Tests for Out of Control

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

0 formulas

10 exam-style examples

~45 min

All Probability and Statistics lectures

Learning objectives

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

This chapter section covers Tests for Out of Control 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 tests for out of control describes physically and when it applies.
State every one of the 0 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. Tests for Out of Control 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: tests for out of control.
Wikimedia Commons, public domain

Probability and Statistics — Tests for Out of Control: 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

1. A single point falls outside the (three sigma) control limits.
2. Two out of three successive points fall on the same side of and more than two sigma units from the center line.
3. Four out of five successive points fall on the same side of and more than one sigma unit from the center line.
4. Eight successive points fall on the same side of the center line.
Engineering Probability and Statistics
Probability and Density Functions: Means and Variances
Variable Equation Mean Variance

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

One-sided test on a mean strength — Tests for Out of Control

A mix must exceed 3715 psi. A sample of 21 cores gives x̄ = 3951 psi, s = 250 psi. Test H₀: μ = 3715 against Hₐ: μ > 3715 at α = 0.05.

Given

μ₀ = 3715 psi
x̄ = 3951 psi
s = 250 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 = 250/√21 = 54.55 psi
Substituting
$z = (3951 - 3715) /54.55 = 4.326$
Compare — 4.33 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 4.33 → reject H₀

Why the other options are there

z = 0.944 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 2

One-sided test on a mean strength — Tests for Out of Control (2)

A mix must exceed 3042 psi. A sample of 44 cores gives x̄ = 3302 psi, s = 359 psi. Test H₀: μ = 3042 against Hₐ: μ > 3042 at α = 0.05.

Given

μ₀ = 3042 psi
x̄ = 3302 psi
s = 359 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 = 359/√44 = 54.12 psi
Substituting
$z = (3302 - 3042) /54.12 = 4.804$
Compare — 4.80 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 4.80 → reject H₀

Why the other options are there

z = 0.724 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 3

One-sided test on a mean strength — Tests for Out of Control (3)

A mix must exceed 4100 psi. A sample of 28 cores gives x̄ = 4308 psi, s = 227 psi. Test H₀: μ = 4100 against Hₐ: μ > 4100 at α = 0.05.

Given

μ₀ = 4100 psi
x̄ = 4308 psi
s = 227 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 = 227/√28 = 42.90 psi
Substituting
$z = (4308 - 4100) /42.90 = 4.849$
Compare — 4.85 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 4.85 → reject H₀

Why the other options are there

z = 0.916 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 4

One-sided test on a mean strength — Tests for Out of Control (4)

A mix must exceed 3728 psi. A sample of 28 cores gives x̄ = 3925 psi, s = 329 psi. Test H₀: μ = 3728 against Hₐ: μ > 3728 at α = 0.05.

Given

μ₀ = 3728 psi
x̄ = 3925 psi
s = 329 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 = 329/√28 = 62.18 psi
Substituting
$z = (3925 - 3728) /62.18 = 3.168$
Compare — 3.17 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 3.17 → reject H₀

Why the other options are there

z = 0.599 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 5

One-sided test on a mean strength — Tests for Out of Control (5)

A mix must exceed 4114 psi. A sample of 44 cores gives x̄ = 4275 psi, s = 156 psi. Test H₀: μ = 4114 against Hₐ: μ > 4114 at α = 0.05.

Given

μ₀ = 4114 psi
x̄ = 4275 psi
s = 156 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 = 156/√44 = 23.52 psi
Substituting
$z = (4275 - 4114) /23.52 = 6.846$
Compare — 6.85 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 6.85 → reject H₀

Why the other options are there

z = 1.032 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 6

One-sided test on a mean strength — Tests for Out of Control (6)

A mix must exceed 3513 psi. A sample of 27 cores gives x̄ = 3687 psi, s = 189 psi. Test H₀: μ = 3513 against Hₐ: μ > 3513 at α = 0.05.

Given

μ₀ = 3513 psi
x̄ = 3687 psi
s = 189 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 = 189/√27 = 36.37 psi
Substituting
$z = (3687 - 3513) /36.37 = 4.784$
Compare — 4.78 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 4.78 → reject H₀

Why the other options are there

z = 0.921 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 7

One-sided test on a mean strength — Tests for Out of Control (7)

A mix must exceed 3657 psi. A sample of 23 cores gives x̄ = 3880 psi, s = 358 psi. Test H₀: μ = 3657 against Hₐ: μ > 3657 at α = 0.05.

Given

μ₀ = 3657 psi
x̄ = 3880 psi
s = 358 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 = 358/√23 = 74.65 psi
Substituting
$z = (3880 - 3657) /74.65 = 2.987$
Compare — 2.99 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.99 → reject H₀

Why the other options are there

z = 0.623 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 8

One-sided test on a mean strength — Tests for Out of Control (8)

A mix must exceed 3727 psi. A sample of 23 cores gives x̄ = 3803 psi, s = 313 psi. Test H₀: μ = 3727 against Hₐ: μ > 3727 at α = 0.05.

Given

μ₀ = 3727 psi
x̄ = 3803 psi
s = 313 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 = 313/√23 = 65.27 psi
Substituting
$z = (3803 - 3727) /65.27 = 1.164$
Compare — 1.16 < 1.645
Conclusion — fail to reject H₀; the evidence is insufficient

Answer: z = 1.16 → fail to reject H₀

Why the other options are there

z = 0.243 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 9

One-sided test on a mean strength — Tests for Out of Control (9)

A mix must exceed 3568 psi. A sample of 22 cores gives x̄ = 3804 psi, s = 244 psi. Test H₀: μ = 3568 against Hₐ: μ > 3568 at α = 0.05.

Given

μ₀ = 3568 psi
x̄ = 3804 psi
s = 244 psi
n = 22
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 = 244/√22 = 52.02 psi
Substituting
$z = (3804 - 3568) /52.02 = 4.537$
Compare — 4.54 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 4.54 → reject H₀

Why the other options are there

z = 0.967 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

Example 10

One-sided test on a mean strength — Tests for Out of Control (10)

A mix must exceed 3855 psi. A sample of 33 cores gives x̄ = 3986 psi, s = 312 psi. Test H₀: μ = 3855 against Hₐ: μ > 3855 at α = 0.05.

Given

μ₀ = 3855 psi
x̄ = 3986 psi
s = 312 psi
n = 33
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 = 312/√33 = 54.31 psi
Substituting
$z = (3986 - 3855) /54.31 = 2.412$
Compare — 2.41 > 1.645
Conclusion — reject H₀; the mean exceeds the requirement

Answer: z = 2.41 → reject H₀

Why the other options are there

z = 0.420 (√n omitted)
Two-tailed critical value 1.96 used

Reference: FE Reference Handbook — Probability and Statistics → Tests for Out of Control

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

Tests for Out of Control contains 0 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.