1、Designation: E122 091An American National StandardStandard Practice forCalculating Sample Size to Estimate, With SpecifiedPrecision, the Average for a Characteristic of a Lot orProcess1This standard is issued under the fixed designation E122; the number immediately following the designation indicate
2、s the year oforiginal adoption or, in the case of revision, the year of last revision. A number in parentheses indicates the year of last reapproval. Asuperscript epsilon () indicates an editorial change since the last revision or reapproval.This standard has been approved for use by agencies of the
3、 Department of Defense.1NOTEEditorial corrections were made to 8.4.1.2 in November 2011.1. Scope1.1 This practice covers simple methods for calculating howmany units to include in a random sample in order to estimatewith a specified precision, a measure of quality for all the unitsof a lot of materi
4、al, or produced by a process. This practice willclearly indicate the sample size required to estimate theaverage value of some property or the fraction of nonconform-ing items produced by a production process during the timeinterval covered by the random sample. If the process is not ina state of st
5、atistical control, the result will not have predictivevalue for immediate (future) production. The practice treats thecommon situation where the sampling units can be consideredto exhibit a single (overall) source of variability; it does nottreat multi-level sources of variability.2. Referenced Docu
6、ments2.1 ASTM Standards:2E456 Terminology Relating to Quality and Statistics3. Terminology3.1 DefinitionsUnless otherwise noted, all statisticalterms are defined in Terminology E456.3.2 SymbolsSymbols used in all equations are defined asfollows:E = the maximum acceptable difference between the truea
7、verage and the sample average.e = E/, maximum acceptable difference expressed as afraction of .f = degrees of freedom for a standard deviation estimate(7.5).k = the total number of samples available from the sameor similar lots. = lot or process mean or expected value of X, the resultof measuring al
8、l the units in the lot or process.0= an advance estimate of .N = size of the lot.n = size of the sample taken from a lot or process.nj= size of sample j.nL= size of the sample from a finite lot (7.4).p8 = fraction of a lot or process whose units have thenonconforming characteristic under investigati
9、on.p0= an advance estimate of p8.p = fraction nonconforming in the sample.R = range of a set of sampling values. The largest minusthe smallest observation.Rj= range of sample j.R=(j 5 1kRj/k , average of the range of k samples, all of thesame size (8.2.2).s = lot or process standard deviation of X,
10、the result ofmeasuring all of the units of a finite lot or process.s0= an advance estimate of s.s =(i 5 1nXi2 X!2/n 2 1!#1/2, an estimate of thestandard deviation s from n observation, Xi, i = 1 to n.s =(j 5 1kSj/k , average s from k samples all of the same size(8.2.1).sp= pooled (weighted average)
11、s from k samples, not all ofthe same size (8.2).sj= standard deviation of sample j.Vo= an advance estimate of V, equal to do/o.v = s/X, the coefficient of variation estimated from thesample.vp= pooled (weighted average) of v from k samples (8.3).1This practice is under the jurisdiction ofASTM Commit
12、tee E11 on Quality andStatistics and is the direct responsibility of Subcommittee E11.10 on Sampling /Statistics.Current edition approved Aug. 1, 2009. Published September 2009. Originallyapproved in 1958. Last previous edition approved in 2007 as E122 07. DOI:10.1520/E0122-09E01.2For referenced AST
13、M standards, visit the ASTM website, www.astm.org, orcontact ASTM Customer Service at serviceastm.org. For Annual Book of ASTMStandards volume information, refer to the standards Document Summary page onthe ASTM website.1Copyright ASTM International, 100 Barr Harbor Drive, PO Box C700, West Conshoho
14、cken, PA 19428-2959, United States.vj= coefficient of variation from sample j.X = numerical value of the characteristic of an individualunit being measured.X=(i 5 1nXi/niaverage of n observations, Xi,i=1ton.4. Significance and Use4.1 This practice is intended for use in determining thesample size re
15、quired to estimate, with specified precision, ameasure of quality of a lot or process. The practice applieswhen quality is expressed as either the lot average for a givenproperty, or as the lot fraction not conforming to prescribedstandards.The level of a characteristic may often be taken as anindic
16、ation of the quality of a material. If so, an estimate of theaverage value of that characteristic or of the fraction of theobserved values that do not conform to a specification for thatcharacteristic becomes a measure of quality with respect to thatcharacteristic. This practice is intended for use
17、in determiningthe sample size required to estimate, with specified precision,such a measure of the quality of a lot or process either as anaverage value or as a fraction not conforming to a specifiedvalue.5. Empirical Knowledge Needed5.1 Some empirical knowledge of the problem is desirablein advance
