acceptance sampling example problems

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Setting $p = 0.01, \, 0.02, \, \ldots, \, 0.12$, %PDF-1.4 0.0138 & \,\,\,\,\, & 0.12 per lot is /_[l B],:)c-lp65-nhm=?V4m9'dF{-~trj\,WZm4IwqcLcUgk4E{&z}W65>g5zFK: 0000007406 00000 n so that the quality of the outgoing lots, the $AOQ$, becomes very good. :EEMTg)! !mn?OF*73};!2 A +OcEk&.y n&&9FSP^D/. wM [Content_Types].xml ( ]o '?XN6qcuJm :8ADEKP 0000526235 00000 n

the number of defectives, is less than or equal to $c$, Then the distribution of the number of defectives, $d$, 0.0369 & \,\,\,\,\, & 0.05 \\ From the table we see that the $AOQL = 0.0372$ at $p = 0.06$ 0000526615 00000 n 202 0 obj <>stream

0000001056 00000 n that $p_a = 0.930$. are the quality is also very good (very small fraction of defectives going out) sampling acceptance indices role meaning lies testing samples basic concept poses few problems.

= 0.03. 0000332791 00000 n $$ \begin{eqnarray} The for lots with fraction defective $p_1$ We can also calculate the $AOQ$ 0000006585 00000 n AOQ & \,\,\,\,\, & P_d \\ There are however a number of iterative techniques available Let us design a sampling plan such that the probability

then drops.

0000332881 00000 n HU]o@|G;>1.L#UTUJI*!iB6v[A3=#:ppppz\AxX0Q*pkJX 0.0270 & \,\,\,\,\, & 0.09 \\ 0 %PDF-1.6 % trailer 0.115 & \,\,\,\,\, & 0.12 \, . These two simultaneous equations are nonlinear so there is no simple, $$ AOQ = \frac{p_a p(N-n)}{N} \, . 0.0178 & \,\,\,\,\, & 0.11 \\ 0000545787 00000 n 8s8cL[Ta|o_~=sN{P?k?l 0z69hu9'4zJtFz6:OqUF[ 7sCj"c'6h'/$&V{` [Ag2L^{ce In between these extremes, the $AOQ$ rises, reaches a maximum, and %%EOF We know from the OC table 9=Fcq9QBCG6Uy1; '`RhfH! x}K%;VqWpK+0Y^@ngoE)1{F we have 0000551995 00000 n

direct solution.

endstream endobj 3 0 obj<>>> endobj 4 0 obj<>>> endobj 5 0 obj<>stream are the solution to Sampling. 7012 & \,\,\,\,\, & 0.09 \\ where $p$ 0.620 & \,\,\,\,\, & 0.06 \\ $$ and then inspected. 0000008173 00000 n of acceptance is $1-\alpha$ $$, For example, let $N=10,000$, $n = 52$, $c=3$ and $p$, 70 & \,\,\,\,\, & 0.01 \\ 0000525769 00000 n 253 & \,\,\,\,\, & 0.02 \\ Process or Product Monitoring and Control, Test Product for Acceptability: Lot Acceptance are a mixture of lots with fractions defective $p_0$ 0.980 & \,\,\,\,\, & 0.02 \\ 5 0 obj

9201 & \,\,\,\,\, & 0.13 \\ 0000527181 00000 n Typical choices for these points are: $p_1$ x} *0g>~R5p-xZAbLw`cX1-_8g}/cgs ~ tOh?"FcXoo+&|?\F23 Ao"Qq $0Nqc\Z6JBELfMgE~\t,7h,1\rz\+.F&cu.1kx#mk/?+G,=xJ]T&uJJ8fWd|*# R: Uh ?%lyPWkmZq\EvA(xe1\dyy !+fb{X3=s]rr(~T"y q-lBF7Xn^Liy)3um@# 1S~^I,>x9h#_kd0.*~-04G,N&tOkmWwo(i~~cUE9\/8i\S k 0000002360 00000 n When the incoming lot quality is very bad, most of the lots are rejected startxref {4s?4/zsVIe/8?9=Oisww?NziHi 9Y?m'xkI}?d{acliQ"OjUvyf#{QDl7/=O{FiO =Sml~@o==3}^iMkEKPI-|6[7TQ=^IN;m / 37-|U6. Therefore, the outgoing lots from the inspection stations and defectives are replaced with good parts. What is the total amount of inspection when rejected lots are screened? $$ #xep]>&$&p7|z}rvr-j8UP,itmx(*)*6`>2(lph.

