(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 172757, 4286] NotebookOptionsPosition[ 162280, 3967] NotebookOutlinePosition[ 163871, 4022] CellTagsIndexPosition[ 163828, 4019] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["Planes de Muestreo", "Title"], Cell[TextData[{ StyleBox["\nGuillermo S\[AAcute]nchez \n", FontSize->12], StyleBox["Actualizado: 2007-09-20 (ejectuado de nuevo con ", FontSize->10], StyleBox["Mathematica", FontSize->10, FontSlant->"Italic"], StyleBox[" 8.0 en Mayo 2012, sin modificar)", FontSize->10] }], "Author", CellChangeTimes->{{3.399363434803706*^9, 3.3993634401652904`*^9}, 3.4180324340485373`*^9, 3.418032480142582*^9, 3.5464999685216246`*^9, { 3.5465001823512*^9, 3.5465002377156973`*^9}}], Cell[CellGroupData[{ Cell["Plan de muestreo simple", "Section"], Cell[CellGroupData[{ Cell["Fundamentos", "Subsection"], Cell[TextData[{ "Nuestro objetivo es establecer un muestreo por atributos \ (aceptaci\[OAcute]n o rechazo) para un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], " del que se toma una muestra ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", aceptandose el lote si la cantidad piezas no conformes en la muestra es \ menor o igual que un valor ", Cell[BoxData[ FormBox[ RowBox[{"r", " "}], TraditionalForm]]], "establecido como criterio de aceptaci\[OAcute]n. " }], "Text"], Cell[TextData[{ "Llamamos ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " a la proporci\[OAcute]n verdadera de uds. disconformes en el lote, que \ nos es desconocida (salvo que se inspeccione el 100%). En un plan de muestreo \ se establece como criterio de aceptaci\[OAcute]n ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " \[LessEqual] ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], " siendo ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], " la fracci\[OAcute]n maxima de unidades defectuosas en el lote que con una \ probabilidad ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], " estamos dispuestos a aceptar. A estos valores de (", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], ") se les denomina usualmente Nivel de Calidad Aceptable (o AQL)-: Por \ tanto, el AQL corresponde a la fracci\[OAcute]n defectuosa, ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", en un lote tal que, para un plan de muestreo determinado, \[EAcute]ste \ ser\[AAcute] rechazado con una probabilidad \[Alpha] = 1 - ", Cell[BoxData[ FormBox[ SubscriptBox["P", "A"], TraditionalForm]]], ". \[Alpha] es conocido como riesgo del fabricante. En la curva OC es de \ inter\[EAcute]s tambien el nivel de calidad rechazable (o RQL) que \ corresponde a la fracci\[OAcute]n de defectuosas de un lote tal que, para un \ plan de muestreo determinado, \[EAcute]ste ser\[AAcute] aceptado con una \ probabilidad \[Beta], conocida como riesgo del consumidor o comprador." }], "Text"], Cell[TextData[{ "En terminos de test de hip\[OAcute]tesis lo anterior podemos expresarlo \ como sigue:\nLa fracci\[OAcute]n ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " real es desconocida. Al tomar una muestra se trata de estimar si ", Cell[BoxData[ FormBox[ RowBox[{"p", " ", "\[LessEqual]", " ", SubscriptBox["p", "A"]}], TraditionalForm]]], " (aceptamos el lote) o ", Cell[BoxData[ FormBox[ RowBox[{"p", ">", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TraditionalForm]]], " (aceptamos el lote). Esto es equivalente a realizar un contraste de hip\ \[OAcute]tesis en el que la hip\[OAcute]tesis nula es ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], ": ", Cell[BoxData[ FormBox[ RowBox[{"p", " ", "\[LessEqual]", " ", SubscriptBox["p", "A"]}], TraditionalForm]]], " y la alternativa ", Cell[BoxData[ FormBox[ SubscriptBox["H", "1"], TraditionalForm]]], ": ", Cell[BoxData[ FormBox[ RowBox[{"p", ">", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TraditionalForm]]], " . Al realizar el contraste podemos cometer dos errores: \n\na) Rechazar ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], " cuando es cierta (error tipo I). Es decir: de la inspecci\[OAcute]n de la \ muestra concluimos que ", Cell[BoxData[ FormBox[ RowBox[{"p", ">", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TraditionalForm]]], " cuando realmente ", Cell[BoxData[ FormBox[ RowBox[{"p", " ", "\[LessEqual]", " ", SubscriptBox["p", "A"]}], TraditionalForm]]], ", por tanto rechazamos un lote que deberiamos aceptar. La probabilidad de \ que esto ocurra es \[Alpha]. A este error se llama riesgo del fabricante pues \ normalmente es \[EAcute]l quien asume sus consecuencias: tirar o \ reinspeccionar producto bueno. Evidentemente el fabricante est\[AAcute]ra \ interesado que este valor sea lo mas peque\[NTilde]o posible, usualmente \ \[Alpha] = 0.05 o 0.01, que quiere decir que en el 5% o 1% de las ocasiones \ se rechar\[AAcute]n lotes buenos." }], "Text"], Cell[TextData[{ "\[Alpha] = {Probabilidad de rechazar ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], " cuando ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], " es cierta} = {probabilidad de rechazar un lote con ", Cell[BoxData[ FormBox[ RowBox[{"p", "=", " ", SubscriptBox["p", "A"]}], TraditionalForm]]], "}" }], "Caption"], Cell[TextData[{ "b) Aceptar ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], " cuando es falsa (error tipo II). Es lo opuesto al caso anterior, \ obtenemos que ", Cell[BoxData[ FormBox[ RowBox[{"p", "\[LessEqual]", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TraditionalForm]]], " cuando realmente ", Cell[BoxData[ FormBox[ RowBox[{"p", ">", SubscriptBox["p", "A"]}], TraditionalForm]]], ", por tanto aceptamos un lote que deberiamos rechazar. La probabilidad de \ que esto ocurra se le llama \[Beta]. A este error se llama riesgo del \ consumidor o comprador pues es normalmente quien asume sus consecuencias: \ tomar por aceptable un producto que no lo es. Natualmente el comprador est\ \[AAcute]ra interesado que este valor sea lo m\[AAcute]s peque\[NTilde]o \ posible, usualmente \[Beta] = 0.05 o 0.1, que quiere decir que en el 5% o \ 10% de las ocasiones se aceptar\[AAcute]n lotes con una fraccion defectiva ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], ". " }], "Text"], Cell[TextData[{ "\[Beta] = {Probabilidad de aceptar ", Cell[BoxData[ FormBox[ SubscriptBox["H", "0"], TraditionalForm]]], " cuando ", Cell[BoxData[ FormBox[ SubscriptBox["H", "1"], TraditionalForm]]], "es cierta} = {probabilidad de aceptar un lote con ", Cell[BoxData[ FormBox[ RowBox[{"p", "=", " ", SubscriptBox["p", "R"]}], TraditionalForm]]], "}. " }], "Caption"], Cell[TextData[{ "Para establecer un plan de muestreo riguroso matematicamente se requiere \ definir previamente \[Alpha], ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", \[Beta] y ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], ". Fijado los parametros anteriores la \[UAcute]nica forma de disminuir \ \[Alpha] y \[Beta] simultanemamente es aumentar el tama\[NTilde]o de la \ muestra. En el l\[IAcute]mite: tama\[NTilde]o muestral = tama\[NTilde]o de la \ poblaci\[OAcute]n \[Alpha] = \[Beta] = 0, pero en la pr\[AAcute]ctica esto no \ suele ser econom\[OAcute]co y a veces inviable (Ej.: an\[AAcute]lisis \ destructivos).\n" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Calculo de \[Alpha] y \[Beta]", "Subsection"], Cell[TextData[{ "Para establecer un plan de muestreo necesitamos relacionar matematicamente \ \[Alpha], ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", \[Beta] y ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], ". Inicialmente utilizamos la distribucion binomial. Esta funci\[OAcute]n es \ valida para muestreo con reemplazamiento y da muy buena aproximaci\[OAcute]n \ cuando el tama\[NTilde]o del lote, ", Cell[BoxData[ FormBox["N", TraditionalForm]]], ", es muy grande con el tama\[NTilde]o de la muestra, ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", situaci\[OAcute]n muy frecuente en la pr\[AAcute]ctica. Mas adelante lo \ extenderemos a la funci\[OAcute]n hipergeom\[EAcute]trica que es la que en \ rigor hay que emplear cuando el muestreo es sin reemplazamiento, aunque la \ diferencias en valores num\[EAcute]ricos van a ser insignificantes con la \ binomial para muestras ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "<<", Cell[BoxData[ FormBox["N", TraditionalForm]]], " (tipicamente ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " \[LessEqual] 0.1 ", Cell[BoxData[ FormBox["N", TraditionalForm]]], "). Adem\[AAcute]s, el m\[EAcute]todo para establecer es similar tanto si se \ emplea la distribuci\[OAcute]n binomial como la hipegem\[EAcute]trica. " }], "Text"], Cell[TextData[{ "La probabilidad de que al tomar una muestra de tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", de una poblaci\[OAcute]n con una proporci\[OAcute]n ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], " defectuosas, haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " defectuosas dada por la funci\[OAcute]n de probabilidad PDF binomial es:" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"PDF", "[", RowBox[{ StyleBox["B", FontWeight->"Bold"], "(", RowBox[{"n", ",", "r", ",", " ", SubscriptBox["p", "A"]}], ")"}], "]"}], "=", RowBox[{ RowBox[{"(", GridBox[{ {"n"}, {"r"} }], ")"}], SuperscriptBox[ RowBox[{ SuperscriptBox[ SubscriptBox["p", "A"], "r"], "(", RowBox[{"1", "-", SubscriptBox["p", "A"]}], ")"}], RowBox[{"n", "-", "r"}]]}]}], TraditionalForm]], "NumberedEquation"], Cell["donde ", "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], "\tproporci\[OAcute]n de muestra defectuosas en el lote (cuando es \ desconocida como es el caso, se supone ", Cell[BoxData[ FormBox[ RowBox[{"p", " ", "=", " ", SubscriptBox["p", "A"]}], TraditionalForm]]], ")\n", Cell[BoxData[ FormBox["n", TraditionalForm]]], "\ttama\[NTilde]o de la muestra\n", Cell[BoxData[ FormBox["r", TraditionalForm]]], "\tn\[UAcute]mero de rechazos en la muestra" }], "Caption"], Cell[TextData[{ "En lenguaje del ", StyleBox["Mathematica", FontSlant->"Italic"], " la funci\[OAcute]n anterior podemos expresar por" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"PB", "[", RowBox[{"n_", ",", "r_", ",", " ", "pA_"}], "]"}], " ", ":=", RowBox[{ RowBox[{"Binomial", "[", RowBox[{"n", ",", "r"}], "]"}], SuperscriptBox["pA", "r"], SuperscriptBox[ RowBox[{"(", RowBox[{"1", "-", "pA"}], ")"}], RowBox[{"n", "-", "r"}]]}]}]], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Por ej.: En una muestra de 100 (", Cell[BoxData[ FormBox["n", TraditionalForm]]], ") uds en una poblaci\[OAcute]n con el 5% de uds. defectuosa (", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ") las probabilidades de que haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " defectuosas se muestran en el gr\[AAcute]fico (la probabilidad de que haya \ m\[AAcute]s de 20 