Estimating, CI's and testing: When do you add √{ (N-n)/(N-1) } to the formula?

Hello,

I'm new to this forum. It seems helpful! I hope I can contribute some myself as well.

But now, I'd be very glad if you could help me with my Statistics test preparation. The test will be about estimating, confidence intervals and hypothesis testing.

My question is when to use what formula (see picture). I don't understand in what case I have to add in √{ (N-n)/(N-1) } to the formula, and when I don't, for estimating, CI's and testing.

This is how it's stated on my formula paper:
http://img15.imageshack.us/img15/7224/vraag1c.png

Thanks in advance.

japzer7 said:

My question is when to use what formula (see picture). I don't understand in what case I have to add in √{ (N-n)/(N-1) } to the formula, and when I don't, for estimating, CI's and testing.

This is how it's stated on my formula paper:
http://img15.imageshack.us/img15/7224/vraag1c.png

Thanks in advance.

Click to expand...

The Sampling Theorem assumes tat the mean and standard deviation of the population are known. In that case, for a sample of size n, the mean of sample means is \(\displaystyle \mu \) and the standard deviation of sample means is \(\displaystyle \sigma / \sqrt{n} \). That is the "usual" formula.

But suppose the population distribution is not known precisely, but has been measured for a total of N cases. The extra factor would account for the uncertainty of \(\displaystyle \sigma \) itself. Note that as N becomes large, the correction factor approaches 1. [I must admit that I have not run across this factor before - I generally just use the simple formula.]