Linked Questions

What is the advantage (if any) to fitting both a linear regression model to a dataset and running an ANOVA test if these are both ultimately identical?

How are they all versions of the same basic statistical method?

I have two categorical / nominal variables. Each of them can take only two distinct values (so, I have 4 combinations in total). Each combination of values comes with a set of numerical values. So, I ...

I am new to statistics and I am trying to understand the difference between ANOVA and linear regression. I am using R to explore this. I read various articles about why ANOVA and regression are ...

On the Mathematics site an OP who is just learning statistics gave his description of the difference between linear regression and ANOVA and asked if his interpretation was correct. I responded that ...

Can I get help completing this tentative (in progress) attempt at getting my bearings on ANOVA's and REGRESSION equivalents? I have been trying to reconcile the concepts, nomenclature and syntax of ...

I found two very useful posts about the difference between linear regression analysis and ANOVA and how to visualise them: Why is ANOVA taught / used as if it is a different research methodology ...

Like, I don't get it. In regression, we say $$Y_i = a + bX_i + \epsilon$$ where $X_i$ is just a real number for each observation $i$, and in analysis of variance with two groups, we say $$Y_i = a + ...

This might be a silly question, but I'm confused why ANOVA is different from a panel data model, with dummy variables for group, and inference on the coefficients on the dummy variables. Is this what ...

I'm working on a study that involves measuring breathing effort of an individual over short periods of time (specifically breath distention). The dependent variable is based on a new therapeutic and ...

ANOVA is usually thought as equivalent to linear regression, which is very interesting. Who first discovered this fact? Did Fisher know about this when he worked on ANOVA?

Show that the cross-product term \hat $y'r$ from $y'y$ is 0. I see that $y'y = ( \hat y + r)'( \hat y + r) = \hat y' \hat y + \hat y'r + r' \hat y + r'r$ But not sure what identity can be used here ...