At my school, linear models is a two year course, regression and anova get their own semester then we do generalized models and other things

👀 good luck!

Great job! I’m sure I only played a small role in that, you’re the one who hustled to learn the material, congratulations!

It's because the assumptions of the Gauss-Markov theorem are used to determine what the standard errors of the coefficient estimators are. So, if those assumptions aren't met, but you still calculate the standard errors in the same way as you would if they were met, then you're going to get incorrect values for the standard errors. Then you use those standard errors to calculate t-statistics and such, so you'll get incorrect values for the t-statistics, and hence incorrect confidence intervals and potentially incorrect results for hypothesis tests.

Thanks! I use Final Cut Pro for editing, Figma and Midjourney for graphics and the manim python library for animations

You’d have a similar conclusion. The ANOVA is only detecting that at least one of them is different, so if that’s the case, there should be some compelling evidence to reject the null hypothesis. But to actually figure out *which* one is different, you’d need to follow up with secondary testing for each of the means

I’m not very well versed it in, but it sounds like it’d be a fancy, high dimensional regression model

You’d prolly get a null result. If you shift all the distributions by the same amount, there wouldn’t be a change in the variance in group means

you could, but most of the time we’re interested in detecting a significant effect, so power is the thing we want to maximize. There’s a trade off between reducing type-I error and power, so we choose to keep alpha constant to signify we tolerate a defined probability of making a wrong decision about rejecting the null

I saw it was really rare, but deep down, I was just looking for an emoji to represent the group lol 😅

You’re right, I just wanted to emphasize that the main assumption is on the residuals. It implies that the outcome is normally distributed, but it’s more of a consequence of the fact that the residuals are normally distributed, rather than an assumption of the model

Yeah I thought about covering multiplicity here, but it deserves its own video

This is not true with ANOVA. It has a type I error rate of 5% for finding *any* difference, not for each particular difference. If you had 1 million populations that were all the same, you would still only have a alpha% chance of finding a fluke. This is the advantage of running an ANOVA and not just running a bunch of two-sample pairwise tests.

It comes from the distribution assumption on the residuals. The residuals were assumed to be normally distributed with some variance, sigma^2. You if you divide the sum of squares by sigma^2, then you get a random variable that’s a standard normal, squaring that gives you a chi-squared distribution. This applies to both the numerator and denominator in the F-statistic.

Yeah for sure, I’ll try my best. This is partially my opinion, so just a heads up. My feeling is that assuming something about the data itself is much stronger than assuming something about the residuals. Very rarely will real-world data follow nice distributions like the Normal, so it’s harder to convince people (read: the statistical referee) that this will hold up. On the other hand, assuming that the residuals is not so bad. It’s like saying, we know there’s an average outcome and people will differ from this average, but they won’t differ too badly from it. In other words, outlier residuals are very rare. It’s confusing because this residual assumption implies that the outcome is also normally distributed in this, but it’s important to note that it’s the residual assumption we make. It’s also important because with stuff like linear regression, we’re looking at how different values of the predictor (i.e. cancer group) shift the distribution of the outcome. If you assume the data itself to have an outcome, it gets more complicated to try to work in how other variables influence it. Assuming the distribution is on the residuals doesn’t come with this baggage. Some people are taught that they should try to transform the outcome so that it “works better” with linear regression or ANOVA. Even though you’re manipulating the outcome, the hope is that this transformations makes the -residuals- look more normal. I hope this helps clarify somewhat. If anyone else sees this and thinks I left something out, please chip in. This is a common question, but even I don’t feel like I get all the nuances.

@@very-normal "it's confusing because this residual assumption implies the outcome is also normally distributed in this" yeah that's the bit that always tripped me up, like I get that you can make one or other the core assumption and build it up from there (it's like picking your axioms in pure maths or something), but in my head the fact that the kinda nebulous residuals assumption implies the much more intuitive distribution assumption meant that I was often fighting between intuition and logic in terms of thinking it through. It also doesn't help that thinking of an example where the residuals are Normal but the distribution _isn't_ is much harder... So it's more about being an assumption of convenience in that it makes the maths much nicer to deal with and is also a weaker and more generalisable assumption, rather than it being anything else like purity or tradition or something. Thanks, I think I get it now! Though no doubt this will be one of those weird bits that'll always feel a little bit of, I feel like I have a much better grasp of the rationale! Much appreciated!

Same question here. In regression I remember them teaching us that you can scale down the data with the different variances in presence of heteroscedasticity. I wonder if that would work here or we have to do some sort of non parametric test

Yeah, common variance is a pretty strong assumption to make. One solution I know of is a variant of the ANOVA called Welch’s ANOVA that can be used when you don’t want to make this assumption. It’s from the same guy behind Welch’s t-test, the version that students learn for two-sample problems when they also can’t assume common variance.

@@very-normal Thank you that's great to know. It seems like Welch's ANOVA is really the way to go, both for small sample size and for no knowledge about the data. (Apparently, it is almost as powerful as the standard ANOVA, even if heterogeneity of variance is fulfilled, so...)

that’s all hypothesis tests tho lol

@@very-normal I meant you need atleast one more group of standard or control to come to any conclusion regarding efficacy

yeahhh he had some L opinions with smoking and eugenics