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I am trying to interpret the SPSS output from a multiple hierarchical regression where the intercept has been eliminated because it is not significant.

I have read previous discussions about inclusion/exclusion of the intercept in this forum and I have seen that the majority of the answers were against the exclusion of the constant from the model, unless we are certain that the intercept is zero.

However, I do not know how I could be sure about that.

All I see is that the intercept is not significant and Eisenhauer (2003)$^{[1]}$ suggests to consider the intercept p value and the model standard error in order to decide whether to include the intercept. Basing on his recommendations I have excluded the intercept.

Now, I see that the R square and multicollinearity indices in this case do not have the same meaning that they have in a model that includes the constant, but what I wonder is: do the standardised and unstandardised beta coefficients have the same meaning of a model that include the intercept or not?

[1] Eisenhauer, J. G. (2003), Regression through the Origin. Teaching Statistics, 25: 76–80.

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The only situations where you might consider excluding the constant are models where it is trivially true that the constant has to be 0. The fact that you looked at the significance level is in itself sufficient evidence that it is not trivially true that the constant is 0. So I would leave it in.

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Assuming you mean unstandardized and standardized (beta) slope coefficients. I'd be interested in seeing exactly how you think $r^2$ and multicollinearity have changed their meaning. I suspect they mean the same thing, just in reference to a model with no intercept term.

At any rate, your raw slope coefficients will also mean the same thing namely: (unit) change in Y is predicted per (unit) change in X. I can see your confusion in the case of standardized slopes because the standardized slope coefficients refer to a Y that is also in standardized units, i.e. one where the intercept actually is 0... and that wasn't the case before - you were just ignoring the fact that it was (presumably) trivially not different from 0. In practice, the interpretation of your standardized slope will be the same, the predicted change in $Z_y$ per $Z_x$. However, to be fair, your prediction should be that $Z_y$ = $Z_x\beta$ + $\bar{y}\over{\sigma_y}$ to account for the extent to which your prediction formula is not estimating that intercept.

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    $\begingroup$ Well, the regression's $R^2$ changes meaning because the denominator (necessarily) changes. In particular, $R^2$ for a regression through the origin is no longer the square of the correlation. The denominator is different from the usual $R^2$, and some packages even refuse to give an $R^2$ at all. $\endgroup$ – Glen_b -Reinstate Monica Jul 2 '13 at 1:00
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    $\begingroup$ @Glen_b: Thanks for clarifying that - I hadn't thought that through all the way and see your point. $\endgroup$ – russellpierce Jul 3 '13 at 2:54

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