Chi-square tests: Goodness of Fit for the Binomial Distribution
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- เผยแพร่เมื่อ 2 ต.ค. 2024
- I work through an example of testing the null hypothesis that the data comes from a binomial distribution. I do this for two tests, one in which the probability of success is specified in the null hypothesis, and one where it is estimated from the data.
Data reference
The Larry Bird free throw data based on information in:
Wardrop, R.L. (1995). Simpson's paradox and the hot hand in basketball. The American Statistician, 49 (1), 24-28.
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THANKYOUUUUU 9 YEARS LATER AND STILL HELPFUL
I'm glad to be of help. I've got lots of good stuff on this channel :)
This is SOOOOOO MUCH easier to understand than a two hour lecture
Thanks Ricky. I'm glad to be of help!
Thank you for this clear and concise explanation.
My teacher gave an assignment with questions about goodness-to -fit procedure without covering the topic in the class 😒
We need you! Thank you so much again. Come back and publish something, please. And if you have a channel on Patreon I'd be happy to support you!
Hi Andrea! Thanks for the kind words. Every day I think about getting back to producing videos. (In fact, just this morning I was mapping some out in my head.) I'm still crushed for time, but I'll force myself to find the time to produce one and go from there. Thanks for the inspiration!
I'd love to do videos on order statistics, and may get around to it at some point in the future, but I've got a lot of other topics lined up before that. Cheers.
A day before my final math exam, I'm going through your video, thinking
WHY COULDN'T I FIND THIS BEFORE D:
it's so clear, it becomes art. Thx for such a great job!!!
I echo this
Why didn't you calculate two-tail p-value?!
Thank you so much this was very very very, simple and straightforward. This video answered all my questions with no waffle added.
These videos are ok. My teacher makes me watch and answer questions about it online which is like deciphering hieroglyphics when it actually comes time to use any of it in R.
Very nice video. Actually, the entire set of videos on statistics is very good. Concerning the video. How would the above methodology generalize if we had multidimensional contingency table. Say, count the elements in 5 boxes, where each box can hold 5 different elements. In this case we would have 16 degrees of freedom. More precisely, how would we compute the theoretical frequencies for the cells, assuming uniform distribution?
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!
Thank you for this but do the 2nd example where the degree of freedom is deducted by 2 works the same for Poisson distribution as well?
Yes. Using the data to estimate lambda would cause a loss of one degree of freedom, and the DF would end up being # cells - 2.
@@jbstatistics Understood thank you!
Why don’t u find the percentiles in chi square table and compare it with the test statistics?
thanks for your very understandable explanation :D
that first example was kinda hard to observe for non-american people
Thanks for your wonderful videos. Love the way you teach concepts.
You are very welcome.
tq for the df point
exam in 2hrs
thanks alot for the help
You are very welcome.
Thank you a lot for the help. Your videos are amazing. But just one question here. In the last part, in the test of the binomial assumption hipotesis, should not the hipotesis be "H0: Larry bird's number of success...a binomial distribution WITH p = CALULATED P". I mean. Why its not mentioned the calculated probability value when it's used to calculate wheter it is a good binomial aproximation? I really got this question. And other doubt is: How can i be sure to use a binomial distribution as aproximation, if the chi-squared test don't prove it's a good aproximation, but only shows evidence that it can't be refused? It is so complex.
legend for getting me to understand degrees of freedom using a basic example. THIS IS HOW TEACHING SHOULD BE DONEEEE
Clear, concise, easily visible to read from a dark screen, what's not to like?
Thank you!!! Amazing video.
When calculating expected values it looks like you have to double the value where x=1 because you can do that 2 ways.
Make/Miss or Miss/Make.
Am I thinking about that right??
Thank you very much.
You are very welcome!
+jbstatistics your new album Purpose is awesome!
Thank you Justin Bieber, you are the very best. I love you and i hope you keep doing your thang. If you can, come to Bogotá, Universidad de Los Andes. People here love you and watch your videos every day.
I'm always happy to help my friends in Colombia! I'm glad to hear you find my videos helpful. Perhaps I'll visit someday!
hol up ur name is justin beiber?
@@realdvgarg lol no :d it is actually Dr. Jeremy Balka, a professor at the University of Guelph
I don't know how I ended up here but boy am glad that i find you. Great content and to the point. Cheers.
Hi ,
How can we perform significance test for non binomial data like ARPU
why P is 0.8
Lets say Larry made a) 1 of 2: 120 times b) 2 of 2: 200 times and of course, c) 0 of 2 18 (338-320) times. The H0 that a the number of makes follows a Binomial distribution would not be rejected (p=0.13). However, when examining only one claim of H0 (p=0.8) by condidering the total number of shots made we would have 520 observed (120 + 2 x 200) (with 156 misses) vs 540.8 and 135.2 expected, the Chi-test now suggests that the prob of each free throw cannot be 0.8 (p-value 0.04). The strange thing for me is that conditioning on independence of each pair of free throws, the two tests should have close p-values however this is not the case. What do I miss? Any intuitive explanation? Thanks a lot!
I don’t quite understand your question but I think the second test is reasonable and isn’t supposed to have a p value closer to the first because you have 5 pairs with 0 made in first with expectation of 13.52 but 18 pairs missed considerably closer to 13.52 as compared with 5 missed pairs of free throws. So is 120 closer to 108.16, even though you are rejecting the bill with a p value below 5%. I don’t know if this is the question you are asking, or it’s something else.
Complicated stuff
Great video. The old value of chi^2 reappeared at 12:30
i see we are doing hit and trial kind of a thing here .....
I don't understand.
very well explained video and clear voice. Thank you
It was really helpful
Thank you .please , i need a book containing this part
I did not get it, why DF-1-1???
Excellent video!!! Very clear...Well done!!!!
Simply GREAT !!! Fantastic Job
how 338 came i didn't understand
beautiful
it'd have been a lot easier if you had shown the process to calculate expected proportion as well!
I don't know what you mean. I go through the expected proportion calculations at 1:52. I don't pull out a calculator and punch the values in, but I show the formulas and the resulting value.
@@jbstatistics at 9:32. there is number of makes, expected proportion and expected number. You've find out expected numbers but i'm clueless about the calculation process there. It will be helpful if you illuminate me in this matter. For e.g in Number of make 1 there is : (2 1).864^1(1-.864)^1 = .2351. How this .2351 came could you please tell me(I mean how to do calculation ?
Super helpful, thank you
im an idiot. i dont understand any of this.. ergghh
That doesn't make you an idiot! :)
what is that software
u r telling about through which i can find p value
Many different software packages can calculate the area under a chi-square distribution. I usually use the (free) statistical software R, and in that software the command pchisq(x,df) yields the area to the left of x under a chi-square distribution with df degrees of freedom.
thanks bro
how did u calculated p-value?
I typically use R to find these areas, but there are a variety of other statistics packages that will do it (e.g. SAS, STATA, Excel, SPSS).
Thank you!
You are very welcome!