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Error Analysis Review

In laboratory work, it is usually necessary to use experimental apparatus to measure physical quantities. The measurements are seldom, if ever, perfect. As a result, imperfections must be taken into account in any set of physical measurements. One of the most important things to be learned in the laboratory is how to make reliable estimates of the uncertainties involved in any physical measurements and how to handle the propagation of these errors, i.e., to know how the uncertainties affect calculated results of an experiment.


Significant Figures

The number of digits required to express the result of an experimental measurement, so that it reflects the accuracy with which the measurement was made, are known as significant figures. Thus, the number of significant figures reflects the limitation of the measuring device and/or the experimenter. Whether the calculations are performed by hand or on a computer, the number of significant figures displayed in the final result must reflect the limitations of the experimental measurement. Often, a computer program will, by default, display either too many or too few digits. If too few digits are displayed, you need to adjust the program setting to show more digits.

For example, if the length of a cylinder is measured as 20.64cm20.64\,\text{cm}, this quantity is said to be measured to four significant figures. If written as 0.0002064km0.0002064\,\text{km}, we still have only four significant figures. The zeros preceding the “2” are used only to indicate the position of the decimal point. The zero between the “2” and the “6” is a significant figure, but the other zeros are not. If the above measurement is made with a meter stick, the last digit recorded is an estimated figure representing a fractional part of a millimeter division. All recorded data should include the last estimated figure in the result, even though it may be zero. If this measurement had appeared to be exactly 20, it should have been recorded as 20.00cm20.00\,\text{cm}, since lengths can be estimated by means of this instrument to about 0.01cm0.01\,\text{cm}. When the measurement is written as 20cm20\,\text{cm} it indicates that the value is known to be somewhere between 19.5cm19.5\,\text{cm} and 20.5cm20.5\,\text{cm}, whereas the value 20.00cm20.00\,\text{cm} indicates a range between 19.995cm19.995\,\text{cm} and 20.005cm20.005\,\text{cm}. Conversely, retaining too many significant figures implies greater accuracy than the figures actually represent.

Rules for Significant Figures

When deciding the number of significant figures to retain, the following rules apply:

Examples --- Addition & Multiplication

The following examples help illustrate these rules.

Addition

When adding these four numbers:

  427.5
   28.03
    0.0654
  396.0
  ------
  851.6

the result is rounded to the first decimal digit (851.6) because the first term (427.5) in the sum is only given to one decimal digit. The sum is expressed to the proper number of significant figures.

Multiplication

In our next example, we will calculate the area of a rectangle. If its length is measured as 1.94cm1.94\,\text{cm} and width as 1.84cm1.84\,\text{cm}, the area as provided by a calculator is

A=1.94×1.84=3.5696cm2.A = 1.94 \times 1.84 = 3.5696\,\text{cm}^2.

Since each factor has three significant figures (as opposed to the five from the calculator), the result should instead be reported as

A=3.57cm2.A = 3.57\,\text{cm}^2.

Error Analysis Procedure

This section is a guid to performing error analysis for an experiment. Steps for performing error analysis are:


Types of Errors

Experimental measurements are characterized by precision and accuracy.

Random errors increase scatter and reduce precision. They arise from uncontrollable variations and inherent limitations of the equipment; they can be estimated and reduced by repeated measurements. Systematic errors arise from calibration issues or faulty methods; they can reduce precision and cannot themselves be reduced by repeated measurements. Often you can estimate precision errors by considering the equipment and its specifications.


The distinction between precision and accuracy.

Figure 1:The distinction between precision and accuracy.

Error Propagation of Measurement Uncertainties

Error analysis can help you recognize if your results are consistent with a known or expected value. Each measurement has some combination of systematic and random errors.

Example: Error Propagation --- Density of a Block

Suppose you are measuring the density of a rectangular block. The block has dimensions length LL, width WW, height HH, mass MM each with their respective measurement uncertainty. The calculated density is mass divided by volume.

Given

The experimental density is

ρexperimental=MV=MLWH=60.0030.00=2.00g/cm3.\rho_\text{experimental} = \frac{M}{V} = \frac{M}{LWH} = \frac{60.00}{30.00} = 2.00\,\text{g/cm}^3.

