I have bought a resistor that works as a heater when a voltage is applied to it, and the seller provided me these $ \left(V (\mathrm{V}), T (\mathrm{°C})\right)$ points: $(6.20$, $200)$, $(7.75$, $250)$, $(9.20$, $300)$, $(10.70$, $350)$, $(13.20$, $400)$, $(14.70$, $450)$, $(16.20$, $500)$, $(17.30$, $550)$. Assuming a linear relationship $T(V)=mV+n$ between voltage and temperature, I performed a linear regression using Excel's LINEST() function, and I got
$$m=30.0439740426118 \mathrm{ °C/V}, \quad SE_m=17.2889340551528 \mathrm{ °C/V}$$
$$n=0.956540771777197 \mathrm{ °C},\quad SE_n=11.9552888753203 \mathrm{ °C}$$
where $SE$ stands for standard error. Here I have a couple of questions:
- Can these standard errors $SE_m$ and $SE_n$ be understood as the uncertainties in $m$ and $n$?
- How could one express the uncertainty of the $T(V)$ obtained from the regression?
Assuming the answer to the first question is positive, my attempt for the second one would be this. If $T(V)=mV+n$, the common uncertainty propagation formula yields $\Delta T(V)=V \Delta m + m\Delta V+\Delta n$. Taking $\Delta V = 0.05 \mathrm{V}$ (experimental error of the voltage, according to the data), the independent variable with its uncertainty will be given by
$$T \pm \Delta T(V) = (m\pm \Delta m)V+m(V \pm \Delta V)\pm \Delta c =\\ mV+n \pm (V\Delta m + m\Delta V + \Delta c) =\\ 30V+20 \pm (V+11.5) $$
Would this be an adequate way to express the uncertainty in the dependent variable of the linear regression?
Best Answer
First, I believe you mixed up the estimates and standard errors. Here is the output of my fit:
Regarding your second question I believe you are interested in the uncertainty of the average temperature at $10V$. Dale's formula assumes that the fit parameters are independent, which is not really true, see here. However, using the above coeffs and your $\sigma_V=0.05V$ I obtain $\sigma_T(10 V) = \sqrt{ (10*0.9565)^2 + (30.0440*0.05)^2 + 11.9553^2} \mathrm{ °C} \approx 15.4 \mathrm{ °C}$. Using the more exact formula yields $\sigma_T(10V) = 4.1 \mathrm{ °C}$. That the uncertainty in the center of the fit is lower than at the edges is also visible in the following plot