Measurements · Module 4 of 10
Probability and Statistics
Repeated measurements never agree exactly. Statistics turns that scatter into a best estimate and a defensible interval, so a set of readings becomes a claim with a number and a confidence.
Readiness check
This module handles scattered data. Tick only what you can do closed-notes.
- Average a list of numbers.
- Square a deviation and sum the squares.
- Take a square root.
- Recall the bell-shaped normal distribution.
- Divide by the square root of a count.
The core idea
A set of readings scatters about a mean, with spread measured by the sample standard deviation s (dividing by n − 1). The mean itself is uncertain by the standard deviation of the means, s/√n, which shrinks with more data. A confidence interval is the mean plus or minus t times that.
mean = (1/n) Σ xis = √[Σ(xi − mean)2 / (n − 1)]CI = mean ± t · s/√nMeasure the same thing repeatedly and the readings will not match, because of random error. Statistics describes this scatter and extracts the best estimate. The readings are treated as a sample from a population that, for random error, is often normal, the familiar bell curve centred on the true mean with a spread set by the population standard deviation. From a finite sample we estimate the centre with the mean and the spread with the sample standard deviation s, computed by dividing the summed squared deviations by n − 1, not n, to correct for using the sample's own mean. Crucially, the mean of the sample is itself a random quantity, and its uncertainty is the standard deviation of the means, s/√n, which falls as the sample grows: averaging more readings pins down the mean even though it does not change the spread of individual readings. To state a result honestly we give a confidence interval, the mean plus or minus t times s/√n, where t is the Student factor for the confidence level and the degrees of freedom. This interval, not a bare average, is the proper output of a set of measurements, and it feeds directly into uncertainty analysis next.
The skills, taught in order
Five skills turn scattered readings into a stated result.
4.1 Populations and the normal distribution
A set of readings is a sample from a larger population. Random error often makes that population normal, a symmetric bell curve described by its mean and standard deviation, which underlies the statistics that follow.
4.2 Mean and standard deviation
The mean estimates the true value; the sample standard deviation s measures the spread, using n − 1 in the denominator. That correction, the degrees of freedom, keeps s from underestimating the true spread when the sample is small.
4.3 The standard deviation of the means
The sample mean has its own uncertainty, s/√n, called the standard deviation of the means. It shrinks as √n, so quadrupling the readings halves it, which is how averaging improves an estimate.
| Quantity | Formula | Describes |
|---|---|---|
| Standard deviation s | √[Σ(xi−mean)2/(n−1)] | spread of readings |
| Std of the means | s/√n | uncertainty of the mean |
| Confidence interval | mean ± t·s/√n | range for the true mean |
The spread of the data and the spread of the mean are different quantities; confusing them is the classic error.
4.4 Confidence intervals with t
A confidence interval brackets the true mean: mean plus or minus t times s/√n. The Student t factor depends on the confidence level and the degrees of freedom n − 1, and it exceeds the normal z for small samples.
4.5 Number of measurements required
Because the interval scales as 1/√n, halving its width needs four times the data. This sets how many readings a target uncertainty demands, and warns that precision bought by repetition gets expensive.
Engineering connection: reporting a tensile strength as a mean with a 95 percent confidence interval, rather than a single number, is what makes the result usable in a design margin.
Worked example 1: mean and standard deviation
Five readings of a length are 9.8, 10.0, 10.2, 10.0, and 10.0 mm. Find the mean and the sample standard deviation.
- ProblemFind the mean and sample standard deviation for the readings in Figure 1.
- Given / findData 9.8, 10.0, 10.2, 10.0, 10.0; n = 5. Find the mean and s.
- AssumptionsIndependent readings with random error.
- Modelmean = (1/n)Σxi; s = √[Σ(xi−mean)2/(n−1)].
- Equationsmean = 50.0 / 5s = √(0.08 / 4)
- Solvemean = 10.0 mm; deviations squared sum to 0.08, so s = √0.02 = 0.14 mm.
- CheckThe readings span 9.8 to 10.2, so a spread of about 0.14 mm is reasonable, and the mean sits centrally.
- ConclusionThe length is 10.0 mm with a spread of 0.14 mm, the inputs the next step needs for a confidence interval.
Worked example 2: a 95 percent confidence interval
For those five readings (mean 10.0 mm, s = 0.14 mm, n = 5), find the 95 percent confidence interval for the true mean. Use t = 2.776 for 4 degrees of freedom.
- ProblemFind the 95 percent confidence interval in Figure 2.
- Given / findmean 10.0 mm, s = 0.14 mm, n = 5, t = 2.776. Find the interval.
- AssumptionsNormal random error; t for 4 degrees of freedom at 95 percent.
- ModelSEM = s/√n; margin = t·SEM; CI = mean ± margin.
- EquationsSEM = 0.1414 / √5 = 0.0632margin = 2.776 × 0.0632
- Solvemargin = 0.18 mm; CI = 10.0 ± 0.18 mm.
- CheckThe margin is larger than the SEM by the t factor, as it must be, and comfortably inside the data spread.
- ConclusionWe are 95 percent confident the true mean lies between 9.82 and 10.18 mm, the proper way to report the length.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Dividing s by n | Standard deviation too small | "Did I use n − 1?" | Use n − 1 for a sample standard deviation. |
| Data spread as mean spread | Interval far too wide | "Spread of readings, or of the mean?" | The mean's uncertainty is s/√n, not s. |
| Using z for small n | Interval too narrow | "Is the sample small?" | Use the Student t factor for small samples. |
| More data shrinks s | Expecting s to fall with n | "Does averaging change the spread?" | More data shrinks s/√n, not s itself. |
Practice ladder
Find the mean and sample standard deviation of 4, 5, and 6.
Show answer
Mean = 15/3 = 5; deviations −1, 0, 1, squares sum 2, s = √(2/2) = 1.
A sample has s = 2 with n = 16. Find the standard deviation of the means.
Show answer
s/√n = 2/√16 = 2/4 = 0.5.
For mean 20, s = 3, n = 9, find the 95 percent confidence interval (t = 2.306 for 8 degrees of freedom).
Show answer
SEM = 3/√9 = 1; margin = 2.306 × 1 = 2.31; CI = 20 ± 2.31.
You need to halve the width of a confidence interval. By what factor must you increase the number of measurements, and why?
What good work looks like
Because the interval scales as 1/√n, halving it needs four times as many measurements. A good answer states the √n dependence and notes the diminishing return of repetition.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take a real set of repeated measurements and report the mean, standard deviation, and a 95 percent confidence interval.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Why n − 1 for a sample?
To correct for using the sample's own mean, giving the degrees of freedom.
2. Write the standard deviation of the means.
s/√n.
3. Write a confidence interval.
mean ± t · s/√n.
4. Why t instead of z?
For small samples the Student t is wider, accounting for the estimated s.
5. Data to halve the interval?
Four times as many, from the 1/√n scaling.
Textbook mapping
This module follows Figliola and Beasley, Theory and Design for Mechanical Measurements, 5th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Describing a population | Figliola and Beasley, Section 4.3, Describing the Behavior of a Population |
| Finite data sets and t | Figliola and Beasley, Section 4.4, Statistics of Finite-Sized Data Sets |
| Number of measurements required | Figliola and Beasley, Section 4.8, Number of Measurements Required |
Section numbers refer to Figliola and Beasley, 5th edition. Any edition with the same chapter titles is equivalent for study.