This might be a silly doubt but at 8:26 its taken the standard deviation for our sample proportion as the standard deviation of binomial random variable X divided by n. Why?

``` 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 ```

0:43, what is the Bernoulli random variable? Was it introduced somewhere else?

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.

could anyone explain the mean of sample proportion? (at 7:56)

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.

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?

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.

Why is the square root of p*p-1 equal to the standard deviation? What makes the formula work?

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)].

Is proportion the same as probability? The proportion is 0.6 at 0:36 . 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?

The sample proportion is the experimental probability (based on data), but might not be the same as the theoretical (true) probability.

Doesn't the sample size have to bigger or equal to 10 and n(1-p) = 10 for us to have a good approximation of normal distribution?

It would be preffered if this condition is met, but you can still do the test with this condition violated.

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AP®︎/College Statistics

Course: AP®︎/College Statistics > Unit 9

Lesson 4: Sampling distributions for sample proportions

Sampling distribution of sample proportion part 1

Name: Sampling distribution of sample proportion part 1
Uploaded: 2017-12-20T21:01:03Z
Description: Formulas for the mean and standard deviation of a sampling distribution of sample proportions.

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

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avishp0
Posted 5 years ago. Direct link to avishp0's post “This might be a silly dou...”
This might be a silly doubt but at
8:26
its taken the standard deviation for our sample proportion as the standard deviation of binomial random variable X divided by n.
Why?
Button navigates to signup pageComment on avishp0's post “This might be a silly dou...”
(32 votes)
Answer
- JorgeMercedes
  Posted 5 years ago. Direct link to JorgeMercedes's post “``` P-hat = X/n Var(P-hat...”
  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
  Comment on JorgeMercedes's post “``` P-hat = X/n Var(P-hat...”
  (18 votes)
Calvin Wang
Posted 4 years ago. Direct link to Calvin Wang's post “0:43, what is the Bernoul...”
0:43
, what is the Bernoulli random variable? Was it introduced somewhere else?
Button navigates to signup pageButton navigates to signup page
(20 votes)
Answer
- Ramen23
  Posted 3 years ago. Direct link to Ramen23's post “IDK if it was introduced ...”
  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.
  Button navigates to signup page
  (16 votes)
Chacko Babu
Posted 5 years ago. Direct link to Chacko Babu's post “could anyone explain the ...”
could anyone explain the mean of sample proportion? (at
7:56
)
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- JorgeMercedes
  Posted 5 years ago. Direct link to JorgeMercedes's post “Let's say you do 99999 ti...”
  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.
  Comment on JorgeMercedes's post “Let's say you do 99999 ti...”
  (23 votes)
eby.annie
Posted 5 years ago. Direct link to eby.annie's post “sd = sqrt(p(1-p)) derived...”
sd = sqrt(p(1-p)) derived here: https://www.khanacademy.org/math/statistics-probability/random-variables-stats-library/binomial-mean-standard-dev-formulas/v/variance-of-binomial-variable
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(9 votes)
Answer
cena.prince
Posted 5 years ago. Direct link to cena.prince's post “Hi Sal...the video is ver...”
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?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Tom Slade
  Posted 3 years ago. Direct link to Tom Slade's post “I've been having a play w...”
  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.
  Button navigates to signup page
  (2 votes)
Emma
Posted 3 years ago. Direct link to Emma's post “Why is the square root of...”
Why is the square root of p*p-1 equal to the standard deviation? What makes the formula work?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Ian Pulizzotto
  Posted 3 years ago. Direct link to Ian Pulizzotto's post “Good question! Actually ...”
  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)].
  Button navigates to signup page
  (3 votes)
hellowKhanLearning
Posted 4 years ago. Direct link to hellowKhanLearning's post “Is proportion the same as...”
Is proportion the same as probability? The proportion is 0.6 at
0:36
. 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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ian Pulizzotto
  Posted 4 years ago. Direct link to Ian Pulizzotto's post “The sample proportion is ...”
  The sample proportion is the experimental probability (based on data), but might not be the same as the theoretical (true) probability.
  Button navigates to signup page
  (4 votes)
peter.dunay.kovacs
Posted 4 years ago. Direct link to peter.dunay.kovacs's post “At 4:40 Sal says, "Well, ...”
At
4:40
Sal 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 ?
Button navigates to signup pageComment on peter.dunay.kovacs's post “At 4:40 Sal says, "Well, ...”
(3 votes)
Answer
Parthiban Rajendran
Posted 5 years ago. Direct link to Parthiban Rajendran's post “And he misses out to expl...”
And he misses out to explain the most important part. How is mu_hat = mu_x/n?
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(3 votes)
Answer
- Bastian Widanski
  Posted 5 years ago. Direct link to Bastian Widanski's post “credit to whom credit is ...”
  credit to whom credit is due: this answer is from JorgeMercedes to a similar question above.
  But I think it perfectly shows why p(1-p)/n is correct, at least it helped me to understand it:
  
  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
  Comment on Bastian Widanski's post “credit to whom credit is ...”
  (0 votes)
Mārtiņš Bružs
Posted 4 years ago. Direct link to Mārtiņš Bružs's post “Doesn't the sample size h...”
Doesn't the sample size have to bigger or equal to 10 and n(1-p) = 10 for us to have a good approximation of normal distribution?
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(2 votes)
Answer
- Saivishnu Tulugu
  Posted 4 years ago. Direct link to Saivishnu Tulugu's post “It would be preffered if ...”
  It would be preffered if this condition is met, but you can still do the test with this condition violated.
  Button navigates to signup page
  (2 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.