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Oct 31, 2022 at 18:06 comment added seanv507 no the approximation is the claim that exp(b) is approximately 1+b, which depends on b being small
Oct 31, 2022 at 18:05 comment added seanv507 y changes by 36% (not 0.36%)
Oct 31, 2022 at 17:35 history edited Dave CC BY-SA 4.0
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Oct 31, 2022 at 17:27 comment added chunguc1004 @seanv507 Also, by looking at exp(𝑧+𝑏)/exp(𝑧)βˆ’1=exp(𝑏)βˆ’1βˆΌπ‘. it seems like you are talking about if the z value is small enough rather than b? which is the coefficient. (My math is rusty.. so sorry if I'm making dumb questions.. but really appreciate any help!)
Oct 31, 2022 at 17:27 comment added chunguc1004 @seanv507 Thanks for your comment! But I meant the raw b value. So if the beta value is 0.36, then you're saying that as long as the coefficient is small enough, I can interpret it as if x changes by 1 unit y changes by 0.36%?
Oct 31, 2022 at 11:09 comment added seanv507 its approximately correct (for small coefficients). if we define $z$ as your log transformed value, then your proportion change is $\exp(z + b)/\exp(z) -1=\exp(b) -1\sim b$ by Taylor expansion of exponential.
Oct 31, 2022 at 6:09 vote accept chunguc1004
Oct 31, 2022 at 6:08 comment added chunguc1004 Thank you for your comment! One other question I have is, if you look at this post (stats.stackexchange.com/questions/549830/…) the answer says that if a dependent variable is logged transform, then b is the % change in y when x changes by 1 unit. However, isn't that incorrect? I thought that we have to exponentiate the coefficient and subtract 1 and multiply by 100 to get the percentage..
Oct 31, 2022 at 5:27 history edited Dave CC BY-SA 4.0
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Oct 31, 2022 at 5:20 history answered Dave CC BY-SA 4.0