Confidence limits of expected odds

  • Thread starter Thread starter Andrusko
  • Start date Start date
  • Tags Tags
    Limits
AI Thread Summary
To convert confidence intervals for proportions into odds, one must first understand the relationship between proportions and odds, which is expressed as p/(1-p). The conversion involves applying this formula to both endpoints of the confidence interval for the proportion. This process results in a new interval that represents the odds corresponding to the original confidence interval. The conversion is straightforward, as it simply changes the notation from proportions to odds. The final outcome provides a clearer understanding of the odds associated with the given proportions.
Andrusko
Messages
42
Reaction score
0
I have two proportions and confidence intervals for each of them. Now the question wants me to convert these to odds.

What does this mean? I sort of understand converting a proportion to odds p/(1-p) but how does one convert an interval into odds?
 
Physics news on Phys.org
It just a different notation. Convert the endpoints of the intervals to odds.
 
Okay, thanks!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top