Inferential statistics
Sampling distribution
None
Sampling distribution of the sample mean
The central limit theorem and the sampling distribution of the sample mean
Discussion and questions for this video
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 In the last video, we learned about what
 is quite possibly the most profound idea in statistics,
 and that's the central limit theorem.
 And the reason why it's so neat is,
 we could start with any distribution that
 has a well defined mean and variance actually,
 I wrote the standard deviation here
 in the last video, that should be the mean,
 and let's say it has some variance.
 I could write it like that, or I could
 write the standard deviation there.
 But as long as it has a well defined
 mean and standard deviation, I don't
 care what the distribution looks like.
 What I can do is take samples in the last video of say,
 size four that means I take literally four
 instances of this random variable, this is one example.
 I take their mean, and I consider
 this the sample mean from my first trial,
 or you could almost say for my first sample.
 I know it's very confusing, because you can consider
 that a sample, the set to be a sample,
 or you could consider each member of the set is a sample.
 So that can be a little bit confusing there.
 But I have this first sample mean,
 and then I keep doing that over and over.
 In my second sample, my sample size is four.
 I got four instances of this random variable,
 I average them, I have another sample mean.
 And the cool thing about the central limit theorem,
 is as I keep plotting the frequency
 distribution of my sample means, it
 starts to approach something that
 approximates the normal distribution.
 And it's going to do a better job of approximating
 that normal distribution as n gets larger.
 And just so we have a little terminology
 on our belt, this frequency distribution
 right here that I've plotted out,
 or here, or up here that I started plotting out,
 that is called and it's kind of confusing,
 because we use the word sample so much that
 is called the sampling distribution of the sample
 mean.
 And let's dissect this a little bit,
 just so that this long description
 of this distribution starts to make a little bit of sense.
 When we say it's the sampling distribution,
 that's telling us that it's being derived
 from it's a distribution of some statistic, which
 in this case happens to be the sample mean
 and we're deriving it from samples
 of an original distribution.
 So each of these.
 So this is my first sample, my sample size is four.
 I'm using the statistic, the mean.
 I actually could have done it with other things,
 I could have done the mode or the range or other statistics.
 But sampling distribution of the sample mean
 is the most common one.
 It's probably, in my mind, the best
 place to start learning about the central limit theorem,
 and even frankly, sampling distribution.
 So that's what it's called.
 And just as a little bit of background and I'll
 prove this to you experimentally,
 not mathematically, but I think the experimental
 is on some levels more satisfying with statistics
 that this will have the same mean
 as your original distribution.
 As your original distribution right here.
 So it has the same mean, but we'll see in the next video
 that this is actually going to start approximating
 a normal distribution, even though my original distribution
 that this is kind of generated from, is completely nonnormal.
 So let's do that with this app right here.
 And just to give proper credit where credit is due,
 this is I think was developed at Rice University this
 is from onlinestatbook.com.
 This is their app, which I think is a really neat app,
 because it really helps you to visualize what a sampling
 distribution of the sample mean is.
 So I can literally create my own custom distribution here.
 So let me make something kind of crazy.
 So you could do this, in theory, with a discrete
 or a continuous probability density function.
 But what they have here, we could take on one of 32 values,
 and I'm just going to set the different probabilities
 of getting any of those 32 values.
 So clearly, this right here is not a normal distribution.
 It looks a little bit bimodal, but it doesn't have long tails.
 But what I want to do is, first just use a simulation
 to understand, or to better understand,
 what the sampling distribution is all about.
 So what I'm going to do is, I'm going
 to take we'll start with five at a time.
 So my sample size is going to be five.
 And so when I click animated, what it's going to do,
 is it's going to take five samples from this probability
 distribution function.
 It's going to take five samples, and you're
 going to see them when I click animated,
 it's going to average them and plot the average down here.
 And then I'm going to click it again,
 and it's going to do it again.
 So there you go, it got five samples from there,
 it averaged them, and it hit there.
 So what I just do?
 I clicked oh, I wanted to clear that.
 Let me make this bottom one none.
 So let me do that over again.
 So I'm going to take five at time.
 So I took five samples from up here, and then it took its mean
 and plotted the mean there.
 Let me do it again.
 Five samples from this probability distribution
 function, plotted it right there.
 I could keep doing it.
 It'll take some time.
 But you can see I plotted it right there.
 Now I could do this 1,000 times, it's going to take forever.
 Let's say I just wanted to do it 1,000 times.
 So this program, just to be clear,
 it's actually generating the random numbers.
 This isn't like a rigged program.
 It's actually going to generate the random numbers according
 to this probability distribution function.
 It's going to take five at a time, find their means,
 and plot the means.
 So if I click 10,000, it's going to do that 10,000 times.
 So it's going to take five numbers from here 10,000 times
 and find their means 10,000 times
 and then plot the 10,000 means here.
 So let's do that.
 So there you go.
