The Sampling Distribution of the Sample Mean

I discuss the sampling distribution of the sample mean, and work through an example of a probability calculation. (I only briefly mention the central limit theorem here, but discuss it in more detail in another video).
The mean and standard deviation of the amount of protein in a quarter pound patty of lean beef was found in the USDA nutrient database at:
ndb.nal.usda.gov/ndb/foods/sho...
For those using R, here is the R code to find the values in the examples:
The probability a randomly selected patty has at least 23.0 grams of protein (mu = 21.4, sigma = 1.9):
1-pnorm(23.0,21.4,1.9)
[1] 0.1998645
or, if we standardize:
1-pnorm((23.0-21.4)/1.9)
[1] 0.1998645
The probability that the mean of 4 randomly selected patties is at least 23.0 grams of protein (sampling from a normal distribution with mu = 21.4, sigma = 1.9):
1-pnorm(23.0,21.4,1.9/sqrt(4))
[1] 0.04607049
or, if we standardize:
1-pnorm((23.0-21.4)/(1.9/sqrt(4)))
[1] 0.04607049

[@emem2060]

thank god for these videos. I learned in 12 min what I didnt understand in 2 days worth of class

[@jbstatistics]

Thanks, I'm glad you like my videos. Thanks for the suggestions. Gamma, Weibull, and lognormal are definitely somewhere on the horizon, but it will be a little while before I can get to them. (I've got a number of other videos lined up right now.) Cheers.

[@dasgomezkanal]

My friend, I want to thank you a lot! for posting this videos. You are an excellent professor and communicator! My statistics profesor is the worse... I can't understand anything at all in class.. I don't even know why it is so expensive to pay for her class 4,500 a semester when all she does is talking and drawing numbers and we are all like 0_o ... Anyway, I really just wanted to thank you a lot for your help. This is my last semester as an undergrad and without these videos I don't think I could pass this subject and graduate. Thank you Thank you Thank you!

[@jbstatistics]

+dasgomezkanal You are very welcome!

[@dinglebeey]

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[@jbstatistics]

Thanks!

[@hessaa3464]

I really like your way in explaining its very simple and understandable. I hope that you create a video tutorial for Gamma, Weibull and Lognormal distributions because I didn't find any good video tutorials for these concepts. Thanks in advance =)

[@alexandra-stefaniamoloiu2431]

Those are simply the best explanations! I'll watch all your videos.

[@jbstatistics]

+alexandra-stefania moloiu Thanks!

[@physiciansmind]

Awesome video. This provided great help for my upcoming stats exam!

[@os7007]

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[@jbstatistics]

Thanks Arian. I'm very glad you've found my videos helpful. Best of luck on your stats exam!

[@homeisthesailor]

@@jbstatistics Having studied Stats at uni and having passed the exam but remained confused Well done! The very opposite of going from the notes of the lecturer to the notes of the student without going through the brain of either.

[@homeisthesailor]

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[@jbstatistics]

Thanks!

[@radhikamamidi9070]

Very informative , crisp and clear.. Watched other videos but understood very well from your videos the concepts well. Thank you.

[@jbstatistics]

Thanks for the compliments!

[@WRONGTURN69]

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[@jbstatistics]

Thanks for the compliments!

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[@jbstatistics]

+Cat Garcia You are very welcome! I'm glad you've found my videos helpful!

[@thecorgiman4118]

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[@jolantabenson9160]

Thank you JB! Excellent tutorials :-)

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[@jbstatistics]

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[@tongyulin6177]

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[@jbstatistics]

Thanks! I'm glad to hear that!

[@farlyn3]

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[@amandafalke7670]

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[@miguelese18]

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[@leahhall7504]

this is the video that made everything make sense in my head

[@jbstatistics]

Thanks!

[@muertaqueen]

the GOAT of stats tutor online

[@jbstatistics]

Thanks!

[@XieQiu]

Very clear explanation.

[@jbstatistics]

Thanks!

[@hiphopsocnroc]

Great explanation.

[@Un_Bison]

Very helpful, Thank you!

[@jbstatistics]

+JustTheRickyshow You're welcome!

[@cameellesuarez298]

A great help! God bless you!

[@jbstatistics]

Thanks!

[@legendarylioness1104]

These are really great!

[@jbstatistics]

Thanks! I'm glad you like them.

