แบบฝึกบท 6
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272
Variables Chapter 6 Control Charts for Variables
Phase II control chart usage Probability limits for control charts Process capability Process capability ratio (PCR) C p R control chart Rational subgroups s control chart 2 s control chart
Shewhart control charts Specification limits Three-sigma control limits Tier chart or tolerance diagram Trial control limits Variable sample size on control charts Variables control charts x control chart
Exercises The Student Resource Manual presents comprehensive annotated solutions to the odd-numbered exercises included in the Answers to Selected Exercises section in the back of this book.
process. the process seem to be inDoes statistical control? If necessary,, revis necessary revisee the trial trial concontrol limits. (b) If specificatio specifications ns on this this diameter are 0.5030 0.5030 ± 0.0010, 0.001 0, find the percentage percentage of nonconfo nonconforming rming bearings produced by this process. Assume that diameter is normally distributed. A high-voltage power supply should have a nominal output voltage of 350 V. A sample of four units is selected each day and tested for process-control purposes. The data shown in Table 6E.2 give the differ-
6.2.
ence between the observed reading on each unit and the nominal voltage times ten; that is, x i = (observed voltage on unit i - 350)10 (a) Se Sett up x an and d R charts on this process. Is the process in statistical control? (b) If specifica specifications tions are at 350 V ± 5 V, V, what can you say about process capability? (c) Is there evidenc evidencee to support support the claim that that voltvoltage is normally distributed?
6.1. The data shown in Table 6E.1 are x and R values for 24 samples of size n = 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, bearin g, with only the last three three decimals recorded recorded (i.e., (i.e., 34.5 should should be 0.50345). (a)) Se (a Sett up x an and d R charts on this
TABLE 6E 1
Bearing Diameter Data
6.3.
The data diameter shown in for Table 6E.3drilled are theindeviations from nominal holes a carbon-fiber composite material used in aerospace manufacturing.
TA B L E 6 E 2
Voltage Data for Exercise 6.2. Sample Number
x3
x4
9 4 8 9
10 6 10 6
15 11 5 13
9 12 16 7 9 15 8 6 16 7 11 15 9
10 11 10 5 7 16 12 13 9 13 7 10 8
7 10 8 10 8 10 14 9 13 10 10 11 12
13 10 9 4 12 13 16 11 15 12 16 14 10
15 8 13
7 6 14
10 9 11
11 12 15
R
Sample Number
1 2 3 4 5 6 7 8 9
34.5 34.2 31.6 31.5 35.0 34.1 32.6 33.8 34.8
3 4 4 4 5 6 4 3 7
13 14 15 16 17 18 19 20 21
35.4 34.0 37.1 34.9 33.5 31.7 34.0 35.1 33.7
8 6 5 7 4 3 8 4 2
10 11 12
33.6 31.9 38.6
8 3 9
22 23 24
32.8 33.5 34.2
1 3 2
18 19 20
R
x2
6 10 7 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Sample Number
x1
Exercises
TA B L E 6 E 3
Hole Diameter Data for Exercise 6.3 Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
6.4.
6.5.
