Test Statistics
Probability and Statistics · FE Reference Handbook section
Learning objectives
What you must be able to do before leaving this section.
This chapter section covers Test Statistics within Probability and Statistics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.
- Explain, in your own words, what test statistics describes physically and when it applies.
- State every one of the 13 relations the handbook lists here and name each symbol with its unit.
- Select the correct relation from the wording of an exam stem within 20 seconds.
- Carry a complete solution from givens to a "most nearly" answer with the correct unit.
- Recognise the distractors generated by the unit trap: probabilities are dimensionless and must land in [0, 1].
Lecture
Why this section exists. Test Statistics is the part of Probability and Statistics that lets you connect a sample of measurements from a construction or materials process to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.
How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.
How it is examined. Items from this page are written as one distribution or one counting rule, then a single probability or interval. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.
The habit that earns the points. Unit discipline. probabilities are dimensionless and must land in [0, 1]. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.
How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.
Photo 1. Where this shows up in practice: test statistics.
Wikimedia Commons, public domain
Probability and Statistics — Test Statistics: reference schematic for orienting the symbols used in this section.
Theory, developed
Read this before the equations — it is what makes them memorable.
The physical situation
Every item from this section describes a sample of measurements from a construction or materials process. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.
The governing principle
The 13 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.
Assumptions and limits of validity
Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.
Solution procedure you should automate
1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.
Photo 2. Probability and Statistics: the physical system the theory above idealises.
Wikimedia Commons, public domain
Notation used in this section
| Zvar | Quantity produced by "Zvar = v" — read its definition and unit from the handbook line directly above the equation. |
|---|---|
| tvar | Quantity produced by "tvar = s" — read its definition and unit from the handbook line directly above the equation. |
| σ | Quantity produced by "σ = standard deviation" — read its definition and unit from the handbook line directly above the equation. |
| µo | Quantity produced by "µo = population mean" — read its definition and unit from the handbook line directly above the equation. |
| X | Quantity produced by "X = hypothesized mean or sample mean" — read its definition and unit from the handbook line directly above the equation. |
| n | Quantity produced by "n = sample size" — read its definition and unit from the handbook line directly above the equation. |
| s | Quantity produced by "s = computed sample standard deviation" — read its definition and unit from the handbook line directly above the equation. |
| Values of Zα/2 | Quantity produced by "Values of Zα/2" — read its definition and unit from the handbook line directly above the equation. |
Handbook notes for this section
Definitions and conditions exactly as the handbook states them.
- The following definitions apply.
- X - no
- X - no
- where
- The Z score is applicable when the standard deviation (s) is known. The test statistic is applicable when the standard deviation
- (s) is computed at time of sampling.
- 80% 1.2816
- 90% 1.6449
- 95% 1.9600
- 96% 2.0537
- 98% 2.3263
- 99% 2.5758
- Engineering Probability and Statistics
- Unit Normal Distribution
- x x x −x x −x x
- x f(x) F(x) R(x) 2R(x) W(x)
- 0.0 0.3989 0.5000 0.5000 1.0000 0.0000
- 0.1 0.3970 0.5398 0.4602 0.9203 0.0797
- 0.2 0.3910 0.5793 0.4207 0.8415 0.1585
- 0.3 0.3814 0.6179 0.3821 0.7642 0.2358
- 0.4 0.3683 0.6554 0.3446 0.6892 0.3108
- 0.5 0.3521 0.6915 0.3085 0.6171 0.3829
- 0.6 0.3332 0.7257 0.2743 0.5485 0.4515
- 0.7 0.3123 0.7580 0.2420 0.4839 0.5161
Core formulas for this FE topic
Definitions, applicability, units, assumptions and worked examples for each relation.
Worked exam-style examples
The four ways this section is written on the real exam — thoughts first, then equations, then substitution.
Nine field density tests give x̄ = 121.4 pcf with s = 3.6 pcf. Construct the 95% confidence interval on the mean (t₀.₀₂₅,₈ = 2.306).
Given
- n = 9
- x̄ = 121.4 pcf
- s = 3.6 pcf
- t = 2.306
Find
95% CI on μ
Start with the thinking
- σ is unknown and n is small — use t, not z.
- Degrees of freedom are n − 1 = 8.
Step-by-step solution
Standard error — SE = s/√n = 3.6/√9 = 1.20 pcf
Margin
Lower limit
Upper limit
Interval — 118.6 pcf ≤ μ ≤ 124.2 pcf
Answer: 118.6 to 124.2 pcf
Why the other options are there
- 119.1 to 123.7 (z = 1.96 used)
- 114.1 to 128.7 (√n omitted)
Reference: FE Reference Handbook — Probability and Statistics — Confidence intervals
A mix must exceed 3890 psi. A sample of 21 cores gives x̄ = 4077 psi, s = 323 psi. Test H₀: μ = 3890 against Hₐ: μ > 3890 at α = 0.05.
