A random, a random, sampling, sampling mean, of p, of population pro, of population proportion, population proportion, we are confident, we are confident, we are confident, we are confident, of population proportion we are confident, confident, and remember this is no, this is not, really not really exact, so we are confident... AAAAAAAAAAAAARRRRRRRRGGGGGGGGHHHHHHH
Amazing video which provides an extremely in depth understanding rather than just throwing the formula at you. If you don't understand it just watch the videos on the central limit theorem and the sampling distribution example problems
For anybody coming here in an introductory stats class, DO not use this since it's an example of when s is unknown. Way too confusing. Just take the sample size 142 and divide by 250 total teachers sampled to obtain the sample mean. The rest is too advanced as s(sample st. deviation) will almost always be provided so you can obtain the t-score value to complete problem. Plug into the formula: t-score ( s/square root of n). t-score is alpha/2 (look up how to obtain alpha value (here it's 1-.99=.01/2=.005) from values of t scores table. In this case, 1st calculate df. df here: 249 (n-1). Any df score above 50 use very bottom of t-score table where it shows 5 separate z-score values for df values above 50. Here since alpha/2 is .005, use z-score of 0.005 which then gives t-score of 2.576. Multiple t-score by (s divided by sq. root of 250) (n). Confidence interval is x bar or sample mean +/- this last answer. Subtract this from sample mean to get lower value and add it to sample mean to get higher value. Write it as Lower value < M < higher value. M stands for population mean.
Tasks like the one in the video are purely algorithmic. You just have to do a bunch of exercises and then you should get everything right 100% of the time. No comprehension needed.
it s from 0.995-0.5 because we can have different z tables, the one sal uses is cumulative z table which tell you the area under the curve all the way to the left. you can get 0.495 if you use a normal standard z table which tells you the area under the graph from a specific data point to the mean i hope this answers your question. Note Think of that whole area from the mean to the left as 50% or 0.5 it s easier if you draw it out
11.34 is quite confusing since some z tables just measure z from the mean. This z table requires you add 0.5 to your .495 and then get your z value that way
Nice videos Sal. Although the way you solve exercises in those videos is more complicated than following the typical formula for confidence intervals, it provides an in depth understanding that I wasn't familiar before
I thought that as well... but, for whatever reason, after watching these videos, I always go back to my texts with much better understanding of the material.
Why not use proportion to calculate the confidence interval? Why do we need the sample std. dev. to solve for the interval, when we could use the the S. E. from sample proportion to get the confidence interval.
I really don't understand why so many people complain about the video being confusing. It's an example based on previous lectures. If you watched those, you wouldn't have any problems following this.
This is correct. I feel that many of the people who are confused haven't watched the previous lessons and just assume that if they skip around, they will understand everything. Unfortunately, math is brutally cumulative.
There was a big confusion on concept of "sample mean" and "sample variance" for me. here is the deal: 1st, the Variance of (sample mean) is not the sample variance, is actually (sample variance) /n (assume iid here). 2nd, in order to unlock/ or start use z or t table, first, you have to have the form of (sample mean - 0 (which is mean of t- distribution or standard-normal distribution )) / sqrt(variance(sample mean)) and above formula has a name called "test-statistic" and this "test-statistic" ,as a whole, is the one going to enjoy the -1.96 , 1.96 quantile from the z or t table. finally, we are interested CI on "sample mean" not the test-statistic (where as the function of sample mean) therefore, after unload all the terms that are not sample mean from test-statistic to the right side the equation, then we get the final result. hope this comment can help someone.
Hi , At 8:25 why do you divide by square root of 250 because the standard deviation is already on 250 samples..not on the whole population...so it should be same. Thanks
Can we just look at the Z table to calculate the interval? On the table, P( Z < 2.34) = 0.9901 Cant we just say that : mean ± (2.34*stderr) Since probability of 2.34 of z score is 99.1%
I have been racking my brain on a statistics question for the last 48 hours; I watched this video and figured it out in 5 minutes. Thank you, Khan Academy!
