6.1 Table of Hole Diameter Data

sample Number

Carbon-fiber composite material1 2 3 4 5 Ri

1 -30 50 -20 10 30 8 802 0 50 -60 -20 30 0 1103 -50 10 20 30 20 6 804 -10 -10 30 -20 50 8 705 20 -40 50 20 10 12 906 0 0 40 -40 20 4 807 0 0 20 -20 -10 -2 408 70 -30 30 -10 0 12 1009 0 0 20 -20 10 2 40

10 10 20 30 10 50 24 4011 40 0 20 0 20 16 4012 30 20 30 10 40 26 3013 30 -30 0 10 10 4 6014 30 -10 50 -10 -30 6 8015 10 -10 50 40 0 18 6016 0 0 30 -10 0 4 4017 20 20 30 30 -20 16 5018 10 -20 50 30 10 16 7019 50 -10 40 20 0 20 6020 50 0 0 30 10 18 50

and 218 1270 and 10.9 63.5

(a) set up and R charts on the process. Is the process in statistical control?

From the problem, it is known that n =5, therefore we use Appendix table VI that D3=0 and D4=2.114.

LC = =63.5

LCL = = 63.5 (0) = 0

UCL = = 63.5 (2.114) =134.239

For chart, it is used A2= 0.577, therefore

UCL = +A2 = 10.9 + 0.577 (63.5) = 47.5395

LC = =10.9

LCL = -A2 = 10.9 - 0.577 (63.5) = -25.7395

Sample

Sam

ple

Range

191715131197531

140

120

100

80

60

40

20

0

_R=63.5

UCL=134.3

LCL=0

R Chart of C1; ...; C5

6.1.1 R chart

Sample

Sam

ple

Mean

191715131197531

50

40

30

20

10

0

-10

-20

-30

__X=10.9

UCL=47.53

LCL=-25.73

Xbar Chart of C1; ...; C5

6.1.2 chart

From the chart and R chart, there aren’t no indication of out-of-control condition. Since they both

exhibit control, we would conclude that the process in in control.

(b) Estimate the process standard deviation using the range method

= = 27.367

(c) if specification are at nominal ±100, what can you say about the capability of this process? Calculate the PCR Cp.

in ten-thousandths of an inch = 10.9 x 10,000 = 109000 inch

USL and LSL= 109,100 and 108,900 or we can say 10.91 and 10.89 (in ten-thousandths of inch).

The process capability:

= = = = 1.218

= x100% = x 100% = 0.821 x 100% ≈ 82%.

The process uses up about 82% of the specification band.

6.7

number of

sample x1 x2 x3 x4 x5 x6 x7 x8 x9 x101 2.5 0.5 2 -1 1 -1 0.5 1.5 0.5 -1.5 0.5 1.3333332 0 0 0.5 1 1.5 1 -1 1 1.5 -1 0.45 0.9264633 1.5 1 1 -1 0 -1.5 -1 -1 1 -1 -0.1 1.1254634 0 0.5 -2 0 -1 1.5 -1.5 0 -2 -1.5 -0.6 1.1737885 0 0 0 -0.5 0.5 1 -0.5 -0.5 0 0 0 0.4714056 1 -0.5 0 0 0 0.5 -1 1 -2 1 0 0.9718257 1 -1 -1 -1 0 1.5 0 1 0 0 0.05 0.8959798 0 -1.5 -0.5 1.5 0 0 0 -1 0.5 -0.5 -0.15 0.8181969 -2 -1.5 1.5 1.5 0 0 0.5 1 0 1 0.2 1.183216

10 -0.5 3.5 0 -1 -1.5 -1.5 -1 -1 1 0.5 -0.15 1.52843411 0 1.5 0 0 2 -1.5 0.5 -0.5 2 -1 0.3 1.20646412 0 -2 -0.5 0 -0.5 2 1.5 0 0.5 -1 0 1.15470113 -1 -0.5 -0.5 -1 0 0.5 0.5 -1.5 -1 -1 -0.55 0.6851614 0.5 1 -1 -0.5 -2 -1 -1.5 0 1.5 1.5 -0.15 1.24833215 1 0 1.5 1.5 1 -1 0 1 -2 -1.5 0.15 1.270389

-0.05 15.99315

-0.00333 1.06621

(a) Set up and s control charts on this process. Does the process exhibit statistical control?

If necessary, construct revised control limits.

The parameters for s chart can be written as:

UCL = B4 = 1.716 x 1.06621= 1.829616

Center line = = 1.06621

LCL = B3 = 0.284 x 1.06621 = 0.302804

The parameters for chart can be written as:

UCL = = = 1.036551

Center line = = -0.00333

LCL = = = -1.04322

Sample

Sam

ple

StD

ev

151413121110987654321

2.0

1.5

1.0

0.5

_S=1.066

UCL=1.830

LCL=0.302

S Chart of 1; ...; 10

6.7.1. s chart

Sample

Sam

ple

Mean

151413121110987654321

1.0

0.5

0.0

-0.5

-1.0

__X=-0.003

UCL=1.040

LCL=-1.047

Xbar Chart of 1; ...; 10

6.7.2 chart

The process exhibit statistical control.

