Quote:
|
Still not right unless the ten rounds he fired are all he's ever going to fire and are therefore the whole population. Otherwise, the best he can hope to do is estimate what the final population (all he ever will fire of that same load from that same gun) standard deviation is going to turn out to be by finding the sample standard deviation. So, he wants to find the sample standard deviation either by calculation or, better yet for small samples, from the lookup table. Calculating the population standard deviation will give him the smallest number and the most likely erroneous.
|
I think we’re talking past each other a bit and if we were sitting across a table we’d realize we agree about this – it’s just difficult to communicate this way.
I see now that we’re looking at the same problem in different ways and it’s the terminology that’s the issue – and I was also correct in my original post on this, just I didn’t explain it as well as I might have. Let me try again.
We start from the understanding that however many shots we chronograph they are only a sample of the theoretically infinite number of shots that we could have taken with the same load under the same conditions. That theoretically infinite number of shots is the population and it’s the statistics of that population that are of interest, not the statistics of the sample itself (except in the unusual circumstance where those are all the shots of that load we ever will shoot and, in effect, the sample equals the population). So, what we are doing is estimating the population variance from the sample variance, and that’s what I meant by using n-1 to calculate the population variance – what would have been better wording is to say that you use n-1 to calculate (estimate) the population variance from the sample data (or the sample sum of squares). If in fact the population and the sample are the same thing – as I said, an unusual circumstance - then the correct divisor is n.