How to deal with these +- values in an equation?

[Martyn Arthur]

TL;DR

How to create chart from an equation with +- values

Hi, I am required to produce a calculation from the following equation which has +- values, y = (0.68+-5.04) +(4.62+-0.14)x.
How do I take into account the +-data to arrive at a final set of figures.
I am required to quantify possible errors, how do I quantify the +-
Am I correct in assuming that the x being at the end of the equation has no alternative meaning and simply requires the second set of figures to be multiplied by x.
Thank you
Martyn

 

[Lnewqban]

Those + and - signs indicate range deviations from a straight line in a x versus y graph of experimental values.

 

[Baluncore]

y = (0.68+-5.04) +(4.62+-0.14)x.
y = a + b * x
a = 0.68 ±5.04 ; you need to check this suspicious value, the error bar swamps the data.
b = 4.62 ±0.14 ; this must be converted to a proportion prior to multiplication.

 

[Martyn Arthur]

Thank you; so do I create the solutions to the equation disregarding them (the forme +- is relatively very large, plot the graph, but then how do I fit these errors into the final outcome?
Seeing the last post I have checked and that is the equation which is a shown, it derives from an Open University Activity "A study by Kearsley et al. (2006) investigated the relationship between particle diameter and crater diameter"
Thanks
martyn

 

[Baluncore]

For each value of x there will be a maximum and a minimum value of y.
Plot the mean and the error bars that give an upper and a lower bound.

 

[Martyn Arthur]

Again thanks, so do I calculate the results of the equation arriving at values for y disregarding the +- such that they then form the same error bars for y for each outcome of the equation calculation?
If so I think I have got it!
Ongoing thanks to you guys.
Martyn

 

[Baluncore]

The error bars increase with x, due to b = 4.62 ±0.14

 

[Martyn Arthur]

Sorry not having a good day, we haven't done a lot on error bars, I have researched the text.
Am I correct that I calculate the values for y disregarding the indicated error figures so that I have a final set of values to plot.
Do I diseragrd the first potential error as it is not asociated with the x element?
If so do I then calculate a further two sets of values for each calculation including the +- values which the are included as the values for the error bars?
So this a table with three columns, the primary value and a top and bottom error value?
Ongoing thanks for your patience
Martyn

 

[Baluncore]

Given x = 1, 2 .
First multiply by b.
b = 4.62 ±0.14 ; min = 4.48 ; mean = 4.62 ; max = 4.76
x = 1 ; min = 4.48 ; mean = 4.62 ; max = 4.76
x = 2 ; min = 2*4.48=8.96 ; mean = 2*4.62=9.24 ; max = 2*4.76=9.52
Finally add a.
a = 0.68 ±5.04 ; min = -4.36 ; mean = +0.68 ; max = +5.72
x = 1 ; min = 4.48-4.36=0.12 ; mean = 4.62+0.68=5.3 ; max = 4.76+5.72=10.48
x = 2 ; min = 8.96-4.36=4.6 ; mean = 9.24+0.68=9.92 ; max = 9.52+5.72=15.24

 

[kuruman]

When you say "I have to produce a calculation" I assume it to mean that you have a set of values for the independent variable $x$ and wish to find corresponding values for $y$ and the uncertainty in your calculated value. If that's the case, then I would
(a) find the value of $y$ for the given value of $x$
(b) find the uncertainty in $y$ using error propagation

Example
Say you have $y=bx+a$ and are given that $b=3.2\pm 0.3$ and $a=2.8\pm 0.5$. You want to calculate $y$ for $x=5.6$.
Step (a): $y=3.2\times5.6+2.8=17.92+2.8=20.72.$
Step (b): The uncertainty in $bx$ due to the uncertainty in $b$ is $$\delta y_{bx}=(bx)\frac{\delta b}{b} =x{\delta b}=5.6\times 0.3=1.68.$$ To this you add the uncertainty in the intercept and get the uncertainty in $y$: $\delta y=1.68+0.5=2.18$. You might be tempted to report $y=20.72\pm2.18$, but such precision is nonsensical. The reported value should be rounded to two significant figures which, in this case, is the accuracy of the input parameters. Thus, you report $y=21\pm 2.$ The uncertainty usually has one but should have no more than two significant figures.