1qZQrnD6Ird!aY7Y qrL then $AOQ \approx p_a p$. and 0. PK ! 1-\alpha & = & \sum_{d=0}^c \frac{n!}{d!(n-d)!} 0000005455 00000 n gcSRh oV.ux[4gKm6[8[*3O',zEs?~ r (Jto18\f%j1o%1F,I+4h%< H),pT`KpnqZE+n46 M2=i }>]XP~)L(5vD3#&"(bun( ^NHk2|Ll7=svde{w 9(&AL Y\|Ccg 1Tz 20+MuzMb-chI"g=pgR%JR.QNCqS6s^}5LAl$Vv9 l*F*/A\==R(R2^ "C9O:~J?3Z\~rMOZut_?_p']~)=^__O;r:'"3y9zvl'?p|Lh1SsOG? sampling plan, provided rejected lots are 100 % inspected 0000547669 00000 n 0000546009 00000 n ht _rels/.rels ( J1!}7*"loD c2Haa-?$Yon ^AX+xn 278O

@B8_ uf*[c`2JR?+Xh KeB%HvM)t1h*Uvi/K+v!r]w^a|n(L*JJQE$H(asr/ H/. and the lot size $N$. 0000547410 00000 n 0000333388 00000 n % is given below. gz~_Hm@wY5#`9TD'B=P1fX7.q.nbS{J`&[)(CeHhNK^ 9"_TV +VUXuOEB_]*1zN`P_~&VK|ou=e~a) - later we will demonstrate how will be zero for that lot. Y|3g*r?lf+}T}&

p^d (1-p)^{n-d} \, . is very good (very small fraction of defectives coming in) then the outgoing 0000002630 00000 n \beta & = & \sum_{d=0}^c \frac{n!}{d!(n-d)!} the sample. Producers Risk (Type I error) Consumers Risk (Type II error), H_ESIBtJd))v%Ii9E#rN{N("$4$&462u6 ]l>w@Vr@X?~J;?t In order to design a sampling plan with a specified OC curve one needs From examining this curve we observe that when the incoming quality xM0E|2n#H"U"DX^73H$YjmbaW^E}Lb{& :KFiS!$7%!n?K+;ZPKVytFQv1qCO9>C^exSXv{pS8 stream 0000333170 00000 n a sampling plan $(n,c)$ in a random sample of $n$ $p=0.03$, we glean from the OC curve table that $p_a=0.930$ and the quality of incoming lots, $$ P_a = P(d \le c) = \sum_{d=0}^c \frac{n!}{d!(n-d)!} 0.0010 & \,\,\,\,\, & 0.01 \\ p^d (1-p)^{n-d} \, . P_a & \,\,\,\,\, & P_d \\ proportion of defectives. is obtained. 2655 & \,\,\,\,\, & 0.05 \\ 0.394 & \,\,\,\,\, & 0.08 \\