x1
x2
x3
x4
−30
+50
−20
0 −50 −10 +20 0 0 +70 0 +10 +40 +30 +30 +30 +10 0
+50
−60 −20
Sample Number
x5
+30
+30
−40
+50 +20 +10
−20
+50
+40
−40
+20
−20 −10
+30
−10
+20
0 +20 −20 +10 +30 +10 +50 +20 0 +20 +30 +10 +40 0 +10 +10 +50 −10 −30 +50 +40 0 +30 −10 0 −20
+20 +20 +30 +30
+10
−20
+50 +30 +10
+50
−10
+40 +20
+50
0
0
x 1
x 2
x 3
+10 +30
−10
TABLE 6E 4
Printed Circuit Board Thickness for Exercise 6.4
+10 +20 +30 +20
0 0 −30 0 +20 0 +20 −30 −10 −10 0
273
+30
0 +10
The values reported are deviations from nominal in ten-thousandths of an inch. (a) Se Sett up x an and d R charts on the process. Is the process in statistical control? (b) Estimat Estimatee the process process standard standard deviation deviation using using the range method. (c) If specifica specifications tions are at nomin nominal al ±10 100, 0, wh what at can can you say about the capability of this process? Calculate the PCR C p. The thickness of a printed circuit board is an important quality parameter. Data on board thickness (in inches) are given in Table 6E.4 for 25 samples of three boards each. (a) Se Sett up x and R control charts. Is the process in statistical control? (b) Estimat Estimatee the process process standard standard deviatio deviation. n. (c) What are the the limits that that you would would expect expect to concontain nearly all the process measurements? (d) If the specific specification ationss are at 0.0630 0.0630 in. ± 0.0015 in., what is the value of the PCR C p? The fill volume of soft-drink beverage bottles is an important quality characteristic. The volume is measured (approximately) by placing a gauge over the crown and comparing the height of the liquid in the neck of the bottle against a coded scale. On this
6.6.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0.0629 0.0630 0.0628 0.0634 0.0619 0.0613 0.0630 0.0628 0.0623 0.0631 0.0635 0.0623 0.0635 0.0645 0.0619
0.0636 0.0631 0.0631 0.0630 0.0628 0.0629 0.0639 0.0627 0.0626 0.0631 0.0630 0.0630 0.0631 0.0640 0.0644
0.0640 0.0622 0.0633 0.0631 0.0630 0.0634 0.0625 0.0622 0.0633 0.0633 0.0638 0.0630 0.0630 0.0631 0.0632
16 17 18 19 20 21 22 23 24 25
0.0631 0.0616 0.0630 0.0636 0.0640 0.0628 0.0615 0.0630 0.0635 0.0623
0.0627 0.0623 0.0630 0.0631 0.0635 0.0625 0.0625 0.0632 0.0629 0.0629
0.0630 0.0631 0.0626 0.0629 0.0629 0.0616 0.0619 0.0630 0.0635 0.0630
scale, a reading of zero corresponds to the correct fill fill height. Fifteen samples of size n = 10 have been analyzed, and the fill heights are shown shown in Table Table 6E.5. (a) Se Sett up x an and d s control charts on this process. Does the process exhibit statistical control? If necessary, construct revised revised control limits. (b)) Se (b Sett up up an R chart, and compare compare with with the s chart in part (a). (c)) Se (c Sett up up an an s2 chart and compare with the s chart in part (a). The net weight (in oz) of a dry bleach product is to be monitored by x and R control charts using a sample size of n = 5. Data for 20 preliminary samples are shown in Table 6E.6. (a) Se Sett up x an and d R control charts using these data. Does the process exhibit statistical control? (b) Estimate the process process mean and and standard standard deviation deviation.. (c) Does fill fill weight seem seem to follow follow a normal distribdistribution?
274
Variables Chapter 6 Control Charts for Variables
TABLE 6E 5
(d) If the the specificat specifications ions are are at 16.2 16.2 ± 0.5, what concluconclusions would you draw about process capability? (e) What fraction fraction of contain containers ers produced produced by this this process is likely to be below the lower specification limit of 15.7 oz? Rework Exercise 6.2 using the s chart. Rework Exercise 6.3 using the s chart. Consider the piston ring data shown in Table 6.3.
