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Thread: What does R, R square , Adjusted R square, STD error of estimates helps for ?

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    Post What does R, R square , Adjusted R square, STD error of estimates helps for ?




    Hello

    After I run out the regression analysis, can someone enlighten me what can I get from R, R square, STD Error estimates and the adjusted r square ?!

    P.S: I am not professional in stats, I have only taken one course in university and that was back 4 years ago so I don't have any knowledge about that except watching couple videos in youtube.

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Well, I'll help you with R. That is, R is the Pearson correlation between the actual values of the dependent variable (Y) and the predicted values of Y (Y-hats) - regardless of the number of predictors (X's) in the regression model. As such, R is an index of the strength of the linear association between Y and Y-hats. Note that R is bounded between 0 and 1.

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Quote Originally Posted by Dragan View Post
    Note that R is bounded between 0 and 1.
    We talking about the same R here?
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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Quote Originally Posted by Dragan View Post
    Note that R is bounded between 0 and 1.
    Between -1 and 1.

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Quote Originally Posted by Miner View Post
    Between -1 and 1.
    Well, no...it is not....rather, the correlation between Y and the Y-hats is bounded between 0 and 1 as I stated above.

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    R^2 is the R value squared, which is bounded between 0 and 1. It is typically interpreted as the amount of variance predicted by a variable (or variables, if there are more than 1 in a model).
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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Based on this...

    Quote Originally Posted by Dragan View Post
    R is the Pearson correlation between the actual values of the dependent variable (Y) and the predicted values of Y (Y-hats)
    ...wouldn't a negative R only occur if your regression line had the wrong slope? How could a least-squares line have the wrong slope?

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Quote Originally Posted by bruin View Post
    Based on this...



    ...wouldn't a negative R only occur if your regression line had the wrong slope? How could a least-squares line have the wrong slope?
    Technically speaking, it is conceivable to obtain a negative value of R^2. This (unusual) case can occur when one conducts a regression without an intercept term i.e. regressing through the origin. However, the OLS estimate of the slope coefficient is still unbiased - unless you force the error terms to sum to zero.

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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?

    Bruin,

    See the first figure for traditional depictations of correlations:

    http://en.wikipedia.org/wiki/Correlation_and_dependence
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    Re: What does R, R square , Adjusted R square, STD error of estimates helps for ?


    Quote Originally Posted by Dragan View Post
    Well, no...it is not....rather, the correlation between Y and the Y-hats is bounded between 0 and 1 as I stated above.
    i thought this could be provable so i decided to give it a try. i haven't done one of this in a while so plz point out potential mistakes:

    to show that 0<Corr(\mathbf{Y},\mathbf{\widehat{Y}})<1 we could show that Cov(\mathbf{Y},\mathbf{\widehat{Y}})>0.

    to do this, it is convenient to use \mathbf{H}, the 'hat matrix' that we know \mathbf{H}=\mathbf{X(X^{t}X)^{-1}X^{t}} so that:

    \mathbf{\widehat{Y}}=\mathbf{X\widehat{\beta}}=\mathbf{X(X^{t}X)^{-1}X^{t}Y}=\mathbf{HY}

    so since we've established \mathbf{\widehat{Y}}=\mathbf{HY}

    we can do:

    Cov(\mathbf{Y,\widehat{Y}})=Cov(\mathbf{Y,HY})=\mathbf{H}Cov(\mathbf{Y,Y})\mathbf{H^{t}}

    now, we know: Cov(\mathbf{Y,Y})=\sigma^{2}\mathbf{I}

    by assumption of independence (or we'd introduce heteroskedasticity and violate the typical assumptions of OLS Regression)

    so we substitute that back in:

    \mathbf{H}Cov(\mathbf{Y,Y})\mathbf{H^{t}}=\mathbf{H}\sigma^{2}\mathbf{I}\mathbf{H}^{t}=\sigma^{2}\mathbf{HH^{t}}

    which is clearly greater than 0 because \sigma^{2} is positive (it's a variance) and \mathbf{HH^{t}} is positive-definite.

    so that (i think) shows Cov(\mathbf{Y},\mathbf{\widehat{Y}})>0 which i'm sure, when standardized, bounds Corr(\mathbf{Y},\mathbf{\widehat{Y}}) to be between 0 and 1.

    comments?
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