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

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To calculate values from the equation y = (0.68±5.04) + (4.62±0.14)x, the ± values represent the uncertainty in the measurements. For each x value, determine the corresponding y values by first calculating the mean and then applying the error propagation method to account for uncertainties. The error bars will increase with x due to the slope (b = 4.62±0.14), and it is essential to include these uncertainties in the final results. After calculating y, report the values with appropriate significant figures, considering the precision of the input parameters. This approach ensures accurate representation of the data and its associated uncertainties.
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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
 
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Those + and - signs indicate range deviations from a straight line in a x versus y graph of experimental values.
 
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.
 
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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
 
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.
 
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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
 
The error bars increase with x, due to b = 4.62 ±0.14
 
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
 
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
 
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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.
 
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