(b) Set up an R chart, and compare with the s chart in part (a)

number of sample

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 R

1 2.5 0.5 2 -1 1 -1 0.5 1.5 0.5 -1.5 42 0 0 0.5 1 1.5 1 -1 1 1.5 -1 2.53 1.5 1 1 -1 0 -1.5 -1 -1 1 -1 34 0 0.5 -2 0 -1 1.5 -1.5 0 -2 -1.5 3.55 0 0 0 -0.5 0.5 1 -0.5 -0.5 0 0 1.56 1 -0.5 0 0 0 0.5 -1 1 -2 1 37 1 -1 -1 -1 0 1.5 0 1 0 0 2.58 0 -1.5 -0.5 1.5 0 0 0 -1 0.5 -0.5 39 -2 -1.5 1.5 1.5 0 0 0.5 1 0 1 3.5

10 -0.5 3.5 0 -1 -1.5 -1.5 -1 -1 1 0.5 511 0 1.5 0 0 2 -1.5 0.5 -0.5 2 -1 3.512 0 -2 -0.5 0 -0.5 2 1.5 0 0.5 -1 413 -1 -0.5 -0.5 -1 0 0.5 0.5 -1.5 -1 -1 214 0.5 1 -1 -0.5 -2 -1 -1.5 0 1.5 1.5 3.515 1 0 1.5 1.5 1 -1 0 1 -2 -1.5 3.5

48

3.2

n =10, therefore we use Appendix table VI that D3=0.223 and D4=1.777. =3.2

LCL = = 3.2 (0.223) = 0.7136

UCL = = 3.2 (1.777) = 5.6864

Sample

Sam

ple

Range

151413121110987654321

6

5

4

3

2

1

0

_R=3.2

UCL=5.686

LCL=0.714

R Chart of 1; ...; 10

6.7.3 R chart6.13

sample number x1 x2 x3 x4 x5 R

1 83 81.2 78.7 75.7 77 79.1 7.3

2 88.6 78.3 78.8 71 84.2 80.2 17.63 85.7 75.8 84.3 75.2 81 80.4 10.54 80.8 74.4 82.5 74.1 75.7 77.5 8.45 83.4 78.4 82.6 78.2 78.9 80.3 5.26 75.3 79.9 87.3 89.7 81.8 82.8 14.47 74.5 78 80.8 73.4 79.7 77.3 7.48 79.2 84.4 81.5 86 74.5 81.1 11.59 80.5 86.2 76.2 64.1 80.2 77.4 22.1

10 75.7 75.2 71.1 82.1 74.3 75.7 1111 80 81.5 78.4 73.8 78.1 78.4 7.712 80.6 81.8 79.3 73.8 81.7 79.4 813 82.7 81.3 79.1 82 79.5 80.9 3.614 79.2 74.9 78.6 77.7 75.3 77.1 4.315 85.5 82.1 82.8 73.4 71.7 79.1 13.816 78.8 79.6 80.2 79.1 80.8 79.7 217 82.1 78.2 75.5 78.2 82.1 79.2 6.618 84.5 76.9 83.5 81.2 79.2 81.1 7.619 79 77.8 81.2 84.4 81.6 80.8 6.620 84.5 73.1 78.6 78.7 80.6 79.1 11.4

and 1586.7 187.0

and 79.3 9.4

(a) Establish and R control charts for compressive strength using these data. Is the process in

statistical control?

For R chart (use D3 = 0 and D4= 2.114 from Apendix table VI because n=5):

UCL = = 9.4 (2.114) =19.8716

Center line = = 9.4

LCL = = 9.4 (0) = 0

For chart (use A2 =0.577 from Apendix table VI because n=5):

UCL = +A2 = 79.3+ 0.577 (9.4) = 84.7568

Center line = =79.3

LCL = -A2 = 79.3 - 0.577 (9.4) = 73.9092

Using minitab we can see the the R Chart and chart as below:

Sample

Sam

ple

Range

191715131197531

25

20

15

10

5

0

_R=9.35

UCL=19.77

LCL=0

1

R Chart of C1; ...; C5

6.13.1 R Chart for 20 subgroups

Sample

Sam

ple

Mean

191715131197531

85.0

82.5

80.0

77.5

75.0

__X=79.33

UCL=84.97

LCL=73.69

Xbar Chart of C1; ...; C5

6.13.2 Chart for 20 subgroups

From the R Chart shown above, there is an out-of-control condition observed. The value which is out-of-control from UCL is R= 22.1 whereas UCL is 19.8716. however, the chart exhibit control.

(b) After establishing the control charts in part (a), 15 new subgroups were collected and the compressive strength are shown in table 6E.8. Plot the and R values against the control units from part (a) and draw conclusions.

sample number x1 x2 x3 x4 x5 xbar R

1 83 81.2 78.7 75.7 77 79.12 7.32 88.6 78.3 78.8 71 84.2 80.18 17.63 85.7 75.8 84.3 75.2 81 80.4 10.54 80.8 74.4 82.5 74.1 75.7 77.5 8.45 83.4 78.4 82.6 78.2 78.9 80.3 5.26 75.3 79.9 87.3 89.7 81.8 82.8 14.47 74.5 78 80.8 73.4 79.7 77.28 7.4

8 79.2 84.4 81.5 86 74.5 81.12 11.59 80.5 86.2 76.2 64.1 80.2 77.44 22.1

10 75.7 75.2 71.1 82.1 74.3 75.68 1111 80 81.5 78.4 73.8 78.1 78.36 7.712 80.6 81.8 79.3 73.8 81.7 79.44 813 82.7 81.3 79.1 82 79.5 80.92 3.614 79.2 74.9 78.6 77.7 75.3 77.14 4.315 85.5 82.1 82.8 73.4 71.7 79.1 13.816 78.8 79.6 80.2 79.1 80.8 79.7 217 82.1 78.2 75.5 78.2 82.1 79.22 6.618 84.5 76.9 83.5 81.2 79.2 81.06 7.619 79 77.8 81.2 84.4 81.6 80.8 6.620 84.5 73.1 78.6 78.7 80.6 79.1 11.421 68.9 81.5 78.2 80.8 81.5 78.18 12.622 69.8 68.6 80.4 84.3 83.9 77.4 15.723 78.5 85.2 78.4 80.3 81.7 80.82 6.824 76.9 86.1 86.9 94.4 83.9 85.64 17.525 93.6 81.6 87.8 79.6 71 82.72 22.626 65.5 86.8 72.4 82.6 71.4 75.74 21.327 78.1 65.7 83.7 93.7 93.4 82.92 2828 74.9 72.6 81.6 87.2 72.7 77.8 14.629 78.1 77.1 67 75.7 76.8 74.94 11.130 78.7 85.4 77.7 90.7 76.7 81.84 1431 85 60.2 68.5 71.1 82.4 73.44 24.832 86.4 79.2 79.8 86 75.4 81.36 1133 78.5 99 78.3 71.4 81.8 81.8 27.634 68.8 62 82 77.5 76.1 73.28 2035 83 83.7 73.1 82.2 95.3 83.46 22.2