 And notice it's already looking a lot
 like a normal distribution.
 And like I said, the original mean of my crazy distribution
 here was 14.45, and after doing 10,000 samples or 10,000
 trials my mean here is 14.42.
 So I'm already getting pretty close to the mean there.
 My standard deviation, you might notice, is less than that.
 We'll talk about that in a future video.
 And the skew and kurtosis, these are
 things that help us measure how normal a distribution is.
 And I've talked a little bit about it in the past,
 and let me actually just diverge a little bit, it's interesting.
 And they're fairly straightforward concepts.
 Skew literally tells so if this
 is let me do it in a different color
 if this is a perfect normal distribution
 and clearly my drawing is very far from perfect
 if that's a perfect distribution,
 this would have a skew of zero.
 If you have a positive skew, that
 means you have a larger right tail than you would otherwise
 expect.
 So something with a positive skew might look like this.
 It would have a large tail to the right.
 So this would be a positive skew,
 which makes it a little less than ideal
 for normal distribution.
 And a negative skew would look like this,
 it has a long tail to the left.
 So negative skew might look like that.
 So that is a negative skew.
 If you have trouble remembering it,
 just remember which direction the tail is going.
 This tail is going towards a negative direction,
 this tail is going to the positive direction.
 So if something has no skew, that
 means that it's nice and symmetrical around its mean.
 Now kurtosis, which sounds like a very fancy word,
 is similarly not that fancy of an idea.
 So once again, if I were to draw a perfect normal distribution.
 Remember, there is no one normal distribution,
 you could have different means and
 different standard deviations.
 Let's say that's a perfect normal distribution.
 If I have positive kurtosis, what's going to happen
 is, I'm going to have fatter tails let
 me draw it a little nicer than that I'm going to have fatter
 tails, but I'm going to have a more pointy peak.
 I didn't have to draw it that pointy,
 let me draw it like this.
 I'm going to have fatter tails, and I'm
 going to have a more pointy peak than a normal distribution.
 So this right here is positive kurtosis.
 So something that has positive kurtosis depending
 on how positive it is it tells you it's
 a little bit more pointy than a real normal distribution.
 And negative kurtosis has smaller tails,
 but it's smoother near the middle.
 So it's like this.
 So something like this would have negative kurtosis.
 And maybe in future videos we'll explore that in more detail,
 but in the context of the simulation,
 it's just telling us how normal this distribution is.
 So when our sample size was n equal 5
 and we did 10,000 trials, we got pretty close
 to a normal distribution.
 Let's do another 10,000 trials, just to see what happens.
 It looks even more like a normal distribution.
 Our mean is now the exact same number,
 but we still have a little bit of skew,
 and a little bit of kurtosis.
 Now let's see what happens if we do the same thing with a larger
 sample size.
 And we could actually do them simultaneously.
 So here's n equal 5.
 Let's do here, n equals 25.
 Just let me clear them.
 I'm going to do the sampling distribution of the sample
 mean.
 And I'm going to run 10,000 trials I'll
 do one animated trial, just so you remember what's going on.
 So I'm literally taking first five samples from up here,
 find their mean.
 Now I'm taking 25 samples from up here, find its mean,
 and then plotting it down here.
 So here the sample size is 25, here it's five.
 I'll do it one more time.
 I take five, get the mean, plot it.
 Take 25, get the mean, and then plot it down there.
 This is a larger sample size.
 Now that thing that I just did, I'm going to do 10,000 times.
 And remember, our first distribution
 was just this really crazy very nonnormal distribution,
 but once we did it whoops, I didn't
 want to make it that big.
 Scroll up a little bit.
 So here, what's interesting?
 I mean they both look a little normal,
 but if you look at the skew and the kurtosis,
 when our sample size is larger, it's more normal.
 This has a lower skew than when our sample size was only five.
 And it has a less negative kurtosis
 than when our sample size was five.
 So this is a more normal distribution.
 And one thing that we're going to explore further
 in a future video, is not only is it more normal in its shape,
 but it's also tighter fit around the mean.
 And you can even think about why that kind of makes sense.
 When your sample size is larger, your odds
 of getting really far away from the mean is lower.
 Because it's very low likelihood,
 if you're taking 25 samples, or 100 samples, that
 you're just going to get a bunch of stuff way out here,
 or a bunch of stuff way out here.
 You're very likely to get a reasonable spread of things.
 So it makes sense that your mean your sample mean is
 less likely to be far away from the mean.
 We're going to talk a little bit more about in the future.
 But hopefully this kind of satisfies you
 that at least experimentally, I
 haven't proven it to you with mathematical rigor, which
 hopefully we'll do in the future.
 But hopefully this satisfies you, at least
 experimentally, that the central limit theorem really
 does apply to any distribution.
 I mean, this is a crazy distribution.
 And I encourage you to use this applet at onlinestatbook.com
 and experiment with other crazy distributions
 to believe it for yourself.