defectuosas es practicamente nula)." }], "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Plot", "[", RowBox[{ RowBox[{"PB", "[", RowBox[{"100", ",", "r", ",", "0.05"}], "]"}], ",", " ", RowBox[{"{", RowBox[{"r", ",", " ", "0", ",", " ", "20"}], "}"}]}], "]"}], " "}]], "Input"], Cell[BoxData[ GraphicsBox[{{}, {}, {Hue[0.67, 0.6, 0.6], LineBox[CompressedData[" 1:eJwVl3c8V/8Xx63swsdokL198BmlNO45SiopUSRJRRkJRTIiQma2ZGQkQlaU ZI/wsSujKC0iJLsi+vb7/P659/F8vO/7vN/ndV7n3vuWsXI2vsDGwsJixrz8 /556w4qTej6FUJnzuSiDHtDUK21wQtoTDqZuX5i54wH/WmoDpKVD4UpB9MHQ KQ9IrY+QWiOdBIlqahuk73nCpd1ujr+l8qDqt6LkEsd16F2xi5+VqoAzjAfz m//zgWsMScpMSSW01Vtoc527AZvi+jp+7K0Gi8Ki9ramG2Cliuzfbevg8JCS 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TraditionalForm]]], " de que haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " o menos piezas defectuosas est\[AAcute] dada por la distribuci\[OAcute]n \ acumulada de probabilidad (CDF) definida como:" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["P", "a"], " ", "=", RowBox[{ RowBox[{"CDF", "[", RowBox[{ StyleBox["B", FontWeight->"Bold"], "(", RowBox[{"n", ",", "r", ",", " ", SubscriptBox["p", "A"]}], ")"}], "]"}], "=", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"x", "=", "0"}], "r"], RowBox[{ RowBox[{"(", GridBox[{ {"n"}, {"r"} }], ")"}], SuperscriptBox[ RowBox[{ SuperscriptBox[ SubscriptBox["p", "A"], "r"], "(", RowBox[{"1", "-", SubscriptBox["p", "A"]}], ")"}], RowBox[{"n", "-", "r"}]]}]}]}]}], TraditionalForm]], "NumberedEquation"], Cell["\<\ donde x\tindice de variaci\[OAcute]n de r\ \>", "Caption"], Cell[TextData[{ "La expresi\[OAcute]n anterior la convertimos al lenguaje del ", StyleBox["Mathematica", FontSlant->"Italic"], " " }], "Text"], Cell[BoxData[ 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Fijado el tama\[NTilde]o de \ la muestra ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " y el el n\[UAcute]mero de rechazos ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " admisibles en la muestra la curva caracteristica (OC) representa la \ probabilidad de aceptaci\[OAcute]n ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], " del lote en funci\[OAcute]n de la fraccion, o porcentaje, de uds \ defectuosas de la poblaci\[OAcute]n ", Cell[BoxData[ FormBox["p", TraditionalForm]]], ". Es decir: se trata de representar ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{"f", " ", RowBox[{"(", RowBox[{"n", ",", " ", "r", ",", " ", "p"}], ")"}]}], TraditionalForm]]], " con ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " y ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " ctes para ello podemos utilizar la funci\[OAcute]n antes definida ", StyleBox["CB[n,r,p]", "Input"], " variando el valor de ", StyleBox["p.", "Input"], " La abscisa de la curva OC representa la fracci\[OAcute]n defectuosa y las \ ordenada la probabilidad de aceptaci\[OAcute]n. 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Es decir " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"alfaB", "[", RowBox[{"100", ",", "2", ",", " ", "0.05"}], "]"}]], "Input"], Cell[BoxData["0.8817370188148795`"], "Output", CellChangeTimes->{3.3993629538197465`*^9, 3.4180324556268005`*^9, 3.546499982655249*^9}] }, Open ]], Cell[TextData[{ "La probabilidad de aceptar un lote que habr\[IAcute]a que rechazar, tomando \ ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], " = 0.06, es muy baja" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"betaB", "[", RowBox[{"100", ",", "2", ",", " ", "0.06"}], "]"}]], "Input"], Cell[BoxData["0.05661277666785658`"], "Output", CellChangeTimes->{3.3993629538665047`*^9, 3.4180324556736755`*^9, 3.5464999827332497`*^9}] }, Open ]], Cell["\<\ El plan de muestreo anterior es claramente favorable al comprador. 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El \[Beta] ser\[AAcute] muy util aplicar en el caso de que estemos \ interesados en protegernos de lotes con baja calidad podemos establecer un \ LTPD (lot tolerance percent defective), -otros nombres sin\[OAcute]nimos son \ RQL o LQL- esto es: el nivel mas porbre de calidad que estamos dispuesto a \ aceptar para un lote individual.\ \>", "Text"], Cell[TextData[{ "Lo dicho puede observarse facilmente representando las OCs para distintos \ valores de {r/n = cte.}: {3/50, , 15/250}, que muestra que en la medida que \ aumentamos ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " va disminuyendo \[Alpha] y \[Beta]." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Plot", "[", RowBox[{ RowBox[{"Evaluate", "[", RowBox[{"Table", "[", RowBox[{ RowBox[{"CB", "[", RowBox[{ RowBox[{"50", " ", "n"}], ",", RowBox[{"3", " ", "n"}], " ", ",", " ", "p"}], "]"}], ",", RowBox[{"{", RowBox[{"n", ",", "1", ",", "5"}], "}"}]}], "]"}], " ", "]"}], ",", " ", RowBox[{"{", RowBox[{"p", ",", " ", "0", 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Cell[TextData[{ "El problema mas frecuente en muestreo es determinar el tama\[NTilde]o de la \ muestra y el n\[UAcute]mero de uds. no conformes en la muestra. Vamos a \ describir como realizarlo en el caso de muestreos simples.