Hi Sal, I have noticed that your handwriting have improved tremendously. I guess you are not using your mouse to write with as before. What device (hardware) do you use to write with? (direct on a screen?) And what software program are you using? I would appreciate your answer so I could use the same tools when teaching. And thanks for all the fantastic videos!! Looking forward to your reply.
why didn't you like so much ?? the video is clear and understandable. English is not my first language. But still, this is the best material for studying statistics!
How does this work when the type of distribution is given? I have a question from an assigment stating that it is a Poisson distribution and the probability of it being zero is (65/600). With those very few arguments given I am having a very hard time figuring it out.
Why is the mean of the current sample (250 people) the same as the mean of the sampling distribution of the sample means? It could well fall on the lower end of the tail of the distribution means, no?
Hi Sal, a minor error correction @ time 17:34: Error: You had mistakenly read the 0.08 as 0.8 thus coming up with the interval (0.488, 0.648) Correction: Confidence 99% interval for .568 +/- 0.08 ought to be (0.560,0.576) Thanks for the great videos!
To find 99% CI if anyway we are good with 99.51 val so what bad what it would do with by taking 99.7% confidence interval i.e. within the 3 standard deviations?
Using the empirical rule mentioned in the earlier videos, if we had to calculate a 99.7% confidence interval, can we directly calculate 3SD values of the distribution without referring the Z table..?
No empirical rule are for intervals that are exact like exactly between 1,2,3 standard deviations from the mean that means exact confidence interval of 68-95-99.7, when you are to look for above or below 68,95,99.7 by any small value use z table e.g 67 or 68.5 or 96 or 99
If anyone wants a good formula for finding the percentage associated with the desired Z-scored, use % - (1 - %) = CI% So for this example: % - (1 - %) = 0.99 2% - 1 = 0.99 % = 0.995 A percentage of 0.995 (0.9951, to be exact) corresponds to a Z-Score of 2.58, which is the number of standard deviations above and below the mean that encompasses the confidence interval of 99%.
Hi Sir, Instead of taking 1 and 0 for sample values can we take something else and if it is going to change the sampling distribution of sample mean and variance.
What a lovely video, Needs quite pre-requisite knowledge to completely understand it.Really appreciate efforts of Salman for Khan Academy. There is no age limit for any of his videos. His concepts are so clear that you won't get confused anywhere.Thanks.
awesome video doesn't it bother you that population size has no say in the calculations? there's no difference between total number of teachers or let's say total population. standard error is free of them
not good description. How did you find the results on the Z table? what is this 0.08 that you add to the 2.5 . It's supposed that you go in depth and you want us to truly understand that intuition behind this, but in this video one needs some prerequisites that you don't explain in order to follow the course
If I were the head of statistician bureau, I could just stop at 0.568. Why introduce confidence interval at all? From what I understood, with 99% confidence, the Proportion of (0.568 +/- 0.08) = (0.488 to 0.648) will favor computer.
That is for population variance. For sample variance you have to multiply the formula with n/(n-1). Or don't use the probability formula. Use the sample variance formula.
I graduated with honors from nursing school and have aced almost every science class I've taken, but this, this is a foreign language. I can't grasp it, I get so lost with where all the numbers come into play and how to keep them straight. Any suggestions?
You don't have to associate the outcomes to zero and one in this problem. It's just a technique of turning a verbal 'yes' or 'no' into a numerical value (called the Bernoulli distribution, he's got a video about that too) , because you can't really add words together. It would make more sense (and be more applicable) if the the teachers were responding on a 1-10 scale versus yes or no. In short, just skip that part and pick up at 142/250=0.568. :)
That's interesting. We estimate, where's the mean, then we estimate precision of our previous estimate, then we can do it again and again. "True" postmodern - there's "never" truth, it's always estimate. Don't believe anything.
