2016
DOI: 10.1088/1681-7575/54/1/59
|View full text |Cite
|
Sign up to set email alerts
|

Sampling for assurance of future reliability

Abstract: Ensuring measurement trueness, compliance with regulations and conformity with standards are key tasks in metrology which are often considered at the time of an inspection. Current practice does not always verify quality after or between inspections, calibrations, laboratory comparisons, conformity assessments, etc. Statistical models describing behavior over time may ensure reliability, i.e. they may give the probability of functioning, compliance or survival until some future point in time.It may not always … Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

0
20
0

Year Published

2018
2018
2024
2024

Publication Types

Select...
5

Relationship

1
4

Authors

Journals

citations
Cited by 5 publications
(20 citation statements)
references
References 4 publications
0
20
0
Order By: Relevance
“…Instead of requiring a large sample to prove the high reliability p=P()false|Xfalse|normalΔ, it is sufficient to apply a smaller sample to prove the smaller reliability P()false|Xfalse|normalΔfalse/γ<p at tighter specifications, while achieving the same statistical type 1 error (ie, the same probability of falsely accepting an unreliable population). For example, when ( n , c ) denotes the sampling plan requiring n devices to be verified and c of these devices to comply, then the sampling plan (80,3) is sufficient instead of (141,1) for a population of size 1201 to 3200, for p =0.973, q =0.92 (table D1 in ISO 2859‐2, and Klauenberg and Elster) and thus γ N =1.263 for Normally (and γ t =1.557 for t 3 ‐) distributed deviations (see Section 4.2 and Equation for the calculation of these factors). That means that, for this particular example, the sample size almost halves and in addition the acceptance number relaxes considerably when the maximum permissible error is decreased to 80% (or 64%, respectively).…”
Section: Theorymentioning
confidence: 99%
See 4 more Smart Citations
“…Instead of requiring a large sample to prove the high reliability p=P()false|Xfalse|normalΔ, it is sufficient to apply a smaller sample to prove the smaller reliability P()false|Xfalse|normalΔfalse/γ<p at tighter specifications, while achieving the same statistical type 1 error (ie, the same probability of falsely accepting an unreliable population). For example, when ( n , c ) denotes the sampling plan requiring n devices to be verified and c of these devices to comply, then the sampling plan (80,3) is sufficient instead of (141,1) for a population of size 1201 to 3200, for p =0.973, q =0.92 (table D1 in ISO 2859‐2, and Klauenberg and Elster) and thus γ N =1.263 for Normally (and γ t =1.557 for t 3 ‐) distributed deviations (see Section 4.2 and Equation for the calculation of these factors). That means that, for this particular example, the sample size almost halves and in addition the acceptance number relaxes considerably when the maximum permissible error is decreased to 80% (or 64%, respectively).…”
Section: Theorymentioning
confidence: 99%
“…In particular, let us assume a population of size N = 3182 and a required reliability of 100 q =97.3% at the confidence level of 90% and at the specification Δ. Then sampling plan ( n , c )=(141,1) is to be applied when sampling without replacement (cf ISO 2859‐2, and Klauenberg and Elster and the hypergeometric distribution). That is, if 2.7% of the devices in the population fail—due to any reason—then the probability of falsely accepting this population is 9.8%.…”
Section: Requirementsmentioning
confidence: 99%
See 3 more Smart Citations