商務(wù)與經(jīng)濟(jì)統(tǒng)計(jì)習(xí)題答案第8版中文版SBE8SM20

1、chapter 20statistical methods for quality controllearning objectives1.learn about the importance of quality control and how statistical methods can assist in the quality control process.2.learn about acceptance sampling procedures.3.know the difference between consumers risk and producers risk.4.be

2、able to use the binomial probability distribution to develop acceptance sampling plans.5.know what is meant by multiple sampling plans.6.be able to construct quality control charts and understand how they are used for statistical process control.7.know the definitions of the following terms:producer

3、's riskassignable causesconsumer's riskcommon causesacceptance samplingcontrol chartsacceptable criterionupper control limitoperating characteristic curvelower control limit13 - 1statistical methods for quality controlsolutions:1.a.for n = 4ucl = m + 3(s / ) = 12.5 + 3(.8 / ) = 13.7lcl = m -

4、 3(s / ) = 12.5 - 3(.8 / ) = 11.3b.for n = 8ucl = m + 3(.8 /) = 13.35lcl = m - 3(.8 /) = 11.65for n = 16ucl = m + 3(.8 /) = 13.10lcl = m - 3(.8 /) = 11.90c.ucl and lcl become closer together as n increases. if the process is in control, the larger samples should have less variance and should fall cl

5、oser to 12.5.2.a.b.ucl = m + 3(s / ) = 5.42 + 3(.5 / ) = 6.09lcl = m - 3(s / ) = 5.42 - 3(.5 / ) = 4.753.a.b.c.ucl = p + 3= 0.0540 + 3(0.0226) = 0.1218lcl = p - 3= 0.0540 -3(0.0226) = -0.0138use lcl = 04.r chart:ucl = = 1.6(1.864) = 2.98lcl = = 1.6(0.136) = 0.22chart:ucl = = 28.5 + 0.373(1.6) = 29.1

6、0lcl = = 28.5 - 0.373(1.6) = 27.905.a.ucl = m + 3(s / ) = 128.5 + 3(.4 / ) = 128.99lcl = m - 3(s / ) = 128.5 - 3(.4 / ) = 128.01b.in controlc.out of control6.process mean = ucl = m + 3(s / ) = 20.01 + 3(s / ) = 20.12solve for s:7.samplenumberobservationsri131422833.6714226183526.3317325303429.679417

7、252121.008538293534.009641423639.676721172922.3312832262828.676941343336.0081029173025.33131126314032.33141223192522.3361317243224.33151443351731.67261518252924.00111630423134.33121728363232.0081840293133.33111918292825.00112022342627.3312= 11.4 and r chart:ucl = = 11.4(2.575) = 29.35lcl = = 11.4(0)

8、 = 0chart:ucl = = 29.17 + 1.023(11.4) = 40.8lcl = = 29.17 - 1.023(11.4) = 17.5r chart:chart:8.a.b.ucl = p + 3= 0.0470 + 3(0.0173) = 0.0989lcl = p - 3= 0.0470 -3(0.0173) = -0.0049use lcl = 0c.process should be considered in control.d.p = .047, n = 150ucl = np + 3= 150(0.047) + 3 = 14.826lcl = np - 3=

9、 150(0.047) - 3 = -0.726thus, the process is out of control if more than 14 defective packages are found in a sample of 150.e.process should be considered to be in control since 12 defective packages were found.f.the np chart may be preferred because a decision can be made by simply counting the num

10、ber of defective packages.9.a.total defectives: 165b.ucl = p + 3= 0.0413 + 3(0.0141) = 0.0836lcl = p - 3= 0.0413 + 3(0.0141) = -0.0010use lcl = 0c. out of controld.p = .0413, n = 200ucl = np + 3= 200(0.0413) + 3 = 16.702lcl = np - 3= 200(0.0413) - 3 = 0.1821e.the process is out of control since 20 d

11、efective pistons were found.10.when p = .02, the probability of accepting the lot iswhen p = .06, the probability of accepting the lot is11.a.using binomial probabilities with n = 20 and p0 = .02.p (accept lot) = f (0) = .6676producers risk: a = 1 - .6676 = .3324b.p (accept lot) = f (0) = .2901produ

12、cers risk: a = 1 - .2901 = .709912.at p0 = .02, the n = 20 and c = 1 plan providesp (accept lot) = f (0) + f (1) = .6676 + .2725 = .9401producers risk: a = 1 - .9401 = .0599at p0 = .06, the n = 20 and c = 1 plan providesp (accept lot) = f (0) + f (1) = .2901 + .3703 = .6604producers risk: a = 1 - .6

