# Determining error in a linear gradient/slope

1. Feb 17, 2014

### 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
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)

Last edited: Feb 17, 2014
2. Feb 17, 2014

### Stephen Tashi

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'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.

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?

3. Feb 17, 2014

### Harry Smith

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
$\Delta r = \sqrt{2 \cdot (0.25)^2}$​
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.

4. Feb 17, 2014

### Stephen Tashi

I don't understand those calculations. Is the accuracy of measuring device given as a "percentage error" instead of an error in meters?

5. Feb 20, 2014

### Harry Smith

It can be given as either. Percentage error just represents the percentage of the deviation from the value.