# Calculating error in fit parameter

1. Jan 19, 2010

### NanakiXIII

I'm working on an experiment and trying to do some data analysis which seems to me should be fairly trivial, but I can't for the life of me figure it out.

I want to know the value of some property $$A$$ and to do so I'm measuring data points $$(x,y)$$, which are related to the value I want by

$$A = \frac{y}{x}$$.

So what I could do, is just calculate all my $$\frac{y_i}{x_i}$$ and find the mean and standard deviation and whatever I want, but I remember once being told that it's a good idea to do the following. I plotted my $$(x,y)$$ as datapoints, then fitted with $$y = A x$$.

I don't remember why this was a good idea (or, why it was a better idea than just averaging), but it makes for nice graphs, anyway. The fitting went fine, but what I'm stuck on is how to calculate the error in $$A$$. I've never been very interested in data analysis, and it's all a little hazy. Is what I'm doing possible and useful? How do I calculate an error in my parameter? I'm sure it's something to do with the sum of the squares of the residuals, but there has to be more to it than that, because that doesn't even have the right units.

If anyone could help untangle my messy knowledge on this subject, I'd be very grateful.

Postscript: I'm not using data analysis software. I'm pretty sure Origin would just do this for me, but I'm just using Mathematica to plot and fit. Since this problem is pretty simple, I want to try to do it by hand, rather than having Origin calculate these things for me without knowing what it actually does.

2. Jan 19, 2010

### rcgldr

A common method for calculating error value is to sum up the squares of the differences between actual and fitted points. The sum of the absolute values of the differences could also be used, but it's not a common practice. The main purpose is to compare the total error between two different fit algorithms.

3. Jan 20, 2010

### NanakiXIII

I'm not interested in comparing fit algorithms, I just want an error of the type $$A = a \pm b$$. Just averaging the squares of the residuals would give me something with the units of $$y^2$$, and a much too big number (since this number would depend on where in the range of the function your datapoints are).

4. Jan 20, 2010

### NanakiXIII

I've worked it out a little more. If I make a least squares fit, what I end up with for $$A$$ is

$$\frac{\partial}{\partial A} \sum_{i} (y_i - A x_i)^2 = -2 \sum_{i} x_i (y_i - A x_i) = 0 \longrightarrow \sum_{i} y_i - \sum_{i} A x_i = 0 \longrightarrow A = \frac{\sum_{i} y_i}{\sum_{i} x_i} = \frac{\bar{y}}{\bar{x}},$$

where the bars indicate the mean. If I just calculate the mean of $$A$$, I get

$$\bar{A} = \frac{1}{N} \sum_{i} \frac{y_i}{x_i} = \texnormal{mean}(\frac{y}{x}).$$

These aren't the same thing, though numerically they're very similar. Which is a better method?