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How do I calculate the error of my data?

  1. Jan 13, 2016 #1
    Hello.


    I have 3 repitions of an experiment. The data consists of mass flow rates vs. time.
    What I want to do is to calculate the mass vs. time, i.e. integrating my data.

    I consider the resolution for each measurement to be exactly 3 minutes. To calculate the area I use the trapetzoid
    rule, i.e. [itex](y(i)+y(i+1))/2*\Delta X= (y(i)+y(i+1))/2*3[/itex]

    Now I want calculate the error. I know using the trapetzoid rule introduce errors, but since I
    have no way of knowing how this affects my result I don't care about that particular error. Rather I want to know
    the error between my 3 repititions.
    My data look like this:
    X # 1 # 2 # 3
    0 y11 y21 y31
    3 y12 y22 y32
    6 y13 y23 y33
    . . . .
    . . . .
    . . . .
    n y1n y2n y3n
    where y is mass flow rates and x is time.

    This is my approach:
    Calculate the mean mass flow rate, i.e
    [itex]yavg(1)=(y11+y21+y31)/3[/itex]
    [itex]yavg(2)=(y12+y22+y32)/3[/itex]
    [itex]yavg(3)=(y13+y23+y33)/3[/itex]
    ...
    [itex]yavg(n)=(y1n+y2n+y3n)/3[/itex]

    and also the standard deviation

    [itex]ystd(1)=std(y11,y21,y31)[/itex]
    [itex]ystd(2)=std(y11,y21,y31)[/itex]
    [itex]ystd(3)=std(y11,y21,y31)[/itex]
    ...
    [itex]ystd(n)=std(y11,y21,y31)[/itex]

    Then use the trapetzoid rule:
    [itex]Area(1)=(yavg(1)+yavg(2))/2*3[/itex]
    [itex]Area(2)=(yavg(2)+yavg(3))/2*3[/itex]
    ...
    [itex]Area(n-1)=(yavg(n-1)+yavg(n))/2*3[/itex]

    But how do I calculate the error? This I what I think:
    [itex]Error(1)=sqrt((ystd(1)^2+ystd(2)^2)/2*3))[/itex]
    [itex]Error(2)=sqrt((ystd(2)^2+ystd(3)^2)/2*3))[/itex]

    or should it be like this:
    [itex]Error(1)=sqrt(ystd(1)^2+ystd(2)^2)[/itex] ??

    [itex]Area(1)+-Error(1)[/itex]
    [itex]Area(2)+-Error(2)[/itex]

    .. etc

    And suppose I want to calculate the total mass, i.e Area(1)+Area(2)...+Area(n-1), then what will the error be?

    Thanks
     
  2. jcsd
  3. Jan 13, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Do you have a good reason to expect the same time dependence of mass flow for all three repetitions?

    You have to be careful to account for the uncertainty exactly once: the uncertainty on area 1 and on area 2 are correlated.
    Luckily, your trapezoid rule can be simplified: total area is 3 minutes * (1/2 yavg(1) + yavg(2) + yavg(3) + ... + 1/2 yavg(n)). The uncertainties on those averages now can* be independent, and you can add them in quadrature.

    *something that would need more investigation: what leads to different measurements between the repetitions?
     
  4. Jan 13, 2016 #3
    Hi, thank you for your answer.

    I am using an instrument (gas chromatograph) to measure concentrations for different chemicals which I then relate to their mass. The chemicals are produced in the same way for all three experiments so I expect the result to be "the same".

    Since I am multiplicating and dividing my areas with constants, should I do the same to the uncertainties also? I.e std=sqrt(3*(var_1*0.5+var_2+...+var_n*0.5)), where var is the variance.

    Random error would be the reason for having different values, although I suspect sample inhomogeneity probably has a larger impact. If this is the case I guess the standard deviation is more a measurement of how good I sample rather than the uncertainty of the measurements :) But I assume that I only have random errors since I normalize my results.
     
  5. Jan 13, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    Sure. But do it outside the square root. Standard deviations scale linearly if you just multiply the uncertain value with some constant, variances scale with the square of that value. This also means your *0.5 should be *0.25.
    That is a possible issue.
    Normalize to what? That could be relevant as well.
     
  6. Jan 14, 2016 #5
    Actually I realized that the data is not normalized, nevermind what I wrote.

    Thank you.

    So if I understand you correctly, I should calculate the error like this:

    ## 3^2*(s1^2*0.25^2+s2^2...+sn^2*0.25^2)=total variance##

    One more question.

    The instrument detect several chemicals. So in addition to calculating the mass for each chemical, I also need to add those masses together to get the total mass. I assume that I just repeat the process for each chemical, and in the end, I add all the variances and square root them to get the total standard deviation.

    E.g.

    Chemical A:
    Average mass of ##A = 3*(A_1*0.5+A_2+...+A_n*0.5)##
    Variance of ##A = 3^2*(S_{A1}^2*0.5^2+S_{A2}^2+...+S_{An}^2*0.5^2)##

    Chemical B:
    Average mass of ##B = 3*(B_1*0.5+B_2+...+B_n*0.5)##
    Variance of ##B = 3^2*(S_{B1}^2*0.5^2+S_{B2}^2+...+S_{Bn}^2*0.5^2)##
    .
    .
    .
    Chemical K:
    Average mass of ##K = 3*(K_1*0.5+K_2+...+K_n*0.5)##
    Variance of ##K = 3^2*(S_{K1}^2*0.5^2+S_{K2}^2+...+S_{Kn}^2*0.5^2)##


    Total mass:
    ##{\text{Average mass of A}}+{\text{Average mass of B}}+...+{\text{Average mass of K}}##
    Total error:
    ##sqrt({\text{Variance of A}}+{\text{Variance of B}}+...+{\text{Variance of K}})##
     
  7. Jan 14, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
    If all those measurements are independent, that works.
     
  8. Jan 14, 2016 #7

    ChrisVer

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    Gold Member

    Watch that your variance now is the same as it would be by error propagation (no correlation between your [itex]x_i[/itex]):
    You have the variable [itex]f(x_1,x_2,...,x_n) = a (bx_1 + x_2 + ... + bx_n)[/itex]
    and its deviation will be [itex]\sigma_f = \sqrt{\Big(\frac{\partial f}{\partial x_1} \sigma_{x1}\Big)^2+ \Big(\frac{\partial f}{\partial x_2 } \sigma_{x2} \Big)^2 + ... + \Big(\frac{\partial f}{\partial x_n} \sigma_{xn} \Big)^2}[/itex]
    or [itex] \sigma_f^2 \equiv Var(f) = a^2 b^2 Var(x_1) + a^2 Var(x_2) + ... +a^2 b^2 Var(x_n) = a^2 \big( b^2 Var(x_1) + Var(x_2) +... + b^2 Var(x_n) \big) [/itex]
     
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