1、1,2,Measures of Process CapabilityStatistical Analysis of Process Capability,3,Process CapabilityRefers to the uniformity of the process.Variability in the process is a measure of the uniformity of the output.- Instantaneous variability is the natural orinherent variability at a specified time- Vari
2、ability over time,4,Process CapabilityA critical performance measure that addresses process results relative to process/product specifications.A capable process is one for which the process outputs meet or exceed expectation.,5,6,Measures of Process CapabilityCustomary to use the Six Sigma spread in
3、 the distribution of the product quality characteristic,7,Key PointsThe proportion of the process output that will fall outside the natural tolerance limits.Is 0.27% (or 2700 nonconforming parts per million) if the distribution is normalMay differ considerably from 0.27% if the distribution is not n
4、ormal,8,9,Measure of Potential Process Capability, CpCp measures potential or inherent capability of the process, given that the process is stableCp is defined as, for two-sided specificationsand , for lower specifications only, for upper specifications only,10,Interpretation of Cpis the percentage
5、of the specification band used up by the process,11,Measure of Potential Process Capability, CpKCpK measures realized process capability relative to action production, given a stable processCp is defined as,12,13,Recommended Minimum Values of the Process Capability RatioTwo-sided One-sidedspecificat
6、ions specifications Existing process 1.33 1.25 New processes 1.50 1.45 Safety, strength, or critical parameter existing process 1.50 1.45 Safety, strength, or critical parameter new process 1.67 1.60,14,Process Fallout (in defective ppm)PCR One-sided specs Two-sided specs 0.25 226628 453255 0.50 668
7、07 133614 0.60 35931 71861 0.70 17865 35729 0.80 8198 16395 0.90 3467 6934 1.00 1350 2700 1.10 1484 967 1.20 159 318 1.30 48 96 1.40 14 27 1.50 4 7 1.60 1 2 1.70 0.17 0.34 1.80 0.03 0.06 2.00 0.0009 0.0018,15,Estimation of Process Capability Ratios,16,Estimation of CpA point estimate of Cp is:where,
8、17,Estimation of Cp - exampleIf the specification limits are LSL = 73.95 and USL = 74.05 and,18,Estimation of CP - exampleand the process usesof the specification band.,19,ExampleTo illustrate the use of the one sided process capability ratios, suppose that the lower specification limit on bursting
9、strength is 200 psi. We will use = 264 and S = 32 as estimates of and , respectively, and the resulting estimate of the one sided lower process capability ratio is,20,21,Uses of Results from a Process Capability Analysis1. Predicting how well the process will hold the tolerances. 2. Assisting produc
10、t developers/designers in selecting or modifying a process. 3. Assisting in establishing an interval between sampling for process monitoring. 4. Specifying performance requirements for new equipment 5. Selecting between competing vendors. 6. Planning the sequence of production processes when there i
11、s an interactive effect on processes or tolerances. 7. Reducing the variability in a manufacturing process.,22,Statistical Analysis of Process Capability,23,The SituationIn many situations, our knowledge is limited to the information that can be obtained from data that has been obtained or that will
12、 be obtained,24,The ProblemThe challenge is to obtain the maximum information from the data and to arrive at the most accurate conclusions,25,Nature of DataMost data are characterized by variation, as opposed to deterministic, due to variation inProcesses and materialsProduct/ManufacturingInspection
13、 & MeasurementOperationEnvironmentetc,26,NeedMethods and techniques are needed for analysis of data that account for Variation in the dataUncertainty in conclusion,27,StatisticsStatistics is the science of analyzing data and drawing conclusionsStatistical methods and techniques that provide tools fo
14、r:- experimental design- analysis of data- making inferences,28,29,30,31,32,33,Example The number of defects per inspected PC-X based on a random sample of 15 from a days production is: 0, 2, 0, 1, 1, 3, 1, 1, 0, 1, 0, 1, 1, 2, 1(i.) Analyze these data and present your results. (ii.) Estimate the pr
15、obability that a randomly selected PC will have at least 3 defects.,34,Example Twenty-five customers selected at random were asked to rate the overall satisfaction, a measure of quality, with the PC-X. Five factors were ranked by each selected customer. Each factor was assigned a rank between 1 and
16、10, with 10 indicating the highest level of customer satisfaction. The ratings were averaged over the five factors for each customer. The results are:9.3 8.5 7.1 9.6 7.35.5 7.7 8.6 5.9 6.97.5 8.8 7.9 6.7 7.87.7 5.2 6.2 7.8 7.66.5 6.0 8.0 8.3 9.1(i.) Analyze the survey data and present your results.
