
#1
Feb1512, 01:16 AM

P: 651

hi, just a simple question
i have a linear graph y=mx+c lets say my y values have an uncertainty of ±1 . my x values don't have uncertainties. so what will my gradient's uncertainty be? PS: can i just use the linear least square fit function from microsoft excel? do they calculate the same uncertainty?i.e is the standard deviation from this method, the same as the uncertainty that i am going to calculate as per above? thanks! 



#2
Feb1512, 04:25 AM

P: 819

Can you simply display your values graphically, along with the error bars in the Y values and then see the variation in M that will fit within the 'envelope'?
Edit: is M held to a single value throughout the range? Or is it allowed to vary with X? 



#3
Feb1512, 06:08 AM

P: 388

The uncertainty in the gradient is characterised by its standard error which Excel's LINEST function can provide using INDEX(LINEST(known_y's, known_x's, const, TRUE), 2,1). Given the standard error you can calculate confidence limits for the gradient in the same way as you presumably have for the data points in order to make the (statistically meaningless) statement that the 'y values have an uncertainty of ±1'. Note that the standard errors in m and c are in general more closely linked to the range in the x values than the errors in the y values.
To correctly interpret the regression statistics I think you should do some reading around the subject: Anscombe's quartet is IMHO a good start to understand some of the limitations of linear regression and the importance of graphical analysis. 



#4
Feb1512, 06:20 AM

P: 819

experimental uncertainty
Thanks for that link, MrAnchovy ! I have not seen that before. An unequivocal demonstration in the power of graphics & the human eyebrain.




#5
Feb1512, 07:25 AM

P: 651

wow, i didn't expect it to be this complicated? i don't think i am at your levels :(
i mean lets say for my yvalues, they are measurements of length of an object. the uncertainty is say ±1cm. so if i do a plot of length against anything(no uncertainty in x), then what will my gradient's uncertainty be? for grad = y/x so (σgrad / grad)^{2} = (σy / y)^{2} + (σx / x)^{2} ??? is it just like that? is this uncertainty propagation? my notes says its just like that? so since i don't have x uncertainties, i end up with σgrad / grad = σy / y ? but thats weird, it says every σgrad is different, because the value σy / y * grad is different for every y value.... 



#6
Feb1512, 07:39 AM

PF Gold
P: 706

(Actually, now that I look at it, I'm not sure this answers your question at all. You've just been given a liner equation, and an uncertainty... Usually linear equations are accompanied by a correlation coefficient, not an uncertainty. )
I was asking some questions along this line some months ago http://www.physicsforums.com/showthread.php?t=525091 The main thing that ought to be distinguished is systematic errors and random errors. You might not have any "random error" in your xvalues, because you are measuring the same xvalue in the same way, and getting the same result. However, there may be a systematic error in your measurement of your xvalue. A quick google search gave me this article which gives some more detail: http://www.ece.rochester.edu/courses...ncertainty.pdf I'm not entirely certain how to account for possible systematic errors... You can really only account for systematic errors you KNOW are present. 



#7
Feb1512, 08:54 AM

P: 388

Ah, I think you are talking about two different things here.
Experimental uncertainty refers to data that are subject to inaccuracy due to one or many random variables affecting the measurement of a quantity. Using certain assumptions, regression analysis can be used to determine the parameters of a 'line of best fit', including the standard errors in these parameters. This analysis in not simple and I have given some hints as to where to start. Error propogation refers to data that are imprecise due to some limitation of the method used to measure (or record) a quantity. For instance an error of ±1cm would be produced by a ruler marked at 2cm intervals if you were unable to interpolate between markings. In this case the analysis is fairly simple. The errors in y can be characterised by y = y' + ε where y is the true value of y, y' is the observed value and ε is the error (in this case, 1cm ≤ ε ≤ 1cm). Taking y = mx + c, we have y' + ε = mx + c and y' = m'x + c where y' are the observed values of y and m' is the observed gradient. These give m' = m + ε / x. So the error in the gradient is 1cm divided by x. Note that the formula σy / y implies that the precision is proportional to y rather than constant; a similar analysis can be performed. For imprecise measurements the simplest way to determine the minimum and maximum gradients that fit the data is as gmax137 says to draw a graph with error bars and plot the steepest/shallowest lines that pass through each bar. 



#8
Feb1512, 10:03 PM

P: 651

but if the grad error is m' = m + ε / x , then wouldn't it mean that for every x, i have a new m'?
lets say i am given x values as 10,20,30,40,50 then wouldn't my m' keep changing for different x? issn't that weird? i didn't measure x by the way, those values were given. but now that you guys talk about error propagation vs uncertainty, i realise that what i am talking about issn't exactly error propgation? i think i am talking about confidence limits? it's not about systematic or maybe random errors. i think it's more about human error. lets say i want to determine the angle of a polarizer which produces the brightest emitted light. but my eyes tells me the brightest light is over a range of angles. i can't pinpoint the angle that is.... so over a range of angles, say 1 degree. so that is my uncertainty in y. so with x fixed/given, how will the gradient's error/uncertainty work out? for that matter, is this called uncertainty or error propagation? or is it a human judgement error? 



#9
Feb1612, 07:16 AM

PF Gold
P: 706

Let's say you have (10,y(10)), (20,y(20)), (30,y(30)), (40,y(40)) You know your xvalues with perfect certainty, but your yvalues are only known to an uncertainty of +/ 1. So you could calculate, one slope by adding 1 to your yvalue on the right, and subtracting 1 from your yvalue to the left. [tex]\frac{\Delta y}{\Delta x}=\frac{(y(40)+1)(y(10)1)}{4010}[/tex] Then calculate another slope, similarly, by subtracting 1 from your yvalue on the right, and adding 1 to your yvalue on the left. Precision uncertainty: If you have a meter marked in millimeters, you can make a guess down to the nearest 10th of a millimeter, but you should make a note that your scale is not as precise as that. Experimental uncertainty: If you perform several trials measuring the same quantity, getting slightly different values each time, you can use statistics to estimate the uncertainty. Population uncertainty: If you perform several trials measuring different quantities which you expect to be near each other, but are not necessarily exactly the same. No, I don't think it is human judgement error, because the light beam really did not land at a point, but in a spread. 



#10
Feb1612, 09:17 AM

P: 388

But it's probably not very relevant either  this is probably more helpful: Estimate the true slope m by drawing a line through two points (x_{1}, y'_{1}) and (x_{2}, y'_{2}). We have m' = (y'_{2}  y'_{1}) / (x_{2}  x_{1}). Inserting the precision errors, we get m' = (y_{2} + ε_{2}  y_{1}  ε_{1}) / (x_{2 } x_{1}) = (y_{2}  y_{1}) / (x_{2}  x_{1}) + (ε_{2} ε_{1}) / (x_{2}  x_{1}) = m + 2ε / (x_{2}  x_{1}). So with your example and using the two extremes of x, the precision error in the gradient m'  m = 2 x 1cm / (50xunit  10xunit) = 0.05cm.xunit^{1} Fitting a straight line to a data set analytically is called linear regression and the most common form is least squares fitting. If you scroll to the bottom of that page you will see expressions for the standard errors (equivalent to the standard deviation from which you can derive confidence limits) in the intercept a and the gradient b. I don't know of any simpler way to show this I am afraid. 



#11
Feb1712, 11:19 PM

P: 651

wow... ok thanks guys!



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