If we 95% confident that unknown parameter lies within our CI (95%, let's say 66-77), can we say that probability of real unknown parameter to catch within 66-77 is 95%? If not, why? Could you explain it clearly and how to get this difference. I can't understand this clearly yet. How then to calculate the probability that unknown parameter lies within our 95% CI? I can't get logic of the next (from wiki) "It is an observed interval (i.e., it is calculated from the observations), in principle different from sample to sample, that potentially includes the unobservable true parameter of interest. How frequently the observed interval contains the true parameter if the experiment is repeated is called the confidence level. In other words, if confidence intervals are constructed in separate experiments on the same population following the same process, the proportion of such intervals that contain the true value of the parameter will match the given confidence level." "Confidence intervals are frequently misunderstood, and published studies have shown that even professional scientists often misinterpret them.[7][8][9][10] A 95% confidence interval does not mean that for a given realised interval calculated from sample data there is a 95% probability the population parameter lies within the interval.[11] Once an experiment is done and an interval calculated, this interval either covers the parameter value or it does not; it is no longer a matter of probability. The 95% probability relates to the reliability of the estimation procedure, not to a specific calculated interval.[12] Neyman himself (the original proponent of confidence intervals) made this point in his original paper:[3]
The population mean is not random. What's the probability that 5 is between 2 and 10? 100%. What's the probability that 13 is between 2 and 10? 0%. Same with any fixed constant like mu or p.
16:20 Can't you simply say that 56,8% (+/-8%) of the teachers see computers as a useful tool, with a confidence level of 99%? 18:06 You mean increase the sample size (from n = 250 to more...)
i am not good at this but i will try my best to expain we want pp percent to be in the middle to find the range i think, and it has to be semetrical but we dont have a function for that, we have a function that tell us a probability from minus infinty to any given point which we can just use a Table remember it has to be semetrical eith the mean in the middle, So 0.99/2 is 0.495, so it has to be 0.495 above the mean the mean is the middle, which gives 0.5 left and right we want 0.495 above the mean WHICH IS 0.5+0.495 = 0.995 we find that on the Table wich tells us how many standard diviation it is From The Mean, now that we have how many standard diviation it is from the mean, we multiply it with the mean which was 0,031 and gives us 0.08 now we know that there i 99 percent that the mean is within 0.08 to the probability i wrote all this, just so it may help people who stoped to understand at that moment correct me if i did a mistake i am still strugling
You don't have to associate the outcomes to zero and one in this problem. It's just a technique of turning a verbal 'yes' or 'no' into a numerical value (called the Bernoulli distribution, he's got a video about that too) , because you can't really add words together. It would make more sense (and be more applicable) if the the teachers were responding on a 1-10 scale versus yes or no. In short, just skip that part and pick up at 142/250=0.568. :)
He thought he was doing it for 95% CI which means if you do 95% of 0.5 (s^2) then you get .475. But the question ask you 99% so with same logic he gets 0.495. (Replying because it threw me off and hoping it will help someone in future)
this video is not clear and i have trouble following this. Also I believe this needs to be redone as there are gaps, like when Sal said 99% he is covering only 2 std deviation whereas it has to be more than that which he has found out from Z table but that was evident
First, somebody isn't necessarily me Second, it seems you only need a to know a small mistake about something to start losing faith in it, so I don't need knowing you :D
From the past were he makes the protein that said computers were a good tool to use A's expressed them as P and then he also repossessed the population as P and then he expressed mean as P too!I just got lost.couldn't understand why he's had to multiply by 0----I found the whole video confusing for me
See its this kinda stuff. For the instructor, if this problem calls for P, DO NOT SAY ANYTHING about it being kinda like a Mean. Let me learn the mechanics first, THEN I'll come back later and make the correlation.
I’m 99% confident that I am going to fail my stats exam on Tuesday.
Did you fail? Mines like 18 hours from now lmao
Lex Luthor mines like 3 days 😭
@@ricegrum5967 Good luck! I managed to pass... But Barely hahaha..
Lex Luthor thank you
you need a bigger sample
Were you okay when you made this? I usually don't have trouble following you but this video was difficult
Yeah I struggled a touch too but with all the confusing terminology, I dunno if there's a better way to explain it.
A random, a random, sampling, sampling mean, of p, of population pro, of population proportion, population proportion, we are confident, we are confident, we are confident, we are confident, of population proportion we are confident, confident, and remember this is no, this is not, really not really exact, so we are confident... AAAAAAAAAAAAARRRRRRRRGGGGGGGGHHHHHHH
Amazing video which provides an extremely in depth understanding rather than just throwing the formula at you. If you don't understand it just watch the videos on the central limit theorem and the sampling distribution example problems
For anybody coming here in an introductory stats class, DO not use this since it's an example of when s is unknown. Way too confusing. Just take the sample size 142 and divide by 250 total teachers sampled to obtain the sample mean. The rest is too advanced as s(sample st. deviation) will almost always be provided so you can obtain the t-score value to complete problem.