17、(ii.) Estimate the proportion of the customer population whose average satisfaction rating is at least 9.,35,Basic ConceptsAnalysis of Location, or Central TendencyAnalysis of VariabilityAnalysis of Shape,36,Population vs. SamplePopulation the total of all possible values (measurement, counts, etc.)
18、 of a particular characteristic for a specific group of objects.Sample a part of a population selected according to some rule or plan.Why sample? - Population does not exist - Sampling and testing is destructive,37,SamplingCharacteristics that distinguish one type of sample from another:the manner i
19、n which the sample was obtainedthe purpose for which the sample was obtained,38,Types of SamplesSimple Random Sample The sample X1, X2, . ,Xn is a random sample if X1, X2, . , Xn are independent identically distributed random variables. Remark: Each value in the population has an equal and independe
20、nt chance of being included in the sample.,39,Analysis of DataData represents the entire populationStatistical analysis is primarily descriptive.Data represents sample from populationStatistical analysis- describes the sample- provides information about the population,40,Analysis of Location or Cent
21、ral TendencySample (Arithmetic) MeanSample MidrangeSample ModeSample MedianSample Percentiles,41,Sample MeanFormula: Remarks: Most frequently used statisticEasy to understandMay be misleading due to extreme values,42,Sample ModeDefinition: Most frequently occurring value in the sampleRemarks: A samp
22、le may have more than one modeThe mode may not be a central valueNot well understood, nor frequently used,43,44,Analysis of VariabilitySample RangeSample VarianceSample Standard DeviationSample of Coefficient of Variation,45,Sample RangeFormula: R = Xmax - Xminwhere Xmax is the largest value in the
23、sampleand Xmin is the smallest sample valueRemarks:Easy to determineEasily understoodDetermined by extreme valuesDoes not use all sample data,46,Sample Variance & Standard DeviationSample VarianceSample Standard Deviations = (sample variance)1/2 RemarksMost frequently used measure of variabilityNot
24、well understood,47,Sample Coefficient of VariationSample VarianceRemarksRelative measure of variationUsed for comparing the variation in two samples of data that are measured in two different units,48,Analysis of ShapeSkewnessKurtosis,49,50,Estimation of SkewnessEstimate of skewness of a distributio
25、n from a random samplewhereand,51,Comparison of Kurtosis,52,Presentation of Data,53,Time Series Graph or Run ChartStem-and-Leaf PlotDigidot PlotBox PlotFrequency DistributionHistogram and Relative Frequency,54,Time Series Graph or Run ChartA plot of the data set x1, x2, , xn in the order in which th
26、e data were obtainedUsed to detect trends or patterns in the data over time,55,Stem-and-Leaf PlotsA quick way to obtain an informative visual representation of the set of data x1, x2, , xn for which each xi consists of at least two digitsSteps for constructing a stem-and-leaf display 1. Select one o
27、r more leading digits for the stem values. The trailing digits become the leaves. 2. List possible stem values in a vertical column. 3. Record the leaf for every observation beside the corresponding stem value. 4. Indicate the units for stems and leaves someplace in the displayThe stem and leaf disp
28、lay does not take the time order of the observed data into account,56,Stem-and-Leaf Plot - ExampleHere are test scores for 25 students:69, 55, 80, 95, 94, 98, 51, 70, 93, 57, 62, 52, 52 58, 61, 51, 64, 67, 78, 68, 69, 68, 96, 73, 71The first step is to place the numbers in order from least to greate
29、st:51, 51, 52, 52, 55, 57, 58, 61, 62, 64, 67, 68, 68, 69, 69, 70, 71, 73, 78, 80, 93, 94, 95, 96, 98,57,Stem-and-Leaf Plot - ExampleNow create the graph:Test Scores5 11225786 124788997 01388 09 34568The numbers on the left side of the vertical line are the stems. The numbers on the right side are t
30、he leaves. In this graph, the stems are the tens digits and the leaves are the unit digits. In this case, 9|3 represents a score of 93.,58,Digidot PlotA combination of the time series graph with the stem and leaf display,59,Box PlotA pictorial summary used to describe the most prominent statistical
