nerdy_hottie said:
Homework Statement
So I'm doing a Least Squares Analysis and I'm wondering about what the 'measured mean value of y for replicate measurements of the unknown' value is supposed to be. I have no idea in the world what it's asking for. The value it is speaking of is not the same as the average value in y. I will post the example so you can see what I'm talking about.
Least-Squares Spreadsheet
X Y
1 2
3 3
4 4
6 5
m 0.615384615 1.346153846 b
sm 0.054392829 0.214144783 sb
R2 0.984615385 0.196116135 sy
n= 4
Mean y= 3.5
Σ(xi-mean x)2 13
*********Measured y= 2.72
k= number of replicate measurements of y= 1
Derived x= 2.2325
sx= 0.373502805
Homework Equations
I'm looking for an equation, or an explanation as to how to obtain the value.
The Attempt at a Solution
I have asterisks (********) next to the measured y value in the spreadsheet. (The value is 2.72). The only reason I know what it is in this case is because this is an example from my textbook. I have no idea where it comes from, but I need it for an equation to be able to do my lab and I don't know how to find the value.
As far as I can make sense of it, I have no means of calculating 'measured mean value of y for replicate measurements of the unknown', as there are no replicate measurements of the y values. Right?
Just in case it helps, this is for an analytical chemistry lab, but it's pertaining to statistics, so I asked it here.
Thanks.
I think that, in principle, you may have a problem that is difficult to solve exactly. You did a least-squares fit of y to x, so the statistical output is valid if the model was of the form
y = \alpha + \beta x +\epsilon, where ##\alpha, \; \beta## are unknown constants and ##\epsilon## is a mean-0 random variable with variance that does not depend on x. If, further, the distribution of ##\epsilon## is NORMAL, you can develop confidence intervals, etc. You have a series of observations
y_i = \alpha + \beta x_i + \epsilon_i, \: i = 1,2, \ldots, n, where the different ##\epsilon_i## are mutually independent and have the same distribution.
You do not know ##\alpha## and ##\beta##, but instead you estimate them as 'a' and 'b' using least-squares formulas. Assuming correctness of the form of statistical model, a and b will be unbiased estimates of the underlying parameters, and the computed total squared error ##S^2## will be related to ##\sigma^2 = \text{Var}(\epsilon)##, via standard formulas.
Standard formulas allow us to give confidence intervals on Ey(x) and on y(x) at some future-measured value of x. These formulas a a bit complicated, but can be found in many sources. However, what you seem to want to do is almost the opposite: you measure y and want to know about x. So, if the original model is valid, what you have is
y = \alpha + \beta x + \epsilon \: \Longrightarrow x = \frac{y - \alpha - \epsilon}{\beta} = \frac{y}{\beta} - \frac{\alpha}{\beta} - \frac{\epsilon}{\beta}. Now the problem you face is that the expected value of ##y/ \beta## is not ##y/b## (although it may be close, sometimes) and that the expected value of ##\alpha / \beta## is not ##a/b##. I suspect that getting exact formulas is somewhere between difficult and impossible, although, of course, one can always resort to Monte-Carlo simulation to get rough estimates.
However, I suspect you are supposed to put ##a## instead of ##\alpha## and ##b## instead of ##\beta## to get an estimate for the mean of x. Whether or not that is of any use is not at all clear.