\n\nSea una poblaci\ \[OAcute]n de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], ", del que desconocemos el n\[UAcute]mero de uds defectuosas. La \ probabilidad ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], " de que haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " o menos defectuosas nos viene dada por la distribuci\[OAcute]n hipergeom\ \[EAcute]trica." }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox["P", "a"], "=", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"x", "=", "0"}], "r"], FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"x"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "x"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}]}], TraditionalForm]], "NumberedEquation"], Cell["donde ", "Text"], Cell[TextData[{ "\n", Cell[BoxData[ FormBox["D", TraditionalForm]]], "\tN\[UAcute]mero de piezas defectuosas en el lote (desconocidas).\n", Cell[BoxData[ FormBox["f", TraditionalForm]]], "\tFraccion de uds defectuosas: ", Cell[BoxData[ FormBox[ RowBox[{"f", " ", "=", " ", RowBox[{"D", "/", "N"}]}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " es desconocida, se toma como valor la fracci\[OAcute]n m\[AAcute]xima, ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", de la poblaci\[OAcute]n que admitiriamos que este formada por uds fuera \ de los l\[IAcute]mites de tolerancia.\n\[Alpha]\t= nivel de significaci\ \[OAcute]n (Probabilidad de cometer un error de tipo I que estamos dispuesto \ a asumir). \n", Cell[BoxData[ FormBox["n", TraditionalForm]]], "\t= tama\[NTilde]o muestral " }], "Caption"], Cell[TextData[{ "Consideremos inicialmente que el tama\[NTilde]o del lote es grande y ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"N", ">>"}], " "}], TraditionalForm]]], Cell[BoxData[ FormBox["n", TraditionalForm]]], " , en este caso en vez de (5) podemos emplear con muy buena aproximacion \ (2), observese que para estas condiciones el resultado es independiente del \ tama\[NTilde]o del lote. " }], "Text"], Cell[TextData[{ "Usualmente \[Alpha], ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], " son fijos. Para cada valor de ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " obtendremos el n\[UAcute]mero m\[AAcute]ximo de rechazos admisibles \ despejando ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " en (3). Esto puede hacerse como sigue:" }], "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{"muestra", "[", RowBox[{"n_", ",", "alfa_", ",", " ", "pA_"}], "]"}], " ", ":=", " ", RowBox[{"Floor", "[", RowBox[{"r", " ", "/.", " ", RowBox[{"FindRoot", "[", RowBox[{ RowBox[{ RowBox[{"alfaB", "[", RowBox[{"n", ",", "r", ",", " ", "pA"}], "]"}], "==", " ", "alfa"}], ",", " ", RowBox[{"{", RowBox[{"r", ",", " ", "1", ",", "2"}], "}"}]}], "]"}]}], "]"}]}]}]], "Input"], Cell[TextData[{ "Por ejemplo, para \[Alpha] = 0.05, ", Cell[BoxData[ FormBox[ RowBox[{" ", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TextForm]]], " = 0.05, y ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " = 100, ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " es " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"muestra", "[", RowBox[{"100", ",", "0.05", ",", " ", "0.05"}], "]"}]], "Input"], Cell[BoxData["8"], "Output", CellChangeTimes->{3.5465000984542522`*^9}] }, Open ]], Cell[TextData[{ "Lo usual que que deseemos tener en cuenta el error de tipo II, es decir \ \[Beta], para ello hemos de fijar \[Beta] y ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], " lo que podemos hacer en funci\[OAcute]n de la experiencia o por que nos \ sean valores dados. En definitiva tenemos cuatro valores fijos: ", Cell[BoxData[ FormBox[ RowBox[{"\[Alpha]", ",", " ", FormBox[ SubscriptBox["p", "A"], TraditionalForm]}], TextForm]]], ", \[Beta] y ", Cell[BoxData[ FormBox[ SubscriptBox["p", "R"], TraditionalForm]]], ", para definir ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " cumpliendo con los cuatro valores anteriores sobre el \[UAcute]nico \ parametro que vamos a poder actuar es sobre el tama\[NTilde]o de la muestra. \ Se trata de resolver el sistema dado por las ecuaciones (3) y (4). Esto puede \ hacerse directamente, que no es facil, o usando m\[EAcute]todos aproximados \ como el que se describe. Este mismo m\[EAcute]todo puede emplearse para \ obtener una primera aproximaci\[OAcute]n que puede ser utilizada para \ posteriormente para resolver el sistema formado por las ecs (3) y (4). " }], "Text"], Cell[TextData[{ "M\[EAcute]todo aproximado.- Consite en aplicar las ecs (6) (requiere que ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " siga aproximadamente una distribucion N(0,1) y que ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"N", ">>"}], " "}], TraditionalForm]]], Cell[BoxData[ FormBox["n", TraditionalForm]]], " para poder asumir que x -n\[UAcute]mero de defectuoso- pueda aproximarse \ por una distribuci\[OAcute]n de Poisson de media \[Lambda] = ", Cell[BoxData[ FormBox[ RowBox[{"n", " ", "p"}], TraditionalForm]]], "). " }], "Text"], Cell[BoxData[{ RowBox[{ RowBox[{"r", " ", "=", " ", RowBox[{ RowBox[{"n", " ", SubscriptBox["p", "A"]}], " ", "+", " ", RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", SqrtBox[ RowBox[{"n", " ", SubscriptBox["p", "A"]}]]}]}]}], " ", ",", " ", RowBox[{ RowBox[{"P", RowBox[{"(", RowBox[{"z", ">", SubscriptBox["z", "\[Alpha]"]}], ")"}]}], "=", "\[Alpha]"}]}], "\n", RowBox[{ RowBox[{"r", " ", "=", " ", RowBox[{ RowBox[{"n", " ", SubscriptBox["p", "R"]}], " ", "+", " ", RowBox[{ SubscriptBox["z", "\[Beta]"], " ", SqrtBox[ RowBox[{"n", " ", SubscriptBox["p", "R"]}]]}]}]}], ",", " ", RowBox[{ RowBox[{"P", RowBox[{"(", RowBox[{"z", ">", SubscriptBox["z", "\[Beta]"]}], ")"}]}], "=", "\[Beta]"}]}]}], "NumberedEquation"], Cell["por