This is a similar question to Intercept significant but not the variables in GLM, but in more detail: my model's dependent variable is change in population density of a state, and the independent variables are various factors that may influence it (i.e. increased access to railways), along with one dummy (categorical) independent variable (if the state borders a coast or not). The results show that only the intercept is statistically significant. In this case, if it's possible for all of my independent variables to be 0, does the significant intercept mean that population density would have increased/decreased regardless of the independent variables? Does having a dummy variable change that interpretation?
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3$\begingroup$ In essence if your model equation is $y = b_0 + b_1 x_1 + \cdots + b_p x_p$ then you're being told that your model is roughly $y \approx b_0$. The intercept is then roughly the mean of $y$, and you're being told it's not zero. $\endgroup$– Nick CoxCommented Jun 26, 2019 at 11:01
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$\begingroup$ To add to @Nick's excellent analysis and interpretation, it is always worth bearing in mind that insignificance of coefficients in such a model does not imply much about the kinds of changes you are exploring. The key issues to bear in mind in every such situation are (a) the distinction between an association and causal relationships; (b) the possibility of nonlinear relationships; (c) the distinction between statistical significance and practical importance; and (d) the possibility that other implicit model assumptions (such as independence of errors) are strongly violated. $\endgroup$– whuber ♦Commented Jun 26, 2019 at 14:54
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2 Answers
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Does the significant intercept mean that population density would have increased/decreased regardless of the independent variables?
Technically, yes. It simply means that the independent variables that you have chosen do not affect your dependent variable. But it does not mean that your dependent variable does not depend on independent variables at all. Example: Try squaring your independent variables. Does that change their coefficient's significance? If yes, then you're simply using the wrong precision of independent variables.
Does having a dummy variable change that interpretation?
The dummy variable simply indicates the presence of some factor and if it's not significant, then it simply means that the dependent variable does not depend on the presence of that factor. It does not change the above interpretation.
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$\begingroup$ Wouldn't it be better to model current population density as a function of initial density, allowing for the possibility that the slope of the second is not 1.0? $\endgroup$ Commented Jun 26, 2019 at 11:07
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1$\begingroup$ (1) Beware of the use of "affect" and "depend on" in this context, because it suggests you are drawing causal inferences where none are warranted. (2) Your remark about "wrong precision" is puzzling: how do you conceive of including squared dependent variables in this regression and how is that related to "precision" of anything? $\endgroup$– whuber ♦Commented Jun 26, 2019 at 14:57
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1$\begingroup$ @whuber (1) Yes you're right, and to clarify, I'm not trying to draw causal inferences. (2) Yes I wrote dependent variable by mistake, Edited it to independent variable now. By precision I mean that the variation in the independent variable might not be much but when you square it, the variation might become more pronounced and may result in a significant coefficient. $\endgroup$– DivyanshCommented Jun 27, 2019 at 12:18
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$\begingroup$ Thank you: that explanation helps. But so far you haven't articulated it clearly in your answer. You appear to suggest including a quadratic term as a way to check the model, but then you go on to characterize it strangely as "you're simply using the wrong precision of independent variables." I'm still unable to make sense of what idea you're trying to convey with that phrase. $\endgroup$– whuber ♦Commented Jun 27, 2019 at 12:41
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$\begingroup$ Oh yes that phrase isn't used very often while discussing stats. What I mean by precision is, suppose you have the independent variable values as
[1,2,3,4,5], so the variation in them is just 1 unit, but when we square them,[1,4,9,16,25], the difference becomes enormous. Therefore, in some cases, it might lead to change in the number of significant digits in data, hence my use of the word precision. $\endgroup$– DivyanshCommented Jul 1, 2019 at 8:42
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You wrote:
The results show that only the intercept is statistically significant.
OK, but be careful not to accept the null and not to confuse lack of significance with lack of meaning.
In this case, if it's possible for all of my independent variables to be 0, does the significant intercept mean that population density would have increased/decreased regardless of the independent variables?
Not necessarily. One or more IVs may have a large parameter estimate, even if it's not significant. There could be colinearity, which can affect significance by inflating the variance. The relationship between the DV and the IV might be nonlinear. There could be an interaction among the IVs.
Does having a dummy variable change that interpretation?
Not really. But nonlinearity isn't going to be an issue.
answered Jun 26, 2019 at 11:25
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$\begingroup$ "OK, but be careful not to accept the null and not to confuse lack of significance with lack of meaning." — could you possibly explain this? For the IVs that were not significant, I looked primarily at the signs of the coefficient estimates as part of my analysis, but is that pretty much the extent of any meaning/information I can get? $\endgroup$ Commented Jun 26, 2019 at 11:50
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$\begingroup$ No. The parameter estimates are just as meaningful if they are not significant. Stat sig is a very narrow question. $\endgroup$ Commented Jun 26, 2019 at 13:33