*Bounty: 50*

*Bounty: 50*

I am looking for someone to just confirm / double-check something for me with regards to errors on measurements.

Let’s say I am trying to determine the slope of a relationship by varying one quantity and measuring another, and then I plot the graph and do a least-squares fit straight line to the data (graph on the left). Then I repeat this procedure twice more, to get the middle and right-most graphs.

Each fit routune will typically give me back a slope and the corresponding 95% confidence interval, so that I obtain $(m_1pmDelta m_1), (m_2pmDelta m_2)$ and $(m_3pmDelta m_3)$. Now I know that the underlying quantity which determines $m$ in each case is the same, so I should be able to quote a best estimate for the slope as their mean

$$

bar{m} = frac{m_1+m_2+m_3}{3}. tag{1}

$$

My question is about the appropriate way to quote the error. We know that for a function $f(x,y)$ with errors in $x$ and $y$ given by $Delta x$ and $Delta y$, respectively, the error on $f$ is given by

$$

Delta f = sqrt{ (Delta x)^2 bigg(frac{partial f}{partial x}bigg)^2 + (Delta y)^2 bigg(frac{partial f}{partial y}bigg)^2 } tag{2}

$$

So I would think I can determine the error in $bar{m}$ to be

$$

begin{align}

Delta bar{m} &= sqrt{ (Delta m_1)^2 bigg(frac{partial bar{m}}{partial m_1}bigg)^2 + (Delta m_2)^2 bigg(frac{partial bar{m}}{partial m_2}bigg)^2 + (Delta m_3)^2 bigg(frac{partial bar{m}}{partial m_3}bigg)^2} tag{3} \

&= frac{1}{3} sqrt{ (Delta m_1)^2 + (Delta m_2)^2 + (Delta m_3)^2 } tag{4}

end{align}

$$

First question, is this correct?

Second question, is it okay to propagate 95% confidence intervals in this way? Should I simply quote now the result as $bar{m} pm Delta bar{m}$ and just explain that $Delta bar{m}$ is the combined 95% confidence interval, or should I convert the 95% number from the fits into standard errors (through the factor of 1.96)?

Thanks in advance,

(I am for now assuming Gaussian errors everywhere.)

**EDIT**

It was suggested in the comments that I first implement weighting in the averaging step before worrying about the errors. This should help to give more weight to slopes which have tighter confidence intervals (and vice versa).

According to this link, the weighted version of the mean would be given by

$$

bar{m}_textrm{w} = frac{sum_i w_i m_i}{sum_iw_i}, hspace{1cm} textrm{where} hspace{0.5cm} w_i = frac{1}{sigma_i^2}tag{5}

$$

and $sigma_i$ is the variance of each slope. Therefore, in my case with the three example slopes, it should be

$$

bar{m}_textrm{w} = frac{m_1/sigma_1^2 + m_2/sigma_2^2 + m_3/sigma_3^2}{1/sigma_1^2 + 1/sigma_2^2 + 1/sigma_3^2}. tag{6}

$$

The variance on the weighted mean slope is given at the above link again by

$$

begin{align}

textrm{Var}(bar{m}_textrm{w}) &= frac{sum_iw_i^2sigma_i^2}{big( sum_iw_ibig)^2}tag{7}\

&= frac{1/sigma_1^2 + 1/sigma_2^2 + 1/sigma_3^2}{big(1/sigma_1^2 + 1/sigma_2^2 + 1/sigma_3^2big)^2}tag{8}\

&= big(1/sigma_1^2 + 1/sigma_2^2 + 1/sigma_3^2big)^{-1}.tag{9}

end{align}

$$

So now my main question remains – these are variances, so should we convert the 95% confidence intervals $Delta m_i$ returned by a fitting algorthm somehow into a variance?

Maybe for a concrete example we could imagine the following values were returned from the fitting routine:

$$

begin{align}

m_1 &= 5.5; (4.9, 6.1)rightarrow Delta m_1 = 0.6\

m_2 &= 5.5; (5.3, 5.7)rightarrow Delta m_2 = 0.2\

m_3 &= 5.2; (4.5, 5.9)rightarrow Delta m_3 = 0.7

end{align}

$$

where the values in brackets represent the 95% confidence intervals. How should the estimate of the slope be reported, including errors?