To find the max/min range, the maximum density consistent with your measurements is the ratio of the maximum mass and the minimum volume:

ρmax=M+δmVδV=M+δm(LδL)(WδL)(HδL)=60.051.99×0.99×4.99=60.0529.69=2.02g/cm3,\rho_{\max} = \frac{M+\delta m}{V-\delta V} = \frac{M+\delta m}{(L-\delta L)(W-\delta L)(H-\delta L)} = \frac{60.05}{1.99\times0.99\times4.99} = \frac{60.05}{29.69} = 2.02\,\text{g/cm}^3,

and similarly, the minimum density is 1.98g/cm31.98\,\text{g/cm}^3. If the expected value lies outside this range, further investigation is required. If the expected density is not in this range, you need to consider possible explanations:

The two types of errors illustrated in Figure 1 require different treatment. The systematic errors illustrated on the target labelled Precise cannot be reduced by taking more measurements, while the average value obtained with the random errors on the target labelled Accurate improves with more measurements. The following sections provide guidance on analyzing systematic and random errors.


Random Errors

Random errors are inherent in nearly all measurements. They arise because of uncontrollable conditions affecting the observer, the measuring device, and the quantity to be measured. On the basis of probability, these errors are as likely to be positive as negative, and more likely to be small than large.

Independent repeated measurements and use of their average in any calculation reduce their effect. This minimization is only achieved if the measurements are independent. A common mistake is biasing a new measurement to agree with the previous measurements. This mistake makes the measurements more precise, but less accurate as shown in Figure 1. Taking truly independent measurements is the best way to reduce random error.

Mean, Standard Deviation, Standard Error

The most well established method to quantify the spread of random errors in a measurement is to compute the standard deviation. Suppose you make NN measurements x1,x2,,xNx_1, x_2, \ldots, x_N of a certain quantity xx. The average value is:

xˉ=x1+x2++xNN=1Ni=1Nxi,\bar{x}=\frac{x_1 + x_2 + \ldots + x_N}{N}=\frac{1}{N}\sum_{i=1}^N x_{i},

and the standard deviation of the individual measurements is

σ=(x1xˉ)2+(x2xˉ)2++(xNxˉ)2N1=1N1i=1n(xixˉ)2.\sigma = \sqrt{\frac{(x_1-\bar{x})^2 + (x_2-\bar{x})^2 + \ldots + (x_N-\bar{x})^2}{N-1}} = \sqrt{\frac{1}{N-1} \sum_{i=1}^{n}\left(x_i-\bar{x}\right)^2}.

In Excel, these operations can be carried out using the functions AVERAGE() and STDEV() or STDEV.S().

If you perform many measurements, then your estimate of the average value improves. The standard deviation of the mean (a.k.a. standard error) for NN measurements is

σmean=σN1.\sigma_{\mbox{mean}} = \frac{\sigma}{\sqrt{N-1}}.

Your best estimate of a measurement xx is then xˉ±σmean\bar{x} \pm \sigma_{\text{mean}}.

Example of Average ± Standard Error

Suppose you take four length measurements (cm): 7.65, 7.61, 7.66, 7.68

The mean value is

xˉ=(7.65+7.61+7.66+7.68)/4cm=7.65cm,\bar{x} = \left(7.65 + 7.61 + 7.66 + 7.68\right)/4\,\text{cm} = 7.65\,\text{cm},

and the standard deviation is

σ=(7.657.65)2+(7.617.65)2+(7.667.65)2+(7.687.65)241cm2=0.029cm.\sigma = \sqrt{\frac{(7.65-7.65)^2 + (7.61-7.65)^2 + (7.66-7.65)^2 + (7.68-7.65)^2}{4-1}\,\text{cm}^2} = 0.029\,\text{cm}.

The standard error (i.e. standard deviation of the mean) of the average value is thus reduced to 0.029cm/3=0.017cm 0.029\,\text{cm} / \sqrt{3} = 0.017\,\text{cm} and therefore the best estimate of the length is xˉ=7.65cm±0.017cm\bar{x} = 7.65\,\text{cm} \pm 0.017\,\text{cm}.

