n. While "intuitively" I find this statement valid/acceptable/../OK, I am wondering: isn't there a (more) formal argument/proof to support it?

Now that I have been thinking some more about this: isn't`p-hat`

, in this context,`X/n`

*by definition*?(3 votes) - what is the Bernoulli random variable?(2 votes)
- A Bernoulli random variable has value 1 with probability p, value 0 with probability 1-p, and no other possible values.

A binomial random variable with n trials can be thought of as a sum of n independent Bernoulli random variables. Also, a binomial random variable with n = 1 trial becomes Bernoulli.

Have a good day!(3 votes)

## Video transcript

- [Instructor] So I have a
gumball machine right over here. It has yellow, and green,
and pink, and blue gumballs. Let me throw a few blue ones in there. And what we're going to
concern ourselves in this video are the yellow gumballs. And let's say that we
know that the proportion of yellow gumballs over here is P. This right over here is a population, population parameter. And for the sake of argument,
just to make things concrete, let's just say that 60% of
the gumballs are yellow, or 0.6 of them. Now let's review some things
that we have seen before. I'm gonna define our
Bernoulli random variable, let's call this capital
Y, which is equal to one if when we take one random
gumball out of that machine we get a, we pick a yellow gumball, and it's equal to zero if when
we pick one random gumball out of that machine we don't pick yellow, so not yellow. From previous videos we
know some interesting things about this Bernoulli random variable. We know its mean. We know the mean of our
Bernoulli random variable. It's going to be the
proportion of yellow balls in this population, so it's
going to be equal to P, which in this particular
case we know is 0.6, and we know what the standard deviation of this Bernoulli random variable is. It is going to be P times one minus P, actually that's the variance. We want to take the square root of that to get the standard deviation. So in this particular scenario, that's going to be the
square root of 0.6 times 0.4. Fair enough. This is all review so far. But now let me define, let me define another random variable X, which is equal to the
sum of 10 independent, independent trials, trials of Y. Now, we have seen random
variables like this before. This is a binomial random variable. Now, what do we know about its
mean and standard deviation? Well, in previous videos we
know that the mean of this binomial random variable is
just going to be equal to n times the mean of each
of the Bernoulli trials right over here, so n times P, which in this particular
situation is going to be, n is 10, we're doing 10
trials, and P is zero, 0.6, which is equal to six. And that makes sense. If 60% of the balls here are yellow and if you were to take a
sample or if you were to take 10 trials, so if you were to
take ten balls one at a time, they have to be independent,
so you keep looking at them and then replacing them,
but if you took 10, then you would expect that
six of them would be yellow. They're not always going to be six yellow, but that would be maybe
what you would expect. All right, now what's the
standard deviation here? So our standard deviation is equal to, and we have proved this in other videos, it's equal to the square root of n times P, times one minus P. Notice you just put an n right over here under the radical sign. And so this is going to get us
in this particular situation to 10 times 0.6 times 0.4 and then the square root of everything. So all of this is review. If it's unfamiliar, I
encourage you to review some of the videos on
Bernoulli random variables and on binomial random variables. But what we're going to do in this video is think about a sampling distribution and it's going to be the
sampling distribution for a sample statistic known
as the sample proportion, which we actually talked
about when we first introduced sampling distributions. So let's say, so let's
just park all of this, this is background right over here. Let's start taking samples of 10. And I didn't pick that randomly. I want to make it
reconcile with what we do with our random variable here. And so let's take a sample of 10 gumballs and let's calculate the
proportion that are yellow. And so we will call that
our sample proportion, and I might as well
just do that in yellow. So we want to calculate
the sample proportion that are yellow. And what is this equivalent to? Well, you could say, Well
this is equivalent to my random variable X. I want to count the number
of gumballs that are yellow and then I'm gonna divide
it by the sample size. So I'm gonna divide it by n. And in this case, it
would be X divided by 10. I know what some of you are thinking. Wait, wait, wait. Hold on for a second. X is sum of 10 independent, independent trials right over here. To be independent, you
can't just take 10 gumballs. You have to take them one at a time and then replace them
back in order for them to be truly independent. But remember, we have our 10% rule. We have our 10% rule, which
tells us that if a sample is less than or equal to
10% of the population, that you can treat each of
the gumballs in this situation as being independent. So let's just say for the sake of argument that there are 10,000 gumballs in here, and so we can feel pretty good that these samples,
that each of the things in the sample are
independent of each other by our 10% rule. And so each of these are, each of these 10 gumballs what we see, they are going to be independent, I'm going to put them in
quotes, by the 10% rule. And so in that situation,
we can make this claim or we can feel good that
this claim is roughly true. And so let's say for that
first sample that we do, our sample proportion is equal to 0.3, so three of our gumballs just happened, three of our 10 gumballs
just happened to be yellow. Then we'd do it again. So we take another sample. We calculate our sample
proportion, this statistic again, remember, it's trying to estimate
our population parameter, and let's say that time it
happens to be seven out of 10, and we just keep doing that. And if we keep doing that,
and we plot it on a dot chart or a dot distribution
I guess we could say, where our possible outcomes
you could have zero out of 10, one out of 10, two, three,
four, five, so half of them, six, seven, eight, nine, 10,
so that would be all of them. And so you could plot okay,
0.3 is one, two, three, that's one scenario where I got, where my sample proportion is 0.3. Then 0.7, that's one
situation where I got a 0.7. And let's say I were to
take another sample of 10 and I were to get 0.7, then
you would plot that over here. And if you kept taking
samples and kept calculating these sample proportions and
you kept plotting it here, you would get a better, and
better, and better approximation for the sampling distribution
of the sampling proportion. But how can we actually
characterize the true sampling distribution for the sample proportion? What is going to be the mean
of this sampling distribution and what is going to be
the standard deviation? Well, we can derive that from
what we see right over here. The mean of our sampling distribution of our sample proportion is
just going to be equal to the mean of our random
variable X divided by n. It's just going to be the
mean of X divided by n, which is equal to what? Well, the mean of X is n times P. This is n times P. You divided it by n,
you're going to get P. And that makes sense. One way to think about
it, the expected value for your sample proportion is going to be the proportion of gumballs
that you actually see. And so this also a good indicator
that this is going to be a reasonably unbiased estimator. Now let's think about
the standard deviation for our sample proportion. Well, we can just view that
as our standard deviation of our binomial random
variable X divided by n. So this is going to be
equal to the square root of n times P, times one minus
P, all of that over n, which is the same thing as if we put this, if we divide by n inside the radical, it'd be the same thing as the square root of nP times one minus P over n squared. Divide the numerator and denominator by n, you will get the square
root of P times one minus P, all of that over n. And so in this particular situation, where our parameter is 0.6, where our population parameter is 0.6, so it's going to be 0.6,
that's the true proportion for our population. And then what would out
standard deviation be for our sample proportion? Well, it's going to be
equal to the square root of 0.6 times 0.4, all of that over 10. And we can get a calculator
out to calculate that. So if we take 0.6 times 0.4
equals, divided by 10, equals, and then we take the square root of that, and we get it's approximately 0.15. Approximately 0.15.