Plug into the formula: t-score ( s/square root of n). t-score is alpha/2 (look up how to obtain alpha value (here it's 1-.99=.01/2=.005) from values of t scores table. In this case, 1st calculate df. df here: 249 (n-1). Any df score above 50 use very bottom of t-score table where it shows 5 separate z-score values for df values above 50. Here since alpha/2 is .005, use z-score of 0.005 which then gives t-score of 2.576. Multiple t-score by (s divided by sq. root of 250) (n). Confidence interval is x bar or sample mean +/- this last answer. Subtract this from sample mean to get lower value and add it to sample mean to get higher value. Write it as Lower value < M < higher value. M stands for population mean.
I really really really really hate statistics
Tasks like the one in the video are purely algorithmic. You just have to do a bunch of exercises and then you should get everything right 100% of the time. No comprehension needed.
Truer words were never, or never will be spoken.
Every problem which I have in the confident interval when I was at the university are solved. Thank you Khan Academy
it s from 0.995-0.5 because we can have different z tables, the one sal uses is cumulative z table which tell you the area under the curve all the way to the left. you can get 0.495 if you use a normal standard z table which tells you the area under the graph from a specific data point to the mean i hope this answers your question. Note Think of that whole area from the mean to the left as 50% or 0.5 it s easier if you draw it out
Can You please explain what did he do when he tried calcuating 0.495.
11.34 is quite confusing since some z tables just measure z from the mean. This z table requires you add 0.5 to your .495 and then get your z value that way
I’m having a Stat Final Exam in about 3 hours, and I’m 99% confident as hell that I will fail the Exam =)
Excellent example video showing the intuition of the subject!
Nice videos Sal. Although the way you solve exercises in those videos is more complicated than following the typical formula for confidence intervals, it provides an in depth understanding that I wasn't familiar before
I thought that as well... but, for whatever reason, after watching these videos, I always go back to my texts with much better understanding of the material.
Why not use proportion to calculate the confidence interval? Why do we need the sample std. dev. to solve for the interval, when we could use the the S. E. from sample proportion to get the confidence interval.
I really don't understand why so many people complain about the video being confusing. It's an example based on previous lectures. If you watched those, you wouldn't have any problems following this.
This is correct. I feel that many of the people who are confused haven't watched the previous lessons and just assume that if they skip around, they will understand everything. Unfortunately, math is brutally cumulative.
they aren't numbered or ordered in any way so it's tough to navigate through the videos as intended.
For all those watching in the future, there is a stats playlist that you can check out. Helped me out quite a bit.
There was a big confusion on concept of "sample mean" and "sample variance" for me. here is the deal:
1st, the Variance of (sample mean) is not the sample variance, is actually (sample variance) /n (assume iid here).
2nd, in order to unlock/ or start use z or t table, first, you have to have the form of (sample mean - 0 (which is mean of t- distribution or standard-normal distribution )) / sqrt(variance(sample mean))
and above formula has a name called "test-statistic"
and this "test-statistic" ,as a whole, is the one going to enjoy the -1.96 , 1.96 quantile from the z or t table.
finally, we are interested CI on "sample mean" not the test-statistic (where as the function of sample mean)
therefore, after unload all the terms that are not sample mean from test-statistic to the right side the equation, then we get the final result.
hope this comment can help someone.
you made it to here just in time for my finals... thanks sal!
Sigh... they really need to redo this.
Last Three videos in statistics playlist with CLt video(4videos) , made me 99% confident in statistics.
Thanks a lot Sal♥️!
Is equal to the standard deviation, is equal to the standard deviation, is equal to the standard deviation, is equal to the standard deviation
Hi ,
At 8:25 why do you divide by square root of 250 because the standard deviation is already on 250 samples..not on the whole population...so it should be same.
Thanks
Can we just look at the Z table to calculate the interval?
On the table,
P( Z < 2.34) = 0.9901
Cant we just say that :
mean ± (2.34*stderr)
Since probability of 2.34 of z score is 99.1%
Can we just do one tailed test instead of two tailed ?