tanto", "Text"], Cell[BoxData[ RowBox[{"n", " ", "=", " ", FractionBox[ RowBox[{ RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", SqrtBox[ RowBox[{" ", SubscriptBox["p", "A"]}]]}], "-", RowBox[{ SubscriptBox["z", "\[Beta]"], " ", SqrtBox[ RowBox[{" ", SubscriptBox["p", "R"]}]]}]}], RowBox[{ SubscriptBox["p", "A"], "-", SubscriptBox["p", "R"]}]]}]], "NumberedEquation"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"Ejemplo", ":", " ", RowBox[{"Sea", " ", SubscriptBox["p", "A"]}]}], "=", " ", RowBox[{ RowBox[{"0.02", " ", "y", " ", SubscriptBox["p", "R"]}], "=", " ", "0.04"}]}], ",", " ", RowBox[{ RowBox[{"con", " ", "\[Alpha]"}], " ", "=", " ", RowBox[{"\[Beta]", " ", "=", " ", "0.05"}]}], ",", " ", RowBox[{ RowBox[{"es", " ", "decir", " ", SubscriptBox["z", "\[Alpha]"]}], " ", "=", RowBox[{ RowBox[{"-", SubscriptBox[ RowBox[{"z", " "}], "\[Beta]"]}], " ", "=", " ", "1.64"}]}], ",", " ", RowBox[{"por", " ", RowBox[{"tanto", ":"}]}]}], TextForm]], "Text"], Cell[BoxData[ RowBox[{"Clear", "[", "n", "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"n", " ", "=", " ", SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{ RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", SqrtBox[ RowBox[{" ", SubscriptBox["p", "A"]}]]}], "-", RowBox[{ SubscriptBox["z", "\[Beta]"], " ", SqrtBox[ RowBox[{" ", SubscriptBox["p", "R"]}]]}]}], RowBox[{ SubscriptBox["p", "A"], "-", SubscriptBox["p", "R"]}]], ")"}], "2"]}], ";"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"n", " ", "=", RowBox[{"n", "/.", RowBox[{"{", RowBox[{ RowBox[{ SubscriptBox["p", "A"], "\[Rule]", "0.02"}], ",", " ", RowBox[{ SubscriptBox["p", "R"], "\[Rule]", "0.04"}], ",", " ", RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", "\[Rule]", "1.64"}], ",", RowBox[{ SubscriptBox[ RowBox[{"z", " "}], "\[Beta]"], "\[Rule]", RowBox[{"-", "1.64"}]}]}], "}"}]}]}]], "Input"], Cell[BoxData["783.8068797358677`"], "Output", CellChangeTimes->{3.3993629662573743`*^9, 3.4180324677050023`*^9, 3.546500098797453*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"(", " ", RowBox[{ RowBox[{"n", " ", SubscriptBox["p", "A"]}], "+", " ", RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", SqrtBox[ RowBox[{"n", " ", SubscriptBox["p", RowBox[{"A", " "}]]}]]}]}], ")"}], "/.", RowBox[{"{", " ", RowBox[{ SubscriptBox["z", "\[Alpha]"], " ", "\[Rule]", "1.64"}], "}"}]}], "/.", RowBox[{ SubscriptBox["p", "A"], "\[Rule]", "0.02"}]}]], "Input"], Cell[BoxData["22.16940639207603`"], "Output", CellChangeTimes->{3.3993629663041325`*^9, 3.418032467751878*^9, 3.5465000988598533`*^9}] }, Open ]], Cell[CellGroupData[{ Cell["Ejemplo", "Subsubsection"], Cell[TextData[{ "Determinar un plan de muestreo para \[Alpha]= ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], "= 0.05 en el que el cliente admite un \[Beta] =0.05, ", Cell[BoxData[ FormBox[ SubscriptBox["P", "R"], TraditionalForm]]], " = 0.1. (Ojo: Como ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " y ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " son enteros los valores de \[Alpha] y \[Beta] no seran justo los \ especificados, sino que ser\[AAcute]n los m\[AAcute]s proximos). " }], "Text"], Cell["El resultado exacto es:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"betaB", "[", RowBox[{"313", ",", "22", ",", " ", "0.1"}], "]"}]], "Input"], Cell[BoxData["0.043623011290468715`"], "Output", CellChangeTimes->{3.3993629778066006`*^9, 3.4180324792988267`*^9, 3.546500151556746*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"alfaB", "[", RowBox[{"313", ",", "22", ",", " ", "0.05"}], "]"}]], "Input"], Cell[BoxData["0.04376826129217115`"], "Output", CellChangeTimes->{3.399362977868944*^9, 3.4180324793144517`*^9, 3.5465001516191463`*^9}] }, Open ]], Cell["\<\ El problema que tiene el caso anterior es que la muestra a tomar es muy \ grande, una alternativa es tomar una muestra mas peque\[NTilde]a manteniendo \ asegurado el riesgo del consumidor. En el ejemplo este valor ser\[IAcute]a\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"betaB", "[", RowBox[{"60", ",", "2", ",", " ", "0.10"}], "]"}]], "Input"], Cell[BoxData["0.05304508181599277`"], "Output", CellChangeTimes->{3.3993629779157023`*^9, 3.418032479345702*^9, 3.546500151665946*^9}] }, Open ]], Cell["\<\ En este caso el fabricante correr\[IAcute]a un riesgo mayor pero ahorrar\ \[IAcute]a gastos de inspecci\[OAcute]n que debe valorar si \ econ\[OAcute]micamente le compensa.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"alfaB", "[", RowBox[{"60", ",", "3", ",", " ", "0.05"}], "]"}]], "Input"], Cell[BoxData["0.352718988398663`"], "Output", CellChangeTimes->{3.3993629779624605`*^9, 3.418032479361327*^9, 3.546500151728346*^9}] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Plan de muestreo en lotes peque\[NTilde]os sin reemplazamiento\ \>", "Section"], Cell["\<\ Cuando el tama\[NTilde]o de la muestra no es mucho menor que el del lote la \ aproximaci\[OAcute]n a la binomial puede no ser suficiente, en este caso es \ necesario operar directamente con la distribuci\[OAcute]n \ hipergeom\[EAcute]trica dada porla ec(5) que podemos escribirla como:\ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"ProbA", "[", RowBox[{"D_", ",", "N_", ",", "n_", ",", "r_"}], "]"}], " ", ":=", " ", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"x", "=", "0"}], "r"], RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"Binomial", "[", RowBox[{"D", ",", "x"}], "]"}], RowBox[{"Binomial", "[", RowBox[{ RowBox[{"N", " ", "-", " ", "D"}], ",", RowBox[{"n", " ", "-", "x"}]}], "]"}]}], ")"}], "/", RowBox[{"Binomial", "[", RowBox[{"N", ",", "n"}], "]"}]}]}]}]], "Input"], Cell[TextData[{ "Vamos a utilizarla para calcular la probabilidad de aceptar el lote fijados \ ", Cell[BoxData[ FormBox["D", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["N", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", y ", Cell[BoxData[ FormBox["r", TraditionalForm]]], ". " }], "Text"], Cell[TextData[{ "Ejemplo.: Sea un lote de 100 (", Cell[BoxData[ FormBox["N", TraditionalForm]]], ") uds en donde exista 5 (", Cell[BoxData[ FormBox["D", TraditionalForm]]], ") defectuosas, tomamos una muestra de 44 (", Cell[BoxData[ FormBox["n", TraditionalForm]]], ") uds y ponemos el criterio de aceptar el lote si en la muestra hay una o \ ninguna defectusas ( x = {0, 1}). " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ProbA", "[", RowBox[{"5", ",", "100", ",", "44", ",", "1"}], "]"}], "//", "N"}]], "Input"], Cell[BoxData["0.2653902798232695`"], "Output", CellChangeTimes->{3.399362978024804*^9, 3.418032479392577*^9, 3.5465001629915657`*^9}] }, Open ]], Cell[TextData[{ "Por lo demas el procedimiento para calcular \[Alpha] y \[Beta] es el mismo \ que el descrito en el apartado anterior para la binomial. Como valor de ", Cell[BoxData[ FormBox[ SubscriptBox["p", "A"], TraditionalForm]]], ", se toma ", Cell[BoxData[ FormBox["f", TraditionalForm]]], "." }], "Text"], Cell[CellGroupData[{ Cell["Muestreo LTPD", "Subsection"], Cell[TextData[{ "Sea una poblaci\[OAcute]n de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], ", del que desconocemos el n\[UAcute]mero de uds defectuosas. Pretendemos \ garantizar que rechazamos con una probabilidad ", Cell[BoxData[ FormBox["P", TraditionalForm]]], " aquellos lotes procedentes de una poblaci\[OAcute]n que contenga una \ fracci\[OAcute]n ", Cell[BoxData[ FormBox["F", TraditionalForm]]], " de la misma est\[AAcute] dentro de unos l\[IAcute]mites de tolerancia \ establecidos. Para ello necesitamos establecer un plan de muestreo en el \ definiremos el tama\[NTilde]o muestral y el n\[UAcute]mero de uds defectuosas \ que estamos dispuesto a admitir en la muestra para aceptar el lote. Este \ criterio est\[AAcute] basado en el propuesto por Dodge-Romig (LTPD, lot \ tolerance percent defective) y se utiliza cuando se pretende ser muy estricto \ en prevenir la aceptaci\[OAcute]n de lotes no conformes.\nEl criterio que se \ seguir\[AAcute] para establecer el plan de muestreo es: Tomaremos ", Cell[BoxData[ FormBox["F", TraditionalForm]]], " = 1 - ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " y estos lotes los rechazaremos con una probabilidad \[Alpha] = 1-", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], ". Observese que ", Cell[BoxData[ FormBox[ SubscriptBox["P", "a"], TraditionalForm]]], " es fijado por el consumidor y corresponde a la fracci\[OAcute]n defectiva \ m\[AAcute]xima que est\[AAcute] dispuesto a aceptar (LTPD), por ello estamos \ limitando el riesgo del consumidor. De (5) se obtiene:" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{"\[Alpha]", "=", " ", RowBox[{ RowBox[{"1", "-", SubscriptBox["P", "a"]}], "=", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"x", "=", "0"}], "r"], FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"x"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "x"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}]}]}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ "Se trata de despejar ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " en la ec. anterior , previamente reescribimos el lado derecha de (3) \ haciendo ", Cell[BoxData[ FormBox[ RowBox[{"D", " ", "=", " ", RowBox[{"f", " ", "N", " "}]}], TraditionalForm]]], " y expresamos el resultado en terminos de la funci\[OAcute]n gamma, con la \ que se puede operar mas facilmente. " }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"eq1", "[", RowBox[{"N_Integer", ",", " ", "n_", ",", " ", "f_", ",", " ", "r_"}], "]"}], " ", ":=", RowBox[{"FunctionExpand", "[", RowBox[{ RowBox[{"ProbA", "[", RowBox[{"D", ",", "N", ",", "n", ",", "r"}], "]"}], "/.", RowBox[{"D", " ", "->", " ", RowBox[{"f", " ", "N"}]}]}], "]"}]}]], "Input"], Cell[TextData[{ "Ahora se trata de fijar el numero de defectos ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " que estamos dispuestos a admitir, y obtener el valor de ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " en la ec. anterior, esto es: resolver eq1[N, n, f,r] = 1-Pa = alfa, \ siendo ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " el valor a calcular." }], "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{"eq2", "[", RowBox[{"N_Integer", ",", "alfa_", ",", " ", "f_", ",", "r_"}], "]"}], " ", ":=", " ", RowBox[{"Ceiling", "[", RowBox[{"n", " ", "/.", " ", RowBox[{"FindRoot", "[", RowBox[{ RowBox[{ RowBox[{"eq1", "[", RowBox[{"N", ",", " ", "n", ",", "f", ",", "r"}], "]"}], " ", "==", " ", "alfa"}], ",", " ", RowBox[{"{", RowBox[{"n", ",", " ", "20"}], "}"}]}], "]"}]}], "]"}]}]}]], "Input"], Cell[TextData[{ "Ejemplo: Si tenemos un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], "=300, y tomamos como criterio de aceptaci\[OAcute]n alfa = 1-0.95 = 0.05 , \ f = 1 - 0.95 = 0.05, y ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " \[LessEqual]se obtiene que el tama\[NTilde]o la muestra es ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " = 83" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"eq2", "[", RowBox[{"300", ",", "0.05", ",", " ", "0.05", ",", "1"}], "]"}], " "}]], "Input"], Cell[BoxData["83"], "Output", CellChangeTimes->{3.3993629781494923`*^9, 3.418032480158207*^9, 3.546500163100766*^9}] }, Open ]], Cell["\<\ En resumen tomariamos una muestra de tama\[NTilde]o 83 y aceptariamos el lote \ si en la muestra hay ninguna o 1 ud. disconforme.