Mathematical Rules of Thumb, illustrated reader · Chapter 16

16Measurement and Uncertainty

Building Trustworthy Results

5 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 16 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-measurement from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 5

Keep the covariance in an uncertainty budget

Does perfect negative correlation make the exact product uncertainty zero?

The same input uncertainty magnitudes combine differently as their correlation changes.

uA2=W2uL2+L2uW2+2LWρuLuW u_A^2=W^2u_L^2+L^2u_W^2+2LW\rho u_Lu_W

Input correlation ρ. First-order propagation; L=10cm, W=5cm, uL=.1cm, uW=.05cm. At cancellation, neglected nonlinear terms matter.

Predict first. Does perfect negative correlation make the exact product uncertainty zero?

Choose an example

Keep the covariance in an uncertainty budget. A=LW with L=10±.1 cm, W=5±.05 cm, and correlation 0 gives first-order u(A)=0.707107 cm². With no correlation, the two contributions add in quadrature.
Input correlation ρ: 0
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Nominal area (cm²)
50
First-order standard uncertainty (cm²)
0.707107
Expanded uncertainty k=2 (cm²)
1.41421

A=LW with L=10±.1 cm, W=5±.05 cm, and correlation 0 gives first-order u(A)=0.707107 cm². With no correlation, the two contributions add in quadrature.

Use the idea

Use rule 16.3.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

First-order propagation; L=10cm, W=5cm, uL=.1cm, uW=.05cm. At cancellation, neglected nonlinear terms matter.

Check your understanding: Does perfect negative correlation make the exact product uncertainty zero?
No. It cancels these first-order contributions; higher-order product terms can remain.

Book source: Rule 16.3.1: Keep covariance terms for correlated uncertainties. Demonstration C16-D01. Worked illustration.

2Demonstration 2 of 5

Watch repetition meet a bias floor

Can more repetitions remove the bias?

Averaging decreases random scatter while a fixed calibration bias persists.

MSE⁡(X‾)=σ2/n+b2 \operatorname{MSE}(\bar X)=\sigma^2/n+b^2

Fixed bias b. Independent unit-SD readings and a constant stipulated bias in common measurement units.

Predict first. Can more repetitions remove the bias?

Choose an example

Watch repetition meet a bias floor. With independent unit-SD noise and fixed bias 0.2, MSE of the mean is 1/n+b². After 100 readings the RMS error is 0.224. Averaging cannot push the error below the bias 0.2; only calibration removes it.
Fixed bias b: 0.2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Single-reading random SD
1
Fixed bias
0.2
RMS error after 100 readings
0.223607

With independent unit-SD noise and fixed bias 0.2, MSE of the mean is 1/n+b². After 100 readings the RMS error is 0.224. Averaging cannot push the error below the bias 0.2; only calibration removes it.

Use the idea

Use rule 16.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Independent unit-SD readings and a constant stipulated bias in common measurement units.

Check your understanding: Can more repetitions remove the bias?
No. Bias correction requires calibration or a better measurement model.

Book source: Rule 16.1.1: Repetition reduces random scatter, not fixed bias. Demonstration C16-D02. Worked illustration.

3Demonstration 3 of 5

Propagate a nonlinear transformation by simulation

Is E[X²] equal to E[X]²?

A seeded histogram makes nonlinear propagation and its mean shift visible. Compare simulated and exact moments.

X∼N(2,σ2),E[X2]=4+σ2 X\sim N(2,\sigma^2),\quad E[X^2]=4+\sigma^2

Input standard deviation σ. 100,000 constructed Gaussian inputs, seed 1603; first-order uncertainty is only a local approximation.

Predict first. Is E[X²] equal to E[X]²?