 But the interesting things are that we're
 approaching a normal distribution,
 but as my sample size got larger,
 it's a better fit for a normal distribution.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

Have something that's not a question about this content? 
This discussion area is not meant for answering homework questions.
Where on the onlinestatbook site is this little software toy?
Thanks, John
Thanks, John
John,
http://onlinestatbook.com/ click on "content" in the upper left "List of Simulations and Demonstrations" scroll to the bottom of the page "Simulations from the Rice Virtual Lab in Statistics" and it's the second one down "Sampling Distribution Simulation". Hope that helps.
http://onlinestatbook.com/ click on "content" in the upper left "List of Simulations and Demonstrations" scroll to the bottom of the page "Simulations from the Rice Virtual Lab in Statistics" and it's the second one down "Sampling Distribution Simulation". Hope that helps.
This is not about the vid (srry), but is your name Firstjohn26 have anything to do with 1 John 2:6 (Whoever claims to live in him must live as Jesus did)?
I have a practice question that I just can't figure out. It is: "Eighteen subjects are randomly selected and given proficiency tests. The mean for this group is 492.3 and the standard deviation is 37.6. Construct the 98% confidence interval for the population standard deviation."
I don't know how to figure out the confidence interval for a standard deviation. Can you please help. Thanks. Katie
I don't know how to figure out the confidence interval for a standard deviation. Can you please help. Thanks. Katie
We already know that:
A range from 1 std.dev. to 1 std.dev. contains 68.3% of outcomes.
A range from 2 std.dev. to 2 std.dev. contains 95.4% of outcomes.
A range from 3 std.dev. to 3 std.dev. contains 99.7% of outcomes.
So the question is, how many Std.Dev's do we have to move away from the mean in both directions on the graph to contain 98% of outcomes. Not 95.4%, Not 99.7%, exactly 98%. Right away you know the answer will be between 2 and 3 std.dev's, as 98% is between 95.4% and 99.7%
To
A range from 1 std.dev. to 1 std.dev. contains 68.3% of outcomes.
A range from 2 std.dev. to 2 std.dev. contains 95.4% of outcomes.
A range from 3 std.dev. to 3 std.dev. contains 99.7% of outcomes.
So the question is, how many Std.Dev's do we have to move away from the mean in both directions on the graph to contain 98% of outcomes. Not 95.4%, Not 99.7%, exactly 98%. Right away you know the answer will be between 2 and 3 std.dev's, as 98% is between 95.4% and 99.7%
To
N=18, mean=492.3, std deviation=37.6. The zscore for a 98% confidence interval is 2.325 (approx). The formula for a confidence interval is X (sample mean) +/ Z*(std deviation)/sqrt(N). Thus, the confidence interval at 98% is 492.3+/(2.325*37.6/sqrt(18)) or 492.3 +/ 20.6. What this really means is that the mean of the population will fall within 471.7 and 512.9 98% of the time.
If I'm wrong please let me know!
If I'm wrong please let me know!
hi...um arent we suppose to use the t.distribution instead of the z because n(the sample size) is less than 30 and one of the properties to using the t.table is when n is less than 30
Well, I'm not sure how exactly what language the answer must be in, but hopefully I can help with the theory:
If we assume a normal distribution, which is a pivotal assumption and is not explicitly stated, (if the distribution is not normal, the question does not give enough info for an answer) the question seems to be asking:
How wide would a range of the distribution, centered on the mean, have to be to contain 98% of outcomes?
We already know that:
A range from 1 std.dev. to 1 std.dev. c
If we assume a normal distribution, which is a pivotal assumption and is not explicitly stated, (if the distribution is not normal, the question does not give enough info for an answer) the question seems to be asking:
How wide would a range of the distribution, centered on the mean, have to be to contain 98% of outcomes?
We already know that:
A range from 1 std.dev. to 1 std.dev. c
Because the distribution is symmetrical, you know that 100  99 = 1% will be below the negative of the std.dev. level Excel gives you. And then you have the std.dev. levels between which 98% of outcomes lie. Then, of course multiply your std.dev level by the std.dev given in the question, subtract it from the mean for the lower end of the range and add it to the mean for the upper end of the range. That's it. That's your 98% confidence interval.
To find the exact answer, pull up Excel and pick a random cell. Type in "=NormsInv(0.99)"
Why not 0.98? NormsInv is a cumulative function, so there is no way to specify only between x std.dev's and x std.dev's. It will only give you the std.dev. below which y% of outcomes fall. So, this will tell you the std.dev. level below which 99% of outcomes fall. Because the distribution is symmetrical, you know that 100  99 = 1% will be below the negative of the std.dev. level Excel gives you. And then
Why not 0.98? NormsInv is a cumulative function, so there is no way to specify only between x std.dev's and x std.dev's. It will only give you the std.dev. below which y% of outcomes fall. So, this will tell you the std.dev. level below which 99% of outcomes fall. Because the distribution is symmetrical, you know that 100  99 = 1% will be below the negative of the std.dev. level Excel gives you. And then
If we know the mean and the standard deviation of the population, then why are we taking samples, if we already have the data?