31、features of the data set, x1, x2, , xn, including its:- Center or location- Spread or variability- Extent and nature of any deviation from symmetry- Identification of outliers,60,Box PlotShows only certain statistics rather than all the data, namely- median- quartiles- smallest and greatest values i
32、n the distributionImmediate visuals of a box plot are the center, the spread, and the overall range of distribution,61,Box PlotGiven the following random sample of size 25:38, 10, 60, 90, 88, 96, 1, 41, 86, 14, 25, 5, 16, 22, 29, 34, 55, 36, 37, 36, 91, 47, 43, 30, 98Arranged in order from least to
33、greatest:1, 5, 10, 14, 16, 22, 25, 29, 30, 34, 36, 36, 37, 38, 41, 43, 47, 55, 60, 86, 88, 90, 91, 96, 98,62,Box PlotFirst, find the median, the value exactly in the middle of an ordered set of numbers.The median is 37Next, we consider only the values to the left of the median: 1, 5, 10, 14, 16, 22,
34、 25, 29, 30, 34, 36, 36We now find the median of this set of numbers. The median for this group is (22 + 25)/2 = 23.5, which is the lower quartile.,63,Box PlotNow consider the values to the right of the median.38, 41, 43, 47, 55, 60, 86, 88, 90, 91, 96, 98The median for this set is (60 + 86)/2 = 73,
35、 which is the upper quartile. We are now ready to find the interquartile range (IQR), which is the difference between the upper and lower quartiles, 73 - 23.5 = 49.549.5 is the interquartile range,64,65,HistogramA graph of the observed frequencies in the data set, x1, x2, , xn versus data magnitude
36、to visually indicate its statistical properties, including- shape- location or central tendency- scatter or variability,66,Guidelines for Constructing HistogramsIf the data x1, x2, , xn are from a discrete random variable with possible values y1, y2, , yn count the number of occurrences of each valu
37、e of y and associate the frequency fi with yi, for i = 1, , k,67,Guidelines for Constructing HistogramsIf the data x1, x2, , xn are from a continuous random variable- select the number of intervals or cells, r,to be a number between 3 and 20, as an initial value use r = (n)1/2, where n is the number
38、 of observations- establish r intervals of equal width, startingjust below the smallest value of x- count the number of values of x withineach interval to obtain the frequency associated with each interval- construct graph by plotting (fi, i) for i = 1, 2, , k,68,Statistical Process Control- Histogr
39、amsPossible answers for a Cliff-like histogramHiding data that should be outside the specificationSupplier is screening the product before shipmentLower specification is a physical limit like zero thickness, but this is not normally the case,upper spec,lower spec,69,Statistical Process Control- Hist
40、ogramsPossible answers for a Bimodal histogramTwo primary sources of process variationThe process is stable, but it has experienced a large shift during the time the data were collected,upper spec,lower spec,70,Statistical Process Control - HistogramsPossible answers for a Comb-like histogramInsuffi
41、cient data collectedToo many classes displayedProcess is unstableProcess is stable but is multimodal,upper spec,lower spec,71,Statistical Process Control - HistogramsPossible answers for a Skewed histogramMay be the natural result of the processFor a machined part, the equipment may be losing tolera
42、nce or tools may be wearing outThe process is shifting slowly to the side with the long tail,upper spec,lower spec,72,Statistical Process Control - HistogramsBy including specification limits on a histogram, the amount of data that falls outside of the specification limits can be easily seen,upper s
43、pec,lower spec,frequency,specification,73,74,Estimation of Lognormal Distribution Estimation of m & sRandom sample of size n, X1, X2, . , Xn from Ln (, )Let Yi = lnXi for i = 1, 2, ., nTreat Y1, Y2, . , Yn as a random sample from N(, )Estimate and using the Normal Distribution Methods,75,Estimation of Weibull Distribution - Estimation of b & qRandom sample of size n, T1, T2, , Tn, from W(, )Point estimates is the solution of g() = 0where,
