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A measurement is performed multiple times for the same object, the student will
\nAdvice is provided including on performing the calculations in Python or spreedsheets together with further reading.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Using a metre rule, you performed serveral repeat measurements of your friend's height :
\nMeasurement | \nHeight [m] | \n
1 | \n$\\var{data[0]}$ | \n
2 | \n$\\var{data[1]}$ | \n
3 | \n$\\var{data[2]}$ | \n
Measurement | \nHeight [m] | \n
1 | \n$\\var{data[0]}$ | \n
2 | \n$\\var{data[1]}$ | \n
3 | \n$\\var{data[2]}$ | \n
4 | \n$\\var{data[3]}$ | \n
5 | \n$\\var{data[4]}$ | \n
When studying a parameter of interest, scientists record many repeat measurements. This is to help account for the imperfect precision and accuracy of measurements.
The best estimate of a parameter, $x$, is the mean of \ud835\udc41 repeat measurements:
\\[ \\bar{x} = \\frac{1}{N}(x_1 + x_2 +\\cdots+ x_N) = \\frac{1}{N}\\sum_{i=1}^N x_i \\]
In this example of measuring the height of your friend, the values of $x_i$ are:
\n$x_1$ = | \n$\\var{data[0]}$ | \n
$x_2$ = | \n$\\var{data[1]}$ | \n
$x_3$ = | \n$\\var{data[2]}$ | \n
Using Python, this can be calculated via:
\n\n\nimport numpy as np
\nx = np.array([$\\var{data[0]}$, $\\var{data[1]}$, $\\var{data[2]}$])
\nmean = np.mean(x)
\n
Or if using a spreadsheet:
\n\n\n=average($\\var{data[0]}$, $\\var{data[1]}$, $\\var{data[2]}$)
\n
To determine how precise the mean is we need to calculate the average deviation of our measurements $x_i$ with the mean value $\\bar{x}$.
\nIn this case, we have repeated measurements of the same object (height of your friend) so we can use the population standard deviation:
\n\\[ \\sigma = \\sqrt{\\frac{1}{N} \\sum_{i=1}^N (x_i - \\overline{x})^2} \\]
\nwhich can be calculated using Python via:
\n\n\nsd = np.std(x)
\n
or via a spreadsheet approach:
\n\n\n=stdevp($\\var{data[0]}$, $\\var{data[1]}$, $\\var{data[2]}$)
\n
We can use this standard deviation to determine the error on the mean value:
\n\\[ \\alpha = \\frac{\\sigma}{\\sqrt{N}} \\]
\nwhich indicates that the more repeat measurements we perform, the smaller the error on the mean becomes.
\nTypically, the error on the mean is only precise to 1 significant figure. It requires 1000's of repeat measurements for an additional significant figure of precision. Hence we round $\\alpha$ to 1 significant figure and then round the mean to the same number of decimal places to produce the final result:
\n\\[ \\bar{x} \\pm \\alpha \\]
\nSee this useful book for more details:
\nHughes, I., Hase, T. , \"Measurements and Their Uncertainties\", Oxford University Press (2010)
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\nNow write the final result and error to the correct precision:
\n[[2]] ± [[3]] m
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