Thanks in advance.
Thanks in advance.
Learning statistics can be a little strange. It almost seems like you're trying to lift yourself up by your own bootstraps. Basically, you learn about populations working under the assumption that you know the mean/stdev, which is silly, as you say, but later you begin to drop these assumptions and learn to make inferences about populations based on your samples.
Once you have some version of the Central Limit Theorem, you can start answering some interesting questions, but it takes a lot of study just to get there!
Once you have some version of the Central Limit Theorem, you can start answering some interesting questions, but it takes a lot of study just to get there!
Is there any difference if I take 1 "sample" with 100 "instances", or I take 100 "samples" with 1 "instance"?
(By sample I mean the S_1 and S_2 and so on. With instances I mean the numbers, [1,1,3,6] and [3,4,3,1] and so on.)
(By sample I mean the S_1 and S_2 and so on. With instances I mean the numbers, [1,1,3,6] and [3,4,3,1] and so on.)
Sal goes over this better than I do in the next video as well!
There is a difference. Your "samples" (random selections of values "x") that are made up of "instances" (referred to as the variable "n") provide what will essentially be the building blocks of your Sampling Distribution of the Sample Mean. Because your "instances" determine the value of the mean of "x", your size of "n" determines the value of "x"'s mean, and the Sampling Distribution of the Sample Mean's standard deviation (Defined as The original dataset's standard deviation divided by the square root of "n").
For example: If you were to take 1 "sample" with 100 "instances", you would get only one piece of data regarding the mean of 100 items [1,1,3,6,3,6,3,1,1,1,1,1...] from your original data. Your sampling distribution of the Sample mean's standard deviation would have a value of ((The original sample's S.D.)/(The square root of 100)), but that wouldn't really matter, because your data will likely be very close to your original data's mean, and you'd only have one sample.
Now if you take 100 samples with 1 instance [3], you'll get many pieces of data, but no change in standard deviation from your first sample: ((The original sample's S.D.)/(The square root of 1)). Functionally, with enough samples taken like this, you'll recreate your original dataset! You won't be creating a useful sampling distribution of the sample mean because "x" will equal the mean of "x". With 100 "samples" of 1 "instance", you're randomly picking 100 values of "x" and replotting them.
I hope that helps.
For example: If you were to take 1 "sample" with 100 "instances", you would get only one piece of data regarding the mean of 100 items [1,1,3,6,3,6,3,1,1,1,1,1...] from your original data. Your sampling distribution of the Sample mean's standard deviation would have a value of ((The original sample's S.D.)/(The square root of 100)), but that wouldn't really matter, because your data will likely be very close to your original data's mean, and you'd only have one sample.
Now if you take 100 samples with 1 instance [3], you'll get many pieces of data, but no change in standard deviation from your first sample: ((The original sample's S.D.)/(The square root of 1)). Functionally, with enough samples taken like this, you'll recreate your original dataset! You won't be creating a useful sampling distribution of the sample mean because "x" will equal the mean of "x". With 100 "samples" of 1 "instance", you're randomly picking 100 values of "x" and replotting them.
I hope that helps.
So if every distribution approaches normal when do I employ say a Poisson or uniform or a Bernoulli distribution? I suppose it's a concept I haven't breached yet but how do I know when or which distribution to employ so I appropriately analyze the data? End goal = solve real world problems!
Not every distribution goes to the Normal. the distribution of the sample mean does, but that's as the sample size increases. If you have smaller sample sizes, assuming normality either on the data or the sample mean may be wholly inappropriate.
In terms of identifying the distribution, sometimes it's a matter of considering the nature of the data (e.g. we might think "Poisson" if the data collected are a rate, number of events per some unit/interval), sometimes it's a matter of doing some exploratory data analysis (histograms, boxplots, some numerical summaries, and the like).
For actually analyzing data: I would suggest hiring someone with more extensive training in Statistics to actually do such. Taking one course in Stats, which is basically what KhanAcademy goes through, isn't really enough to prepare someone to be a data analyst. I see the primary goal of taking one or two stats courses as giving you enough information to allow you to understand the results of statistical analyses. You can better tell the statistician what you want in his/her own terms, and you can better understand what s/he gives back to you.
In terms of identifying the distribution, sometimes it's a matter of considering the nature of the data (e.g. we might think "Poisson" if the data collected are a rate, number of events per some unit/interval), sometimes it's a matter of doing some exploratory data analysis (histograms, boxplots, some numerical summaries, and the like).
