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Simple error analysis

  1. Dec 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A series of 5 measurements of Y are recorded with increasing X. (i.e. gradient has units Y/X)
    The error in measurement of Y is ±0.05
    The error in measurement of X is ±0.005.
    How is the error in the gradient calculated?
    2. Relevant equations - measurements
    Y | X
    0.2 | 0.23
    0.3 | 0.27
    0.5 | 0.31
    0.6 | 0.38
    0.8 | 0.42

    3. The attempt at a solution
    I assumed that the error in the gradient a would given by the equation.
    εa = √((εY)2 + (εX)2)

    I was unsure however if it was given by the addition of the individual error in each measurement.
    Thanks
     
  2. jcsd
  3. Dec 17, 2014 #2

    haruspex

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    I really dislike questions like this. The distribution of the errors in the inputs is not stated. It seems that those who pose these questions expect you to interpret the given errors as some (unspecified but fixed) number of standard deviations in normal distributions. In practice, errors induced by reading instruments are closer to uniform distributions. That being the case, the error in Y/X will be a different distribution, and it is not clear what an answer of the form ±ε would mean.
    Putting that aside, the formula you quote is for fractional errors. That is, if the actual error in Xi is εi then the fractional error is εi/Xi. Of course, this is only an approximation, and only works when εi is small compared with Xi.
     
  4. Dec 18, 2014 #3

    BvU

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    I understand Haru's dislike, but since this is something that probably happens to many students, perhaps a few questions and comments may be useful:

    The big question is: what's this about? If Sal has more context, he/she can get more and better assistance. What are you varying, what are you measuring, where do the error estimates come from ? Are they statistical or are they systematic ?

    The comments:

    Error estimates are usually pretty rough. 0.05 and 0.005 are testimony. There is a chance they are one half the least significant digit of some measuring instrument, in which case see Haru's comment.

    Notation:
    It's good practice to state errors (or rather: error estimates) to one decimal place (unless the first digit is 1, or unless there is a good reason to state more).
    it's good practice to state measurement to the same number of decimal places as the estimated error. So here X, Y = 0.230, 0.20 etc. That the trailing digit is always zero is suspicious.

    Experiments:
    usually you vary something (the independent variable) and you measure something (the dependent variable). The varied quantity is probably measured too, like in this case. A simple plot shows the independent variable on the x-axis and the dependent variable on the y-axis. If there is a reason to expect linear behaviour you do a linear least squares analysis to find a slope and an intercept. Other theoretically expected behaviours may bring you to plot calculated values (square root, logarithm, etc.) to obtain an expected linear relationship.

    Analysis: With the given information, I would do a linear least squares fit to estimate the slope and the error therein. I would be ignoring the error in X and I would feel justified to do so: it's ten times smaller than the error in Y so when things add up quadratically the one half percent wrong is much smaller than the error in the estimated errors: " √ ( 0.052 + 0.0052 ) = 0.05 "
     
  5. Dec 18, 2014 #4
    I made these values and data up to custom to my actual problem.
    The experiment was zeeman splitting using a CCD. The data was recorded by the program, with ΔE (Y from before) recorded to 1 decimal place and B (X from before) recorded to 2 decimal places. Gradient was automatically plotted. I want to know what the error in it is and so just need to know what method
     
  6. Dec 18, 2014 #5

    BvU

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    Method is linear regression. Expressions are e.g. here (for error see here).
    Otherwise, Excel has a Regression entry under Data | Data Analysis

    What does "gradient was automatically plotted" mean ? You got a plot of the data and it showed Y = 3.0 * X - 0.5 ?
     
  7. Dec 18, 2014 #6

    haruspex

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    ... as fractions of the X and Y values respectively.
     
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