# Uncertainty for repeated measurements

I am having some trouble calculating the percent uncertainty of obtained cell potential. My cell potentials were 1.788, 1.787, and 1.792 volts with the uncertainty of voltmeter $$\pm$$ 0.001 volts. Here was my attempt:

The average cell potential: \begin{align*} \frac{1.788+1.787+1.792}{3}=1.789 \end{align*} Percent uncertainty when measurements are repeated was found by the following formula: \begin{align*} \text{% E_{cell} uncertainty}=\frac{\frac{\text{range}}{2}}{\text{mean}}\times 100\%. \end{align*} Therefore: \begin{align*} \text{% E_{cell} uncertainty}=\frac{\frac{1.792-1.787}{2}}{1.789}\times 100\%\approx\text{0.1\% (1 s.f.)}. \end{align*} But, my calculation doesn't involve voltmeter's uncertainty of $$\pm$$ 0.001 V. What should I do? If I were to use standard error, then,

Standard deviation: \begin{align} \sqrt{(1.788-1.789)^2+(1.787-1.789)^2+(1.792-1.789)^2}=0.002 \end{align}

Standard error of average measurement: \begin{align} \frac{0.002}{\sqrt{3}}=0.0012 \end{align}

Total uncertainty: \begin{align} 0.0012 + 0.001 = 0.0022 \end{align}

Total percentage uncertainty: \begin{align} \frac{0.0022}{1.789}\times100=0.1\% \end{align}

• I'm sorry, I didn't get it. May I ask you to 'answer the question' instead of 'add a comment' for a more detailed explanation?
– qmoi
Jun 4, 2023 at 11:19
• 1. where did sqrt(12) come from? 2. what is meant by s^2_1meas,Corr = s^2_1meas + s^_device? I am not familiar with this formatting.
– qmoi
Jun 4, 2023 at 12:20
• As you have few measurements you need to use Student's t test because so few samples may not represent a normal distribution. With $t=n-1=2$, calculate $\pm t\;s/\sqrt(3)$. At $90$% confidence $t=2.9$ and 4.3 at $95$%. If this is bigger than your instrumental error you can ignore the latter. Jun 4, 2023 at 14:20
• @porphyrin, thanks for the comment! Can you explain it to me some more?
– qmoi
Jun 4, 2023 at 16:13
• @Poutnik, thanks for the link! I'll check it out!
– qmoi
Jun 4, 2023 at 16:14

Imagine you measured 10 values:

• 6 times 1.790 V
• 2 times 1.789 V
• 2 times 1.791 V

Estimation of standard deviation of 1 measurement is $$s_\mathrm{meas} = \sqrt{\frac {\sum_{i=1}^{n} {(x_i - \bar{x})^2}}{n-1}} = \\ \sqrt {\frac{2 \cdot (1.789-1.790)^2 + 2 \cdot (1.791-1.790)^2 }{ 10-1}} \pu{V} = \\ \sqrt {\frac{\pu{4e-6}}{ 9}} \pu{V} =\pu{0.0006667 V} \approx \pu{0.0007} V$$

With the voltmeter resolution $$\ce{0.001 V}$$, the related estimation of the standard deviation for the uniform distribution is $$s_\mathrm{resol} = \frac{\pu{0.001 V}}{\sqrt{12}}=\pu{0.0002887} V \approx \pu{0.0003 V}$$

The overall corrected estimation of the measurement standard deviations is:

$$s_\mathrm{meas, corr} = \sqrt{s^2_\mathrm{meas} + s^2_\mathrm{resol}}=\sqrt{0.0007^2 + 0.0003^2} V \approx \pu{0.0008 V}$$

For you case, you need first to do the Student test of values that are off, as porphyrin suggested. Note that 3 values is not exactly many to work with.

For a small number of measurements,it is often recommended to estimate the standard deviation from the data range: $$s = R \cdot \frac{1}{d_n}$$.

|  n | 1/d_n |
|----+-------|
|  2 | 0.886 |
|  3 | 0.591 |
|  4 | 0.486 |
|  5 | 0.430 |
|  6 | 0.395 |
|  7 | 0.370 |
|  8 | 0.351 |
|  9 | 0.337 |
| 10 | 0.325 |


For the measurement data $$\pu{1.788 V}$$, $$\pu{1.787 V}$$, and $$\pu{1.792 V}$$,

$$s(n=3)_\mathrm{meas} \approx (\pu{1.792 V} - \pu{1.787 V} ) \cdot 0.591 = \pu{0.002955 V} \approx \ce{0.003 V}$$

$$s_\mathrm{meas, corr} = \sqrt{s^2_\mathrm{meas} + s^2_\mathrm{resol}}=\sqrt{0.003^2 + 0.0003^2} V \approx \pu{0.003 V}$$

Note that the reason for keeping $$s$$ with just the single valid digit is this: (Standard deviation of standard deviation)

$$\frac{\mathrm{SD}(s)}{s} = \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot \sqrt{\frac{n-1}{2} - \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } \approx \frac{1}{\sqrt{2(n-1)}}$$

For $$n=51$$, $$\frac{\mathrm{SD}(s)}{s} \approx 0.1$$

what well illustrates why uncertainties with two valid digits are rare and must be well justified and supported by much of data.

• It helped me so much!! Thank you!!
– qmoi
Jun 10, 2023 at 6:49
• Consider then accepting the answer, unless you need further clarification. Jun 10, 2023 at 6:51