For actually analyzing data: I would suggest hiring someone with more extensive training in Statistics to actually do such. Taking one course in Stats, which is basically what KhanAcademy goes through, isn't really enough to prepare someone to be a data analyst. I see the primary goal of taking one or two stats courses as giving you enough information to allow you to understand the results of statistical analyses. You can better tell the statistician what you want in his/her own terms, and you can better understand what s/he gives back to you.
Me and my friend Callum have been experimenting with sampling distribution progran on online stat book used by Sal (http://onlinestatbook.com/stat_sim/sampling_dist/index.html). However we found a result we cannot explain nor rationalise: When we ask for a sample size of 2 for the median disribution of any population it aproximates the population distribution and not a 'bell curve'. I am very disturbed by this because surely the median of 2 numbers is the same as the mean of 2 numbers and according to the central limit theorem should approximate a normal distribution. Is this assumption correct? Is the programme wrong? Or is there something we fail to understand?
The distribution of the sample median is not normal even if you take a larger sample size, such as n=5,10,or 25. The distribution of the sample median seems to be more related to the distribution of the population.
But I don't know why.
But I don't know why.
Thanks Jilarra, I agree. Unfortunately I cannot access the program again. However I would like to see if the simulation of n=2 for both the median and the mean are similar (or the same or completely different). If anyone happens to test this please post here to let me know the results :)
when n = 2 (n being sample size) the central limit theorem is not going to give a very good approximation to the normal
why can we say that the sampling distribution of mean follows a normal distribution for a large enough sample size even though the population is may not be normally distributed?
Properly, the sampling distribution APPROXIMATES a normal distribution for a sufficiently large sample (sometimes cited as n > 30). A coin flip is not normally distributed, it is either heads or tails. But 30 coin flips will give you a binomial distribution that looks reasonably normal (at least in the middle).
@ 9:15 two distributions are shown and compared (N=5 and N=25) and Sal explains in terms of skew and Kurzweillosis (or something) that the N=25 distribution is more normal. But wait... it does not LOOK more normal to me. Specifically, it looks a lot lumpier... as if it were composed of less data. Each bin is fatter and there are less bins. Am I making sense? Can someone explain?
The lumpier look you're seeing is exactly because of the fewer number of bins. If we wanted, we could go in and specify how we wanted the bins formed, but typically there's just a computer algorithm that chooses the bins in some fashion. If we chose a few more bins there, it would looks much more smooth.
The bottom histogram looks more normal because of the general behavior of the distribution. The one for n=5 is like a normal distribution that was smashed down a bit. It's too short in the middle and has too "fat" of tails. If you think back to, say, the Empirical Rule, the top one would probably have less than 68% of the data within 1 standard deviation of the mean.
p.s. the word is "kurtosis," it's a way to describe the "peakedness" of the graph. A graph with high kurtosis will have much sharper peak (picture 1 below), a graph with low kurtosis will have much more of a rolling hill look to it.
Picture 1:
http://commons.wikimedia.org/wiki/File:Orographic_lifting_of_the_air__NOAA.jpg
Picture 2:
http://en.wikipedia.org/wiki/File:FoothillsCO.JPG
The bottom histogram looks more normal because of the general behavior of the distribution. The one for n=5 is like a normal distribution that was smashed down a bit. It's too short in the middle and has too "fat" of tails. If you think back to, say, the Empirical Rule, the top one would probably have less than 68% of the data within 1 standard deviation of the mean.
p.s. the word is "kurtosis," it's a way to describe the "peakedness" of the graph. A graph with high kurtosis will have much sharper peak (picture 1 below), a graph with low kurtosis will have much more of a rolling hill look to it.
Picture 1:
http://commons.wikimedia.org/wiki/File:Orographic_lifting_of_the_air__NOAA.jpg
Picture 2:
http://en.wikipedia.org/wiki/File:FoothillsCO.JPG
I'm a little confused about what you're doing at 04:40. Lets say the PDF represents the 32 species of animals on a small island. So that application selects 5 types of animals lets say zebras, goats, penguins, gorillas and porcupines and plots their mean on the graph below. How the hell can you get the mean of a set of 5 species of animals? I don't get it.
@cnidoblast, selecting 5 types of animals invalidates the CLT. One of the assumptions of the most common CLT (there are actually many versions, this one is the most common) is that the observations, what Mr. Khan calls samples, are independent and identically distributed instances of a random variable. A random variable is a function that converts an observation from a random process in to a number. Your animals are not numbers, so it's meaningless to sum them much less find the mean. If you're talking about averaging their weights then it still fails the CLT assumptions because the weights that you're averaging do not come from an identical distribution. That is, the distribution of weights of zebras is very different from the distribution of weights of goats. Hope this helps! :)
Could you define a measure of skewness as (meanmedian)/standard deviation? An advantage of this would be that it is easier to calculate, and it can only take values between 1 and 1
I'm having some issues with this question.