and 2778 456.8 and 79.37143 13.05143

For R chart (use D3 = 0 and D4= 2.114 from Apendix table VI because n=5):

UCL = = 13.05143 (2.114) = 27.60

Center line = = 13.05143

LCL = = 13.05143 (0) = 0

For chart (use A2 =0.577 from Apendix table VI because n=5):

UCL = +A2 = 79.37143 + 0.577 (13.05143) = 86.90

Center line = =79.37143

LCL = -A2 = 79.37143 - 0.577 (13.05143) = 71.84

Using minitab we can see the the R Chart and chart as below:

Sample

Sam

ple

Range

343128252219161310741

30

25

20

15

10

5

0

_R=13.05

UCL=27.60

LCL=0

11

R Chart of 1; ...; 5

6.13.3 R Chart for 35 subgroups

Sample

Sam

ple

Mean

343128252219161310741

88

86

84

82

80

78

76

74

72

70

__X=79.37

UCL=86.90

LCL=71.84

Xbar Chart of 1; ...; 5

6.13.4 Chart for 35 subgroupsConclusions:

6.19 Control charts for and R are maintained for an important quality characteristic. The sample size is n= 7; and R are computed for each sample. After 35 samples, we have found that:

=7805 and =1200

(a) Set up and R charts using these data

Control limits for the R Chart (since n=7, the value of D3 is 0.076 and D4 is 1.924 ):

UCL = D4 = 1.924 x 34.286 =65.966

Center line = = = = 34.286

LCL = D3 = 0.076 x 34.286 =2.605

Control limits for the Chart (since n=7, the value of A2=0.419 and is = 223):

UCL = +A2 = 223 + 0.419 (34.286) =237.3658

Center line = =223

LCL = -A2 = 223 - 0.419 (34.286) = 208.6342

(b) Assuming that both charts exhibit control, estimate the process mean and standard deviation

The process mean = µ ≈ = 223

The estimator of is = = = 12.679

(c) If the quality characteristic is normally distributed and if the spesifications are 220± 35, can the process meet the specifications? Estimate the fraction nonconforming.

USL = 220 +35 = 255 and LSL = 220-35=185

P= P{x<185} + P{x>255}

= Φ +1- Φ

= Φ(-2.996) +1 – Φ(2.5237)

= 0.001368 + 1 - 0.994194 = 0.007174

It is about 0.7174 percent [7174 parts per millon(ppm)] of the products produced will be aoutside of the specifications.

(d) Assuming the variance to remain constant, state where the process mean should be located to minimize the fraction nonconforming. What would be the value of fraction nonconforming under these conditions?

6.25 Suppose that the following the construction of the and R control charts in exercise 6.23, the process engineers decided to change the subgroup size to n = 2. Table 6E.11 contains 10 new subgroups of tickness data. Plot this data on the control charts from exercise 6.23 (a) based on the new subgroup size. Is the process in statistical control?

subgroup x1 x2 x3 x4 x bar R1 459 449 435 450 448.25 242 443 440 442 442 441.75 33 457 444 449 444 448.5 134 469 463 453 438 455.75 315 443 457 445 454 449.75 146 444 456 456 457 453.25 137 445 449 450 445 447.25 58 446 455 449 452 450.5 99 444 452 457 440 448.25 17

10 432 463 463 443 450.25 3111 445 452 453 438 447 1512 456 457 436 457 451.5 2113 459 445 441 447 448 1814 441 465 438 450 448.5 2715 460 453 457 438 452 2216 453 444 451 435 445.75 1817 451 460 450 457 454.5 1018 422 431 437 429 429.75 1519 444 446 448 467 451.25 2320 450 450 454 454 452 4

and 8973.75 333 and 448.6875 16.65

From exercise 6.23, we know that

n old =4 , = 16.65

and from Appendix Table VI we have

d2 (old) =2.059 d2 (new)= 1.128,

therefore, the new control limits on the chart are found :

UCL = = 448.69 + 1.88(0.548)16.65= 465.84

LCL = = 448.69 - 1.88(0.548)16.65= 431.54

For the R chart, the new parameters are :

UCL = D4 = 3.267 (0.548)16.65 = 29.808

CL = = = 9.1242

LCL = max {0, D3 }= 0

From here, we can make the chart to se whether the process in statistical control or not.

6.31.Specifications on a cigar lighter detent are 0.3220 and 0.3200 in. Samples of size 5 are taken every 45 min with the results shown in Table 6E.13 (measured as deviations from 0.3210 in 0.0001 in).