[@mutalemalama]

The terminology really fucks me over. I don't know when to implement certain formulas so that's a real hinderance although, this has sort of widened my perspective so cheers mate

[@pomegranate8593]

good explaining!!!!!!!!!

[@ilose21]

Thank you, apparently you are a better instructor than my online applied statistics instructor...

[@jbstatistics]

You are very welcome.

[@superkamiguru6510]

My teacher never teaches in class so i'm forced to scour the internet and search for problems similar to the work he gives out to help me understand it better. After watching this video, I still am at a lost for what to do

[@hakuyihnew499]

Thank you, very interesting

[@chumpapii]

Hey man what if thre are 2 given sample mean like between 445 and 485?

[@mrx42]

Thank you master

[@jietang118]

Great video!

[@jbstatistics]

Thanks!

[@xitrumkc]

So Good!

[@user-cm4tu5jj5n]

Thanks a lot. It is helpful to solve my problems.

[@jbstatistics]

I'm glad to be of help!

[@rosemarieguira1042]

It helps me a lot. Thank you so much po.

[@shefalisingh4675]

I really like your way of explaining concepts. But if in a real case scenario where we don't have mu value(suppose a very large sample) how do we even calculate the probability??

[@jbstatistics]

Thanks for the compliment. As you know, we don't (generally) know mu in real life. So then we can't calculate a probability this way. The primary purpose of the discussion in this video is that we later turn this problem on its head, and use a known value of X bar to say something about the unknown value of mu. Working with the mathematical concepts discussed in this video (and some others), we can come up with an appropriate formula for a confidence interval for mu, and an appropriate test statistic to use in hypothesis testing. Statistical inference is built on concepts related to the sampling distribution of the estimator of a parameter. Edit: Which is what I allude to at the end of the video.

[@katieandrews3005]

Thanks for the videos! Like another commenter, I can't seem to figure out how you got 0.200 at the time mark of 7:20 in your video.

[@jbstatistics]

Jenny FromDaBloc It's an area under the standard normal curve, and it can be found with software or a standard normal table. I have videos that show how to use the standard normal table. Cheers.

[@CorginShep]

Take the Z value (at 0.842) and subtract it from 0.5. Because we want the probability above 23grams (at least) and the Z value 0.842 represents the area from 0 to 0.842 you need to subtract it from the whole area (which is 0.5 on the normal distribution table).

[@cesar.vasconcelos]

Can somebody please explain how to properly estimate the population parameters: true mu and true sigma, when we don't have them? Can I use the sample mean as a good point estimator for the population mean? Can I use the sample standard deviation as a good point estimator for the true sigma?

[@lucaslau8379]

perfection!

[@jbstatistics]

Thanks!

[@ItsASoccerTing]

how did he get from 0.8 to 0.2 in the first example

[@overlordfury996]

search for statistical table for normal distribution and see 0.8 exact in the table

[@jeeruajay2111]

I have one doubt , x' mean is equal to population mean when we take mean of the sampling means otherwise not ? but your told that only one sample mean is equal to the population mean how tell me ?

[@ThePeacefulResistor]

thank you so much!

[@jbstatistics]

You are very welcome Nunya!

[@himayunrashid2864]

thankyou sir..............

[@Oingoboingo710]

Can someone please refer me to the video wherein Mr jbstats prove that E({x bar}) is equal to the population mean?

[@elvinjafarli6257]

Man, I wish you was my stats professor

[@jbstatistics]

I'm glad I can still be of help!

[@ShuusakuSama]

Is this only for for n>=30? I remember you need to use the t distribution is n

[@jbstatistics]

I talk about this in detail in my other videos. When sampling from a normally distributed population, the random variable Z = (X bar - mu)/(sigma/sqrt(n)) has the standard normal distribution. When sigma is replaced with the sample standard deviation S, the quantity T = (X bar - mu)/(S/sqrt(n)) has a t distribution with n-1 degrees of freedom. This is always true when we are sampling from a normally distributed population, regardless of the sample size.

[@letseconomics2938]

@8:39 why do we choose the middle data? We have to draw a sample of size 4 so we can take four values from the left or right as well? And will the mean will be the same as 21.4 for all samples we draw from this distribution?

[@jbstatistics]

I don't know what you mean by "why do we choose the middle data". We're randomly picking 4 values from that distribution; they'll be whatever they'll be. It's 4 random draws from that distribution. The distribution in green is the distribution of the mean of 4 randomly picked values from the distribution in white. Yes, each of the 4 has exactly the same distribution (the distribution in white).

