1
$\begingroup$

I'm attempting to figure out how the equation works for the scores of hospitals in the US News & World Report 2018-2019.$^\dagger$ Specifically, I'm having trouble with the log transformation formula that they're giving on page 38-40.

Transformation is applied to the weighted reputation data using the formula $\log(R_X + 10) – 1$, where $R_X$ is the weighted reputation score for hospital X.

I'm assuming $\log$ refers to the natural logarithm, although I tried that, and a logarithm with base 10 and that didn't work either.

The article claims that:

An untransformed score of 1% has a transformed value of 4 (4 times greater)

An untransformed score of 10% has a transformed value of 29 (2.9 times greater),

An untransformed score of 60% has a transformed value of 81 (1.35 times greater)

But I've tried $\log(60+10)-1$ and that doesn't equal 81. I'm not sure where I'm going wrong.


$\dagger$: Olmsted et al. (2018): U.S. News & World Report 2018-19 Best Hospitals: Specialty Rankings

$\endgroup$
3
$\begingroup$

The transformation

The part you quote from the original paper is missing one important detail:

The transformed data are then scaled to a minimum of 0 and maximum of 100.

Using this, we can obtain their new scores as follows:

  1. $f(x) = \ln(x+10) - 1$
  2. $g(x) = \frac{f(x) - f(0)}{f(100) - f(0)} \cdot 100\%$

Plugging their numbers

Now to check that:

An untransformed score of 1% has a transformed value of 4 (4 times greater)

An untransformed score of 10% has a transformed value of 29 (2.9 times greater),

An untransformed score of 60% has a transformed value of 81 (1.35 times greater)

  • $g(1) \approx 3.98$
  • $g(10) \approx 28.91$
  • $g(60) \approx 81.15$

So that seems to work.


Implementation in R

f <- function(x){log(x + 10) - 1}
g <- function(x){(f(x) - f(0)) / (f(100) - f(0)) * 100}

Which gives:

> round(g(c(1, 10, 60)))
[1]  4 29 81

Commentary

Going slightly off-topic, but this transformation seems a bit arbitrary. Perhaps I'm missing something, but it comes across as changing the data to match what the author thinks it should be.

Also note that it can be misleading to say that $x\%$ is $\frac{x}{y}$ times greater than $y\%$.

For example, let's say I have two choices of medicine:

  • Drug A, which cures $90\%$ of patients;
  • Drug B, which cures $99\%$ of patients.

Yes, drug B would indeed be expected to cure a mere $\frac{99}{90} = 1.1\times$ more patients than drug A, but you could also say it results in $10\times$ less uncured patients!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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