Fill Height Data for Exercise 6.5 Sample Number x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 1
2.5
0 .5
2.0
−1 .0 .0
1 .0 .0
−1.0
0.5
1.5
0.5 −1.5
2
0.0
0.0
0.5
1.0
1.5
1.0
−1.0
1.0
1.5 −1.0
3
1.5
1 .0
1.0
.0 −1 .0
0 .0 .0
−1.5 −1.0 −1.0
1.0 −1.0
4
0.0
0.5
.0 −2 .0
5
0.0
0 .0
0.0
−0.5
0.5
1.0
−1.5 0.0 −2.0 −1.5 −0.5 −0.5 0.0 0.0
6
1.0 −0.5
0.0
0.0
0.0
0.5
−1.0
1.0
.0 −2 .0
1. 0
7
1.0 −1.0
−1.0 −1.0
0.0
1.5
0.0
1.0
0.0
0.0
8
0.0 −1.5
−0.5
1.5
0.0
0.0
0.0
−1.0
9
−2.0 −1.5
1.5
1.5
0.0
0.0
0.5
1.0
0.0
1.0
10
−0.5
3.5
0.0
−1.0 −1.5 −1.5 −1.0 −1.0
1.0
0.5
11
0.0
1.5
0.0
12
0.0 −2.0
−0. 5
13
0 .0 .0 −1.0
0.0
2.0
−1.5
0.5
−0.5
2.0 −1.0
2.0
1.5
0.0
0.5 −1.0
0.5
0.5
0.5
1.0
15
1.0
0.0
0.0
−1.0 −0.5 −2.0 −1.0 −1.5 1.5
1.5
1.0
−1.0
0.0
−1.5 −1.0 −1.0 0.0 1.0
1.5
1.5
−2.0 −1.5
Data for Exercise 6.6.
x1
x2
6.10.
x and R chart developed in Exercise 6.9. Is the process in control? 6.11. Control charts on x and s are to be maintained on the torque readings of a bearing used in a wingflap actuator assembly. Samples of size n = 10 are to be used, and we know from past experience that when the process is in in control, control, bearing torque torque has has a normal
TABLE 6E 6
Sample Number
Assume that the specifications on this component are 74.000 + 0.05 mm. (a) Se Sett up x an and d R control charts on this process. Is the process in statistical control? (b) Note that the contro controll limits on the x chart in part (a) are identical to the control limits on the x chart in Example Example 6.3, 6.3, where the the limits limits were based on s. Will this always happen? (c) Estimat Estimatee process capabili capability ty for the the piston-ring piston-ring process. Estimate the percentage of piston rings produced that will be outside of the specification. spec ification. Table 6E.7 shows 15 additional samples for the piston ring process process (Table (Table 6.3), 6.3), taken after after the initial control charts were established. Plot these data on the
0.5 −0.5
0 .0 .0 −0.5
−1.0 −0.5 −0.5 −1.0
14
1.5
6.7. 6.8. 6.9.
x3
x4
x5
1 2 3 4 5
15.8 16.3 16.1 16.3 16.1
16.3 15.9 16.2 16.2 16.1
16.2 15.9 16.5 15.9 16.4
16.1 16.2 16.4 16.4 16.5
16.6 16.4 16.3 16.2 16.0
6 7 8 9 10 11 12 13 14 15 16 17 18
16.1 16.1 16.2 16.3 16.6 16.2 15.9 16.4 16.5 16.4 16.0 16.4 16.0
15.8 16.3 16.1 16.2 16.3 16.4 16.6 16.1 16.3 16.1 16.2 16.2 16.2
16.7 16.5 16.2 16.4 16.4 15.9 16.7 16.6 16.2 16.3 16.3 16.4 16.4
