Determining error in a linear gradient/slope

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Discussion Overview

The discussion revolves around determining the error in the gradient of a linear fit from experimental data. Participants explore concepts related to error analysis, specifically focusing on standard deviation and the interpretation of measurement uncertainties in the context of a lab report.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks guidance on determining the error in the gradient of their graph, noting the maximum and minimum gradients derived from 'worst' lines of fit.
  • Another participant questions the meaning of "error" and the interpretation of notation like "9.35 +/- 2.61," suggesting it could represent a mean value and standard deviation or known limits.
  • A clarification is provided that "worst line of fit" refers to the maximum and minimum gradients achievable within the bounds of measurement error.
  • Participants discuss the absence of x values in the data set and inquire whether multiple measurements exist for the y values.
  • One participant explains their calculation of error based on the accuracy of the measuring device, using standard deviation formulas.
  • Another participant expresses confusion regarding the calculations and asks if the accuracy is expressed as a percentage error.
  • A response indicates that accuracy can be represented as either a fixed error in meters or as a percentage error.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of error and the calculations involved. There is no consensus on the correct approach to determining the error in the gradient, and the discussion remains unresolved.

Contextual Notes

Participants have not fully clarified the assumptions behind their calculations, and there is uncertainty regarding the definitions of error and measurement accuracy. The discussion also lacks specific details about the x values corresponding to the y data points.

Harry Smith
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Hi, so I'm currently writing a lab report and need to determine the error in the value of my graphs gradient, m. I have found the maximum and minimum gradient from the 'worst' lines of fit but am lost where to go from there. My data analysis/statistics lecture notes are no help on this topic, neither is Google. It seems like it should be an easy topic!

m = 11.20 m2
mmin = 9.79 m2
mmax = 12.06 m2

and I've found the mean gradient and its deviation (though this is little use)

mavg = (11.20 ± 1.14) m2

The data in y is
11.39 ± 1.67
21.39 ± 2.29
33.06 ± 2.85
47.27 ± 3.40
56.25 ± 3.71
66.02 ± 4.02
(all in m2)

Thanks in advance!
 
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Harry Smith said:
Hi, so I'm currently writing a lab report and need to determine the error in the value of my graphs gradient, m.


What you mean by "error"? One sees quantities in reports given in notation like "9.35 +/- 2.61", but unfortunately, I know of no standard interpretation for this notation. Such notation often means that the measurements have a mean value of 9.35 and a standard deviation of 2.61. However, sometimes the notation means that the measured quantity is known (with certainty) to be between the stated limits.

I have found the maximum and minimum gradient from the 'worst' lines of fit but am lost where to go from there.

I'm lost as to how you would define a "worst" line of fit. However, your remark at least reveals that you are fitting linear graphs to the data.
The data in y is
11.39 ± 1.67
21.39 ± 2.29
33.06 ± 2.85
47.27 ± 3.40
56.25 ± 3.71
66.02 ± 4.02
(all in m2)

Why didn't you state the x values?

Does your data consist of many individual measurements of (x,y)? For example, assuming the y value 21.39 is for x = 1, do you have many different measurements like (1,22.28), (1,18.33)... ? Or do you have just one measurement (1,21.39) and are getting the +/- 2.29 from some manufacturers specification for the measuring device?
 
By error I mean standard deviation. Apologies.

By worst line of fit I mean the maximum mmax or minimum mmin possible gradient achievable from the set of data within the bounds of the error.

The x values are 1, 2, 3, 4, 5, 6 with no error so it needn't be considered.

The ± 2.29 originates from the datum it corresponds with. The accuracy of the measuring device was to the nearest 0.25 m. The datum were then averaged, with error Δr of
[itex]\Delta r = \sqrt{2 \cdot (0.25)^2}[/itex]​
This was then squared, with error Δr2 of
[itex]\Delta r ^2 = r ^2 \cdot \sqrt{2 (\frac{\Delta r}{r}) ^2}[/itex]​
where r2 = y.

Hope that cleared it up.
 
Harry Smith said:
The accuracy of the measuring device was to the nearest 0.25 m. The datum were then averaged, with error Δr of
[itex]\Delta r = \sqrt{2 \cdot (0.25)^2}[/itex]​
This was then squared, with error Δr2 of
[itex]\Delta r ^2 = r ^2 \cdot \sqrt{2 (\frac{\Delta r}{r}) ^2}[/itex]​
where r2 = y.

I don't understand those calculations. Is the accuracy of measuring device given as a "percentage error" instead of an error in meters?
 
Stephen Tashi said:
I don't understand those calculations. Is the accuracy of measuring device given as a "percentage error" instead of an error in meters?
It can be given as either. Percentage error just represents the percentage of the deviation from the value.