(a) Set up an R chart and examine the process for statistical control.

sample number x1 x2 x3 x4 x5 x bar R

1 1 9 6 9 6 6.2 82 9 4 3 0 3 3.8 93 0 9 0 3 2 2.8 94 1 1 0 2 1 1 25 -3 0 -1 0 -4 -1.6 46 -7 2 0 0 2 -0.6 97 -3 -1 -1 0 -2 -1.4 38 0 -2 -3 -3 -2 -2 39 2 0 -1 -3 -1 -0.6 5

10 0 2 -1 -1 2 0.4 311 -3 -2 -1 -1 2 -1 512 -16 2 0 -4 -1 -3.8 1813 -6 -3 0 0 -8 -3.4 814 -3 -5 5 0 5 0.4 1015 -1 -1 -1 -2 -1 -1.2 1

and -1 97 and -0.06667 6.466667

Control limits for the R Chart (since n=5, the value of D3 is 0 and D4 is 2.114 ):

UCL = D4 = 2.114 x 6.47 =13.67

Center line = = = 6.47

LCL = D3 = 0 x 6.47 = 0

Sample

Sam

ple

Range

151413121110987654321

20

15

10

5

0

_R=6.47

UCL=13.67

LCL=0

1

R Chart of C1; ...; C5

From the picture, there is a sample that is out of control (sample 12). Then, we can eliminate the extrim data and recalculate the parameter.

(c) What parameters would you recommend for an R chart for on-line control?

The parameters recommended in this case are:

UCL = D4 new= 2.114 x5.643 =11.929

Center line = new= = 5.643

LCL = D3 new= 0 x5.643 = 0

The chart is given below:

Sample

Sam

ple

Range

1413121110987654321

12

10

8

6

4

2

0

_R=5.64

UCL=11.93

LCL=0

R Chart of C1; ...; C5

The R chart above indicate that there’s no out-of-control sample. Therefore, since the R chart exhibit control, we would conclude that the parameters we would reccomend are UCL = 11.929, Center line = 5.643 and LCL = 0

(d) Estimate the standard deviation of the process.

= = 2.426

(e) What is the process capability?

Cp= = = 1.374

P = = = 72.78%

6.37 Thirty samples each of size 7 have been collected to establish control over a process. The following

data were collected: and

(a) calculate trial control limits for the two charts

Trial control limit for chart:

UCL = +A2 = 90 +0.419(4)=91.676

CL = = =90

LCL = -A2 = 90 - 0.419(4)=88.324

Trial control limit for R chart:

UCL = D4 = 1.92 4(4)=7.696

CL= = =4

LCL = = D3 =0.076(4) = 0.304

(b) On the assumption that the R chart is in control, estimate the process standard deviation.

= = 1.479

(c) Suppose an s chart were desired. What would be the appropriate control limits and center line?

Suppose that is given above

UCL = B6 =1.806 (1.479)= 2.671

CL = = 1.479

LCL =B5 = 0.113 (1.479)= 0.167

6.43. and R charts with n=4 and the control parameters below both exhibit control.

Chart R ChartUCL = 815 UCL=46.98

Center Line = 800 Center Line=20.59LCL= 785 LCL=0

What is the probability that a shift in the process mean to 790 will be detected in the first sample following the shift?

First, calculate the probability of not detecting this shift that is:

Since N(µ,̴ /n), then :

We suppose that L=3 (the usual three-sigma limits), k=(µ1-µ0)/σ with σ can be estimated by =

= =10 →k=(790-800)/10= -1

then we have:

= (5)- (-1)

= 1- 0.158655 = 0.841344

Last, the probability that such a shift will be detected in the first subsequent is 1-β= 1- 0.841344 = 0.158655

6.49 The following and s charts based on n=4 have shown statistical control:

Chart S ChartUCL = 710 UCL=18.08

Center Line = 700 Center Line=7.979LCL= 690 LCL=0

(a) Estimate the process parameters µ and σ

For µ, the estimate parameter is = 700

For σ, the estimate parameter is =8.66

(b) If the specifications are at 705 ± 15, and the process output is normally distributed, estimate the fraction nonconforming.

= = = 0.577

= = =1.733x100% =173.3%. the process uses up more than 100%

of the tolerance band. In this case the process is very yield-sensitive, and a large number of nonconforming units will be produced.

(c) For the Chart, find the probability of type I error, assuming σ is constant

(d) Suppose the process mean shifts to 693 and the standard deviation simultaneously shifts to 12. Find the probability of detecting this shift on the Chart on the first subsequent sample.

(e) For the shift of part(d), find the average run length.

6.55. Michelson actually made 100 measurements on the velocity of light in five trials of 20 observations each. The second set of 20 measurements is shown in table 6E.20.

(a) plot these new measurement on the control charts constructed in exercise 6.56. are these new measurements in statistical control? Give a practical interpretation of the control charts.

(b) is there evidence that the variability in the measurement has decreased between trial 1 and trial 2?

6.61.The vane heights for 20 of the castings from Fig. 6.25 are shown in table 6E.23. Construct the “between/within “ control charts for these process data using a range chart to monitor the within-castings vane height. Compare these to the control charts shown in fig.6.27

7.1 The data in table 7E.1 represent the results of inspecting all units of a personal computer produced for the past ten days. Does the process appear to be in control?

day units inspected

non conforming

units

fraction non-

conforming1- stdev LCL UCL

1 80 4 0.05 0.06 0.94 0.026552 -0.01966 0.1396562 110 7 0.064 0.06 0.94 0.022643 -0.00793 0.127933 90 5 0.056 0.06 0.94 0.025033 -0.0151 0.13514 75 8 0.107 0.06 0.94 0.027423 -0.02227 0.1422685 130 6 0.046 0.06 0.94 0.020829 -0.00249 0.1224876 120 6 0.05 0.06 0.94 0.021679 -0.00504 0.1250387 70 4 0.057 0.06 0.94 0.028385 -0.02516 0.1451558 125 5 0.04 0.06 0.94 0.021241 -0.00372 0.1237249 105 8 0.076 0.06 0.94 0.023176 -0.00953 0.129529

10 95 7 0.074 0.06 0.94 0.024366 -0.0131 0.133097Total : 1000 Total: 60 Total: 0.62 0.132398

Because UCL and LCL can’t be negative, then the value of LCL = 0. From the calculation above, we

can se from the p-chart below that there’s no indication of out-of-control condition, therefore the

process appear to be in control.

