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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:

  1. Can these standard errors $SE_m$ and $SE_n$ be understood as the uncertainties in $m$ and $n$?
  2. 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?

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  • $\begingroup$ Yes to basically everything $\endgroup$ Apr 21, 2023 at 10:39
  • $\begingroup$ @naturallyInconsistent therefore, no to something? $\endgroup$ Apr 21, 2023 at 10:45
  • $\begingroup$ At least the uncertainty propagation has square roots and squarings. The correct thing to do is not to overestimate the uncertainties, but rather as most likely as possible. $\endgroup$ Apr 21, 2023 at 10:57
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    $\begingroup$ Might Cross Validated be better suited for this question? $\endgroup$
    – Kyle Kanos
    Apr 21, 2023 at 20:29
  • $\begingroup$ @Kyle Kanos I didn't know it, but yes, it could be better suited. $\endgroup$ Nov 6, 2023 at 12:56

2 Answers 2

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First, I believe you mixed up the estimates and standard errors. Here is the output of my fit:

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  17.2889    11.9553   1.446    0.198    
x            30.0440     0.9565  31.409 6.92e-08 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.29 on 6 degrees of freedom
Multiple R-squared:  0.994, Adjusted R-squared:  0.9929 
F-statistic: 986.5 on 1 and 6 DF,  p-value: 6.919e-08

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

enter image description here

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the common uncertainty propagation formula yields $\Delta T(V)=V \Delta m + m\Delta V+\Delta n$.

There is a little mistake in this formula, and some non-standard notation. The non-standard notation is that usually the standard deviation of $m$ would be represented as $\sigma_m$ rather than $\Delta m$. So, correcting the mistake and using the standard notation, this should be $$\sigma_T (V) = \sqrt{V^2 \ \sigma_m{}^2 + m^2 \ \sigma_V{}^2+\sigma_n{}^2}$$

Other than that it seems correct. So with your information listed above for $V=10$ I get $\sigma_V (10)=173.3$

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