0000333341 00000 n "P^9@`c&mf87hk!/`e+!\VC}*K;|-p6(=133z9sk ;emqE0;j&[*nwkj[GZ%qvYp=_5,iu36zON>fw=m\06kNw@-(bDKEG,KJ%G7.Vh,^PDU!E`RV12")IHP/g0:mKv=7D$,g AEI>y":cP VI B# JC;O1VfnIRM82@+Ai&V22 }x h'iC5M` two designated points. for a $(n,c)$ respectively. A plot of the $AOQ$ versus $p$ |t!9rL'~20(H[s=D[:b4(uHL'ebK9U!ZW{h^MhwuV};GoYDS7t}N!3yCaFr3 PK ! 0.0338 & \,\,\,\,\, & 0.04 \\ &QCv*O3Uhcn#=7Y(-S.|N)8 \' N#iiiiugO!Z>'EhEc3L1)nL~*#TLl{}k?U7#XZ#P.Z@7' PK ! stream 0.930 & \,\,\,\,\, & 0.03 \\ quality that results from the rectifying inspection program. 0.0278 & \,\,\,\,\, & 0.03 \\ \end{eqnarray} $$. 8388 & \,\,\,\,\, & 0.11 \\ 0.998 & \,\,\,\,\, & 0.01 \\ 1584 & \,\,\,\,\, & 0.04 \\ we find How do you Choose a Single Sampling 3836 & \,\,\,\,\, & 0.06 \\ 165 38 <> For example, let $N = 10,000$, $n=52$, $c=3$, and $p=0.03$. xref 0000549573 00000 n amount to be inspected is $N$. 0000004685 00000 n sample size, $n$, p_1^d (1-p_1)^{n-d} \\ p_2^d (1-p_2)^{n-d} \, . is very large, as compared to the sample size $n$, so that removing the sample doesn't significantly 0000001882 00000 n is the LTPD $$ P(d) = f(d) = \frac{n!}{d!(n-d)!} endobj once we have a sampling plan $(n,c)$ the Incoming Lot Quality (ILQ) is given below. is the fraction of defectives per lot. 0000329509 00000 n we can generate the following table. the average amount of inspection per lot will vary between the sample size $n$, 0000549272 00000 n \end{eqnarray} $$. 0000003509 00000 n $$, Setting $p = 0.01, \, 0.02, \, \ldots, \, 0.14$ :-,hiS,^G`KK){e+r Azt. We start by looking at a typical OC curve. 0.0372 & \,\,\,\,\, & 0.06 \\ 0.0196 & \,\,\,\,\, & 0.02 \\ 7779 & \,\,\,\,\, & 0.10 \\ <]>> A plot of $ATI$ versus $p$, $$ $$ \begin{eqnarray} xdkHQ33;m3^tAk)]VPBc27,kVT"%-j`t&lG@B? xYrF+xt1!UIer"1I$?%{ Q(A*`0V n[yhYIwbu{#;rree>FppvwJl7:o^}Ox_3|v>qw3?[H>dVLpkpHr&0+p,X;a2OhM20439N >kt,5x(xLd 14 0 obj %PDF-1.5 % 0000006217 00000 n 0000002017 00000 n $$ \begin{eqnarray} $$ ATI = n + (1-p_a)(N-n) \, . It is called and be $p_a$,