Given
- μ₀ = 3890 psi
- x̄ = 4077 psi
- s = 323 psi
- n = 21
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 323/√21 = 70.48 psi
Substituting
Compare — 2.65 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 2.65 → reject H₀
Why the other options are there
- z = 0.579 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3780 psi. A sample of 44 cores gives x̄ = 4010 psi, s = 269 psi. Test H₀: μ = 3780 against Hₐ: μ > 3780 at α = 0.05.
Given
- μ₀ = 3780 psi
- x̄ = 4010 psi
- s = 269 psi
- n = 44
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 269/√44 = 40.55 psi
Substituting
Compare — 5.67 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 5.67 → reject H₀
Why the other options are there
- z = 0.855 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3173 psi. A sample of 34 cores gives x̄ = 3320 psi, s = 255 psi. Test H₀: μ = 3173 against Hₐ: μ > 3173 at α = 0.05.
Given
- μ₀ = 3173 psi
- x̄ = 3320 psi
- s = 255 psi
- n = 34
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 255/√34 = 43.73 psi
Substituting
Compare — 3.36 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 3.36 → reject H₀
Why the other options are there
- z = 0.576 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3037 psi. A sample of 45 cores gives x̄ = 3182 psi, s = 341 psi. Test H₀: μ = 3037 against Hₐ: μ > 3037 at α = 0.05.
Given
- μ₀ = 3037 psi
- x̄ = 3182 psi
- s = 341 psi
- n = 45
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 341/√45 = 50.83 psi
Substituting
Compare — 2.85 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 2.85 → reject H₀
Why the other options are there
- z = 0.425 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3390 psi. A sample of 41 cores gives x̄ = 3546 psi, s = 308 psi. Test H₀: μ = 3390 against Hₐ: μ > 3390 at α = 0.05.
Given
- μ₀ = 3390 psi
- x̄ = 3546 psi
- s = 308 psi
- n = 41
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 308/√41 = 48.10 psi
Substituting
Compare — 3.24 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 3.24 → reject H₀
Why the other options are there
- z = 0.506 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3950 psi. A sample of 28 cores gives x̄ = 4082 psi, s = 260 psi. Test H₀: μ = 3950 against Hₐ: μ > 3950 at α = 0.05.
Given
- μ₀ = 3950 psi
- x̄ = 4082 psi
- s = 260 psi
- n = 28
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 260/√28 = 49.14 psi
Substituting
Compare — 2.69 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 2.69 → reject H₀
Why the other options are there
- z = 0.508 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 4118 psi. A sample of 26 cores gives x̄ = 4221 psi, s = 158 psi. Test H₀: μ = 4118 against Hₐ: μ > 4118 at α = 0.05.
Given
- μ₀ = 4118 psi
- x̄ = 4221 psi
- s = 158 psi
- n = 26
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 158/√26 = 30.99 psi
Substituting
Compare — 3.32 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 3.32 → reject H₀
Why the other options are there
- z = 0.652 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 3900 psi. A sample of 42 cores gives x̄ = 4050 psi, s = 202 psi. Test H₀: μ = 3900 against Hₐ: μ > 3900 at α = 0.05.
Given
- μ₀ = 3900 psi
- x̄ = 4050 psi
- s = 202 psi
- n = 42
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 202/√42 = 31.17 psi
Substituting
Compare — 4.81 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 4.81 → reject H₀
Why the other options are there
- z = 0.743 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
A mix must exceed 4050 psi. A sample of 37 cores gives x̄ = 4114 psi, s = 198 psi. Test H₀: μ = 4050 against Hₐ: μ > 4050 at α = 0.05.
Given
- μ₀ = 4050 psi
- x̄ = 4114 psi
- s = 198 psi
- n = 37
- z₀.₀₅ = 1.645
Find
Test statistic and conclusion
Start with the thinking
- One-sided alternative → one critical value.
- Compare the statistic, not the raw difference.
Step-by-step solution
Statistic — z = (x̄ − μ₀)/(s/√n)
Standard error — s/√n = 198/√37 = 32.55 psi
Substituting
Compare — 1.97 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement
Answer: z = 1.97 → reject H₀
Why the other options are there
- z = 0.323 (√n omitted)
- Two-tailed critical value 1.96 used
Reference: FE Reference Handbook — Probability and Statistics → Test Statistics
Self-check
Answer these without notes before moving on.
- Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
- Which assumption, if violated, makes the main relation of this section invalid?
- Given a sample of measurements from a construction or materials process, what is the first quantity you would compute, and why that one first?
- Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
- Rework Example 1 above from the givens alone, without reading the solution lines.
Chapter summary
- Test Statistics contains 13 relations; you must be able to find this page in under 15 seconds.
- Exam style: one distribution or one counting rule, then a single probability or interval.
- Unit rule: probabilities are dimensionless and must land in [0, 1].
- Work the 10 examples until the solution path, not the answer, is automatic.
Common traps in this section
- probabilities are dimensionless and must land in [0, 1]
- Answering the intermediate quantity instead of the quantity requested.
- Rounding intermediate values before the final step.
- Using a relation from an adjacent handbook section that shares a symbol.
- Skipping the sketch — most lost points on this page start with a misread geometry.