Sample

Pro

port

ion

10987654321

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00

_P=0.06

UCL=0.1331

LCL=0

P Chart of nonconforming unit

Tests performed with unequal sample sizes

7.7. A control chart indicates that the current process fraction nonconforming is 0.02. If fifty items are inspected each day, what is the probability of defecting a shift in the fraction nonconforming to 0.04 on the first day after the shift? By the end of the third day following the shift?

From the problem above it is known that , n=50,

P{D=x} =

7.13.

(a) set up a control chart for the number nonconforming in samples of n=100

sample number

sample size

number nonconforming

sample fraction of

nonconforming1 100 10 0.12 100 15 0.153 100 31 0.314 100 18 0.185 100 24 0.246 100 12 0.127 100 23 0.238 100 15 0.159 100 8 0.08

10 100 8 0.08 Total: 164 = 0.164

The np Control Chart:

= 0.164, n=100

UCL = n + 3 = 100(0.164) +3 = 27.508 ~28

Center line = n = 100 x 0.164=16.4

LCL= n - 3 = 100(0.164) -3 = 5.292 ~ 5

(b) For the chart established in part (a), what is the probability of detecting a shift in the process fraction nonconforming to 0.30 on the first sample after the shift has occured?

δ= the magnitude of the process shift = 0.3 – 0.164 =0.136

L= 3 (if it is assumed that three-sigma limits are used)

n = 100

7.19 A fraction nonconforming control chart has center line 0.01, UCL=0.0399, LCL=0, and n=100. If three-sigma limits are used, find the smallest sample size that would yield a positive lower control limit.

The sample size must be

= 891

Thus, if n ≥ 892 units, the control chart will have a positive lower control limit.

7.25 A fraction nonconforming control chart is to be established with a center line of 0.01 and two-sigma control limits.

(a) How large should the sample size be if the lower control limit is to be nonzero?

The sample size must be

= 396

Thus, if n ≥ 397 units, the control chart will have a positive lower control limit.

(b) how large should the sample size be if we wish the probability of detecting a shift to 0.04 to be 0.5?

→ n= = = 1.63 ≈2

7.31.Consider an np chart with k-sigma control limits. Derive a general formula for determining the minimum sample size to ensure that the chart has a positive lower control limit.

Based on the LCL for fraction nonconforming, the minimum sample size to ensure that the chart has a positive lower control limit can be obtained:

LCL = p - L > 0

, therefore the LCL for np chart should be:

LCL = np - L > 0

7.37 Consider the data in Exercise 7.35. Suppose a new inspection unit is defined as 2500 m of wire.

Based on exercise 7.35 we can get the information below:

sample number

number of nonconformities

sample number

number of nonconformities

1 1 12 62 1 13 93 3 14 114 7 15 155 8 16 86 10 17 37 5 18 68 13 19 79 0 20 4

10 19 21 911 24 22 20

Total: 22 Total : 189

= = 8.59

Sample

Sam

ple

Count

21191715131197531

25

20

15

10

5

0

_C=8.59

UCL=17.38

LCL=0

1

1

1

C Chart of C1

(a) What are the center line and control limits for a control chart for monitoring future production based on the total number of nonconformities in the new inspection unit?

The center line = = 8.59

UCL = +3 = 8.59 +3 =17.38

LCL = - 3 = 8.59 -3 = - 0.2, because these calculations yield a negative value for the LCL, then set LCL=0.

(b) What are the center line and control limits for a control chart for average nonconformities per unit used to monitor future production?

sample number

number of nonconformities

sample size average

1 1 2500 0.00042 1 2500 0.00043 3 2500 0.00124 7 2500 0.00285 8 2500 0.00326 10 2500 0.0047 5 2500 0.0028 13 2500 0.00529 0 2500 0

10 19 2500 0.007611 24 2500 0.009612 6 2500 0.002413 9 2500 0.003614 11 2500 0.004415 15 2500 0.00616 8 2500 0.0032

17 3 2500 0.001218 6 2500 0.002419 7 2500 0.002820 4 2500 0.001621 9 2500 0.003622 20 2500 0.008

189 0.0756

= = =0.0034, therefore, the parameters for the control chart:

The center line = =0.0034

UCL = +3 = 0.0034 +3 =0.0069

LCL = -3 = 0.0034 -3 ≈0

7.43. Find 0.900 and 0.100 probability limits for a c chart when the process average is equal to sixteen nonconformities.

7.49. A textile mill wishes to establish a control procedure on flaws in towels it manufactures. Using an inspection unit of 50 units, past inspection data show that 100 previous inspection units had 850 total flaws. What type of control chart is appropriate? Design the control chart such that it has two-sided probability control limits of α=0.06, approximately. Give the center line and control limits.

The appropriate control chart for this case is control charts for nonconformities because each product contain more than one nonconformities. We can use the u Chart

From the problem above, it is known that:

n=50, then the past inspection data shows 100 sample number, number of nonconformities= 850, α=0.06→β= 1-0.06 = 0.94

= =8.5

7.55. A production line assembles electric clocks. The average number of nonconformities per clock is estimated to be 0.75. The quality engineer wishes to establish a c chart for this operation, using an inspection unit of six clocks. Find the three-sigma limits for this chart.

It is known that =0.75

Inspection unit= 6

UCL = +3 = 0.75+3 = 3.348

LCL = - 3 = 0.75-3 =-1.848, because these calculations yield a negative value for the LCL, then set LCL=0.