13、604 = .3396for a given sample size, the producers risk decreases as the acceptance number c is increased.13.a.using binomial probabilities with n = 20 and p0 = .03.p(accept lot)= f (0) + f (1)= .5438 + .3364 = .8802producers risk: a = 1 - .8802 = .1198b.with n = 20 and p1 = .15.p(accept lot)= f (0)

14、+ f (1)= .0388 + .1368 = .1756consumers risk: b = .1756c.the consumers risk is acceptable; however, the producers risk associated with the n = 20, c = 1 plan is a little larger than desired.14.cp (accept)p0 = .05producersrisk ap (accept)p1 = .30consumers risk b(n = 10)0.5987.4013.0282.02821.9138.086

15、2.1493.14932.9884.0116.3828.3828(n = 15)0.4633.5367.0047.00471.8291.1709.0352.03522.9639.0361.1268.12683.9946.0054.2968.2968(n = 20)0.3585.6415.0008.00081.7359.2641.0076.00762.9246.0754.0354.03543.9842.0158.1070.1070the plan with n = 15, c = 2 is close with a = .0361 and b = .1268. however, the plan

16、 with n = 20, c = 3 is necessary to meet both requirements.15.a.p (accept) shown for p values below:cp = .01p = .05p = .08p = .10p = .150.8179.3585.1887.1216.03881.9831.7359.5169.3918.17562.9990.9246.7880.6770.4049the operating characteristic curves would show the p (accept) versus p for each value

17、of c.b.p (accept)cat p0 = .01producers riskat p1 = .08consumers risk0.8179.1821.1887.18871.9831.0169.5169.51692.9990.0010.7880.788016.a.b.ucl = m + 3(s / ) = 95.4 + 3(.50 / ) = 96.07lcl = m - 3(s / ) = 95.4 - 3(.50 / ) = 94.73c.no; all were in control17.a.for n = 10ucl = m + 3(s / ) = 350 + 3(15 / )

18、 = 364.23lcl = m - 3(s / ) = 350 - 3(15 / ) = 335.77for n = 20ucl = 350 + 3(15 / ) = 360.06lcl = 350 - 3(15 / ) = 339.94for n = 30ucl = 350 + 3(15 / ) = 358.22lcl = 350 - 3(15 / ) = 343.78b.both control limits come closer to the process mean as the sample size is increased.c.the process will be decl

19、ared out of control and adjusted when the process is in control.d.the process will be judged in control and allowed to continue when the process is out of control.e.all have z = 3 where area = .4986p (type i) = 1 - 2 (.4986) = .002818.r chart:ucl = = 2(2.115) = 4.23lcl = = 2(0) = 0chart:ucl = = 5.42

20、 + 0.577(2) = 6.57lcl = = 5.42 - 0.577(2) = 4.27estimate of standard deviation:19. = 0.665 = 95.398 chart:ucl = = 95.398 + 0.577(0.665) = 95.782lcl = = 95.398 - 0.577(0.665) = 95.014r chart:ucl = = 0.665(2.115) = 1.406lcl = = 0.665(0) = 0the r chart indicated the process variability is in control. a

21、ll sample ranges are within the control limits. however, the process mean is out of control. sample 11 (= 95.80) and sample 17 (=94.82) fall outside the control limits.20.= .053 = 3.082 chart:ucl = = 3.082 + 0.577(0.053) = 3.112lcl = = 3.082 - 0.577(0.053) = 3.051r chart:ucl = = 0.053(2.115) = 0.112

22、1lcl = = 0.053(0) = 0all data points are within the control limits for both charts.21.a.warning: process should be checked. all points are within control limits; however, all points are also greater than the process proportion defective.b.warning: process should be checked. all points are within con

23、trol limits yet the trend in points show a movement or shift toward ucl out-of-control point.22.a.p = .04ucl = p + 3= 0.04 + 3(0.0139) = 0.0817lcl = p - 3= 0.04 - 3(0.0139) = -0.0017use lcl = 0b.for month 1 = 10/200 = 0.05. other monthly values are .075, .03, .065, .04, and .085. only the last month

24、 with = 0.085 is an out-of-control situation.23.a.use binomial probabilities with n = 10.at p0 = .05, p(accept lot)= f (0) + f (1) + f (2)= .5987 + .3151 + .0746 = .9884producers risk: a = 1 - .9884 = .0116at p1 = .20,p(accept lot)= f (0) + f (1) + f (2)= .1074 + .2684 + .3020 = .6778consumers risk:

25、 b = .6778b.the consumers risk is unacceptably high. too many bad lots would be accepted.c.reducing c would help, but increasing the sample size appears to be the best solution.24.a.p (accept) are shown below: (using n = 15)p = .01p = .02p = .03p = .04p = .05f (0).8601.7386.6333.5421.4633f (1).1303.2261.2938.3388.3658.