Determining error in a linear gradient/slope

[Harry Smith]

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
m_min = 9.79 m²
m_max = 12.06 m²

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

m_avg = (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!

 

[Stephen Tashi]

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?

 

[Harry Smith]

By error I mean standard deviation. Apologies.

By worst line of fit I mean the maximum m_max or minimum m_min 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

\Delta r = \sqrt{2 \cdot (0.25)^2}​

This was then squared, with error Δr² of

\Delta r ^2 = r ^2 \cdot \sqrt{2 (\frac{\Delta r}{r}) ^2}​

where r² = y.

Hope that cleared it up.

 

[Stephen Tashi]

The accuracy of the measuring device was to the nearest 0.25 m. The datum were then averaged, with error Δr of

\Delta r = \sqrt{2 \cdot (0.25)^2}​

This was then squared, with error Δr² of

\Delta r ^2 = r ^2 \cdot \sqrt{2 (\frac{\Delta r}{r}) ^2}​

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?

 

[Harry Smith]

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.