7.61.(a) Set up a c chart for the total number of errors. Is the process in control?

day record 1 record 2 record 3 record 4 record 5Number of

nonconformities

1 8 7 1 11 17 442 11 1 11 2 9 343 1 1 8 2 5 174 3 2 5 1 4 155 3 2 13 6 5 296 6 3 3 3 1 167 8 8 2 1 5 248 4 10 2 6 4 269 1 6 1 3 2 13

10 15 1 3 2 8 2911 1 7 13 5 1 2712 6 7 9 3 1 2613 7 6 3 3 1 2014 2 9 3 8 7 2915 6 14 7 1 8 3616 2 9 4 2 1 1817 11 1 1 3 2 1818 5 5 19 1 3 3319 6 15 5 6 6 3820 2 7 9 2 8 2821 7 5 6 14 10 4222 4 3 8 1 2 1823 4 1 4 20 5 3424 15 2 7 10 17 5125 2 15 3 11 2 33

Total : 698

= =27.92

UCL = +3 = 27.92 +3 = 43.77

LCL = - 3 = 27.92 -3 = 12.068

Sample

Sam

ple

Count

252321191715131197531

50

40

30

20

10

_C=27.92

UCL=43.77

LCL=12.07

1

1

C Chart of C1

From the c chart above, it is indicate that the process is out of control because sample 24 is plotted outside the UCL.

(c) Set up a t chart for the total number of errors, assuming a geometric distribution with a=1. Is the process in control?

(d) discuss the findings from parts (a) and (b). Is the poisson distribution a good model for the customer error data? Is there evidence of this in the data?

9.1. A machine is used to fill cans with motor oil additive. A single sample can is selected every hour and the wight of the can is obtained. Since the filling process is automated, it has very stable variability, and long experience indicates that σ=0.05 oz. The individual observations for 24 hours of operation are shown in table 9E.1

(a) Assuming that the process target is 8.02 oz, set up tabular cusum for this process. Design the cusum using the standardized values h=4.77 and k=1/2.

µ0= 8.02, n= 1, σ=0.05, h=4.77 , k=1/2

=max , with =0

=max , with =0

Calculation (for i=1, for the rest is ilustrated from the table):

=max

= max

= max

=0

=max

= max

= max

=0

sample number x

a b

xi-8.52 Ci+ N+ 7.52-xi Ci- N-

1 8 -0.52 0 0 -0.48 0 02 8.01 -0.51 0 0 -0.49 0 03 8.02 -0.5 0 0 -0.5 0 04 8.01 -0.51 0 0 -0.49 0 05 8 -0.52 0 0 -0.48 0 06 8.01 -0.51 0 0 -0.49 0 07 8.06 -0.46 0 0 -0.54 0 08 8.07 -0.45 0 0 -0.55 0 09 8.01 -0.51 0 0 -0.49 0 0

10 8.04 -0.48 0 0 -0.52 0 011 8.02 -0.5 0 0 -0.5 0 012 8.01 -0.51 0 0 -0.49 0 013 8.05 -0.47 0 0 -0.53 0 014 8.04 -0.48 0 0 -0.52 0 015 8.03 -0.49 0 0 -0.51 0 016 8.05 -0.47 0 0 -0.53 0 017 8.06 -0.46 0 0 -0.54 0 018 8.04 -0.48 0 0 -0.52 0 019 8.05 -0.47 0 0 -0.53 0 020 8.06 -0.46 0 0 -0.54 0 021 8.04 -0.48 0 0 -0.52 0 022 8.02 -0.5 0 0 -0.5 0 023 8.03 -0.49 0 0 -0.51 0 024 8.05 -0.47 0 0 -0.53 0 0

(b) Does the value of σ=0.05 seem reasonable for this process?

If σ=0.05 , then H= 5(0.05) = 0.25 It seems to be reasonable because there’s no out-of-control process.

9.7. Set up a tabular cusum scheme for the flow width data used in example 6.1 (see Tables 6.1 and 6.2). when the procedure is applied to all 45 samples, does the cusum react more quickly than the chart to the shift in the process mean? Use σ=0.14 in setting up the cusum, and design the procedure to quickly detect a shift of about 1σ.

sample number

wafersxbar=xi1 2 3 4 5

1 1.3235 1.4128 1.6744 1.4573 1.6914 1.51192 1.4314 1.3592 1.6075 1.4666 1.6109 1.49513 1.4284 1.4871 1.4932 1.4324 1.5674 1.48174 1.5028 1.6352 1.3841 1.2831 1.5507 1.47125 1.5604 1.2735 1.5265 1.4363 1.6441 1.48826 1.5955 1.5451 1.3574 1.3281 1.4198 1.44927 1.6274 1.5064 1.8366 1.4177 1.5144 1.58058 1.419 1.4303 1.6637 1.6067 1.5519 1.53439 1.3884 1.7277 1.5355 1.5176 1.3688 1.5076

