12
$\begingroup$

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.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ 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
    Commented Jul 15, 2013 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
    Commented Jul 15, 2013 at 12:32
  • 1
    $\begingroup$ @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
    Commented Jul 15, 2013 at 15:02
  • 1
    $\begingroup$ @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
    Commented Jul 15, 2013 at 22:54
  • 1
    $\begingroup$ @Glen_b I totally agree and voted for reopening. $\endgroup$
    – COOLSerdash
    Commented Jul 15, 2013 at 22:58

1 Answer 1

Reset to default
4
$\begingroup$

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.
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.