Choose an example

Propagate a nonlinear transformation by simulation. For Gaussian x with mean 2 and SD 0.5, E[x²]=4+σ²=4.25. The constructed Monte Carlo result differs from squaring the mean. Straight-line estimate 4σ=2 versus simulated SD 2.04: close enough here.
Input standard deviation σ: 0.5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Seed
1603
Monte Carlo mean of x²
4.23575
Exact mean of x²
4.25
Monte Carlo standard deviation
2.04046
Straight-line estimate of SD (4σ)
2

For Gaussian x with mean 2 and SD 0.5, E[x²]=4+σ²=4.25. The constructed Monte Carlo result differs from squaring the mean. Straight-line estimate 4σ=2 versus simulated SD 2.04: close enough here.

Use the idea

Use rule 16.3.5 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

100,000 constructed Gaussian inputs, seed 1603; first-order uncertainty is only a local approximation.

Check your understanding: Is E[X²] equal to E[X]²?
Not with nonzero variance. E[X²]=Var(X)+E[X]².

Book source: Rule 16.3.5: Use Monte Carlo propagation when linear uncertainty rules bend. Demonstration C16-D03. Worked illustration.

4Demonstration 4 of 5

Spend effort on the biggest contributor

Which halving cuts the total most?

Three independent contributions of 4, 2 and 1 mm. Halve one and watch the combined uncertainty.

uc=u12+u22+u32 u_c=\sqrt{u_1^2+u_2^2+u_3^2}

Contributor halved. Independent contributions, unit sensitivity, root-sum-square combination.

Predict first. Which halving cuts the total most?

Choose an example

Spend effort on the biggest contributor. Contributions 4, 2 and 1 mm combine as √(16+4+1)=4.583 mm. Halving contributor 2 gives 4.243 mm, a 7.42% cut. Squaring makes the 4 mm term dominate, so effort spent here barely moves the total.
Contributor halved: 2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Combined before (mm)
4.58258
Combined after (mm)
4.24264
Reduction (%)
7.41799

Contributions 4, 2 and 1 mm combine as √(16+4+1)=4.583 mm. Halving contributor 2 gives 4.243 mm, a 7.42% cut. Squaring makes the 4 mm term dominate, so effort spent here barely moves the total.

Use the idea

Use rule 16.3.4 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Independent contributions, unit sensitivity, root-sum-square combination.

Check your understanding: Which halving cuts the total most?
Halving the 4 mm term: 4.58 mm drops to 3 mm. Halving the 1 mm term saves under 2%.

Book source: Rule 16.3.4: Improve the dominant uncertainty contributors first. Demonstration C16-D04. Worked illustration.

5Demonstration 5 of 5

See when k=2 really means 95%

Does ±2u from three repeat readings give 95% coverage?

The interval ±2u reaches about 95% coverage only when the uncertainty itself is well estimated. With few degrees of freedom the Student t tails are heavier.

P(|Tν|≤2),Tν∼tν P(|T_\nu|\le 2),\quad T_\nu\sim t_\nu

Effective degrees of freedom ν. Normal measurement errors, uncertainty estimated with ν effective degrees of freedom, coverage computed from the t distribution.

Predict first. Does ±2u from three repeat readings give 95% coverage?

Choose an example

See when k=2 really means 95%. With ν=10 effective degrees of freedom, the interval ±2u covers about 92.7% of a Student t distribution. Reaching the normal 95.45% would need k≈2.28. The shortfall is now small, but report ν so a reader can check it.
Effective degrees of freedom ν: 10
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Degrees of freedom
10
Coverage of ±2u (%)
92.6612
Factor needed for 95.45% coverage
2.28368

With ν=10 effective degrees of freedom, the interval ±2u covers about 92.7% of a Student t distribution. Reaching the normal 95.45% would need k≈2.28. The shortfall is now small, but report ν so a reader can check it.

Use the idea

Use rule 16.3.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Normal measurement errors, uncertainty estimated with ν effective degrees of freedom, coverage computed from the t distribution.

Check your understanding: Does ±2u from three repeat readings give 95% coverage?
No. With ν=2 or 3 it covers only about 82% to 86%; a factor near 3.3 is needed at ν=3.

Book source: Rule 16.3.3: Coverage factor two is an approximate 95% convention. Demonstration C16-D05. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.