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Plan de muestreo con reemplazamiento o para muestras grandes", "Section"], Cell[TextData[{ "En el caso de que la inspecci\[OAcute]n sea con reemplazamiento, o que el \ tama\[NTilde]o de la muestra sea peque\[NTilde]a comparado con el del lote la \ funci\[OAcute]n de probabilidad (PDF) de que en un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], " , con ", Cell[BoxData[ FormBox["D", TraditionalForm]]], " piezas defectuosas, tomada una muestra de tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " aparezcan ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " defectuosas esta dada por la distribuci\[OAcute]n hipergeom\[EAcute]trica: \ un lote con una fracci\[OAcute]n ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " de piezas defectuosas est\[AAcute] dada por la distribuci\[OAcute]n \ hipergeom\[EAcute]trica (HG):" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"PDF", "[", RowBox[{"HG", "(", RowBox[{"D", ",", "N", ",", "n", ",", "r"}], ")"}], "]"}], "=", FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"r"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "r"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}], TraditionalForm]], "NumberedEquation"], Cell["\<\ donde N \ttama\[NTilde]o del lote D\tn\[UAcute]mero de piezas defectuosas en el lote n\tTama\[NTilde]o de la muestra r\tn\[UAcute]mero de rechazos en la muestra\ \>", "Caption"], Cell[TextData[{ "En lenguaje del ", StyleBox["Mathematica", FontSlant->"Italic"], " la funci\[OAcute]n anterior podemos expresar por" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"Pa", "[", RowBox[{"D_", ",", "N_", ",", "n_", ",", "r_"}], "]"}], " ", ":=", RowBox[{ RowBox[{"Binomial", "[", RowBox[{"D", ",", "r"}], "]"}], RowBox[{ RowBox[{"Binomial", "[", RowBox[{ RowBox[{"N", " ", "-", " ", "D"}], ",", RowBox[{"n", " ", "-", "r"}]}], "]"}], "/", RowBox[{"Binomial", "[", RowBox[{"N", ",", "n"}], "]"}]}]}]}]], "Input"], Cell[TextData[{ "Por ej.: Un lote de 100 (", Cell[BoxData[ FormBox["N", TraditionalForm]]], ") uds en donde exista 5 (", Cell[BoxData[ FormBox["D", TraditionalForm]]], ") defectuosas, si tomamos una muestra de 44 (", Cell[BoxData[ FormBox["n", TraditionalForm]]], ") uds la probabilidad de que no haya ninguna ud defectuosa en la muestra \ es ( ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " = 0): " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Pa", "[", RowBox[{"5", ",", "100", ",", "44", ",", "0"}], "]"}], " ", "//", "N"}]], "Input"], Cell[BoxData["0.05073637702503682`"], "Output", CellChangeTimes->{3.399362978180664*^9, 3.4180324802050824`*^9, 3.5465001631475663`*^9}] }, Open ]], Cell[TextData[{ "Asumiendo las mismas condiciones, la probabilidad de que haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " o menos piezas defectuosas est\[AAcute] dada por la distribuci\[OAcute]n \ acumulada de probabilidad (CDF) definida como:" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"CDF", "[", RowBox[{"HG", "(", RowBox[{"D", ",", "N", ",", "n", ",", "r"}], ")"}], "]"}], "=", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{"x", "=", "0"}], "r"], FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"x"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "x"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}]}], TraditionalForm]], "NumberedEquation"], Cell[TextData[{ "donde \nx\tindice de variaci\[OAcute]n de r\nProbA\tProbabilidad de que \ tomada ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " uds haya ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " o menos defectuosas " }], "Caption"], Cell[TextData[{ "La expresi\[OAcute]n 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" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ProbA", "[", RowBox[{"5", ",", "100", ",", "44", ",", "1"}], "]"}], "//", "N"}]], "Input"], Cell[BoxData["0.2653902798232695`"], "Output", CellChangeTimes->{3.3993629782274227`*^9, 3.418032480236333*^9, 3.5465001632099667`*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Planes de muestreo (por atributos) ", "Section", CellChangeTimes->{{3.5464998654834433`*^9, 3.5464998781662655`*^9}, 3.5464999490371904`*^9}], Cell[BoxData[ RowBox[{"Quit", "[", "]"}]], "Input", CellChangeTimes->{{3.3993632223665266`*^9, 3.399363225094076*^9}}], Cell[TextData[{ "En este apartado vamos a establecer un plan de muestreo por atributos \ (aceptaci\[OAcute]n o rechazo) secuencial consistente en hacer una primera \ inspecci\[OAcute]n para un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], " del que se toma una muestra ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " , si en esta se produce cero rechazos se acepta el lote, en el caso de que \ se produzca 1 o 2 rechazos se toma una muestra adicional del mismo lote, si \ se dan mas de 2 se rechaza el lote. Si en esta muestra ampliada contiene cero \ rechazos el lote se acepta, y si contiene dos se rechaza. En el caso de que \ contenga un rechazo puede ocurrir:\na) Que en el primer muestreo se tuviera \ solo un rechazo en cuyo caso se vuelve a tomar otra muestra adicional. Si en \ ella aparecen cero rechazos se acepta el lote, en caso contrario se rechaza \ definitivamente. \nb) Que en el primer muestreo se dieron dos rechazos en \ este caso se rechaza el lote.