Which statistics formula to use in Excel?

I have the following problem for my stats class and I am trying to figure out which statistical formual I need to use to get the answer. Please advice! Any help will be much appreciated.

A sample of salaries finds sample mean $35 thousand and sample variance 81. If the population mean is to be estimated within $2 thousand (that is, $2 thousand above and $2 thousand below), how many employee salaries must be sampled? Assume 95% confidence and t-multiple = 2.

Hi,

You need to use power of analysis to determine the sample size. Excel doesn't have a built in function to do this, you'll need an add-in, or a stat package such as minitab.

There's an explaination of power here:

http://www.indiana.edu/~statmath/sta...wer/power.html

Thank you for the link. I have a statistical program that I use for class assignments but I am not sure which function to use NORMDIST or NORMINV? Can anyone lead me in the right direction. I don't need the answer, just help on how to interpret the problem.

Originally Posted by Peace_love

I have the following problem for my stats class and I am trying to figure out which statistical formual I need to use to get the answer. Please advice! Any help will be much appreciated.

A sample of salaries finds sample mean $35 thousand and sample variance 81. If the population mean is to be estimated within $2 thousand (that is, $2 thousand above and $2 thousand below), how many employee salaries must be sampled? Assume 95% confidence and t-multiple = 2 and round to the nearest integer.

I think you need to try with NORMDIST because I think this is normal or Gaussian distribution (so google for it, maybe I didn't solve it correctly)

you have situation that formula is:

P = [-A/B < (x-m)/(o/sqrt(n)) < A/B]

where A=2000/35000 and B=1/sqrt(81) and o is standard deviation

so.. it's P=[-0,5143 < (something) < 0,5143] and because graph of Gaussian distribution is mirror image you get:

P=2*Y(0,5143)-1

Y(0,5143) = (from table of Gaussian distribution), if I find correct one, is 0,69672

so P= 2*0,69672-1=0,3934 = 40%

So, I guess 40% of employees is the answer...

however, I didn't find where to put 95% confidence and t-multiple = 2, therefore I doubt in my result, but I hope I lead you to the right path

Good luck, and hope someone more skilled than me can help you more

Well, the answer that I am supposed to find should the number of people needed, not the percent. I am very confused by this problem. At first I thought I should use the NORMINV function but that did not lead me to the correct answer. I am trying to understand that problem because I don't. If I understand it I'm sure I can solve it.

Yes, you are right.. Number should be answer... Was my answer due to forgotteness or due to wine

I read table from a wrong side

So for 95% you get from table 1,96.
But I was focused on that t-multiple = 2 and only thing I can figure is that it's part of Student distribution...

However, no matter how I put numbers in... I can't get any of solutions ... by intuition.. the number must be small... 324 max...