3. For the general population, mean IQ is 100 with a standard deviation of 15. A sample of 100 people is selected at random from the population, with a sample mean of 102. This sample mean comes from a distribution of sample means with the following properties:
a. a mean of 100 and a standard error of 1.5
b. a mean of 102 and a standard error of 1.5
c. a mean of 100 and a standard error of 15
d. a mean of 102 and a standard error of 15
I think that the answer is either a or b, because you would divide the SD 15 by the square root of the original mean 10, which gives 1.5. But I have no idea what to do about the mean 100/102? Can anyone explain why it is one or the other?
3. For the general population, mean IQ is 100 with a standard deviation of 15. A sample of 100 people is selected at random from the population, with a sample mean of 102. This sample mean comes from a distribution of sample means with the following properties:
a. a mean of 100 and a standard error of 1.5
b. a mean of 102 and a standard error of 1.5
c. a mean of 100 and a standard error of 15
d. a mean of 102 and a standard error of 15
I think that the answer is either a or b, because you would divide the SD 15 by the square root of the original mean 10, which gives 1.5. But I have no idea what to do about the mean 100/102? Can anyone explain why it is one or the other?
THe general population is known to have a mean IQ of 100. That means that the distribution of sample means also has a mean of 100.
At 513 pm: For some reason, I understand this when it comes to means but in Sampling distribution of the sample proportion Using population (4,5,9), sample size n = 2 I am struggling to construct a table that represents the sampling distribution of the sample proportion of odd numbers. Can you please explain?
How would one answer a question such as "what is the sampling distribution of the sample mean? Explain." after being given a problem where the only info given is the mean of a (normal) distribution and its standard deviation? There is also a number that is being randomly computed and averaged. Is the sample mean the mean of the normal distribution?
The sampling distribution of a normal distribution is itself normally distributed. The mean of the sampling distribution is the mean of the original distribution (by symmetry there is no other possible result), and the standard deviation of the sampling distribution shrinks by the square root of the sample size.
This derives from the properties of the variance. When you add two random variables, the variance of the sum adds. Thus when you add n identical random variables, the variance of the sum is n times the original variance and the standard deviation (square root of the variance) is sqrt(n) times the original standard deviation. Divide this by n, to AVERAGE n identical random variables, and you get the above result.
This derives from the properties of the variance. When you add two random variables, the variance of the sum adds. Thus when you add n identical random variables, the variance of the sum is n times the original variance and the standard deviation (square root of the variance) is sqrt(n) times the original standard deviation. Divide this by n, to AVERAGE n identical random variables, and you get the above result.
how do distributions provide a link between probabilities and statistical tests
Statistical tests are generally trying to compute the probability of something. Most often, there is an assumption (hypothesis), and we find the probability of the observed results assuming that hypothesis is true.
The probabilities can be calculated in a few different ways, but a very common method is through a distribution. So, we think that the data or a function of it, like a test statistic, has a particular distribution (this is generally _proven_, so it's not just a guess), and we can use that distribution to calculate probabilities.
The probabilities can be calculated in a few different ways, but a very common method is through a distribution. So, we think that the data or a function of it, like a test statistic, has a particular distribution (this is generally _proven_, so it's not just a guess), and we can use that distribution to calculate probabilities.
what is the relationship between M, meu, and meu with subscript m?
I have a question that I dont quite understand and it goes like this: "Assume the weights of eggs produced on an egg farm have a normal distribution with mean 64 grams and standard deviation 7 grams. and it also says "describe the distribution of weights of 12 (randomly chosen) mixed grade eggs?
9:08, how do you get five samples from the nonnormally distributed probability function? How do you get a set of data from the probability function?
Computers can quite easily simulate uniform distributions (for example the rand() function in matlab that gives a number between 0 and 1 accordingly to an uniform distribution). With that number you can simulate all sorts of other distributions.
For example if you want to simulate a fair dice you do :
x = rand(1)
if (x<1/6) then y = 1
elseif (x<2/6) then y = 2
elseif (x<3/6) then y = 3
elseif (x<4/6) then y = 4
elseif (x<5/6) then y = 5
else y=6
This is how you can simulate easily discrete distributions.
For example if you want to simulate a fair dice you do :
x = rand(1)
if (x<1/6) then y = 1
elseif (x<2/6) then y = 2
elseif (x<3/6) then y = 3
elseif (x<4/6) then y = 4
elseif (x<5/6) then y = 5
else y=6
This is how you can simulate easily discrete distributions.
My professor said the answer to the problem is "NOT" 0. I take meticulous notes, record lectures, online research, etc. Why can't I figure this out. Do I need to somehow calculate a sample proportion? Not sure what else to do. If the sample proportion is not given, how do I find it. The problem is the Z scores are above 3 and our Standard Normal Distribution Table stops at 3. Again, he said the answer is not 0. Below are some problems directly pasted here:
1) Given a normal distribution with a µ = 100 and σ = 10, if you select a random sample of n = 25, what is the probability that the sample mean is between 90 and 97.5?
