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# 421 Note 16 for STAT 401 at PSU

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Date Created: 02/06/15

STAT 401 Ch 16 Quality Control In this chapter we describe some graphical methods that help monitor the quality of manufactured products We will concen trate on methods for monitoring quality when quality is quan ti ed in terms of the mean average value of some trait char acteristic of the product We will describe the X chart and the CUSUM chart Fchart Think of the simple context where we want to monitor the mean strength of plastic sheets Suppose we have in mind some target value no If the production process shifts for some rea son and the plastic sheets produced have more or less average strength we want to interrupt and make adjustments To deter mine if and when the production process shifts samples of size n are taken at time points l1t2t3 Let X1X2X3 denote the sample means at times l1l2l3 The fcharts plots the points 111 022 133 ifa point falls above the UCL Upper Control Limit or below the LCL Lower Control Limit line the process is declared out of control and corrective measures are taken Performance Characteristics Similarly to hypothesis testing we can commit two types of errors It is possible for a point liX to fall outside the control limits when in fact the true mean equals the target value 0 and it is possible for a point to fall within the contrail limits when in fact the true mean has shifted away from 0 In quality contrail it is customary to use i w in which case assuming normality 6 L 3 LCLzyo 3 or Pa point lift falls outside the control limits when in fact y 0 X m X m W w m PZgt 3 orZlt 3 PZ gt 3PZlt 3 1 c1gt3 c1gt 3 0026 and B P 292 falls outside the control limits when y yo Ac 6 C Plty0 3 ltXilto4 n n f f M3 7A ltIgtlt 3 M which means that the probability of this type of error depends on the shift A measured in standard deviation units and the sample size n mma A l 25 50 100 200 300 Bforn4l 9936 9772 8413 1587 0013 In the area of quality control it is common to eXpress the quality characteristics in terms of the eXpected or average run length 3 ARL needed to observe an outof control point instead of 01 and When the process is in control ie y go we would like t9 have a long ARL and when the process shifts out of control we would like to have small ARL To determine the ARL let Y denote a run length ie Y rst i for which falls outside control limits If you think of Y as the number of trials until the rst success countering the trial that results in success we see that Y is 1a geometric rV so see p 127 EU 2 1 1p l P P where pprobof success here success is when falls outside control limits Thus the process is in control 1 1 1 EY 3842 p 01 0026 incontrol ARL When the process is out of control E Y i outofcontrol ARL and the table of Bvalues we saw gives A l 25 50 100 200 300 ARLforn4l 15625 4386 630 119 10013 Unknown parameters The previous discussion of fcharts assumed for simplic ity that both the target value yo and the standard deviation 6 are known In practice both 6 and go may be unknown The target value no is typically unknown when the process is rst subjected to quality control It is often the case that the man agement wants to keep the process at its present state ie sales may be good but the mean value of the primary quality char acteristic is unknown Both yo and 6 are estimated using k samples gathered when the process is assumed to be in control Let thz fk de note the k sample means It is recommended that k 2 20 with sample size n 34 5 or 6 Then use 1 k A X X kl instead of 0 Instead of 6 we use a biasadjusted average of the sample standard deviations Recall that the sample standard deviation S is a biased estimate of 6 If the sample has come from a normal population it can be shown that W ans where a M mm see p167 for de nition of the gamma function As the fol lowing table indicates an 2 l for large values of n n l 3 4 5 6 7 8 an for n 4 886 921 940 952 959 965 1 Let now S 2le S be the average of the standard dev1a tions computed from each of the k samples Then if the sample size for each sample is n it follows by the properties of eXpec tation that S ES zanc orE gt 6 an if all samples are normal with the same variance 6 Thenin stead of 6 use its unbiased estimator San Using X and S an instead of no and 6 the control limits are S elm13 S UCL X 3 7 elm13 LCL 3 Remark 1 Recomputing Control Limits The control lim its using S an and S an are based on the assumption that the k sample from which f1 fk and S1 Sk were computed were taken when the process is in control It is possible how ever that one or more of X 1 Xk falls outside the control lim its If this is the case and if an assignable cause can be found it is recommended that control limits be recalculate without the corresponding samples Remark 2 Supplemental Rules The large Bvalues correspondingly large ARL values for small deviations from the target value yo led to additional rules for declaring the process out of contrail The following supplemen tal rules were proposed by the Western Electric Co Additional ones are implemented by Minitab 1 Two out of three successive fall outside the 26 limits on the same side 2 Four out of ve successive fall outside the 16 limits on the same side 3 Eight