Errors during the measurement process

 

Standard error of the mean

The foregoing analysis has examined the way in which measurements with random errors are distributed about the mean value. However, we have already observed that some error remains between the mean value of a set of measurements and the true value, i.e. averaging a number of measurements will only yield the true value if the number of measurements is infinite. If several subsets are taken from an infinite data population, then, by the central limit theorem, the means of the subsets will be distributed about the mean of the infinite data set. The error between the mean of a finite data set and the true measurement value (mean of the infinite data set) is defined as the standard error of the mean, α. This is calculated as:

                                                    α = α /n^1/2                                                      (3.18)

α tends towards zero as the number of measurements in the data set expands towards infinity. The measurement value obtained from a set of n measurements, x₁, x₂, … x_n,

can then be expressed as:

                                                          x = x_mean ± α

For the data set C of length measurements used earlier, n = 23, α = 1.88 and α = 0.39. The length can therefore be expressed as 406.5 ± 0.4 (68% confidence limit). However, it is more usual to express measurements with 95% confidence limits (±2 α boundaries). In this case, 2 α = 3.76, 2α = 0.78 and the length can be expressed as 406.5 ± 0.8 (95% confidence limits).

Estimation of random error in a single measurement

In many situations where measurements are subject to random errors, it is not practical to take repeated measurements and find the average value. Also, the averaging process becomes invalid if the measured quantity does not remain at a constant value, as is usually the case when process variables are being measured. Thus, if only one measurement can be made, some means of estimating the likely magnitude of error in it is required. The normal approach to this is to calculate the error within 95% confidence limits, i.e. to calculate the value of the deviation D such that 95% of the area under the probability curve lies within limits of ±D. These limits correspond to a deviation of ±1.96. Thus, it is necessary to maintain the measured quantity at a constant value whilst a number of measurements are taken in order to create a reference measurement set from which α can be calculated. Subsequently, the maximum likely deviation in a single measurement can be expressed as: Deviation = ±1.96. However, this only expresses the maximum likely deviation of the measurement from the calculated mean of the reference measurement set, which is not the true value as observed earlier. Thus the calculated value for the standard error of the mean has to be added to the likely maximum deviation value. Thus, the maximum likely error in a single measurement can be expressed as:

                                         Error = ±(1.96 α + α )                                                          (3.19)

Example 3.4

Suppose that a standard mass is measured 30 times with the same instrument to create a reference data set, and the calculated values of α and ˛ are α = 0.43 and α = 0.08. If the instrument is then used to measure an unknown mass and the reading is 105.6 kg, how should the mass value be expressed?

Solution Using (3.19), 1.96 α + α = 0.92. The mass value should therefore be expressed as: 105.6 ± 0.9 kg.

Before leaving this matter, it must be emphasized that the maximum error specified for a measurement is only specified for the confidence limits defined. Thus, if the maximum error is specified as ±1% with 95% confidence limits, this means that there is still 1 chance in 20 that the error will exceed ±1%.