18、.5.1.1 We may have some idea about the standard deviationof the characteristic.5.1.2 If we have not had enough experience to give a preciseestimate for the standard deviation, we may be able to state ourbelief about the range or spread of the characteristic from itslowest to its highest value and po
19、ssibly about the shape of thedistribution of the characteristic; for instance, we might be ableto say whether most of the values lie at one end of the range,or are mostly in the middle, or run rather uniformly from oneend to the other (Section 9).5.2 If the aim is to estimate the fraction nonconform
20、ing,then each unit can be assigned a value of 0 or 1 (conforming ornonconforming), and the standard deviation as well as theshape of the distribution depends only on p8, the fractionnonconforming in the lot or process. Some rough idea con-cerning the size of p8 is therefore needed, which may bederiv
21、ed from preliminary sampling or from previous experi-ence.5.3 More knowledge permits a smaller sample size. Seldomwill there be difficulty in acquiring enough information tocompute the required size of sample. A sample that is largerthan the equations indicate is used in actual practice when theempi
22、rical knowledge is sketchy to start with and when thedesired precision is critical.5.4 The precision of the estimate made from a randomsample may itself be estimated from the sample. This estima-tion of the precision from one sample makes it possible to fixmore economically the sample size for the n
23、ext sample of asimilar material. In other words, information concerning theprocess, and the material produced thereby, accumulates andshould be used.6. Precision Desired6.1 The approximate precision desired for the estimate mustbe prescribed. That is, it must be decided what maximumdeviation, E, can
24、 be tolerated between the estimate to be madefrom the sample and the result that would be obtained bymeasuring every unit in the lot or process.6.2 In some cases, the maximum allowable sampling error isexpressed as a proportion, e, or a percentage, 100 e. Forexample, one may wish to make an estimate
25、 of the sulfurcontent of coal within 1 %, or e = 0.01.7. Equations for Calculating Sample Size7.1 Based on a normal distribution for the characteristic, theequation for the size, n, of the sample is as follows:n 5 3so/E!2(1)The multiplier 3 is a factor corresponding to a low probabil-ity that the di
26、fference between the sample estimate and theresult of measuring (by the same methods) all the units in thelot or process is greater than E. The value 3 is recommendedfor general use. With the multiplier 3, and with a lot or processstandard deviation equal to the advance estimate, it is practi-cally
27、certain that the sampling error will not exceed E. Wherea lesser degree of certainty is desired a smaller multiplier maybe used (Note 1).NOTE 1For example, multiplying by 2 in place of 3 gives a probabil-ity of about 45 parts in 1000 that the sampling error will exceed E.Although distributions met i
28、n practice may not be normal, the followingtext table (based on the normal distribution) indicates approximateprobabilities:Factor Approximate Probability of Exceeding E3 0.003 or 3 in 1000 (practical certainty)2.56 0.010 or 10 in 10002 0.045 or 45 in 10001.96 0.050 or 50 in 1000 (1 in 20)1.64 0.100
29、 or 100 in 1000 (1 in 10)7.1.1 If a lot of material has a highly asymmetric distribu-tion in the characteristic measured, the sample size as calcu-lated in Eq 1 may not be adequate. There are two things to dowhen asymmetry is suspected.7.1.1.1 Probe the material with a view to discovering, forexampl
30、e, extra-high values, or possibly spotty runs of abnor-mal character, in order to approximate roughly the amount ofthe asymmetry for use with statistical theory and adjustment ofthe sample size if necessary.7.1.1.2 Search the lot for abnormal material and segregate itfor separate treatment.7.2 There
31、 are some materials for which s varies approxi-mately with , in which case V (=s/) remains approximatelyconstant from large to small values of .7.2.1 For the situation of 7.2, the equation for the samplesize, n, is as follows:n 5 3 Vo/e!2(2)If the relative error, e, is to be the same for all values
32、of ,then everything on the right-hand side of Eq 2 is a constant;hence n is also a constant, which means that the same samplesize n would be required for all values of .E122 09127.3 If the problem is to estimate the lot fraction noncon-forming, then so2is replaced by po(1po) so that Eq 1becomes:n 5
33、3/E!2po1 2 po! (3)7.4 When the average for the production process is notneeded, but rather the average of a particular lot is needed, thenthe required sample size is less than Eq 1, Eq 2, and Eq 3indicate. The sample size for estimating the average of thefinite lot will be:nL5 n/1 1 n/N!# (4)where n