2) Given a normal distribution with a µ = 50 and σ = 8, if you select a random sample of n = 100, what is the probability that the sample mean is between 47 and 49.5?
3) Given a normal distribution with a µ = 50 and σ = 5, if you select a random sample of n = 100, there is a 35% chance that the sample mean is above what value?
I'm really struggling here with the Z's being greater than 3. working on this for three days. Not just trolling for answers and being lazy. I desperately want to know the techniques and steps to calculate situations like this. Thank you very very much.
1) Given a normal distribution with a µ = 100 and σ = 10, if you select a random sample of n = 25, what is the probability that the sample mean is between 90 and 97.5?
2) Given a normal distribution with a µ = 50 and σ = 8, if you select a random sample of n = 100, what is the probability that the sample mean is between 47 and 49.5?
3) Given a normal distribution with a µ = 50 and σ = 5, if you select a random sample of n = 100, there is a 35% chance that the sample mean is above what value?
I'm really struggling here with the Z's being greater than 3. working on this for three days. Not just trolling for answers and being lazy. I desperately want to know the techniques and steps to calculate situations like this. Thank you very very much.
The key to all of these questions is using the standard error of the mean which is described in one of the next videos in the section.
Briefly, the SE (standard error) = standard deviation / sqr (sample size).
For 1) the SE = 10 (standard deviation) / 5 (sqr of 25) = 2. If you use a z table, we are looking for the probability of z between 5 {(90100)/2} and 1.25 {(97.5100) / 2}. This is .1056 using this online table (http://www2.fiu.edu/~millerr/Normal%20Table.pdf).
The other problems are solved similarly
Briefly, the SE (standard error) = standard deviation / sqr (sample size).
For 1) the SE = 10 (standard deviation) / 5 (sqr of 25) = 2. If you use a z table, we are looking for the probability of z between 5 {(90100)/2} and 1.25 {(97.5100) / 2}. This is .1056 using this online table (http://www2.fiu.edu/~millerr/Normal%20Table.pdf).
The other problems are solved similarly
I need help putting together the formula to anser the question, "A population is bimodal with a variance of 5.77. One hundred samples of size 30 are randomly selected and the 100 sample means are calculated. The standard deviation of the sample means is approximately:
I have a question that I cant figure out please help:
Identify the class width, class midpoints, & class boundaries for the given frequency distribution
Daily low temp (F) Frequency Daily low temp (F) Frequency
3235 1 4851 7
3639 3 5255 7
4043 5 5659 1
4447 11
Identify the class width, class midpoints, & class boundaries for the given frequency distribution
Daily low temp (F) Frequency Daily low temp (F) Frequency
3235 1 4851 7
3639 3 5255 7
4043 5 5659 1
4447 11
The class widths are the width of each interval which in this case is 4 (e.g. {32, 33, 34, 35} has 4 items),
the mid points are the mid point of each class, (top + bottom)/2, 33.5 in the case of the first one.
The boundaries between the ranges except you want to include the data that gets rounded up or down, so you add 0.5 to the top boundary or subtract it from the lower. So they would be 31.5, 35.5, 39.5, ..., 59.5
the mid points are the mid point of each class, (top + bottom)/2, 33.5 in the case of the first one.
The boundaries between the ranges except you want to include the data that gets rounded up or down, so you add 0.5 to the top boundary or subtract it from the lower. So they would be 31.5, 35.5, 39.5, ..., 59.5
I have a question m failing to solve. ' A population has a mean of 200 and a standard deviation of 50. A simple random sample of size 100 will be taken and the sample mean x will be used to estimate the population mean. Show the sampling distribution of the sample mean
at 8:45, it has been said that even for single samples the central limit theorem is true. It is not so, central limit theorem is applicable only for sample MEANS. For example, out of a population of 5000 if I have taken the sample of n=50, central limit theorem does NOT apply to that. It applies only when I have taken (e.g.)40 samples of n=50. However, this is as per my understanding. Please correct me if I am wrong.
What I don't understand is when you have a large Binary distribution for example, and you approximate it using Normal distribution.. If you only have one sample consisting of x values, you haven't got a standard deviation really.. we always have those kinds of questions on the exam but i always get the formula wrong then..
As long as you know all the values in the sample, you can do the series of calculations described under "basic examples" here http://en.wikipedia.org/wiki/Standard_deviation to figure out what the sample's standard deviation is. Of course, you have to divide by N1 with samples like the wikipedia article (as well as Sal's video on standard deviation) explains, otherwise it's exactly the same. Perhaps you are limiting your definition of "standard deviation" to "standard deviation of population", which you of course can't figure out with just one sample of values? If it's not specified that the population's SD is asked for in the exam question you're describing, it's safe to assume that they are asking for the sample's SD.