\nEl procedimiento podr\[IAcute]a extenderse \ para contemplar un n\[UAcute]mero indefinido de inspecciones seceuenciales, \ pero por razones practicas se ha optado por limitarlo a tres." }], "Text"], Cell[TextData[{ "El plan de muestreo se establecer\[AAcute] definiendo como requisito de \ aceptacion que cumpla el criterio ", Cell[BoxData[ FormBox[ RowBox[{"P", "/", "F"}], TraditionalForm]]], " que definimos como aquel que para un lote cualquiera procedente de una \ poblaci\[OAcute]n que contenga una fracci\[OAcute]n ", Cell[BoxData[ FormBox["F", TraditionalForm]]], " dentro de los l\[IAcute]mites de aceptaci\[OAcute]n (o su equivalente: una \ fracci\[OAcute]n ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " de uds. defectuosas, ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " = 1 - ", Cell[BoxData[ FormBox["F", TraditionalForm]]], ") lo rechazaremos con una probabilidad ", Cell[BoxData[ FormBox["P", TraditionalForm]]], ". El m\[EAcute]todo que describimos permite definir este plan de muestreo \ para cualquier valor de ", Cell[BoxData[ FormBox["P", TraditionalForm]]], " y ", Cell[BoxData[ FormBox["F", TraditionalForm]]], ", no obstante numericamente se resuelve el caso part\[IAcute]cular de que ", Cell[BoxData[ FormBox[ RowBox[{"P", "/", "F"}], TraditionalForm]]], " sea 95/95.\n\nPara los c\[AAcute]lculos que siguen vamos a utilizar el \ programa de c\[AAcute]lculo simb\[OAcute]lico y num\[EAcute]rico ", StyleBox["Mathematica.", FontSlant->"Italic"], " " }], "Text"], Cell[CellGroupData[{ Cell["Fundamentos", "Subsection"], Cell[TextData[{ "En el caso de que la inspecci\[OAcute]n sea sin reemplazamiento, la funci\ \[OAcute]n de probabilidad (PDF) de que en un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], " , con ", Cell[BoxData[ FormBox["D", TraditionalForm]]], " piezas defectuosas, tomada una muestra de tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " aparezcan ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " defectuosas esta dada por la distribuci\[OAcute]n hipergeom\[EAcute]trica: \ un lote con una fracci\[OAcute]n ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " de piezas defectuosas est\[AAcute] dada por la distribuci\[OAcute]n \ hipergeom\[EAcute]trica (HG):" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"PDF", "[", RowBox[{"HG", "(", RowBox[{"D", ",", "N", ",", "n", ",", "r"}], ")"}], "]"}], "=", FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"r"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "r"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}], TraditionalForm]], "NumberedEquation"], Cell["\<\ donde N \ttama\[NTilde]o del lote D\tn\[UAcute]mero de piezas defectuosas en el lote n\tTama\[NTilde]o de la muestra r\tn\[UAcute]mero de rechazos en la muestra\ 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" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ProbA", "[", RowBox[{"5", ",", "100", ",", "44", ",", "1"}], "]"}], "//", "N"}]], "Input"], Cell[BoxData["0.2653902798232695`"], "Output", CellChangeTimes->{3.3993629782897663`*^9, 3.3993632359419327`*^9}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Inspeccion con reemplazamiento o para muestras grandes", "Subsection"], Cell[BoxData[ RowBox[{"Quit", "[", "]"}]], "Input", CellChangeTimes->{{3.3993633852246404`*^9, 3.3993633879677763`*^9}}], Cell[TextData[{ "En el caso de que la inspecci\[OAcute]n sea con reemplazamiento, o que el \ tama\[NTilde]o de la muestra sea peque\[NTilde]a comparado con el del lote la \ funci\[OAcute]n de probabilidad (PDF) de que en un lote de tama\[NTilde]o ", Cell[BoxData[ FormBox["N", TraditionalForm]]], " , con ", Cell[BoxData[ FormBox["D", TraditionalForm]]], " piezas defectuosas, tomada una muestra de tama\[NTilde]o ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " aparezcan ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " defectuosas esta dada por la distribuci\[OAcute]n hipergeom\[EAcute]trica: \ un lote con una fracci\[OAcute]n ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " de piezas defectuosas est\[AAcute] dada por la distribuci\[OAcute]n \ hipergeom\[EAcute]trica (HG):" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"PDF", "[", RowBox[{"HG", "(", RowBox[{"D", ",", "N", ",", "n", ",", "r"}], ")"}], "]"}], "=", FractionBox[ RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"D"}, {"r"} }], ")"}], Binomial, Editable->False], " ", TagBox[ RowBox[{"(", GridBox[{ { RowBox[{"N", "-", "D"}]}, { RowBox[{"n", "-", "r"}]} }], ")"}], Binomial, Editable->False]}], TagBox[ RowBox[{"(", GridBox[{ {"N"}, {"n"} }], ")"}], Binomial, Editable->False]]}], TraditionalForm]], "NumberedEquation"], Cell["\<\ donde N \ttama\[NTilde]o del lote D\tn\[UAcute]mero de piezas defectuosas en el lote n\tTama\[NTilde]o de la muestra r\tn\[UAcute]mero de rechazos en la muestra\ \>", 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