Chi-square Tests for One-way Tables
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- čas přidán 10. 09. 2024
- I introduce the chi-square test for one-way tables (sometimes called a goodness-of-fit test), and work through an example.
The data used here is from a classic 1905 genetics experiment by William Bateson and Reginald Punnet.
Thank you, very much! The evidence is strong in favor of this video being very effective at teaching the chi-squared test for One-way tables! =)
this absolutely saved my life bruh, i had no clue wtf i was doing trying to do a two way table chi squared test for a one way table and i got so frustrated but this cleared everything up
Why is the third video in this playlist set to private?
You are very welcome, and thanks for the compliment!
You're welcome Craig. I'm glad that it helped.
what is the min sample for this one way table? can i work with 20 sample?
I just wish I found your channel sooner, your videos are great! Subscribed!
+LuvNotH8 Thanks!
So the story of discovery of Mendelian inheritance
was used. Cool!
Professor~ I don't understand why there is minus 1 in getting DF. In yr playlist "Continuous Probability Distributions", with the Chi intro video, you said DF = k square variables, no minus 1. --- Yes, I watched almost all of yr playlists. They are concise but suuuuper clear!
I think it might be to do with the fact that the total number of samples is used to calculate the expected number in each category. If the total number of samples has been fixed, then once the number of samples in the first three categories is determined then the number in the fourth category is also determined, i.e. only three degrees of freedom.
you are the best
Thanks!
GREAT VIDEO!
I'm doing one sample Chi-square test to compare the frequencies of two classrooms to see if there is a significant different or not. The thing is the data I have is frequencies of 8 and 0. I know people use Fisher's exact test for small frequency data. But mine isn't 2*2 matrix which means I can't use Fisher's exact test. What can I do?
Is there any way to do a pairwise comparison? (test to see if the percentages are statistically significant)
at 7:56, when DF=3, shouldn't the chi-square distribution look like increasing from 0 first, and then fall after x=1? The figure in your video seems like the chi-square distribution of DF=1... am I getting something wrong?
It does increase and then decrease. In order to put the observed value of the test statistic in the plot, I had to stretch the limit on the x axis way out. If it was a chi-square distribution with 1 DF, say, then there wouldn't be that seemingly vertical line on the left extreme. It's not actually vertical, it just might seem that way, with that effect exaggerated by the choice of limits on the x axis.
what if your one way table is not about phenotypes and is not 4 numbers?
how do you do the first chart? I was really hoping to get that one solved but we just skipped over it :( i needed that the most ;-;
I find its so useful,although taking a master economics course,still can’t follow the professor ,but after watching the video,i got a lot of missing points
+jing qian Good to hear! I'm glad I could be of help.
Very very good and thanks for the video.
Thanks Professor Balka :) This was a big help.
Great channel man, great job! Thanks!
You are very welcome! And thanks for the compliment!
i liked it
Thanks...
I'm confused. Please tell me where i've gone wrong in my logic...
P value is the evidence AGAINST the null --- P value is low, therefore little evidence against the null --- Therefore null is likely true --- Therefore true ratio is likely to be 9:3:3:1 .... but that obviously doesn't make sense?
The lower the p-value, the stronger the evidence against the null hypothesis. In the example the p-value is tiny, thereby giving extremely strong evidence against the null hypothesis.
You're my hero.
Did Gregor Mendel know these methods?
I had a doubt. if we had failed to reject, then that means we dont have enough evidence to say the null is false, right? So in that case the null may or may not be true and hence its inconclusive?
I think so, yes.
what if 9331 ratio is not given then how do I calculate expected results ???
The expected count is the count we would expect to get, on average, if the null hypothesis were true. So there needs to be a null hypothesis in order to get expected counts. Are you asking how we'd get the expected counts *in this situation* if the 9:3:3:1 ratio were not given? If so, a google search of 9:3:3:1 would yield plenty of information about how that ratio arises under independent inheritance.
Hey Ive been watching your videos throughout my class so just a heads up thanks for all this. I have a question on the chi -square table. How do I know the probability of interest for the p-value?
If the question don't give the proportion 9:3:3:1....how can I find it??
In order to carry out the test, the null hypothesis probabilities must be given in some way. This might be given in numerical values, or a worded description (e.g. "equally likely"). The 9:3:3:1 ratio specifically comes up in 2nd generation independent assortment in genetics, and a google search of 9:3:3:1 is very informative.
@@jbstatistics....can I give the proportion as 1/4 if number of classes are 4??
If there are 4 classes, and we wish to test the null hypothesis that those classes are equally likely, then yes. But we are not always testing that the classes are equally likely; there may be a different question of interest.
@@jbstatistics..do you mean the proportion should be given as information...if not equally likely?
Yes. In order to carry out a hypothesis test, there must first be a hypothesis to test. That hypothesis is not based on the sample data, but on the nature of the problem at hand.
Hi, The lectures on statistics are great. Can I share them on my blog. Awaiting your permission.
is this the same thing as a one dimensional chi square test?
Yes, these tests are sometimes called one-dimensional chi-square tests.
if you have 10 times the counts, the x2 will be 1347. it does not make any sense. the X2 is propotional to the total count. the more the counts, the more the x2.
i do think that's the point of squaring the difference from the expected, to augment the bigger difference. When the sample size gets bigger, if the null hypothesis is true (independence inheritance between the two genes), we would expect the difference is not too big in larger sample size compared to smaller sample size. Percentage-wise is the diff probably the same, but sample size matters too
So you have 3 degrees of freedom, but how do you use that to find a p-value? In this case the test statistic is so large you can just assume it is very close to zero. But in order to make the claim that there is a significant difference, are you comparing the p-value to some alpha? I feel the end of this problem left out how to actually come to a conclusion.
+camsully4 The p-value is the area to the right of the observed value of the test statistic under a chi-square distribution with 3 degrees of freedom. It's found using software (or a chi-square table, if you must).
I'm not of the school of thought that one must always pick a significance level, so I don't teach statistics that way. (You might note that I did not used the the phrase "significant difference" in this video.) The p-value is very near 0. It is, for all intents and purposes, impossible to obtain the observed data if the null hypothesis were in fact true. Thus there is extremely strong evidence against the null hypothesis. If you need to pick a significance level for whatever reason, then go ahead and carry out your test that way, but it's very reasonable to simply state that a minuscule p-value gives strong evidence against the null hypothesis. Cheers.
binomial distribution at 3:09
or the multinomial dis if you like