Estimation through Variation or Instrumentation

Frequently, a set of measurements has to be taken in an experiment under a prescribed condition. It may be difficult to judge exactly when this condition is satisfied. In this case, it may be necessary to vary the physical quantities that produce this condition and to note how much each may be varied without appreciably altering the prescribed condition. This variation may be used to determine the error.

Finally, the measuring devices (meter stick, voltmeter, and so on) are not perfectly accurate even if one could make a precise reading with them. The manufacturer specifies the accuracy in the measuring device; for example, a particular type of voltmeter may be guaranteed to give readings accurate to ±1%\pm 1\% of full scale reading. Part of this tolerance is random error that can average out by taking multiple readings. As discussed in the next section, another part of the instrument error is systematic. An ammeter may always read 0.5%0.5\% high because of the value of a resistor internal to the device.


Systematic Errors

Systematic errors occur in an experiment because of a defective measuring apparatus, faulty methods, or incomplete working equations or assumptions. They are definite in sign and magnitude and cannot be reduced by averaging because the same error is included in each measurement.

These errors are often more important than the random errors. Calibrating the measuring apparatus, modifying the method, or correcting the equations may reduce them. These are generally described as corrections that are made in the performing of the experiment.

Examples of Systematic Errors

Report these sources of error as Vernier caliper error or reaction time error. The expression human error is too ambiguous to be useful in describing measurement errors.


Difference Between Experimental and Expected Values

If the true value of a quantity is known, then the systematic error can be estimated as difference between the experimental and expected values. It is important to remember that this is an estimate of the systematic error. The difference between your experimental value and the expected value is not an estimate of the random experimental error.

To report this difference as a percentage (percent difference), divide the difference by the true, expected, or actual value and multiply by 100%100\%. In Excel, you can use percent format instead of multiplying by 100%100\%.

The percent difference between experimental and actual values is

Experimental ValueActual ValueActual Value×100%.\frac{\text{Experimental Value} - \text{Actual Value}}{\text{Actual Value}} \times 100\%.

Sometimes, rather than a difference to actual values, you will be looking at how some quantity changes from an initial to final state. Similar to percent difference, you can report a percent change (when applicable) to help contexualize the process with

Final Experimental ValueInitial Experimental ValueInitial Experimental Value×100%.\frac{\text{Final Experimental Value} - \text{Initial Experimental Value}}{\text{Initial Experimental Value}} \times 100\%.

Example of % Difference

Suppose a student measures the value of gravity and will compare to the standard accepted value.

The systematic relative difference, expressed as a percentage, is

9.409.8039.803×100%=0.4039.803×100%=4.11%\frac{9.40 - 9.803}{9.803} \times 100\% = \frac{-0.403}{9.803} \times 100\% = -4.11\%

It is important to maintain the proper sign of the error as it indicates whether your experimental value was larger or smaller than the actual value (or increased/decreased in the context of % change). There is no definite allowable difference between the experimental and expected values in this course’s experiments. However, all measurements should be made with the greatest care, so as to reduce the error as much as possible.

It should be noted that the percent difference is not an estimate of experimental error. Any measurement has random and systematic errors even if the percent difference happens to be small.

Often, scientists do not know the actual value of the quantity they are measuring and they give their best estimate with error. Comparing the difference between the experimental and actual values with your best estimate of the experimental error (e.g. using standard deviation σ\sigma, standard error σmean\sigma_\text{mean}, or direct measurement uncertainties δ\delta) will help determine if there is a systematic error in your results.


Propagation of Systematic Errors through 1. Brute Force & 2. Calculus

Through indirect measurement, the effect of systematic errors can be followed through the equations used. Two simple examples will be discussed.

Calculation from a Single, Direct Measurement (Sphere)

Consider the volume of a sphere VV as calculated from a direct measurement of its diameter, DD. We would use

V=16πD3V = \frac{1}{6} \pi D^3

where π=3.1415926\pi = 3.1415926\ldots. Suppose the value of DD is measured as 3.04cm3.04\,\text{cm} and VV computed to be 14.71cm314.71\,\text{cm}^3. It is later discovered that the value of DD has a systematic error ΔD=+0.01cm\Delta D = +0.01\,\text{cm}, (unnecessarily large, in all probability). What is the error ΔV\Delta V in VV and the corrected value of VV?