10 1.4039 1.6697 1.5089 1.4627 1.522 1.513411 1.4158 1.7667 1.4278 1.5928 1.4181 1.524212 1.5821 1.3355 1.5777 1.3908 1.7559 1.528413 1.2856 1.4106 1.4447 1.6398 1.1928 1.394714 1.4951 1.4036 1.5893 1.6458 1.4969 1.526115 1.3589 1.2863 1.5996 1.2497 1.5471 1.408316 1.5747 1.5301 1.5171 1.1839 1.8662 1.534417 1.368 1.7269 1.3957 1.5014 1.4449 1.487418 1.4163 1.3864 1.3057 1.621 1.5573 1.457319 1.5796 1.4185 1.6541 1.5116 1.7247 1.577720 1.7106 1.4412 1.2361 1.382 1.7601 1.506021 1.4371 1.5051 1.3485 1.567 1.488 1.469122 1.4738 1.5936 1.6583 1.4973 1.472 1.539023 1.5917 1.4333 1.5551 1.5295 1.6866 1.559224 1.6399 1.5243 1.5705 1.5563 1.553 1.568825 1.5797 1.3663 1.624 1.3732 1.6887 1.526426 1.4483 1.5458 1.4538 1.4303 1.6206 1.499827 1.5435 1.6899 1.583 1.3358 1.4187 1.514228 1.5175 1.3446 1.4723 1.6657 1.6661 1.533229 1.5454 1.0931 1.4072 1.5039 1.5264 1.415230 1.4418 1.5059 1.5124 1.462 1.6263 1.509731 1.4301 1.2725 1.5945 1.5397 1.5252 1.472432 1.4981 1.4506 1.6174 1.5837 1.4962 1.529233 1.3009 1.506 1.6231 1.5831 1.6454 1.531734 1.4132 1.4603 1.5808 1.7111 1.7313 1.579335 1.3817 1.3135 1.4953 1.4894 1.4596 1.427936 1.5765 1.7014 1.4026 1.2773 1.4541 1.4824

37 1.4936 1.4373 1.5139 1.4808 1.5293 1.491038 1.5729 1.6738 1.5048 1.5651 1.7473 1.612839 1.8089 1.5513 1.825 1.4389 1.6558 1.656040 1.6236 1.5393 1.6738 1.8698 1.5036 1.642041 1.412 1.7931 1.7345 1.6391 1.7791 1.671642 1.7372 1.5663 1.491 1.7809 1.5504 1.625243 1.5971 1.7394 1.6832 1.6677 1.7974 1.697044 1.4295 1.6536 1.9134 1.7272 1.437 1.632145 1.6217 1.822 1.7915 1.6744 1.9404 1.7700

9.13.Consider the velocity of light data introduced in exercises 6.56 and 6.55. use only the 20 obseervations in Exercise 6.56 to set up a cusum with target value 734.5. plot all 40 observations from both exercises 6.56 and 6.55 on this cusum. What conclusions can you draw?Cusum with target value 734.5 for 20 observations:

sample number

xa b

xi-735 Ci+ N+ 734-xi Ci- N-1 850 115 115 1 -116 0 02 1000 265 380 2 -266 0 03 740 5 385 3 -6 0 04 980 245 630 4 -246 0 05 900 165 795 5 -166 0 06 930 195 990 6 -196 0 07 1070 335 1325 7 -336 0 08 650 -85 1240 8 84 84 19 930 195 1435 9 -196 0 0

10 760 25 1460 10 -26 0 011 850 115 1575 11 -116 0 012 810 75 1650 12 -76 0 013 950 215 1865 13 -216 0 014 1000 265 2130 14 -266 0 015 980 245 2375 15 -246 0 016 1000 265 2640 16 -266 0 017 980 245 2885 17 -246 0 018 960 225 3110 18 -226 0 019 880 145 3255 19 -146 0 020 960 225 3480 20 -226 0 0

Plot of Cusum with target value 734.5 for all observations:

Sample

Cum

ula

tive S

um

37332925211713951

4000

3000

2000

1000

0 0

UCL=359

LCL=-359

CUSUM Chart of C1

It indicates that the process is out of control.

9.19.Rework Exercise 9.4 using an EWMA control chart with λ= 0.1 and L=2.7. Compare your results to those obtained within the cusum.

Based on the exercise 9.4, the tabular cusum for target value=1050 and σ=25 is described below:

sample number

xa b

xi-1062.5 Ci+ N+ 1037.5-xi Ci- N-1 1045 -17.5 0 0 -7.5 0 02 1055 -7.5 0 0 -17.5 0 03 1037 -25.5 0 0 0.5 0.5 14 1064 1.5 1.5 1 -26.5 0 05 1095 32.5 34 2 -57.5 0 06 1008 -54.5 0 0 29.5 29.5 17 1050 -12.5 0 0 -12.5 17 28 1087 24.5 24.5 1 -49.5 0 09 1125 62.5 87 2 -87.5 0 0

10 1146 83.5 170.5 3 -108.5 0 011 1139 76.5 247 4 -101.5 0 012 169 -893.5 0 0 868.5 868.5 113 1151 88.5 88.5 1 -113.5 755 214 1128 65.5 154 2 -90.5 664.5 315 1238 175.5 329.5 3 -200.5 464 416 1125 62.5 392 4 -87.5 376.5 517 1163 100.5 492.5 5 -125.5 251 618 1188 125.5 618 6 -150.5 100.5 719 1146 83.5 701.5 7 -108.5 0 0

20 1167 104.5 806 8 -129.5 0 0

The calculations for EWMA control chart with λ = 0.1 and L=2.7 are summarized in the table below:

To ilustrate the calculation, consider the first observation is:

UCL= = =1056.75

Center line = µ0 =1050

LCL = = =1043.25

z1 = λx1 +(1- λ)z0 = 0.1(1045)+0.9(1050) = 1049.5

sample number

x EWMA sample number

x EWMA

1 1045 1049.5 11 1139 1058.92 1055 1050.5 12 169 961.93 1037 1048.7 13 1151 1060.14 1064 1051.4 14 1128 1057.85 1095 1054.5 15 1238 1068.86 1008 1045.8 16 1125 1057.57 1050 1050 17 1163 1061.38 1087 1053.7 18 1188 1063.89 1125 1057.5 19 1146 1059.6

10 1146 1059.6 20 1167 1061.7

Sample

EWM

A

191715131197531

1100

1080

1060

1040

1020

1000

__X=1066.3

UCL=1083.4

LCL=1049.2

EWMA Chart of C1

9.25.Derive the variance of the exponentially weighted moving average zi

9.31.An EWMA control chart uses λ=0.4. How wide will the limit be on the Shewhart control chart, expressed as a multiple of width of the steady-state EWMA limits?

