Pearson's chi square test (goodness of fit) | Probability and Statistics | Khan Academy
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Pearson's Chi Square Test (Goodness of Fit)
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you make me want to burn my statistics book. I just spent 2.5 hours trying to understand a chapter and came out clueless, and then you made everything perfectly clear in 15 minutes (i paused a few times)
YOU ARE AMAZING
thank you so so so much for making this. you don't understand how much this helped me; my textbook could not have done a poorer job explaining it
most statistic books should be burned anyway
also statistic teachers
He is really amazing
The ASA Statement on p-Values: Context, Process, and Purpose
www.tandfonline.com/doi/full/10.1080/00031305.2016.1154108
"Failing Grade: 89% of Introduction-to-Psychology Textbooks That Define or Explain Statistical Significance Do So Incorrectly"
journals.sagepub.com/doi/10.1177/2515245919858072
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I was dumb lost my password. I would appreciate any tips you can give me
literally the only video on youtube that gets right on the point! No bullshit intro, just right to point on what i need to know, Love it!
you are the reason im getting a degree at harvard
Did u get that degree bro it's been 7 years can I get a update?
hes also the reason your degree is overpriced
I wonder if you had to also watch his videos to study for stat tests at Harvard ;)
@@bakhodiryakubov3981 I'm doing a Master in Physics at Oxford and I've found these videos more useful then my statistics lectures lol
@@randomman1000tweeny1 wow didnt expect that
Hi there, thank you Khan Academy this really helped me and its already 5 years since this was uploaded. I am so glad to be born into this generation of freedom of information online.
It's 6 years since you made this comment.
Did you find ur role in this world?!
What do you do now?
But my dad didn't let me have access
When the AP test is in 2 hours and you're finally trying to study 😬
Statistics is fucking impossible to study for. I've never taken such a dry class in my life
i study the time im taking the test
Why do you think it is boring? personally I quite enjoy it.
I’m actually cramming for a university exam in like 4hrs...
@@stoirmdraodih6810 same right now
This video was suggested by my bio professor & he was 101% right. This video makes it super easy :) thanks
Its Funny how I pay my professors/school thousands of dollars per semester and yet Khan always teaches me better for free.
Pay khan sir as well than if you feel like this
Man, Sal, i thought i was done with khanacademy in grade 12 and here i am in my final semester of engineering :')
I'm still a bit confused about the conclusion.
The calculated value is higher than the critical value.
The smaller the P-value, the more significant the result.
The smaller the P-value, the higher the cricital value.
Thus I'd assume the hypothesis should be accepted, rather than rejected.
Otherwise I'd use a P-Value of 0.01 rather than 0.05 and the calculated value would be lower than the critical value and thus be accepted which makes no sense to me.
Realize the following fact. The more the observed data deviates from the expected data, the higher the differences between those values and so a higher Chi-squared value.
Yes you would accept the null hypothesis if you had used a P-Value of 0.01. Lower P-value means you need stronger evidence for you to change your default assumption / null hypothesis. There is a greater than 1% chance the owner is right given what you saw. In the video he rejects the owner's distribution because it has a less than 5% chance.
1st year AP Stat teacher loves Khan videos... Thank you Sal :)
I have massive respect for your work, man! Thanks for making people throughout the world more educated, that´s one of the essential things this world needs. And it´s one of the essential things I need haha.
A clarification for anyone arriving directly at this video like me: the definition of the chi squared statistic for a sample is the sum of (x-E(x))^2/s^2, where s^2 is the variance of the data. When the variable is Poisson-distributed, like in this case, then you are in the particular case where s^2=E(x), as appears in the video
helpful, thank you!
so the p-value would be 0.043293 which is < 0.05 so reject the null hypothesis - same result as using critical values 11.07 < 11.44.
Please correct if I am wrong but it will be a two tailed tests and the P value to consider would be for 0.975% (12.83). So as 11.44 is within 12.83, we would accept the hypothesis.