Very Nicely Explained . Such insight explanation on the topic is hard to find .
The repeating makes it unwatchable to me :(
ขอบคุณมากครับสำหรับซับไตเติ้ลไทย
Makes perfect sense...thanks
I have been racking my brain on a statistics question for the last 48 hours; I watched this video and figured it out in 5 minutes. Thank you, Khan Academy!
Good vs not good? The question asked to the teachers was whether or not it was an ESSENTIAL tool for their teaching.
Hi Sal, I have noticed that your handwriting have improved tremendously. I guess you are not using your mouse to write with as before. What device (hardware) do you use to write with? (direct on a screen?) And what software program are you using? I would appreciate your answer so I could use the same tools when teaching. And thanks for all the fantastic videos!! Looking forward to your reply.
Since we didn't know the exact standard deviation and used the sample deviation shouldn't we have used the t table as opposed to the z table?
You would do that with a smaller sample size.
If the sample size is smaller than 30, then we use t table. If its larger, then z table.
why didn't you like so much ?? the video is clear and understandable. English is not my first language. But still, this is the best material for studying statistics!
Great vid.
How does this work when the type of distribution is given? I have a question from an assigment stating that it is a Poisson distribution and the probability of it being zero is (65/600). With those very few arguments given I am having a very hard time figuring it out.
Whoa perfect timing, thanks!
Why is the mean of the current sample (250 people) the same as the mean of the sampling distribution of the sample means? It could well fall on the lower end of the tail of the distribution means, no?
Hi Sal, a minor error correction @ time 17:34:
Error: You had mistakenly read the 0.08 as 0.8 thus coming up with the interval (0.488, 0.648)
Correction: Confidence 99% interval for .568 +/- 0.08 ought to be (0.560,0.576)
Thanks for the great videos!
Sal's calculation is correct. It is 0.08, not 0.008
Even I am prone to such errors while multiplying decimals. don't know why. cheers anyway.
To find 99% CI if anyway we are good with 99.51 val so what bad what it would do with by taking 99.7% confidence interval i.e. within the 3 standard deviations?
So I'm not the only one lost watching this... AP Stat exam tomorrow and I'm cramming y'all!!!!
good video
Using the empirical rule mentioned in the earlier videos, if we had to calculate a 99.7% confidence interval, can we directly calculate 3SD values of the distribution without referring the Z table..?
Exactly what I was saying lol
No empirical rule are for intervals that are exact like exactly between 1,2,3 standard deviations from the mean that means exact confidence interval of 68-95-99.7, when you are to look for above or below 68,95,99.7 by any small value use z table e.g 67 or 68.5 or 96 or 99
If anyone wants a good formula for finding the percentage associated with the desired Z-scored, use
% - (1 - %) = CI%
So for this example:
% - (1 - %) = 0.99
2% - 1 = 0.99
% = 0.995
A percentage of 0.995 (0.9951, to be exact) corresponds to a Z-Score of 2.58, which is the number of standard deviations above and below the mean that encompasses the confidence interval of 99%.
Hi Sir,
Instead of taking 1 and 0 for sample values can we take something else and if it is going to change the sampling distribution of sample mean and variance.
It will change the numerical value, but the meaning of the value won't change.
(1-0.99)/2=0.005....0.5-0.005=0.495
What a lovely video, Needs quite pre-requisite knowledge to completely understand it.Really appreciate efforts of Salman for Khan Academy.
There is no age limit for any of his videos. His concepts are so clear that you won't get confused anywhere.Thanks.
awesome video
doesn't it bother you that population size has no say in the calculations? there's no difference between total number of teachers or let's say total population. standard error is free of them
At 8 seconds, that should be sigma sub of x bar, not sigma of x. It's the standard deviation of x bar, not the standard deviation of x.
He makes this way too complicated, I'm lost watching this
All those who have their stats tests soon, what discipline are you studying in and where?
not good description. How did you find the results on the Z table? what is this 0.08 that you add to the 2.5 . It's supposed that you go in depth and you want us to truly understand that intuition behind this, but in this video one needs some prerequisites that you don't explain in order to follow the course
If I were the head of statistician bureau, I could just stop at 0.568. Why introduce confidence interval at all?
