What is the Standard Deviation of the Mean for Lab Homework Data?

In summary, when calculating the mean and standard deviation of a set of measurements with unequal uncertainties, it is necessary to use the weighted average formula for the mean and the error propagation formula for the variance. This involves taking into account the individual uncertainties of each measurement in order to determine the overall mean and variance. It is also important to note that when calculating the variance, the derivative should be evaluated using the equation for the mean with unequal uncertainties.
  • #1
jumi
28
0

Homework Statement



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.

Homework 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]

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.
 
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  • #2
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]
 
  • #3
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.)
 
  • #4
How would calculate the s.d. from a single number representing the distribution?
 
  • #5
tms said:
How would calculate the s.d. from a single number representing the distribution?
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.
 
  • #6
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.
 
  • #7
Just realized it makes no difference. So long as the error ranges are proportional to the s.d.s... Doh!
 
  • #8
Thanks for the replies.

So what exactly should I do? Are the formulas tms posted correct?
 
  • #9
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.
 
  • #10
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:

What is error analysis for lab?

Error analysis for lab is the process of identifying and quantifying the uncertainties or errors in a scientific experiment or measurement. It involves evaluating the sources of error and their potential impact on the accuracy and reliability of the results.

Why is error analysis important in a lab setting?

Error analysis is important in a lab setting because it allows scientists to understand the limitations and potential biases of their experiments. This helps to ensure the validity and reproducibility of their results, and allows for improvements to be made in future experiments.

What are the types of errors that can occur in a lab?

There are three main types of errors that can occur in a lab: random errors, systematic errors, and human errors. Random errors are unpredictable and can be caused by factors such as equipment limitations or environmental conditions. Systematic errors are consistent and can be caused by faulty equipment or incorrect experimental procedures. Human errors are mistakes made by the experimenter, such as misreading measurements or recording incorrect data.

How can errors be minimized in a lab experiment?

Errors can be minimized in a lab experiment by using precise and calibrated equipment, following standardized procedures, and repeating measurements multiple times. It is also important to identify potential sources of error and take steps to control or eliminate them.

What is the difference between accuracy and precision?

Accuracy refers to how close a measured value is to the true or accepted value, while precision refers to how close multiple measurements of the same quantity are to each other. A measurement can be precise but not accurate if it consistently misses the true value by the same amount. A measurement can also be accurate but not precise if it is far from the true value, but the repeated measurements are close to each other.

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