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Error Analysis for Lab

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Using 5 different ammeters, you get the following data (all measured in Amps):
    [itex]I_{A} = 128 ± 2[/itex]
    [itex]I_{B} = 121 ± 1[/itex]
    [itex]I_{C} = 114 ± 8[/itex]
    [itex]I_{D} = 120 ± 3[/itex]
    [itex]I_{E} = 122 ± 4[/itex]

    Calculate the mean current and the standard deviation of the mean.

    2. Relevant equations

    Standard Deviation of the mean: [itex]σ_{mean} = \frac{σ_{s}}{\sqrt{N}}[/itex]

    where [itex]σ_{s}[/itex] is the standard deviation.

    Quadrature sum for error propagation: [itex]Total Error = \sqrt{σ_{1}^{2} + σ_{2}^{2} + σ_{3}^{2} + ...}[/itex]

    Error propagation formula: [itex]σ_{p} = \sqrt{(\frac{\partial f}{\partial a})^2σ_{a}^2 + (\frac{\partial f}{\partial b})^2σ_{b}^2 + (\frac{\partial f}{\partial c})^2σ_{c}^2 + ...}[/itex], where [itex]p = f(a,b,c,...)[/itex]

    3. The attempt at a solution

    To get the mean, I added all the data (using the quadrature formula, too) so I could divide by 5.

    I got: [itex]\overline{I} = \frac{605 ± 9.6954}{5}[/itex]

    To divide that by 5, I used the error propagation formula and got: [itex]\overline{I} = 121 ± 2[/itex]

    To get the standard deviation, I would normally get the variance, [itex]σ_{s}^2 = \frac{1}{N - 1} \sum (y_{i} - \overline{y} )^2[/itex], and take the square root.

    But subtracting and squaring each data point with the error propagation would require a lot of arithmetic, and that doesn't seem like the right path...

    Is there something I'm doing wrong or an easier method to do this? Or do I just have to grit my teeth and do all the arithmetic...?

    Thanks in advance.
     
  2. jcsd
  3. Jan 20, 2013 #2

    tms

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    When the errors in the separate measurements are not equal, you do not use the usual equations for mean and standard deviation. Instead, the mean is given as
    [tex]\mu \approx \frac{\Sigma (x_i/\sigma_i^2)}{\Sigma (1/\sigma_i^2)},[/tex]
    and the variance is
    [tex]\sigma_\mu^2 \approx \frac{1}{\Sigma(1/\sigma_i^2)}.[/tex]
     
  4. Jan 20, 2013 #3

    haruspex

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    tms, how would you calculate the σi values from error ranges? Do you assume the error ranges represent standard deviations, or do you take the ranges to represent uniform distributions and compute each s.d. on that basis? (The second sounds right to me.)
     
  5. Jan 21, 2013 #4

    tms

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    How would calculate the s.d. from a single number representing the distribution?
     
  6. Jan 21, 2013 #5

    haruspex

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    I don't understand your response. Your advice (which seems reasonable to me) was to obtain a mean by taking a weighted average of the readings, where the weights are derived from the uncertainties in the readings. But in your formula you express those uncertainties as variances, whereas the given uncertainties are in the form ±error. I'm merely asking what procedure you regard as appropriate to derive the variances from the error ranges.
     
  7. Jan 21, 2013 #6

    tms

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    I was trying to be Socratic. You suggested assuming the [itex]\sigma_i[/itex]s represented uniform distributions and calculating the s.d.s from them. I was trying to suggest that the most reasonable way a single number could represent a distribution is if it were the s.d.

    In fact, the [itex]\sigma_i[/itex]s are the uncertainties in the measurements. I should have made that explicit.
     
  8. Jan 21, 2013 #7

    haruspex

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    Just realised it makes no difference. So long as the error ranges are proportional to the s.d.s... Doh!
     
  9. Jan 22, 2013 #8
    Thanks for the replies.

    So what exactly should I do? Are the formulas tms posted correct?
     
  10. Jan 22, 2013 #9

    haruspex

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    tms' formula for the mean is certainly valid.
    I'm not sure I understand the one for the variance. You have two indicators of variance: the individual error ranges and the scatter of the individual 'central' readings. But tms' formula only seems to involve the former.
     
  11. Jan 22, 2013 #10

    tms

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    The derivation is in Bevington. Very briefly,
    [tex]\sigma_\mu^2 = \sum \left[ \sigma_i^2 \left( \frac{\partial \mu}{\partial x_i }\right)^2\right].[/tex]
    When the uncertainties are unequal, evaluate the derivative using the equation for the mean given above.
     
    Last edited: Jan 22, 2013
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