Is it possible to do a linear regression without the individual samples Just curious if I have only categorical predictors and a continuous response variable, is it possible to do a regression using only summary statistics of each possible combination of my indicators (e.g, sum, sum of squares, n, etc)?
 A: As whuber points out, I may well have misinterpreted the question; consequently I'll edit to answer it two ways. [I'm still making some assumptions here relating to the form of the model, so there's still some potential room for misinterpretation even now.]


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*If "in each category" means "at each combination of factor levels" (cells in a k-way table, for k categorical predictors), then you actually only need the counts and the sum of the y's at each combination (in each cell) to get the coefficients. (Sums of squares of y's would be needed for some other things, as suggested below.)

*If by "in each category" you mean "for each categorical variable" then that would consist just of marginal summaries, and 
marginal summaries (sums and sums of squares) are not sufficient.
Here you need sums of cross products as well; $x_i'x_j$ (across all pairs of predictors) and $x_i'y$. (This is equivalent to finding the counts and sum of y's in each cell in 1. above)
With those as well as sums and sums of squares (which are just $x_0'x_j$ and $x_0'y$ for $x_0=\mathbf{1}$, in order to fit a constant term), you can form $X'X$ (a matrix whose $i,j$ element is $x_i'x_j$) and $X'y$ (a column vector with $i$-th element $x_i'y$, in both cases indexing from $0$ rather than $1$) and hence attempt to solve the regression problem (i.e. solve the system $X'X \hat{\beta}=X'y$).
This could then be done, for example, by forming the Cholesky decomposition of $X'X$ (don't attempt to directly invert $X'X$):
If $A=X'X$ and $z = X'y$ you're trying to solve $A\hat{\beta}=z$ for symmetric (and hopefully positive definite) $A$.


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*Form $A=LL'$ (other versions of Cholesky form $LDL'$ or $R'R$ or $R'DR$, but the basic approach is similar)

*Solve $L \zeta=z$ for $\zeta$ by forward substitution

*Solve $L'\hat{\beta}=\zeta$ for $\hat{\beta}$ by back substitution
You then have the parameter estimates
If you need things like $R^2$ or $s^2$ you'll also need $y'y$.
