3 ms·

Log-odds difference from contingency table: ln (4×11) - ln (1×8) = ln (44/8) = ln 5.5 ~= 2.3 × lg 5.5 ~= 2.3 × 0.75 = 1.75. Standard error of log-odds differen

by kqr 1y ago

Log-odds difference from contingency table: ln (4×11) - ln (1×8) = ln (44/8) = ln 5.5 ~= 2.3 × lg 5.5 ~= 2.3 × 0.75 = 1.75. Standard error of log-odds difference: √(1/1 + 1/11 + 1/4 + 1/8) ~= √(1 + 0.09 + 0.25 + 0.13) = √(1.47) = 1.47^0.5 = 10^(0.5 × lg 1.47) ~= 10^(0.5 × 0.16) = 10^0.08 ~= 1.2. The effect is 1.75/1.2 ~= 1.5 standard errors from zero, thus not significant by common thresholds. You would find Fisher's exact test in agreement, but I like the (vepry slow) party trick of doing this all in my head. Techniques: - Poor man's logistic regression: https://entropicthoughts.com/contingency-table-for-poor-mans-logistic-regression https://entropicthoughts.com/contingency-table-for-poor-mans... - Mental logarithms: https://entropicthoughts.com/learning-some-logarithms https://entropicthoughts.com/learning-some-logarithms

FabHK 1y ago
Very neat! Just one comment: For square root of 1.47 you don't need to take the log, half it, and exp it again. It's just sqrt(147/100), or a bit more than sqrt(144/100) = 12/10 = 1.2. (If you like mental math you can then run an iteration of Newton's method: you take the remainder between the number and the square of your guess, 1.47-1.44 = 0.03, and divide it by twice your current guess, and add it to the current guess: 0.03 /(2x1.2) = 3/2.4 / 100 = 1.25 / 100 = 0.0125, so the next guess would be 1.2 + 0.0125 =1.2125. And the actual sqrt of 1.47 is 1.2124355653..., so that's pretty close. Newton's method roughly doubles the number of accurate digits at each iteration.)
- kqr 1y ago
  Oh. That's actually very helpful. I hadn't realised Newton--Raphson could be that easy to use in practice. The closeness between 1.47 and 144 did strike me.
  - FabHK 1y ago
    Indeed - taking the square root of numbers iteratively is very satisfying, because of the super fast convergence. Sqrt 10: Guess 3, then residual is 10 - 33 = 1, so next guess add 1/(23) = 1/6, so now we're at 3.166..., and the square of that is 10.0277..., so you're very close already.
  - jacquesm 1y ago
    I did this in high school when we weren't allowed a calculator during tests and got into an argument with a teacher who told me that it is impossible to do this in your head so I must have cheated. A short demonstration on random numbers sufficed. I've taught my kids the method and they are now set up for similar arguments ;) The hardest part if you want to generate more digits is to keep track of the intermediary sum during the multiplication, that's why I never got past three significant digits. That would be an interesting little contest though!

by kqr 1y ago

FabHK 1y ago

Very neat! Just one comment: For square root of 1.47 you don't need to take the log, half it, and exp it again. It's just sqrt(147/100), or a bit more than sqrt(144/100) = 12/10 = 1.2. (If you like mental math you can then run an iteration of Newton's method: you take the remainder between the number and the square of your guess, 1.47-1.44 = 0.03, and divide it by twice your current guess, and add it to the current guess: 0.03 /(2x1.2) = 3/2.4 / 100 = 1.25 / 100 = 0.0125, so the next guess would be 1.2 + 0.0125 =1.2125. And the actual sqrt of 1.47 is 1.2124355653..., so that's pretty close. Newton's method roughly doubles the number of accurate digits at each iteration.)

kqr 1y ago
Oh. That's actually very helpful. I hadn't realised Newton--Raphson could be that easy to use in practice. The closeness between 1.47 and 144 did strike me.
- FabHK 1y ago
  Indeed - taking the square root of numbers iteratively is very satisfying, because of the super fast convergence. Sqrt 10: Guess 3, then residual is 10 - 33 = 1, so next guess add 1/(23) = 1/6, so now we're at 3.166..., and the square of that is 10.0277..., so you're very close already.
- jacquesm 1y ago
  I did this in high school when we weren't allowed a calculator during tests and got into an argument with a teacher who told me that it is impossible to do this in your head so I must have cheated. A short demonstration on random numbers sufficed. I've taught my kids the method and they are now set up for similar arguments ;) The hardest part if you want to generate more digits is to keep track of the intermediary sum during the multiplication, that's why I never got past three significant digits. That would be an interesting little contest though!