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SSREG is calculated with $\sum_{i=1}^{n} (\hat Y_i - \bar{Y})^2$ which is the regression sum of squares

SSRES is calculated with $\sum_{i=1}^{n} ( Y_i - \hat Y_i)^2$ which is the residual sum of squares

If we are given

$$\sum_{i=1}^{n} X_i = 323.02, \sum_{i=1}^{n} Y_i = 333.15, \sum_{i=1} ^{n} X_i^2 = 1021.487, \hat\beta_0 = 1.097, \hat\beta_1 = 0.674 $$

I cant seem to find relevant formulas to help me compute it with given info the best I found was

$SSREG = \hat\beta^2_{1} \sum_{i=1}^{n} (X_i - \bar{X})^2 $

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Is there an easier one?

Tree Garen
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1 Answers1

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You should be given the sample size $n$, then For the SSReg you are right, you have all its components, just plug it in the formula $$ SSreg = \hat{\beta}_1^2(\sum X_i^2 -n \bar{X}_n^2)=0.0674^2(1021.487 - 323.02^2/n ) $$ and $$ SSres=SST-SSreg = \sum Y_i^2-n \bar{Y}_n^2-SSreg $$ where for the SST you need $\bar{Y}_n$ that you can derive from $$ \hat{\beta}_0=\bar{Y}_n-\bar{X}_n\hat{\beta}_1. $$

V. Vancak
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  • Many thanks! Quick question though: $SSreg = \hat{\beta}_1^2(\sum X_i^2 -n \bar{X}_n^2)=0.0674^2(1021.487 - 323.02^2/n )$ for this you put $n \bar{X}^2$ but did $323.02^2 / n$ in the equation. So is it $SSreg = \hat{\beta}_1^2(\sum X_i^2 - \bar{X}_n^2 / n)=0.0674^2(1021.487 - 323.02^2/n )$ – Tree Garen May 20 '19 at 20:49
  • $\sum X_i = n\bar{X}_n$, $(\sum X_i ) ^2 = n^2\bar{X}_n^2$ – V. Vancak May 20 '19 at 21:12
  • So $\bar{X_n}^2 = 323.02^2$ right? Then how is $n \bar{X_n}^2 = 323.02^2/n$ ? – Tree Garen May 20 '19 at 21:18
  • $ (\sum X_i )^2/n =n^2 \bar{X}_n^2/n = n \bar{X}_n^2 $ – V. Vancak May 20 '19 at 21:33