Is it possible to calculate a Type II error for a geometric distribution? I can't see how it can be since the Type II error calculations use phi which is for the normal distributions. I could approximate the geometric to the normal - if there is a way to do this, even if there is it's not on my syllabus and so I'm not meant to use it.

Anyone who can tell me for certain one way or another would be a great help.

Ewwww! Just reading that brought back nightmares of college stats.

Then it made me laugh as I remembered the instructor, his problem actually saying the word "statistics" and his tendancy to quiz the class on the results of flipping a coin a certain number of times by asking "how much head do you get?"

What was Type II error again? IIRC is it the probably of concluding something true is false?

Not surprisingly, I think I forgot how to do all this stuff. If you could outline how you usually do it with the normal distribution, I'm sure you can make a few adjustments for it to handle the geometric distribution. Eg: instead of say

f(x) = Gaussian function (you know, the exp -0.5 *((x-mean)/standard deviation)^2 / root 2 pie times std dev)

replace f(x) with the geometric one: P(k) = p*(1-p)^(k-1) or something like that.

And about approximating the normal with a geometric - it might work. You should pick the probabilities (p and q=1-p in the above) to be around half and half or else your geometric distribution won't really be approximately normal.

Thanks for the help but I've handed the project in now so it's too late for any more adjustments. Just hope I got a good mark. Thanks anyway though!