From what I understood, with 99% confidence, the Proportion of (0.568 +/- 0.08) = (0.488 to 0.648) will favor computer.
Mr. Sal, why did you sum .495 & .5?
When we use the z table and t table to find the confidence interval
i thought the variance of Bernoulli Distribution is p(1-p), which is 0.568*(1-0.568) = 0.245376, no?
That is for population variance. For sample variance you have to multiply the formula with n/(n-1). Or don't use the probability formula. Use the sample variance formula.
I graduated with honors from nursing school and have aced almost every science class I've taken, but this, this is a foreign language. I can't grasp it, I get so lost with where all the numbers come into play and how to keep them straight. Any suggestions?
watch all the video starting from basic descriptive statistics
I don't get the idea of associating two outcomes with 0 and 1. WHat video should I watch to get it?
You don't have to associate the outcomes to zero and one in this problem. It's just a technique of turning a verbal 'yes' or 'no' into a numerical value (called the Bernoulli distribution, he's got a video about that too) , because you can't really add words together. It would make more sense (and be more applicable) if the the teachers were responding on a 1-10 scale versus yes or no. In short, just skip that part and pick up at 142/250=0.568. :)
which playlist is this in?
Anthony Salazar Statistics
+Anthony Salazar Go to www.KhanAcademy.org and search for confidence intervals
0.99 divided into 2 = 0.495
Not sure if you'll see this but thank you for clarifying i was so lost lol
I COULD KISS YOU. This is all I wanted to know. Now I can work on the rest of my Homework. Phew...
i watch every single video in this playlist then end up not understanding 1% of it because khan speeds through it. oof. at least i can get 99% of it
That's interesting. We estimate, where's the mean, then we estimate precision of our previous estimate, then we can do it again and again. "True" postmodern - there's "never" truth, it's always estimate. Don't believe anything.
just a quick question how do you record and write the stuff on the black ground?
If we 95% confident that unknown parameter lies within our CI (95%, let's say 66-77), can we say that probability of real unknown parameter to catch within 66-77 is 95%? If not, why? Could you explain it clearly and how to get this difference. I can't understand this clearly yet. How then to calculate the probability that unknown parameter lies within our 95% CI?
I can't get logic of the next (from wiki)
"It is an observed interval (i.e., it is calculated from the
observations), in principle different from sample to sample, that
potentially includes the unobservable true parameter of interest. How
frequently the observed interval contains the true parameter if the
experiment is repeated is called the confidence level. In other words,
if confidence intervals are constructed in separate experiments on the
same population following the same process, the proportion of such
intervals that contain the true value of the parameter will match the
given confidence level."
"Confidence intervals are frequently misunderstood, and published studies
have shown that even professional scientists often misinterpret
them.[7][8][9][10]
A 95% confidence interval does not mean that for a
given realised interval calculated from sample data there is a 95%
probability the population parameter lies within the interval.[11] Once
an experiment is done and an interval calculated, this interval either
covers the parameter value or it does not; it is no longer a matter of
probability. The 95% probability relates to the reliability of the
estimation procedure, not to a specific calculated interval.[12] Neyman
himself (the original proponent of confidence intervals) made this point
in his original paper:[3]
The population mean is not random. What's the probability that 5 is between 2 and 10? 100%. What's the probability that 13 is between 2 and 10? 0%. Same with any fixed constant like mu or p.
16:20 Can't you simply say that 56,8% (+/-8%) of the teachers see computers as a useful tool, with a confidence level of 99%?
18:06 You mean increase the sample size (from n = 250 to more...)
Can someone please please explain to me why he added 0.5 to 0.495??? Why is it 0.5?? Please!!! Thank you in advance
why must we use n as 250 and not 6250 to calculate the std dev of sampling distribution??
@08:50
Why in previous video we remove one class with multiplying it with zero.
Not in this case.
Why didn't we remove it with multiplying with zero.
