2
$\begingroup$

In a previous question I was trying to figure out how to compute the confidence interval of a ratio. Having come to the conclusion that the correct way to do this is Fieller's Theorem, I realize I'm very unsure about my interpretation of the degrees of freedom.

I'm trying to compute confidence intervals for a speedup -- computed as a ratio:

$$ \frac{\text{time mean}_{\text{1 thread}}}{\text{time mean}_{n \text{ threads}}} $$

Each thread count has $k$ samples.

I want to say I need 1 degree of freedom here... but I also have been known to be wrong in these things before...

$\endgroup$
  • $\begingroup$ Online documentation indicates the degrees of freedom will be $a+b-2,$ where $a$ is sample size of numerator and $b$ is sample size of denominator. Why do you think it should be 1 degree of freedom? $\endgroup$ – soakley Aug 23 '13 at 20:20
  • $\begingroup$ A day later, I now... don't really recall. :S I grabbed my stats book and agree that $a + b - 2$ is reasonable (though, it doesn't have Fieller's Theorem). $\endgroup$ – Matthew G. Aug 23 '13 at 21:44

Your Answer

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

Browse other questions tagged or ask your own question.