That's what I thought too, have you found an answer to this?
yup i agree it should be a two tailed tests because the alternative hypothesis is H1 = not correct.
Chi square only has 1 tail for all those seeing this comment.. so the P value remains 5% as the video, and the null hypotheses therefore remains rejected as the video shows the solving.
the concept of chi square can only do 1 tail test is incorrect because chi square goodness of fit can do 2 tail test as well. It all depends on how you state your hypothesis.
@@eXcelMathS Hey you might be right, this exercise didn't really apply well to my bio-statistics course, I had to see two different outcomes for each exposure for my exercises in the Chi-square, but it's the same concept, so you may certainly be right about the 2-tail testing
Thank you so much
Thank you!!
I think Channels like khan academy is a boon for people who want to understand things in inexpensive or Frugal way.
Wow, that was a very nice example and intro to chi-square! Great job, Sir Khan!
it's funny how you always repeat yourself but that makes your video's so easy to understand! thanks for your help!
At 8:30 Khan justefies the further work by saying after he has calculated the X^2 value, calculated, "What is the probability of getting a result this extreme"? By talking of probablity he indicates that there are more options. The expected values are constant. So does that mean that what we're testing is the probability of getting an observed set similar to those that we have? How (on earth) can this probability be calculated on bases of the area beaneath a random graph? Hope someone can answe
This was so helpful! Thank you Khan Academy and you guy who made this!
David Hager The guy who made the video is Khan, it's one guy that made evrything basically.
Thanks for this test. Made it much clearer.
It is very unfair to dislike this tutorial.
Thank you so much! You just taught me how to calculate Chi-squared and my finals are tomorrow. :)
how was the exam? 4 years :D
How was the exam 8 years ago?
@@thewilmarvelfan 12 years?
One thing I don't entirely understand is why do we almost always choose a significance level of 5%? And what exactly would we gain or lose if we make it more or less than 5%? If we make it higher than 5%, doesn't it then make the hypothesis more difficult to accept, since it would require more accuracy to accept it? This would seem to make it more easy to reject a true hypothesis, but at the same time, doesn't it also make it more likely that we won't accept a false hypothesis?
Edit: I read that medical experiments have a tendency to choose a significance level of 1%, but wouldn't it be safer to choose a higher significance level, so that a false hypothesis won't be accepted, or did I misunderstand something?
The choice of a significance level will be appropriate, not only to the industry but the precision of available techniques. If your measuring/quantification techniques have an accuracy of micrometres choosing a significance that equates to an accuracy of nanometres is not only inappropriate but impractical.
Can you please elaborate what 'extreme' means? It's used a lot in this video. Thanks!
When do you use n-2 for your degrees of freedom? We were taught something like this.
If the total number of customers is not given upfront, then we need data about all the days for calculating chi-square. In such a case will degrees of freedom be still five?
Thank youuuu it really helped 👍🏻💙
This material helped me out a lot! I know about khan academy since about 5 years ago and i just want to tell you guys to keep doing what you are doing, you probably account for more than 40% of my education
Thanks for this!! My exams are in 3 days time.
To Mcgraw-Hill Connect: THIS IS HOW YOU MAKE AN INSTRUCTIONAL VIDEO
I feel this initial presumption is wrong. Null Hypotheses should always start with "negative".
As null hypothesis means that attributes are not related so how owners distribution is correct? Please explain I'm confused about that statement
Awesome explanation
Thank U!!
it's relieving to read others in higher levels of school similarly benefiting from these vids. They're a lifesaver. Thanks Sal!
How amazing that I pay 3.400 euro's at university and get bored to death with useless info when Khan explains this in about 11 minutes...
Sal, another clear video. However, may be you could have talked a little bit as to why chi-sq is appropriate for this particular type of problem (or similar type) - how would a student know when to select the same distribution for a problem at hand.
amazing
Good luck students taking the AP statistics tmr! Pray for me guys
I thought to get the degrees of freedom for interaction in a contigency table is d.f. = (rows - 1) x (columns - 1)
Khan just did d.f. = n - 1
Whats up?