[@letseconomics2938]

@@jbstatistics I mean if we have to choose randomly 4 values. Why didn't we choose the extreme 4 values of the distribution?

[@jbstatistics]

@@letseconomics2938 You're not understanding what's happening. We're not choosing any specific values, or letting our judgement decide which side we want to grab them from. We're randomly picking 4 values from the distribution in white. If we do that, then the distribution of the mean of those 4 randomly picked values is the green distribution. That's just how the math works out. The green distribution has a lower variance for the reasons I discuss in the earlier part of the video.

[@jbstatistics]

I'll just add that sure, we might end up randomly picking 4 extreme values from one tail of the distribution from which we're sampling, that will have non-zero probability of occurring. But the probability of that is very small and is reflected in the sampling distribution of the sample mean.

[@sadiaafrin6240]

Thank you so much

[@jbstatistics]

You are very welcome!

[@beabittenbender8373]

super helpful thanks

[@jbstatistics]

You are very welcome.

[@edwardtification]

which is the other video you derived the formula?

[@ezquerzelaya]

+e t wondering the same thing

[@CarlosAugusto-yr3bn]

Same

[@nutritionbyelsa]

wooohoooooo thank you

[@MannonCannon]

I get that the SD decreases as N increase. But I'm stuck on the idea that variance increases as sample size increases. I dont get it. Please explain

[@jbstatistics]

I am not sure what you are asking me. The standard deviation of the sampling distribution of the sample mean is just the square root of the variance of the sampling distribution of the sample mean. The standard deviation of the sample mean and the variance of the sample mean both decrease as the sample size increases.

[@MannonCannon]

That's ok I realised I read my lecture slides wrong. Was confused as to why they were saying the SD and variance increased with sample size, however they were saying the opposite. Thanks for actually replying though :)

[@2foolish]

beautiful

[@jbstatistics]

Thanks!

[@bushranahvi2263]

how did you calculate P(z>= 1.648)=0.046?

[@jbstatistics]

That area can be found by using software or a standard normal table.

[@bushranahvi2263]

thankyou!

[@nickyclement]

P(Z>1.648) can be found by looking up Z of 1.648 on standard normal table. Area under the standard normal curve to the RIGHT of Z is ~0.954. Area to the LEFT of Z is what we are looking for, so P(Z>1.648)= 1(total area under std. norm. curve) - 0.954 = 0.046

[@bigfatjuice]

Where is the video where you derive the properties of X bar?

[@jbstatistics]

I derive the mean and variance of the sample mean in this video: kzbin.info/www/bejne/bZ68dXuYp7F4h9E. I discuss the central limit theorem in this video: kzbin.info/www/bejne/hqbNoJ9nrqhklaM

[@randallkim5903]

If sigma is not given, is it possible to do any probability calculations?

[@jbstatistics]

In short, no. We might conceivably have an estimate of sigma, which could give us a rough estimate of the probabilities, but without knowledge of the value of sigma the exact probabilities can't be calculated.

[@randallkim5903]

Thank you!

[@trishayvonnefabular7119]

im confused about the square root of n.. i didnt get the answer.. pls. help

[@jbstatistics]

What part were you having trouble with? What value were you getting?

[@etlogfactory6253]

I hope someone still sees this, but can someone explain how 0.842 becomes 2 😭

[@jbstatistics]

The area to the right of 0.842 under the standard normal curve is 0.200. That's found with any statistical software, such as R, or by using a standard normal table.

[@corinalohrman5764]

why is there no sound

[@jbstatistics]

Hi Corina. There is sound when I play the video. I'm not sure what the problem is, but I believe it's something on your end. Cheers.

[@corinalohrman5764]

Yup it was thanks. I appreciate your good work

[@nimas1638]

i am the 1000th like.... what was the probability of that?

[@jimallysonnevado3973]

Wheres the proof

[@jbstatistics]

Of what part? The mean and variance? The fact that if we're sampling from a normally distributed population, then the sample mean is normally distributed? Of the central limit theorem? I derive the mean and variance of the sampling distribution here: kzbin.info/www/bejne/bZ68dXuYp7F4h9E. I don't have video proofs of the other parts yet.

[@musicmanxii]

This is like chinese to me. God help me.

[@randomrandom316]

Really useful, thanks a lot!

[@jbstatistics]

+Shailendra Pandey You are very welcome!

[@jbstatistics]

Thanks!