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I have a linear regression model where the dependent variable is logged and an independent variable is linear. The slope coefficient for a key independent variable is negative: $-.0564$. Not sure how to interpret.

Do I use the absolute value then turn it into a negative like this: $(\exp(0.0564)-1) \cdot 100 = 5.80$

or

Do I plug in the negative coefficient like this: $(\exp(-0.0564)-1) \cdot 100 = -5.48$

In other words, do I use the absolute figure and then turn that into a negative or do I plug in the negative coefficient? How would I phrase my findings in terms of a one-unit increase in X is associated with a __ percent decrease in Y? As you can see, these two formulas produce 2 different answers.

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    $\begingroup$ Could you add more details about your model? That would help us to answer the question. Here are some comments: Normally, you would just exponentiate the regression coefficient, so just $\exp{(\beta)}$. If the coefficient is negative, $\exp{(\beta)}<1$ and if the coefficient is positive, then $\exp(\beta)>1$. I think the interpretation is like this: the exponentiated coefficient is the multiplicative term to use to calculate the estimated dependent variable when the independent variable increases by 1 unit. In this case, the multiplicative term is $0.945$. See also here. $\endgroup$ – COOLSerdash Jul 15 '13 at 10:38
  • $\begingroup$ Thanks @Glen_b for the clarification. I will delete my comment and wait until the OP provides additional information about his goals. How would one calculate the mean? $\endgroup$ – COOLSerdash Jul 15 '13 at 12:32
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    $\begingroup$ @COOLSerdash Sorrt, somehow I missed the question about calculating the mean. If it's normal on the log scale, then conditioning on knowing the parameter values, you'd be computing the mean of a lognormal ($\exp(\mu + \frac{1}{2} \sigma^2)$). If you don't condition on at least the variance-parameter, the exponentiated estimate is instead log-t ... and then it doesn't have a mean. $\endgroup$ – Glen_b Jul 15 '13 at 15:02
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    $\begingroup$ @COOLSerdash Yeah, I agree that normally statisticians would use log-linear model refer to a model whose linear predictor has a log-link (which is natural in the Poisson regression case), but as you note, the question says "where the dependent variable is logged", clearly suggesting modelling $\log(y) = \alpha + \beta x+\varepsilon$. Needless to say, I don't think it's a duplicate of a Poisson regression question, which would model $\log(\text{E}(y))$ as linear in $x$, not $\text{E}(\log(y))$. $\endgroup$ – Glen_b Jul 15 '13 at 22:54
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    $\begingroup$ @Glen_b I totally agree and voted for reopening. $\endgroup$ – COOLSerdash Jul 15 '13 at 22:58
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You should not take the absolute value of the coefficient--although this would let you know the effect of a 1-unit decrease in X. Think of it this way:

Using the original negative coefficient, this equation shows the percentage change in Y for a 1-unit increase in X:

(exp[−0.0564*1]−1)⋅100=−5.48

Your "absolute value" equation actually shows the percentage change in Y for a 1-unit decrease in X:

(exp[-0.0564*-1]−1)⋅100=5.80

You can use a percentage change calculator to see how both of these percentages map onto a 1-unit change in X. Imagine that a 1-unit change in X were associated with a 58-unit change in linear Y:

  • Our linear version of Y going from 1,000 to 1,058 is a 5.8% increase.
  • Our linear version of Y going from 1,058 to 1,000 is a 5.482% decrease.
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