Hypothesis Testing

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

13 formulas

10 exam-style examples

~60 min

All Probability and Statistics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Hypothesis Testing 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 hypothesis testing 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. Hypothesis Testing 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.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: hypothesis testing.
Wikimedia Commons, public domain

Probability and Statistics — Hypothesis Testing: 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.

Notation used in this section

yi•	Quantity produced by "yi• = / yij and y•• / / yij" — read its definition and unit from the handbook line directly above the equation.
j	Quantity produced by "j=1 i=1 j=1" — read its definition and unit from the handbook line directly above the equation.
i	Quantity produced by "i=1 j=1 i=1 i=1i=1" — read its definition and unit from the handbook line directly above the equation.
SStotal	Quantity produced by "SStotal = SStreatments + SSerror" — read its definition and unit from the handbook line directly above the equation.
If N	Quantity produced by "If N = total number observations" — read its definition and unit from the handbook line directly above the equation.
SSerror	Quantity produced by "SSerror = SStotal − SStreatments" — 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.

Let a "dot" subscript indicate summation over the subscript. Thus:
n a n
One-Way Analysis of Variance (ANOVA)
Given independent random samples of size ni from k populations, then:
k ni k k ni
k ni y2
k y2 y2
Montgomery, Douglas C., and George C. Runger, Applied Statistics and Probability for Engineers, 4 ed., New York: John Wiley and Sons, 2007.

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.

Example 1

95% confidence interval on mean density

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
$E = t \cdot S E = 2.306 (1.20) = 2.77 p c f$
Lower limit
$121.4 - 2.77 = 118.6 p c f$
Upper limit
$121.4 + 2.77 = 124.2 p c f$
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

Example 2

One-sided test on a mean strength — Hypothesis Testing

A mix must exceed 3516 psi. A sample of 36 cores gives x̄ = 3772 psi, s = 284 psi. Test H₀: μ = 3516 against Hₐ: μ > 3516 at α = 0.05.

Given

μ₀ = 3516 psi
x̄ = 3772 psi
s = 284 psi
n = 36
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 = 284/√36 = 47.33 psi
Substituting
$z = (3772 - 3516) /47.33 = 5.408$
Compare — 5.41 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 5.41 → reject H₀

Why the other options are there

z = 0.901 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 3

One-sided test on a mean strength — Hypothesis Testing (2)

A mix must exceed 3116 psi. A sample of 31 cores gives x̄ = 3250 psi, s = 325 psi. Test H₀: μ = 3116 against Hₐ: μ > 3116 at α = 0.05.

Given

μ₀ = 3116 psi
x̄ = 3250 psi
s = 325 psi
n = 31
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 = 325/√31 = 58.37 psi
Substituting
$z = (3250 - 3116) /58.37 = 2.296$
Compare — 2.30 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.30 → reject H₀

Why the other options are there

z = 0.412 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 4

One-sided test on a mean strength — Hypothesis Testing (3)

A mix must exceed 4085 psi. A sample of 40 cores gives x̄ = 4205 psi, s = 235 psi. Test H₀: μ = 4085 against Hₐ: μ > 4085 at α = 0.05.

Given

μ₀ = 4085 psi
x̄ = 4205 psi
s = 235 psi
n = 40
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 = 235/√40 = 37.16 psi
Substituting
$z = (4205 - 4085) /37.16 = 3.230$
Compare — 3.23 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 3.23 → reject H₀

Why the other options are there

z = 0.511 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 5

One-sided test on a mean strength — Hypothesis Testing (4)

A mix must exceed 3732 psi. A sample of 29 cores gives x̄ = 3862 psi, s = 248 psi. Test H₀: μ = 3732 against Hₐ: μ > 3732 at α = 0.05.

Given

μ₀ = 3732 psi
x̄ = 3862 psi
s = 248 psi
n = 29
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 = 248/√29 = 46.05 psi
Substituting
$z = (3862 - 3732) /46.05 = 2.823$
Compare — 2.82 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.82 → reject H₀

Why the other options are there

z = 0.524 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 6

One-sided test on a mean strength — Hypothesis Testing (5)

A mix must exceed 3764 psi. A sample of 24 cores gives x̄ = 3971 psi, s = 353 psi. Test H₀: μ = 3764 against Hₐ: μ > 3764 at α = 0.05.

Given

μ₀ = 3764 psi
x̄ = 3971 psi
s = 353 psi
n = 24
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 = 353/√24 = 72.06 psi
Substituting
$z = (3971 - 3764) /72.06 = 2.873$
Compare — 2.87 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.87 → reject H₀

Why the other options are there

z = 0.586 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 7

One-sided test on a mean strength — Hypothesis Testing (6)

A mix must exceed 3962 psi. A sample of 42 cores gives x̄ = 4172 psi, s = 209 psi. Test H₀: μ = 3962 against Hₐ: μ > 3962 at α = 0.05.

Given

μ₀ = 3962 psi
x̄ = 4172 psi
s = 209 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 = 209/√42 = 32.25 psi
Substituting
$z = (4172 - 3962) /32.25 = 6.512$
Compare — 6.51 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 6.51 → reject H₀

Why the other options are there

z = 1.005 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 8

One-sided test on a mean strength — Hypothesis Testing (7)

A mix must exceed 3109 psi. A sample of 27 cores gives x̄ = 3358 psi, s = 354 psi. Test H₀: μ = 3109 against Hₐ: μ > 3109 at α = 0.05.

Given

μ₀ = 3109 psi
x̄ = 3358 psi
s = 354 psi
n = 27
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 = 354/√27 = 68.13 psi
Substituting
$z = (3358 - 3109) /68.13 = 3.655$
Compare — 3.65 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 3.65 → reject H₀

Why the other options are there

z = 0.703 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

Example 9

One-sided test on a mean strength — Hypothesis Testing (8)

A mix must exceed 3450 psi. A sample of 23 cores gives x̄ = 3585 psi, s = 266 psi. Test H₀: μ = 3450 against Hₐ: μ > 3450 at α = 0.05.

Given

μ₀ = 3450 psi
x̄ = 3585 psi
s = 266 psi
n = 23
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 = 266/√23 = 55.46 psi
Substituting
$z = (3585 - 3450) /55.46 = 2.434$
Compare — 2.43 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.43 → 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 → Hypothesis Testing

Example 10

One-sided test on a mean strength — Hypothesis Testing (9)

A mix must exceed 4161 psi. A sample of 34 cores gives x̄ = 4257 psi, s = 190 psi. Test H₀: μ = 4161 against Hₐ: μ > 4161 at α = 0.05.

Given

μ₀ = 4161 psi
x̄ = 4257 psi
s = 190 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 = 190/√34 = 32.58 psi
Substituting
$z = (4257 - 4161) /32.58 = 2.946$
Compare — 2.95 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.95 → reject H₀

Why the other options are there

z = 0.505 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Hypothesis Testing

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

Hypothesis Testing 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.