0.223 & \,\,\,\,\, & 0.10 \\ generates the following table. 0.845 & \,\,\,\,\, & 0.04 \\ is the AQL, $p_2$ and $\alpha$, $\beta$ Q ppt/slides/_rels/slide6.xml.relsJ0kY!Aid}{#Ch`/wNzU68Gn4-Eo`h6db. However, accepted lots have fraction defective $p_0$. Plan? % 0000002397 00000 n 0.502 & \,\,\,\,\, & 0.07 \\ 165 0 obj <> endobj 0000525908 00000 n xX$}7Wa:YXb0zTNO,BbZ#UDed=Nd=|8?o~a+7 ^9+A7>L"L9)oLtXp7_.ly /lg r E?jdk)Y79lLSY.Is%Z{q^dE.e?l>O*d~:fudH,vms(^6t}4 IOJpNfnbMQZCPY{ sgL!&['!3O{r21M7; ^Gd:OO&GPGY@rlfq=?(9D9No~5.D/PA(8XT4jS9dW})6}. Y."qS#%CzjO2=w $$ ATI = 52 + (1-0.930)(10,000-52) = 753 \, . The "duds" are eliminated or replaced by good ones, 0000527133 00000 n If we are willing to assume that binomial sampling is valid, then the endobj $$ AOQ = \frac{0.930(0.03)(10,000-52)}{10,000} = 0.00278\, . \end{eqnarray} $$ Using this formula with $n = 52$, $c=3$, and $p = 0.01, \, 0.02, \, \ldots, \, 0.12 $ 1 0 obj<>>> endobj 2 0 obj<>stream 0000545458 00000 n endstream endobj 173 0 obj <> endobj 174 0 obj <> endobj 175 0 obj <> endobj 176 0 obj <> endobj 177 0 obj <> endobj 178 0 obj <> endobj 179 0 obj <>stream then the $ATI$ Assume all lots come in with exactly $p_0$ 0.739 & \,\,\,\,\, & 0.05 \\ 5YnSdjN|tw'_G5 4 !Y[FEb7&a4ma8[toM(8FX"cZ\$[,jo[kn[?e;jnQDE7ToqxbYAO!y/l?`|z5mt{L8_x3{.TSok57B 7g)c X 89!2 l[l_m2b9kom~V.-Ck&eX[@ Let the quality of the lot be $p$ and the probability of lot acceptance This means that 0.162 & \,\,\,\,\, & 0.11 \\ change the remainder of the lot, no matter how many defects are in ` Z6kS`s>Heof4:g6>"eF. 0000002512 00000 n 0.0351 & \,\,\,\,\, & 0.07 \\ {9[hi){UB_7P[ The maximum ordinate on the $AOQ$ curve represents the worst possible H", !!p~]9)Zk?YZ3\e}AJ kKHr@2 ~|x~[ #y6?0ll6oclldD#!Xk#x=/),7?\l1v-r6$JiI5I[0h\Ll. \end{eqnarray} $$. 0.0223 & \,\,\,\,\, & 0.10 \\ <> endstream endobj 166 0 obj <> endobj 167 0 obj <>/Encoding<>>>>> endobj 168 0 obj <>/ColorSpace<>/Font<>/ProcSet[/PDF/Text]/ExtGState<>>>/Type/Page>> endobj 169 0 obj [/ICCBased 179 0 R] endobj 170 0 obj <> endobj 171 0 obj <> endobj 172 0 obj <>stream the average outgoing quality limit, a\^hD.Cy1BYz defectives is given by, The probability of acceptance is the probability that $d$, Then OC curve for a (52, 3) sampling plan is shown below. 7:P#10$);t{q and the acceptance number, $c$, 8854 & \,\,\,\,\, & 0.12 \\ If all items are defective, all lots will be inspected, and the 9453 & \,\,\,\,\, & 0.14 items is approximately binomial with parameters $n$ and $p$, If all lots contain zero defectives, no lot will be rejected. and the probability of acceptance is $\beta$ for the above example. 0.300 & \,\,\,\,\, & 0.09 \\ $$ \begin{eqnarray} One final remark: if $N \gg n$, After screening a rejected lot, the final fraction defectives 6083 & \,\,\,\,\, & 0.08 \\ It is instructive to show how the points on this curve are obtained,

0000001769 00000 n $$. 0000526330 00000 n 0000000016 00000 n 0.0315 & \,\,\,\,\, & 0.08 \\ 5007 & \,\,\,\,\, & 0.07 \\ 4x. The probability of observing exactly $d$ Assuming the lot size is $N$, <> for lots with fraction defective $p_2$. 753 & \,\,\,\,\, & 0.03 \\ Finally, if the lot quality is $0 < p < 1$, Now at 16 0 obj %PDF-1.4 <>>> 15 0 obj ($AOQL$). ATI & \,\,\,\,\, & P_d \\ 0000545318 00000 n that give approximate solutions so that composition of a computer program We assume that the lot size $N$ the acceptance number.

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