1. Direct, brute-force method

2. Method based on calculus

Find the derivative of VV with respect to DD,

dVdD=36πD2dV=36πD2dD\frac{{\rm d}V}{{\rm d}D} = \frac{3}{6} \pi D^2 {\rm d}V = \frac{3}{6} \pi D^2 {\rm d}D

Note that this shows a direct dependence of a change in VV on an infinitesimal change in DD. Now divide through by V=16πD3V = \frac{1}{6} \pi D^3

dVV=3dDD.\frac{{\rm d}V}{V} = 3 \frac{{\rm d}D}{D}.

If a finite change in DD, such as the systematic error ΔD\Delta D, is small compared to DD, then dV{\rm d}V and dD{\rm d}D can be approximated by ΔV\Delta V and ΔD\Delta D, respectively, and so the following can be written:

ΔVV3ΔDD.\frac{\Delta V}{V} \approx 3 \frac{\Delta D}{D}.

This means that the fractional or percent error in VV is very nearly 3 times as large as that in DD. Another way to say, a 1% error in diameter produces a 3% error in volume. Therefore,

ΔVV+30.013.041%.\frac{\Delta V}{V} \approx + 3 \frac{0.01}{3.04} \approx 1 \%.

Hence,

ΔV0.15cm3\Delta V \approx 0.15\,\text{cm}^3

and

Vcorrected=VΔV=14.56cm3.V_{\text{corrected}} = V - \Delta V = 14.56\,\text{cm}^3.

Calculation from Two or More Direct Measurements (Cylinder)

Consider the volume VV of a right, circular cylinder as calculated from its diameter DD and length LL. For convenience, let DD be measured as 2.00cm2.00\,\text{cm} and LL as 5.00cm5.00\,\text{cm}. Then,

V=14πLD2=15.71cm3.V = \frac{1}{4} \pi L D^2 = 15.71\,\text{cm}^3.

If DD has an error of +0.01cm+0.01\,\text{cm} and LL an error of 0.02cm-0.02\,\text{cm}, what is the error ΔV\Delta V in VV and what is the corrected value of VV?

1. Direct, brute-force method

To obtain VcorrectedV_{\text{corrected}}

2. Method based on calculus

Find the partial derivatives of VV

VL=14πD2Vd=12πDL\frac{\partial V}{\partial L} =\frac{1}{4} \pi D^2 \hspace{2 cm} \frac{\partial V}{\partial d} =\frac{1}{2} \pi D L
dV=VLdL+VDdD{\rm d} V = \frac{\partial V}{\partial L} {\rm d} L + \frac{\partial V}{\partial D} {\rm d} D

so that

ΔVVLΔL+VDΔD=π4D2ΔL+π2DLΔD.\Delta V \approx \frac{\partial V}{\partial L} \Delta L + \frac{\partial V}{\partial D} \Delta D = \frac{\pi}{4} D^2 \Delta L + \frac{\pi}{2} D L \Delta D.

Now dividing through by V=14π,D2,LV = \frac{1}{4}\pi , D^2 , L

ΔVVΔLL+2ΔDD.\frac{\Delta V}{V} \approx \frac{\Delta L}{L} + \frac{2 \Delta D}{D}.

This shows how fractional errors in each measured quantity contribute to the total error in the calculated result; i.e. the fractional (or percent) systematic error in VV is very nearly equal to the fractional systematic error in LL plus twice the fractional error in DD. In our case,

ΔLL=0.025.00=0.4%\frac{\Delta L}{L} = \frac{-0.02}{5.00} = -0.4 \%

and

2ΔDD=+0.022.00=+1.0%\frac{2\Delta D}{D} = \frac{+0.02}{2.00} = +1.0 \%

so that

ΔVV+0.6%.\frac{\Delta V}{V} \approx +0.6 \%.

Thus ΔV=+0.09cm3\Delta V = +0.09\,\text{cm}^3, i.e. the systematic error, is +0.09cm3+0.09\,\text{cm}^3; the correction is 0.09cm3-0.09\,\text{cm}^3 and Vcorrected=+15.62cm3V_{\text{corrected}} = +15.62\,\text{cm}^3.