# Not sure what statistical analysis to perform

• Sep 29th 2010, 12:58 AM
Pookabear
Not sure what statistical analysis to perform
Hi guys,

This may be an easy question and apologies if it's in the wrong forum, but it's uni-level stats so I figured I'd put it in here.

I'm really stumped as to how to approach the following: I have a data set and it says "it is believed that the true mean value for one concentration is given by a weighted average of the true concentrations found at other locations".

I have alot more info but I'm trying to do this myself so I'm just after a gentle nudge in the right direction, rather than a step-by-step guide.
I'm thinking that since the two concentration sets have different n's, I need to take this into account to work out the weighted average. So say 1 is the first concentration discussed, 2 is one of the true concentrations with n=6 and 3 is the other true concentration where n=5, will it be something like the mean of 1 = (6/11)*mean of 2 + (5/11)*mean of 3? So I'm weighting those concentrations based on how many I have in the sample? And then just do a t-test where H0 = that being true and H1 = that being false but two-sided.

I really hope this makes sense, the more I read the question the more confused I get!
• Sep 29th 2010, 01:17 AM
pickslides
Quote:

Originally Posted by Pookabear
I'm thinking that since the two concentration sets have different n's, I need to take this into account to work out the weighted average. So say 1 is the first concentration discussed, 2 is one of the true concentrations with n=6 and 3 is the other true concentration where n=5, will it be something like the mean of 1 = (6/11)*mean of 2 + (5/11)*mean of 3?

I have not seen anything like this before.

It looks like you are are trying test proportions here not the mean like asked in the question.

Quote:

Originally Posted by Pookabear
I'm thinking that since the two concentration sets have different n's,

so you have 2 samples? Can you work it to be a difference in means formulation? i.e $\displaystyle H_0: \mu_1 = \mu_2, H_a: \mu_1 \neq \mu_2$

The fact that $\displaystyle n_1 \neq n_2$ won't stop you from performing a t-test.
• Sep 29th 2010, 09:26 AM
Pookabear
Thanks for the reply. I did reply but seems my answer isn't showing so I'll give it another bash!

What I have is concentrations taken at 3 different depths; 400m (n=8), 600m (n=4) and 1000m (n=4). It's asking to test whether the weighted mean of 400m and 1000m = mean of 600m. Originally I thought it'd be weighted due to the different sample sizes but now I'm thinking that it's weighted because if 400m is closer to 600m than 1000m is, those results will carry more weight in the analysis. It's an odd question but I think I'm slowly nutting it out :)