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

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

$\mu_{0} = 3715 psi$
$x̄ = 3951 psi$
$s = 250 psi$
$n = 21$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3042 psi$
$x̄ = 3302 psi$
$s = 359 psi$
$n = 44$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 4100 psi$
$x̄ = 4308 psi$
$s = 227 psi$
$n = 28$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3728 psi$
$x̄ = 3925 psi$
$s = 329 psi$
$n = 28$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 4114 psi$
$x̄ = 4275 psi$
$s = 156 psi$
$n = 44$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3513 psi$
$x̄ = 3687 psi$
$s = 189 psi$
$n = 27$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3657 psi$
$x̄ = 3880 psi$
$s = 358 psi$
$n = 23$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3727 psi$
$x̄ = 3803 psi$
$s = 313 psi$
$n = 23$
$z_{0}._{05} = 1.645$

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 \to fail to reject H_{0}

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

$\mu_{0} = 3568 psi$
$x̄ = 3804 psi$
$s = 244 psi$
$n = 22$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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

$\mu_{0} = 3855 psi$
$x̄ = 3986 psi$
$s = 312 psi$
$n = 33$
$z_{0}._{05} = 1.645$

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 \to reject H_{0}

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