This isn't contingency table
this is a one-way goodness-of-fit chi-square test, not a two-or-more way test of independence. it's one-way goodness-of-fit test (which tests randomness or preference) because there is only one column of one category or group (that is, weekdays of 6 days). so df=k-1 or 6-1=5. you are asking about a situation where there is more than one column and/or row of categories/groups to test independence, not goodness-of-fit.
I have an exam tomorrow you saved me
Thank you so much for your great lecture videos! I have one question if I may. I followed from Chi Square Distribution video 1 to here. And just confused that is that (30-20)^2/ 20 an x^2 in your Chi Square distribution definition? Basically, is x = (one_observe - mean ) / sqrt(mean) a standard normal distribution? shouldn't it be (one_sbserve - mean) / standard_deviation a standard normal?
Thanks for the video, this just explains how to, but is there a video on why the difference between observed and expected follows Chi-squared test? And also does the expected distri (e.g. normal vs binomal vs possion) matter? what is the proof for that?
The Chi-Square distribution is a sum of squares of standard normal variables Zi. What this test is doing is assuming that the squared errors between the observed and expected distributions are distributed ~ Chi-Square. Makes it a bit easier to understand the test when you understand the motivation behind using the Chi-Square distribution.
Why is it a problem when the probability is less than the significance level?
if they teach like that in schools, we would all be statisticians.
Hi ! In the chi square test, each random variable follows a normal distribution N(0,1). However, when you compute the sample chi square statistic, you add terms like ( (Oi - Ei) ** 2 ) / Ei. I don't understand how those terms can be seen as samples from N(0,1)
Did you finally understand them?
Thank you got an exam in 2hrs
why are we calculating ((error)^2/expected) what is the meaning of this value?
Hi! thanks for the amazing content. Just a little question on the assumed null hypothesis: don't we usually take the "NO news" kind of thing in null. I mean would it be more conventional if we state the null hypothesis to be: "Owner's distribution is incorrect"
Thats true, in his other video on "Chi square test for association (independence)", he mentions null hypothesis is "Negative". Seems like a contradiction. Did you find the answer?
@@2828jordan I have the same question here. Anyone find anything?
The null hypothesis here should be that there is no statistical significance between the expected frequency and observed frequency (owner is right). As proven in the video, the null hypothesis is rejected. Hence, there IS a statistical significance between the expected frequency (provided by the owner) and the observed frequency. In conclusion, the owner is wrong.
Thank you!! This example helped me a lot in understanding other purposes of Goodness-of-fit test. Really appreciate it!!
This example assumes that the confidence level is 95% and an alpha of 0.05 or (1 - .95), that is the risk of making a Type 1 error (a false positive - rejecting the null hypothesis when it shouldn't have been rejected). As the vid shows, the chi-square statistic of 11.44 > critical value of 11.07, so we reject the null hypothesis and accept the alternative hypothesis. We get the same result by looking at p of 0.04 < alpha of 0.05. Imagine the example used a confidence level of 99% instead of 95%, we'd have an alpha of 0.01 instead of 0.05. With an alpha of 0.01, the chi-square statistic would still be 11.44, but it would < a new critical value of 15.09, so we could fail to reject the null hypothesis. Again, we would get the same result by looking at p of 0.04 > alpha of 0.01.
I feel like an actual idiot. I took AP Stats last year, my freshman year, and everything clicked, I got a 5, and I thought “hooray!! End of story.” But now I’m taking AP Bio, and lo and behold unit 0 (why is that a thing?) is Chi-Squared. I threw away all my notes. I somehow completely wiped it from my memory. I’m just hoping this will help, I have a quiz tomorrow. 😢
Thanks a lot for the explanation. What would happen if the random variables were not independent? Would it still follow Chi square distribution? How would it affect the degree of freedom?
i suppose usually for simplicity sake we always assume random variables are independent
I can't stand universities.
