Finding uncertainty in the slope/intercept for a non-linear least squares fit I have the following function:
$$M = a(\log_{10}W-2.5)+b$$
I also have a set of data with actual measured values of $W$ and $M$ (each have individual $\pm$ errors). Here's a small sampling of the data:
W(x)    M(y)        M_model
245     -19.59      -19.05  
155     -16.64      -17.76
314     -20.26      -19.75
351     -20.78      -20.07
192     -17.96      -18.37
...     ...         ...

Using $W$ and $M$ values from the dataset, I did a non-linear least squares fit (using Matlab's optimtool)and a $y = ax + b$ form where $x = \log_{10}(W)-2.5$. Found values of $a$ and $b$, and then calculated $M_{model}$ in the table above.
$$a = -6.359$$
$$b = -19.83$$
How do I find what uncertainty there might be in $a$ (slope) and $b$ (intercept)? I need the function in the format:
$$M = (-6.359\pm c)(\log_{10}W-2.5)+(-19.83\pm d)$$
Update
I have tried using LINEST() in Excel which provides me with the errors, but it also provides me with slightly different $a$ and $b$ values too. Is there a more robust way of calculating these uncertainties?
Update
I have tried Michael's solution below, and that gives me one figure, substituting his $x$ for my $M$. I think that gives me the $\pm slope$, but I am not sure if that is correct. Also not sure how to get $\pm intercept$ as I only have one set of residuals (for $M$).
 A: Supposing (unobservable) errors (as opposed to observable residuals) are independent and identitcally distributed, and normally distributed with expectation $0$ and variance $\sigma^2$, then the least-squares estimate $\hat a$ of the slope $a$ satisfies
$$
\hat a \sim N\left(a, \frac{\sigma^2}{\sum_i (x_i-\bar x)^2} \right)
$$
where $\bar x=(x_1+\cdots+x_n)/n$.
Now let $\hat\sigma^2$ be the estimate of $\sigma^2$ that comes from dividing the sum of squares of observed residuals by the degrees of freedom, in this case $n-2$.  Then
$$
(n-2) \frac{\hat\sigma^2}{\sigma^2}\sim \chi^2_{n-2},
$$
and (believe it or not!) this random variable is independent of $\hat a$.
Therefore
$$
t = \frac{\hat a - a}{\hat\sigma/\sqrt{\sum_i(x_i-\bar x)^2}}
$$
has a $t$-distribution with $n-2$ degrees of freedom.
So
$$
\hat a \pm A \frac{\hat\sigma}{\sqrt{\sum_i(x_i-\bar x)^2}}
$$
are the endpoints of a confidence interval for $a$.  The number $A$ you get from one of the standard tables for the $t$ distribution, and depends on what confidence level you want.
Two thoughts on the updated version of the question:


*

*The estimates of slope and intercept are correlated with each other (except when the mean of the $x$-values is $0$.  If the line is $y=ax+b$, then we'd have a confidence region for the pair $(a,b)$ that would be an ellipse whose axes are not parallel to the $x$- and $y$-axes.  So it would be somewhat misleading to express them separately as $\hat a\pm\text{something}$ and $\hat b\pm\text{something}$.  However, if the line is expressed point-slope form, as $y-\bar y=a(x-\bar x)$, where $\bar y$ and $bar x$ are the mean values, then $\bar y$ would be uncorrelated with $\hat a$.

*All this assumes the $y$s are random variables and the $x$s are fixed.  That can make some sense if all results are regarded as conditional expected values, given the $x$ values.  It's pretty conventional to do things that way, but possibly in some situations it's not appropriate.