successiveJZ fall on the same side CUSUM Chart The CUmulative SUM procedure is also designed to im prove the performance of the f chart The basic idea is that the Fchart uses information only from the current sample ig noring information contained in previous samples However a small deviation from the target is more easily detected by com bining information in all samples Thus the CUSUM chart plots 11S1 12S2 where S1i1u0a SzSii2 0a S3SzX3 0aSiSi1C 0a 7 However note that the variance of S is VarS 16211 where n is the sample size at each inspection time Thus it is not possible to have the simple control limits we have for the J chart The control limits take the form of V mask which is superimposed on the plot of the points tiSi 12 A different V mask is used for each point Here we illustrate the V mask for the point ti Si If the arms of the V mask do not include all previous points the process is declared out of control If a point falls below the lower arm as in the gure the mean g has shifted to a value higher than no and if a point falls above the upper arm g has shifted to a lower value We next eXplain how the V mask is 8 constructed though in practice this is done by computer e g Minitab We need to know the slope of the lower arm ie of the angle 9 and h These and the appropriate sample size n are determined by specifying the size of the shift A from the target value that is of particular concern the ARL or B at the speci ed shift and the incontrol ARL or X as follows A 1 The slope k of the lower arm is E ii For the sample size connect the speci ed in and outof control ARL s in the Kemp nomogram p689 and read A 2 cm to obtain the number in the k scale Then solve k n round up iii Connecting again the in and outofcontrol ARL s read 6 the number in the h scale and set h h w Illustration of a CUSUM Chart Vmask as Minitab produces it Given a sequence of k sample means here k 20 you can ask Minitab to superimpose a V mask at any sample mean The above V mask corresponds to the 15th inspection point In practice a new V mask is constructed for each new sample mean as soon as it becomes available 10 A wood products company manufactures charcoal briquettes for barbecues It packages these briquettes in bags of various sizes the largest of which is supposed to contain 40 lb Table 164 displays the weights of bags from 16 different samples each of size n 4 The rst 10 of these were drawn from a normal distribution with y yo 40 and 6 5 Starting with the eleventh sample the mean has shifted upward to y 403 Table 164 Observations 2s and cumulative sums for Example 168 Sample Number Observations x 209 40 1 4077 3995 4086 3921 4020 20 2 3894 3970 4037 3988 3972 08 3 4043 4027 4091 4005 4042 34 4 3955 4010 3939 4089 3998 32 5 4191 3907 3985 4032 4006 38 6 3906 3990 3984 4022 3976 14 7 3963 3942 4004 3950 3965 21 8 4195 4074 4043 3940 4041 20 9 4028 4089 3961 4048 4032 52 10 3928 4049 3888 4072 3984 36 11 4057 4004 4085 4051 4049 85 12 3990 4067 4051 4053 4040 125 13 4070 4054 4073 4045 4061 186 14 3958 4090 3962 3983 3998 184 15 4016 4069 4037 3969 4023 207 16 4046 4021 4009 4058 4034 241 11 Figure 168 J control chart for the data of Example 168 12 Figure 169 shows CUSUM plots with a particular V mask superimposed The plot in Figure 169a is for current time r 12 A11 points in this plot lie inside the arms of the mask However the plot for r 13 displayed in Figure 169b gives the mean has shifted to higher value 13 Equivalent to the graphical form of the CUSUM procedure ie V mask is the Computational Form of CUSUM Procedure Set do e0 0 and calculate 611612 and e1e2 from the following recursive formulas 61 maX 00111 0 10 e maX 0611 0 10 A for i 12 where kslope of lower arm as before Rule If dt gt h or er gt h where h is found from the Kemp nomogram as describe before then the process is declared out of control at inspection time ti If this happens due to 61 gt h we conclude the mean has shifted to a higher value If it happens due to e gt h we conclude the mean has shifted to a lower value Ex Cont Suppose that the shift to be quickly detected is A 3 Note IN the description of CUSUM A is the shift in ordinary units not in standard deViation units We specify ARL when A 3 ie when y 403 org 2 397 to be 7 so B l 2 857 Carry out the computational form of the CUSUM procedure 14 S01 Here yo 40 but for the computational form we must A also know k and h As always k E 15 and the h is found from the Kemp nomogram However now we have different information we know that n 4 we know A 3 and we know the desired ARL as shift A 3 which is 7 But from this info we can nd I 15 cm 5 M Then we can connect k 6 with the outCSofcontrol ARL7 to nd h 38 from which we nd h nh 95 We also nd an incontrol ARL of 500 Now we are ready to proceed with the computational form of CUSUM do 0 I 15 d1 max max d2 max max 613 max max 0610 i1 4015 00 4020 4015 05 0611 2 4015 0 05 3972 4015 0 0d23 4015 00 4042 4015 27 etc To illustrate the calculations for the e we have e0 0 e1 max 0e0 i1 3985 max 0 0 4020 3985 0 e2 max 0e1 i2 3985 max 00 3972 3985 13 etc The rst outof control signal comes at the 13th sample and is 16 due to 03913 117 being greater than h 95 We conculate that the mean has shifted to a higher value which is indeed the case 17

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