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One SD is the SD of Time, the other SD of Rate = 50/Time. The division means that you need to go through the CV's or use the delta method if you prefer. In short, this turns out to be SD(Rate) = SD(Time)*50/mean(Time)^2 so all is well. As to the preferred method I would have thought that the SD...

It sounds like you are trying to locate a point (an animal perhaps?) which has emitted a sound. This sound is detected at 5 stations, and the arrival time referenced against a "zero" station. You then pick a point at random, find the distance to each station, work out what the times would have...

This sounds like an interesting problem, but you will need to explain it in much simpler terms for those of us who aren't familiar with the situation.

If I am interpreting you correctly, the subject is just guessing. Of the 80 TRUE answers they will get 40 right by guessing yes. Of the 20 FALSE answers they will get 10 right by guessing no. Total 50 right by guessing. I think your friend is right.

Are you still interested in an exact solution? I think one would be possible using dynamic programming which is just a fancy way of saying that you can build up a table of smaller values to calculate the final value.

One observation may not influence another but they both may be influenced in the same way by ecological conditions at the site.

Since the year to year differences don't seem to be of particular interest, why not just combine the 4 years for each season for each of the 47 sites and use the Wilcoxon signed rank test suggested by Karabiner on the 47 matched pairs. (Or even the matched pairs t test. You have plenty of sites...

I can't think of any practical way to get an exact solution. In principle, there will be an exact answer for any given set of conditions, but it will probably involve tabulating all the possibilities, calculating the probability of each, and then collating the results. That's 2^100 for 100 days...

Would you be happy with an accurate approximation? or must it be exact?

Y = rho*X + sqrt(1-rho^2)*Z is true for X, Y and Z standard normal and so could be roughly true (I think) for uniform. The second version is an attempt to convert the normal formula into uniform distributions.

New version, in Excel formulas Y =NORMSDIST(rho*NORMSINV(X)+SQRT(1-rho^2)*NORMSINV(RAND())) Y and X are correlated with correlation rho and both lie on (0,1). You can scale Y to (0,5).

This sort of works, if it is what you are looking for. X is the variable you start with on (0, 1) and rho is the desired correlation If rho>0 then make Y = rho*X + sqrt(1-rho^2)*Z where Z is uniform (0, 1) Y is now correlated with X with correlation rho. Y is on (0, sqrt(2)) and can be scaled...

Cheers, kat

If the simulated samples do in fact come from a uniform distribution, and if you test the fit against a uniform distribution using 20 bins, then no matter how big the sample, the various X2 values you get will be distributed as Chi Square with 19 df. Some X2 values will be large, and some small...

The same j for every pair, or a new j for each pair?

Your mean is about 19.7 The mean of Chi square with 19df is 19. Very close. Your SD is about 5.3. The SD of Chi square with 19df is sqrt(2*19) is about 6.2, so things look OK to me.

If they are two different groups, I would do them separately. How many in each group? How will you test them?

Most of the theory is based on straight lines and equal errors. Often a situation is not like this, but it becomes so if we log the y's. Then we can use the standard techniques on the transformed data.

Sample P2 is about 35% and P1 is about 50%, so no way is P2 - P1 >= 10% a suitable Ho. Do you have to do a hypothesis test? They can be very confusing. Most people would just find a suitable confidence interval for the difference and see if 10% lies in it or not.

Ah ha! I see it now. Well done. You snuck in a post while I was away looking up the Fisher-Box quote. kat