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How should I interpret Exp (B) = 3.661E-11 in negative binomial regression? I am stumped when it comes to these E-11, E-10, and E-12 values. I got these values on interaction terms consisting of categorical variables. Can anyone assist?

I think I am clear on how to interpret it when Exp (B) is greater than 1. For Exp (B) values greater than one, the criterion variable (Y) is expected to be the value of the Exp (B) times higher as a result of a one unit change in the (X) predictor. One would apply the following formula: (Exp (B) – 1.00) * 100. If Exp(B) = 1.5 then: (1.5 – 1.00) * 100 = 0.5 * 100 = 50%. One could then say that for every one standard deviation increase in the predictor (X) one would expect the mean criterion variable (Y) to increase by 50%.

For Exp (B) values that are less than one, the same approach is used. If the Exp (B) value of a predictor (X) were 0.75 it would indicate that with one unit increase in the predictor (X) the predicted score of the criterion variable would be expected to be a factor of 0.75 times the mean criterion variable (Y). To determine the percentage of decrease in the mean rate of the criterion variable, one would use the same formula: (Exp (B) – 1.00) * 100 such that (0.75 – 1) * 100 = - 0.25 * 100 = - 0.25%. One could say that the mean criterion variable (Y) would be expected to decrease by 25% for every one standard deviation increase in the predictor variable (X).

When the Exp(B) value is so small like 3.661E-11 then applying the formula results in 1 - 0.00000000003661 = .9999 * 100 = 99.99% so the mean criterion variable would be 99.99% lower? I am just having a difficult time wrapping my brain around that. Could it be that although these Exp(B) values are statistically significant, that they really could be indicating a problem with the data? The frequency counts of the interaction terms that generated these values are very low (between 1 and 4).

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    $\begingroup$ Just to be clear, are you aware of scientific (exponential) notation? $\endgroup$ – usεr11852 says Reinstate Monic Jan 27 '18 at 21:13
  • $\begingroup$ I think so. Wouldn't 3.661E-11 be 0.00000000003661? $\endgroup$ – Laura Jan 27 '18 at 22:10
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    $\begingroup$ What, then, are you trying to ask? You highlight "E-11" as being something special: in what way? $\endgroup$ – whuber Jan 27 '18 at 22:46
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    $\begingroup$ It would be much better to edit all your explanations into your original question as it is quite hard to follow the thread in comments. At first reading I think you are on the right lines. $\endgroup$ – mdewey Jan 28 '18 at 13:43
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    $\begingroup$ One possible problem is the scale of the variable. You didn't tell us what the variables are, but if you were predicting (say) the probability of voting for a Republican and you had an IV of "income in billions of dollars" you might get something like this. So, check that the scales are reasonable (and tell us more about the data, the variables, the research questions and so on). $\endgroup$ – Peter Flom - Reinstate Monica Jan 28 '18 at 15:51
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I think your last sentence hits the nail on the head here. What is happening is that for some combination of your predictor variables there is a zero count. So the software is trying to generate a coefficient which will send the intercept (which is presumably positive) to zero. But there is no multiplier which does that so it stops when it reaches a very small predicted value. If you look at the coefficients on the log scale it is about -24 which is trying to go to $-\infty$ but stopping there. For advice on what to do you could try browsing the tag or looking at How to deal with perfect separation in logistic regression? Note that most of those are about logistic regression but similar issues arise for you here.

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The outcome for a negative binomial is a count. The coefficient B means that when the predictor changes by 1 unit, then you expect a B difference in the log of the outcome, so log(y|x) - log(y|¬x). This is the same as log((y|x) /(y|¬x)), which gives you a factor to multiply.

Say your expected count goes from 5 to 10. 10/5=2. Log(2) is .301. You expect the outcome variables to increase by .301 times, or 30%. If instead your expected count goes from 10 to 5, log(5/10) is -.301, so you expect it to decrease by .301 times, or decrease by 30%. That's the case whether it's a small number or not. Exp(.00000005) is 1.00000005, so you expect a 1.00000005 times increase, which is a very small increase (depending on your unit of analysis). Small coefficient means small change.

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You're interpreting the coefficient correctly, assuming this is a log-link model. If the exp of the coefficient is > 1, the average response value is higher on average. If it's less than 1, it's lower on average.

Your statement that every increase of 1 standard deviation results in a mean response x% higher is only correct if you've scaled your input variables to have standard deviation 1.

In the level of the categorical variable associated with this coefficient you probably have no examples of positive counts, or the positive counts are all also in other low-count categorical levels.

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  • $\begingroup$ I have not scaled my input variables to have a standard deviation of 1. So that being the case, would it be correct to say that every one point increase on the scale predictor variable results in a mean response x% higher on the criterion count variable? $\endgroup$ – Laura Jan 28 '18 at 15:49

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