34、 is the value computed from Eq 1, Eq 2, or Eq 3. Thisreduction in sample size is usually of little importance unless nis 10 % or more of N.7.5 When the information on the standard deviation islimited, a sample size larger than indicated in Eq 1, Eq 2, andEq 3 may be appropriate. When the advance est
35、imate s0isbased on f degrees of freedom, the sample size in Eq 1 may bereplaced by:n 5 3s0/E!21 1 =2/f! (5)NOTE 2The standard error of a sample variance with f degrees offreedom, based on the normal distribution, is =2s4/f . The factor1 1 =2/f! has the effect of increasing the preliminary estimate s
36、02byone times its standard error.8. Reduction of Empirical Knowledge to a NumericalValue of so(Data for Previous Samples Available)8.1 This section illustrates the use of the equations inSection 7 when there are data for previous samples.8.2 For Eq 1An estimate of socan be obtained fromprevious sets
37、 of data. The standard deviation, s, from any givensample is computed as:s 5 (i 5 1nXi2 X!2/n 2 1!#1/2(6)The s value is a sample estimate of so. A better, more stablevalue for somay be computed by pooling the s values obtainedfrom several samples from similar lots. The pooled s value spfor k samples
38、 is obtained by a weighted averaging of the kresults from use of Eq 6.sp5 (j 5 1knj2 1!sj2/(j 5 1knj2 1!#1/2(7)8.2.1 If each of the previous data sets contains the samenumber of measurements, nj, then a simpler, but slightly lessefficient estimate for somay be made by using an average ( s)of the s v
39、alues obtained from the several previous samples. Thecalculated s value will in general be a slightly biased estimateof so. An unbiased estimate of sois computed as follows:so5sc4(8)where the value of the correction factor, c4, depends on thesize of the individual data sets (nj)(Table 13).8.2.2 An e
40、ven simpler, and slightly less efficient estimate forsomay be computed by using the average range ( R) takenfrom the several previous data sets that have the same groupsize.so5Rd2(9)The factor, d2, from Table 1 is needed to convert the averagerange into an unbiased estimate of so.8.2.3 Example 1Use
41、of s.8.2.3.1 ProblemTo compute the sample size needed toestimate the average transverse strength of a lot of bricks whenthe value of E is 50 psi, and practical certainty is desired.8.2.3.2 SolutionFrom the data of three previous lots, thevalues of the estimated standard deviation were found to be215
42、, 192, and 202 psi based on samples of 100 bricks. Theaverage of these three standard deviations is 203 psi. The c4value is essentially unity when Eq 1 gives the followingequation for the required size of sample to give a maximumsampling error of 50 psi:n 5 3 3 203!/5025 12.225 149 bricks (10)8.3 Fo
43、r Eq 2If s varies approximately proportionatelywith for the characteristic of the material to be measured,compute the average, X, the standard deviation, s, and thecoefficient of variation v for each sample. The pooled V valuevpfor k samples, not necessarily of the same size, is obtainedby a weighte
44、d average of the k results. Then use Eq 2.vp5 (j 5 1knj2 1!vj2/(j 5 1knj2 1!#1/2(11)8.3.1 Example 2Use of V, the estimated coefficient ofvariation:8.3.1.1 ProblemTo compute the sample size needed toestimate the average abrasion resistance (that is, averagenumber of cycles) of a material when the val
45、ue of e is 0.10 or10 %, and practical certainty is desired.8.3.1.2 SolutionThere are no data from previous samplesof this same material, but data for six samples of similarmaterials show a wide range of resistance. However, the valuesof estimated standard deviation are approximately proportionalto t
46、he observed averages, as shown in the following text table:Lot No.SampleSizeAvgCyclesStandardDeviationCoefficientof Varia-tion, %110 9013142 10 190 32 173 10 350 45 134 10 450 71 165 10 1000 120 126 10 3550 680 19Pooled 15.43ASTM Manual on Presentation of Data and Control Chart Analysis, ASTMMNL 7A,
47、 2002, Part 3.TABLE 1 Values of the Correction Factor C4and d2for SelectedSample Sizes njASample Size3,(nj) C4d22 .798 1.134 .921 2.065 .940 2.338 .965 2.8510 .973 3.08AAs njbecomes large, C4approaches 1.000.E122 0913The use of the pooled coefficient of variation for Voin Eq 2gives the following for
48、 the required size of sample to give amaximum sampling error not more than 10 % of the expectedvalue:n 5 3 3 15.4!/1025 21.322 test specimens (12)8.3.1.3 If a maximum allowable error of 5 % were needed,the required sample size would be 86 specimens. The datasupplied by the prescribed sample will be
49、useful for the studyin hand and also for the next investigation of similar material.8.4 For Eq 3Compute the estimated fraction nonconform-ing, p, for each sample. Then for the weighted average use thefollowing equation:p 5total number nonconforming in all samplestotal number of units in all samples(13)8.4.1 Example 3Use of p:8.4.1.1 ProblemTo compute the size of sample needed toestimate the fraction nonconforming in a lot of alloy steel trackbolts and nuts when the value of E is 0.04, and practicalcertainty