From the control limit of EWMA control chart for steady-state value, then the wide is:

= = =

10.1.Discuss how you would use a cusum in the short production run solution. What advantages would it have relative to a Shewhart chart, such as a DNOM version of the chart?

10.7.Reconsider the data in Exercise 10.4 and 10.6. Suppose the process measurement are individual data values, not subgroup averages.

sample number

head1 2 3 4

xbar R xbar R xbar R xbar R1 53 2 54 1 56 2 55 32 51 1 55 2 54 4 54 43 54 2 52 5 53 3 57 24 55 3 54 3 52 1 51 55 54 1 50 2 51 2 53 16 53 2 51 1 54 2 52 27 51 1 53 2 58 5 54 18 52 2 54 4 51 2 55 29 50 2 52 3 52 1 51 3

10 51 1 55 1 53 3 53 511 52 3 57 2 52 4 55 112 51 2 55 1 54 2 58 213 54 4 58 2 51 1 53 114 53 1 54 4 50 3 54 215 55 2 52 3 54 2 52 616 54 4 51 1 53 2 58 517 53 3 50 2 57 1 53 118 52 1 49 1 52 1 49 219 51 2 53 3 51 2 50 320 52 4 52 2 50 3 52 2

(a) Use observations 1-20 in Exercise 10.4 to construct appropriate group of control charts.

chart:

CL= 52.988

UCL = = 52.988+ = 58.727

LCL= = 52.988- = 47.248

R chart:

CL=2.158

UCL = D4MR= 3.267x2.158= 7.05

LCL=D3MR = 0 x 2.158 = 0

= = =52.988

MR= = 2.158

(b) Plot observations 21-30 from Exercise 10.6 on the charts from part (a). Discuss your findings.

sample number

head1 2 3 4

x bar R

x bar R

x bar R

x bar R

21 50 3 54 1 57 2 55 522 51 1 53 2 54 4 54 323 53 2 52 4 55 3 57 124 54 4 54 3 53 1 56 225 50 2 51 1 52 2 58 426 51 2 55 5 54 5 54 327 53 1 50 2 51 4 60 128 54 3 51 4 54 3 61 429 52 2 52 1 53 2 62 330 52 1 53 3 50 4 60 1

(c) Using observations 1-20, construct an individual chart using the average of the readings on all four heads as an individual measurement and an s control chart using the individual measurements on each head. Discuss how these charts function relative to the group control chart.

10.13. Specification on a bearing diameter are established at 8.0 ± 0.01 cm. Sample of size n=8 are used, and a control chart for s shows statistical control, with the best current estimate of the population standard deviation S=0.001. If the fraction of nonconforming product that is barely acceptable is 0.135%, find the three-sigma limits on the modified control chart for this process.

UCL=USL-

USL= 8.01 and LSL = 7.99

Z0.00135= 3

n= 8

σ= 0.001

UCL=8.01- = 8.008061

LCL= 7.99+ = 7.991939

10.19. Set up a moving center line EWMA control chart for the concentration data in Exercise 10.16. Compare it to the residuals control chart in Exercise 10.16 part (c).

=0.1, λ=0.7055, =3.227. observations 8, 56 and 90 exceed control limits

10.25.

(a) Discuss the use of the moving range method to estimate the process standard deviation when the data are positively autocorrelated.

(b)Discuss the use of the sample variance s2 with positively autocorrelated data. Specifically, if the observations at lag ρi, is s2 still an unbiased estimator for σ2?

(c) Does your answer in part (b) imply that s2 would really be a good way (in practice) to estimate σ2

in constructing a control chart for autocorrelated data?

11.1. A product has three quality characteristics. The nominal values of these quality characteristics and their sample covariance matrix have been determined from the analyzis of 30 preliminary samples of size n=10 as follows:

S=

The sample means for each quality characteristic for 15 additional samples of size n=10 are shown in table 11E.1. Is the process in statistical control?

sample number

1 3.1 3.7 32 3.3 3.9 3.13 2.6 3 2.44 2.8 3 2.55 3 3.3 2.86 4 4.6 3.5

7 3.8 4.2 38 3 3.3 2.79 2.4 3 2.2

10 2 2.6 1.811 3.2 3.9 3

12 3.7 4 3

13 4.1 4.7 3.214 3.8 4 2.915 3.2 3.6 2.8

11.7. rework exercise 11.6, assuming that the subgroup size is n=5

(a) find the phase II control limits assuming that α=0.005.

UCL= = =21.3

(b) Compare the control limits from part (a) to the chi-square control limit. What is the magnitude of the difference in the two control limits?

Chi-square control limit = = =18.548

(c) how many preliminary samples would have to be taken to ensure that the exact phase II control limit is within 1% of the chi-square control limit?

11.13. consider all 30 observations on the first two process variables in Table 11.6. Calculate an estimate of the sample covariance matrix using both estimators S1 and S2 discussed in Section 11.3.2. Are the estimates very different? Discuss your findings.

11.19. Consider the p=9 process variables in Table 11.5.

(a) perform a PCA on the first 30 observations. Be sure to work with the standardized variables.

(b) how much variability is explained if only the first r=3 principal components are retained?

(c) Construct an appropriate set of pairwise plot of the first r=3 principal component scores.

(d) Now consider the last 10 observations. Obtain the principal component scores and plot them on the chart in part (c). Does the process seem to be in control?