At 12:33, it's "0.9902" and not "0.992"
To me, it looks like you're treating "'p" as a random variable by saying there's a 99% chance it lies within 0.08 of p-hat.
why did you add 0 in the numerator finding x bar
Sal is a force for good
a bit fast. but great vid
11:15 lost inside nothingness
.995 from where? Why not 99?
i am not good at this but i will try my best to expain
we want pp percent to be in the middle to find the range i think, and it has to be semetrical
but we dont have a function for that, we have a function that tell us a probability from minus infinty to any given point which we can just use a Table
remember it has to be semetrical eith the mean in the middle, So 0.99/2 is 0.495,
so it has to be 0.495 above the mean
the mean is the middle, which gives 0.5 left and right
we want 0.495 above the mean WHICH IS 0.5+0.495 = 0.995
we find that on the Table wich tells us how many standard diviation it is From The Mean,
now that we have how many standard diviation it is from the mean, we multiply it with the mean which was 0,031 and gives us 0.08
now we know that there i 99 percent that the mean is within 0.08 to the probability
i wrote all this, just so it may help people who stoped to understand at that moment
correct me if i did a mistake
i am still strugling
I wish I thought math was more interesting. Would learn so much quicker.
12:39 i can't find this problem in comment.., 0.4951 * 2 = 0.9902 not 0.992
18.35 mins really? Not necessary at all
why they times 1*142(=good tool )+0*108(=bad tool)??????where is the sentence we define 1 and 0 for these?......dont get it....... is that rule?
You don't have to associate the outcomes to zero and one in this problem. It's just a technique of turning a verbal 'yes' or 'no' into a numerical value (called the Bernoulli distribution, he's got a video about that too) , because you can't really add words together. It would make more sense (and be more applicable) if the the teachers were responding on a 1-10 scale versus yes or no. In short, just skip that part and pick up at 142/250=0.568. :)
10:18 where did he get the .475 from??????????
He thought he was doing it for 95% CI which means if you do 95% of 0.5 (s^2) then you get .475. But the question ask you 99% so with same logic he gets 0.495. (Replying because it threw me off and hoping it will help someone in future)
Why did you divide the standard deviation of the sample with n? this doesn't make sense to me.
Watch Standard error of mean video
@@praveen7028 thanks bro, got that figured out
Is it ironic that Khan Academy is now doing this same exact thing??
this video is not clear and i have trouble following this. Also I believe this needs to be redone as there are gaps, like when Sal said 99% he is covering only 2 std deviation whereas it has to be more than that which he has found out from Z table but that was evident
How did you get 0.495??
Patricia Canonizado 99%/2=49.5%
Why did you subtract 1 from the denominator in 4:30?
Watch Statistics:Standard deviation
First, somebody isn't necessarily me
Second, it seems you only need a to know a small mistake about something to start losing faith in it, so I don't need knowing you :D
I just had a midterm on this X___X why just now khan... why x__X
From the past were he makes the protein that said computers were a good tool to use A's expressed them as P and then he also repossessed the population as P and then he expressed mean as P too!I just got lost.couldn't understand why he's had to multiply by 0----I found the whole video confusing for me
P is used for Population Proportion... which in essence is the mean, but different.
See its this kinda stuff. For the instructor, if this problem calls for P, DO NOT SAY ANYTHING about it being kinda like a Mean. Let me learn the mechanics first, THEN I'll come back later and make the correlation.
Salman stated the correct result later in the video. You just were not paying attention.
Where did the 0.495 come from?
It has some mean.
এতদিন কোথায় ছিলেন?
If I play it on 1.5 speed then it's actually watch-able.
Absolutely Watchable and understandable👌☄
you didn't need to use parenthesis when using the calculator
I didn't get it why he multiplied 2.58 * estimator in the end(around 16:00)
The z-value for a 99% confidence interval is 2.58. The formula in a stats book will show this clearly.
hmm i dont understand how to get the variance at 3:44
were just using the formula and multiplying by 142 and 108 respectively
Isn't this Wrong? Shouldn't 99% be at 3 Standard deviations. In this example it is 2 Standard deviations which would only be 95%.
who else is watching this 20 mins before their STATS final?
i still dont get shit
IOC
please do this from a script. I can't follow your train of thought when you are repeating yourself every few words.
99 percent /2
You watched the whole video in hope. But in the end.... I Am Not gonna do it😅