Errors in measurement

Error is the difference between the measured value and the true value of the quantity: $e = V_m - V_t$. Absolute error is $e$; relative (percentage) error is $e/V_t \times 100$.

Types of error

1. Gross errors: caused by human mistakes such as wrong reading, wrong recording, using the instrument wrongly or forgetting to correct zero error. They can be avoided by care and by repeating the measurement.
2. Systematic (fixed) errors: they follow a definite law and have the same magnitude and sign for repeated readings under the same conditions, so they can be predicted and corrected.
   • Instrumental: worn anvils, wrong calibration, zero error, backlash.
   • Environmental: temperature, humidity, pressure, vibration, dust.
   • Observational: parallax, wrong estimation of the fraction of a division.
3. Random (accidental) errors: small variations in repeated readings that cannot be predicted individually, caused by small changes in temperature, friction, operator judgement and so on. They are treated by statistics: the mean of many readings is taken and the scatter is described by standard deviation.

Sources of error

Reducing errors

Measure at standard temperature $20^\circ$C or allow the work and instrument to soak, apply corrections for known systematic errors, calibrate regularly, use correct measuring force (ratchet), follow Abbe's principle, and take the mean of several readings to reduce random error.

(a) Error propagation

For $V = \dfrac{\pi d^2 h}{4}$ the relative error depends on each variable as

$$\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta h}{h} \quad (\text{maximum}), \qquad \frac{\delta V}{V} = \sqrt{\left(2\frac{\delta d}{d}\right)^2 + \left(\frac{\delta h}{h}\right)^2} \quad (\text{probable})$$ $$V = \frac{\pi (25)^2 (60)}{4} = 29\,452\ \text{mm}^3$$

• $2\,\delta d/d = 2(0.02/25) = 0.0016$
• $\delta h/h = 0.05/60 = 0.000833$

Maximum error: $\delta V = 29452 \times (0.0016 + 0.000833) = 71.7\ \text{mm}^3$ (0.243 %).

Probable error: $\delta V = 29452\sqrt{0.0016^2 + 0.000833^2} = 29452 \times 0.001804 = 53.1\ \text{mm}^3$ (0.180 %).

The diameter contributes most of the error because it is squared.

Answer: $V = 29\,452 \pm 72$ mm³ (maximum), $\pm 53$ mm³ (probable).

(b) Averaging

Number of readings $n = 8$.

$$\bar d = \frac{25.02+24.98+25.01+25.03+24.99+25.00+25.02+25.01}{8} = \frac{200.06}{8} = 25.0075\ \text{mm}$$

Sum of squares $= 1950\times10^{-6}$ mm².

$$s = \sqrt{\frac{1950\times10^{-6}}{n-1}} = \sqrt{278.6\times10^{-6}} = 0.0167\ \text{mm}$$ $$\sigma_{\bar d} = \frac{s}{\sqrt{n}} = \frac{0.0167}{\sqrt 8} = 0.0059\ \text{mm}$$

Since the standard error varies as $1/\sqrt n$, halving it needs $n$ four times larger: $4 \times 8 = 32$ readings. Averaging reduces random error but not systematic error.

Answer: mean = 25.0075 mm, $s$ = 0.0167 mm, standard error = 0.0059 mm; 32 readings for half the standard error.

The method of least squares chooses $a$ and $b$ so that $\sum (y_i - a - b x_i)^2$ is minimum. This gives the normal equations

$$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}, \qquad a = \frac{\sum y - b\sum x}{n}$$

$n = 6$:

$$b = \frac{6(222.86) - (30)(30.42)}{6(220) - 30^2} = \frac{1337.16 - 912.6}{1320 - 900} = \frac{424.56}{420} = 1.0109$$ $$a = \frac{30.42 - 1.0109(30)}{6} = 0.0157\ \text{mm}$$

Fitted line: $y = 0.0157 + 1.0109\,x$.

The residuals add to zero (as they must) and $\sum e^2 = 0.0021\ \text{mm}^2$.

Estimate at $x = 5$ mm: $y = 0.0157 + 1.0109(5) = 5.070$ mm.

• Intercept $a = 0.016$ mm is the zero error of the gauge.
• Slope $b = 1.011$ means the gauge reads about 1.1 % high (scale error); the calibration factor is $1/b = 0.989$.

Answer: $y = 0.0157 + 1.0109x$ (mm); reading at 5 mm = 5.070 mm.