I'm trying to picture skew and kurtosis, but I have no idea how much the numbers actually mean. Is there a video that gives a good idea of how much skew is, say: 0.1, 0.5, 1, and 10? Same thing with kurtosis. I like having a feel for what the value means in my brain.
only the mean follows the CLT ?
What would be the difference between the distribution of a sample variable and the sampling distribution of the mean?..? I'm so confused between these two terms
Sal repeats "well defined mean" and "well defined variance" a couple of times at the very beginning of the video. When are these quantities not well defined?
what does the ! following a number mean
7! means 7 factorial which is the same as 7*6*5*4*3*2*1 or 5040
This seems to be a simple question to answer, but I'm actually not 100% certain about it:
say there's a population that's normally distributed with mean u and standard deviation s. An independent sample of N observations is drawn from the population. What is the distribution of the sample mean? I think it's still a normal distribution, but I'm not sure if this is correct and sufficient, because I'm still in the process of getting comfortable with all this stat lingo.
Thanks!
say there's a population that's normally distributed with mean u and standard deviation s. An independent sample of N observations is drawn from the population. What is the distribution of the sample mean? I think it's still a normal distribution, but I'm not sure if this is correct and sufficient, because I'm still in the process of getting comfortable with all this stat lingo.
Thanks!
Yes. If X is normally distributed, then the sample mean xbar will also be normally distributed regardless of the sample size. If X is _not_ normally distributed, then we have to make sure the sample size is large enough for the Central Limit Theorem to kick in.
There were two cases talked about; n=5 and n=25. It was said that after 10,000 samples the n=25 was a closer fit to the normal distribution than the n=5 case. What I want to know is, if there were infinite samples, would the n=5 and the n=25 cases both be a perfect normal distribution?
If this is so: As the number of samples tends to infinity, does the n=25 case converge to the normal distribution faster than the n=5 case?
If this is so: As the number of samples tends to infinity, does the n=25 case converge to the normal distribution faster than the n=5 case?
This is answered in the next video in the series.
2:20 Can you really do it with the mode? It seems like there would be some distributions in which no matter how many samples you take, the mode would not be normally distributed.
In this example Sal took 10,000 samples of 5 for a total of 50,000 samples in the first example. Why not just take 50,000 samples of the original distribution and calculate the mean and SD?
Every random variable has some sort of probability distribution. When we have a lot of data, we can plot them in a histogram and "see" the probability distribution. This is what we often do to see the distribution of the raw data. But the sampling distribution o the sample mean is trickier business. When we calculate the sample mean (xbar) , we have 1 value. Xbar is still a random variable, but for a given dataset, we have only 1 value of xbar, and using just 1 value is not going to provide a very useful plot. We'd really like to see how the sampling distribution of xbar behaves, but for that we need to have a lot of xbar's.
So we can do some experiments like Sal has done. He decided how to generate some data (according to that very strange population he was making on the top panel), and then he can drew 5 observations from that distribution. By calculating the mean, we get 1 observation from the sampling distribution of the sample mean. If we do this over and over again, that lets us get 10,000 observations from the sampling distribution of the sample mean. Plotting all of these together lets us see how the sampling distribution of the sample mean behaves  at least for the distribution Sal specified.
If we had put all 50,000 observations that we drew together and calculated the sample mean and SD, that would just be 1 observation from the sampling distribution of the sample mean (with n=50,000 instead of n=5). If we plotted all 50,000 together, that would be plotting the distribution of the raw data, not the distribution of the sample mean.
Let me know if this helps. It's a pretty tricky concept to grasp, I've had college students struggle to understand this, and that was when I was there explaining it in person.
So we can do some experiments like Sal has done. He decided how to generate some data (according to that very strange population he was making on the top panel), and then he can drew 5 observations from that distribution. By calculating the mean, we get 1 observation from the sampling distribution of the sample mean. If we do this over and over again, that lets us get 10,000 observations from the sampling distribution of the sample mean. Plotting all of these together lets us see how the sampling distribution of the sample mean behaves  at least for the distribution Sal specified.
If we had put all 50,000 observations that we drew together and calculated the sample mean and SD, that would just be 1 observation from the sampling distribution of the sample mean (with n=50,000 instead of n=5). If we plotted all 50,000 together, that would be plotting the distribution of the raw data, not the distribution of the sample mean.
Let me know if this helps. It's a pretty tricky concept to grasp, I've had college students struggle to understand this, and that was when I was there explaining it in person.
video wont play plese fix
maybe there's a bug or something.
are sample mean and population mean the same? while solving ques for confidence intervals why do we always subtract the sample mean from the value when the formula includes population mean?
P(1.28 < z < 1.75)
there are less videos on econometrics..:(
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.6 years, and standard deviation of 0.5 years.
If you randomly purchase 25 items, what is the probability that their mean life will be longer than 3 years?
If you randomly purchase 25 items, what is the probability that their mean life will be longer than 3 years?
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