Interpret coefficients in Log-linear model with variable in percentage

[Nelson Joseph]

Excuse me, I have this equation LnY = 5.733−0.001𝑋1 + 0.075𝑋2, where Y is the number of patients in a province and my variable X2 is the percentage of homes that have water in a certain province that goes from 0 to 100 (If I wanted Know the effect of increasing my variable X2 by 1%, 5% and 10%, how would it be interpreted with the coefficient? I know that it is multiplied by 100 or it can also be exp 100*(exp(0.075X2)-1) and a similar value should come out. But my variable It is already independent and is in percentage, is the same procedure followed?

And finally, if the province that I want to evaluate already had a value of 4% and I want to know the effect of reaching 10% (a variation of 6%) it is advisable to make a transformation of my equation to eliminate the Ln and have the number of patients by adding the value of 10 in this new equation. Thanks

[Clyde Schechter]

You just have to work with the regression equation. It's not that hard--it's high school algebra. The fact that X2 represents a percentage of something is really of no importance.

Now, you have to be clear on the statement of your problem. When you refer to increasing X2 by 1%, 5%, and 10% is that what you really mean? For example, if a typical value of X2 is, say, 50%, increasing that by 1%, 5%, and 10% would mean changing X2 to 50.5%, 52.5%, and 55% respectively. That's OK if it's what you want. But my hunch is you really want to see the effect o changing X2 from 50% to 51%, 55%, and 60% respectively. That's not changing them by 1%, 5%, and 10%. That's changing them by 1 percentage point, 5 percentage points, and 10 percentage points, respectively. The failure to distinguish between percentage change and percentage point change is a constant source of confusion in these situations. Sloppy language leads to sloppy thinking, which leads to misleading results. I'll assume you really meant percentage points.

LnY(X1, X2) = 5.733 - 0.001X1 + 0.075X2
LnY(X1, X2+ 1 percentage point) = 5.733 - 0.001X1 + 0.075(X2+1)

Subtracting the top equation from the bottom one, many of the terms cancel out:
LnY(X1, X2+1) - LnY(X1, x2) = 0.075*(X2+1) - 0.075*X2 , which simplifies to just 0.075.

Now, the left hand side of this equation can also be simplified since the difference of logarithms is the logarithm of the ratio:
Ln [Y(X1, X2+1)/Y(X1, X2)] = 0.075.

Exponentiating both sides
Y(X1, X2+1)/Y(X1,X2) = exp(0.075), which is 1.079 to three decimal places.
So Y(X1, X2+1) = 1.079 * Y(X1, X2).

In words, a 1 percentage point (not percent) increase in X2 is associated with a 7.9 percent (not percentage point) increase in Y.

Working this out for 5 and 10 percentage point increases in X2 is left as a homework exercise. (The results may surprise you if you are accustomed to using rules of thumb about multiplying a coefficient by 100 or the like. Those rules of thumb only work for small differences in X2 and are way off base for moderate to large differences.)

[Nelson Joseph]

Thanks Clyde for the very specific explanation. You are right this is easy, in my case I saw so many examples and I confused the background of this. Now I understand it much better, do you think it is possible that you would recommend a bibliography to me?
Thanks for your time