volume 1, page 395 of Curriculum, problem 20

can anyone possibly explain me, how have we calculated that multiple R-squared is 0.36

i feel im mixing a lot of concepts…. *sigh*

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I think It’s the sum of the weighted correlations of each independent variable to the dependent ‘squared’. So you’d start with a correlation matrix.

And the weights would be the regression coefficients of each respective independent variable.

I think!

The multiple r2 is the explained variation / total variation.  So that would be rss/sst (51.433/142.869) = 0.36

the multiple r2 is not the correlation, so you take the sqrt and the corr = 0.60

Thats why answers b&c don’t work.

stunnerrunner wrote:

The multiple r2 is the explained variation / total variation.  So that would be rss/sst (51.433/142.869) = 0.36

the multiple r2 is not the correlation, so you take the sqrt and the corr = 0.60

Thats why answers b&c don’t work. Apart from this you also just need to remember that -

sst (total sum of squares) = rss (regression sum of squares) + sse (sum of squared errors).

Best regards,

Sooraj.

I think he was asking about how to actually get the ‘multiple r-square’. Aren’t the formulas for the coefficient of determination above for ‘single linear regression’ only?

No mate I believe its for multiple regression as well.

And as to why I wrote the formulae.. The general format is such that for regression, you could be given RSS and SSE, but they need not give SST. You could be asked to calculate R2.

Please correct me if am wrong.

Best regards,

Sooraj.