I spent 8 hours trying to understand the math of what my professor was doing in R, and didn't understand anything.
I SPENT 12 MINUTES watching this video.
Boom. Done.
Can someone please explain...
As it defines: "A random variable has a Chi-square distribution if it can be written as a sum of squares of independent standard NORMAL variables."
So, why is it still valid to use Pearson's chi square test on variables with not normal distribution obviously ? I see it everywhere and I can't understand why it's correct...
*subscribes as a thank you for cutting some of the unending stress I am exuding from my stats final in 5 hours* *nervous smile*
Why is H0: Distribution not correct and H1: Distribution is correct. Usually H0 is the hypothesis that we are suspicious or something that is against the research. May Thanks in advance if someone can help.
1:53 What is trying to be said with"a result more extreme than this". Is it a distribution that fits the Owners Distribution less ?
I may be stupid, but this makes little sense to me.
- What are the random variables here? The day of the week is non-numerical. Are there 6 random variables corresponding to the number of customers on each day of the week?
- Based on the last part, where there is a chi-squared distribution there is a normal distribution being squared? What are our standard normal distributions here?
If the probability of getting this result or less likely than this is less than 5% then we shouldn’t be rejecting Null Hypothesis. In effect what owner of store saying is correct and the observed values are just by chance.
it amazing how well he writes with the mouse. It honestly amazes me.
Probably a digital pen
@@CrunchyDark 8 years ago I didn't know about this technology FFS looool
@@TheNotSoOriginal It's all good :) it happens to all of us. Sometimes I have ideas for new technology but then I check online and it already exists :P
@@TheNotSoOriginal surprised that you replied ! Do you get notifications of all your comments , I mean the ones which are like 3-4 yrs old ?
thank you so much :)
question; do they always give you the percentage for the expected? because in all that i have done i had to either multiply the row total by the column total divided by the grand total or either dividing the total amount of scores by the amount of levels (the no. of days in this case) was i doing it wrong?
its not a random graph... he doesn't go into the probabilitstic proof of how a chi-sq distribution is dervied and the validity of it but I'm sure you can find it if you dig. But the probability is the p(finding the data, given the suggested frequencies). i.e. how likely is it to find the observed data IF the suggested distribution is TRUE. If that likelihood is very small (
why does the X² test divides to the expected value and not the variance?
We don't have direction or we are not calculate increase or decrease so don't you think we should go for 2 tailed test 0.025 both side please explain
W
O
W
Is chi-square a one-tailed or a two-tailed test? i read some articles where it says that it is essentially a 1 tailed test..why so?? i am not very much clear about it?? in goodness of fit, why we always take it to be one-tailed?? please help..
what if you had 3 replicates for each variable (say i have variables a, b, c, d and 3 replicate values for each (eg variable a has values 0.1, 0.2, 0.3) can I do the chi squared test on the 4 variables? Do I have to find the mean of each variable then do the chi squared test? or is there a better test for that? Thanks!
@neoaeonian- uh, that's actually proof that YOU are wrong. The higher the numbers, the smaller the relative errors from the expected value that are expected. So, since in your case they stay the same when you've increased your sample size tenfold, you are clearly doing something terribly, terribly wrong. Which is also obvious from your ((30-20)/20)^2 = 10000/200. I don't even.
This might be so old I can't even reply to it directly, but it's truly something special.
When you get a chi2 t-statistic, you only get obviously positive values. So how can I know which values are greater o lower than the mean, I mean, which are on the right and left tail? This info is lost!
My stats teacher taught this well but I thought I'd come here anyway
finally i understood it properly :D
So if my chi squared value worked out at 1.86 and my S.L (at 5%) was 3.84, then i would accept my null? How can i word that using the words 'chance' and 'probability'?
I don't understand why you divided every term by the expected value before adding. Why?
@CogitoErgoCogitoSum this test is a non-parametric statistical test therefore you would use it when you cannot obtain the population parameters μ and σ
Your voice sounds like Jeremy Allen White Lol
btw Thanks a lot dear Sal
Thanks a bunch. It helped me with the homework.
Running Pearson's chi-square test on restaurant attendance by days of the week is like trying to predict which day you'll feel like eating out. It's a statistical puzzle as mysterious as trying to figure out why Monday feels like a salad day, but Saturday suddenly screams 'pizza party!' Chi-square might tell us there's a correlation, but it won't explain the 'Taco Tuesday' phenomenon or why Fridays always seem to lead to fries. It's a statistical adventure where the only certainty is that any day can be a 'cheat day' when it comes to dining out!
so if the chi squared statistic i got from the sum is smaller than that from the table, what does that mean?
Excellent thank you GK. \\ For myself only \\ X is a multinomial RV with 6 events with p_1 = .1, p_2 = .1, etc. What is the probability P(X1=30, X2=14, etc) = 200! / 30!14! etc p_1 ^ 30 p_2 ... \\ this will be an extremely low #. Also is the shop owners distribution really a mulitnomial? That would be if like 200 people each chose a day independently of whwat ot go to. hmmm... to come back to later
I actually need help here. I have a study (quantitative) where the respondents need to choose 10 out of the 20 characteristics of their preferred leadership behavior. These respondents are classified into 4 Generations namely: Boomers, Generation X, Y, and Z. What statistical tool/method should I employ to gather reliable data. I badly need serious responses here.
I remember when learning about this test that if the normalized value for one of the columns is less than a particular value, that you are supposed to group it with the next column until your combined normalized value is greater or equal to the specified value. If I'm not mistaken that number is 0.05. Am I correct? For example, if (Observed-Expected)/Expected = 0.02 for one column and the column to the right is 0.34. Then you would add 0.02 + 0.34 to get 0.36 and you'd lose a degree of freedom.
Currently studying for the USMLE STEP 3, which is heavy on biostats. As a physician, I will NEVER USE THIS KNOWLEDGE! WHY MUST I LEARN IT!?!?! :((((
Sir I have encountered a rare case which I am unable to source for answer ... What if the test of associate is the same for every Column ? What will you conclude for this case . (O-e)^2/E also obtain an answer of 0.01 for every result for example how will you conclude for this case ?
Maybe someone can help me: There is a group of 26 students. They are asked a question and two possible variants are suggested. 12 of them choose the first answer, and 14 of them choose the second variant.
What test should I use to statistically determine whether students favor one answer over another? SHould I use the t-test or some kind of other test and how do I calculate it?
Over 7 years and this is still very useful :)
12 years now
and its still helpful
Is anyone equiped to explain the history or logical background behind df = 5 and if we had 5 sums, we would be able to find out the 6th sum? Does it have something to do with fitting the summed values to the Chi squared curve and somehow knowing the 6th point from that?
I just was just wondering why you expect the data and not assume it has a Chi-square distribution? Is it because you have to little data points? I would be really gratefull if someone could explain this to me :)!
Wow you explained this so well, I actually got it! Thank you :)
is the significance level ALWAYS 5% no matter what for every problem?? How did he get 5%?
Saved. My. Life.
Well, did you end up buying the restaurant or not?!?!?! 😛
So if our calculated number was like, 9.something, then it being smaller than 11.07 means we accept the null hypothesis???
Don't forget everyone! The chi-squared statistic is actually a measure of badness of fit, not goodness of fit, like it would seem obvious to be.
Can someone calculate this?
(14-20)/20 ?
You don't have to do the chi square test to see that for 200 customers a week is not worth buying :)) just kidding, congrats for the video!
What is the t value u enter when creating a table. And the a= . Is a=.05. So a chi sqaure test was conducted for goodness of fit (a=??)