16.6 16.1 16.1 16.3 16.1 16.3 16.2 16.4 16.3 16.2 16.3 16.3 16.5
16.4 16.5 16.3 16.5 16.5 16.4 16.5 16.1 16.4 16.2 16.2 16.2 16.1
26
7 4.0 4.01 1 2 74 74 .0 .0 15 15
7 4.0 4.03 3 0 7 3.9 3.98 8 6 7 4.0 4.00 00
7 4.0 4.00 09
0 .0 .0 44 44
27
7 3.9 3.99 9 5 74 74 .0 .0 10 10
7 3.9 3.99 9 0 7 4.0 4.01 1 5 7 4.0 4.00 01
7 4.0 4.00 02
0 .0 .0 25 25
28
7 3.9 3.98 8 7 73 73 .9 .9 99 99
7 3.9 3.98 8 5 7 4.0 4.00 0 0 7 3.9 3.99 90
7 3.9 3.99 92
0 .0 .0 15 15
29
7 4.0 4.00 0 8 74 74 .0 .0 10 10
7 4.0 4.00 0 3 7 3.9 3.99 9 1 7 4.0 4.00 06
7 4.0 4.00 04
0 .0 .0 19 19
30
7 4.0 4.00 0 3 74 74 .0 .0 00 00
7 4.0 4.00 0 1 7 3.9 3.98 8 6 7 3.9 3.99 97
7 3.9 3.99 97
0 .0 .0 17 17
31
7 3.9 3.99 9 4 74 74 .0 .0 03 03
7 4.0 4.01 1 5 7 4.0 4.02 2 0 7 4.0 4.00 04
7 4.0 4.00 07
0 .0 .0 26 26
32
7 4.0 4.00 0 8 74 74 .0 .0 02 02
7 4.0 4.01 1 8 7 3.9 3.99 9 5 7 4.0 4.00 05
7 4.0 4.00 06
0 .0 .0 23 23
33
7 4.0 4.00 0 1 74 74 .0 .0 04 04
7 3.9 3.99 9 0 7 3.9 3.99 9 6 7 3.9 3.99 98
7 3.9 3.99 98
0 .0 .0 14 14
34
7 4.0 4.01 1 5 74 74 .0 .0 00 00
7 4.0 4.01 1 6 7 4.0 4.02 2 5 7 4.0 4.00 00
7 4.0 4.01 11
0 .0 .0 25 25
35
7 4.0 4.03 3 0 74 74 .0 .0 05 05
7 4.0 4.00 0 0 7 4.0 4.01 1 6 7 4.0 4.01 12
7 4.0 4.01 13
0 .0 .0 30 30
36
7 4.0 4.00 0 1 73 73 .9 .9 90 90
7 3.9 3.99 9 5 7 4.0 4.01 1 0 7 4.0 4.02 24
7 4.0 4.00 04
0 .0 .0 34 34
37
7 4.0 4.01 1 5 74 74 .0 .0 20 20
7 4.0 4.02 2 4 7 4.0 4.00 0 5 7 4.0 4.01 19
7 4.0 4.01 17
0 .0 .0 19 19
38
7 4.0 4.03 3 5 74 74 .0 .0 10 10
7 4.0 4.01 1 2 7 4.0 4.01 1 5 7 4.0 4.02 26
7 4.0 4.02 20
0 .0 .0 25 25
19 20
16.4 16.4
16.0 16.4
16.3 16.5
16.4 16.0
16.4 15.8
39
7 4.0 4.01 1 7 74 74 .0 .0 13 13
7 4.0 4.03 3 6 7 4.0 4.02 2 5 7 4.0 4.02 26
7 4.0 4.02 23
0 .0 .0 23 23
40
7 4.0 4.01 1 0 74 74 .0 .0 05 05
7 4.0 4.02 2 9 7 4.0 4.00 0 0 7 4.0 4.02 20
7 4.0 4.01 13
0 .0 .0 29 29
TABLE 6E 7
Piston Ring Diameter Data for Exercise 6.10 Sample Number, i
Observations
x¯ i
Ri
275
Exercises
6.12.
distribution with mean m = 80 inch-pounds and stan = 10 inch-pounds. Find the center dard deviation s = line and control limits for these control charts. Samples of n = 6 items each are taken from a process at regular intervals. A quality characteristic is measured su red,, and x and and R values are calculated for each sample. After 50 samples, samples, we have 50
xi i∑ =1
Data for Exercise 6.14 Sample Number x1 x2
50 =
Ri 2000 and i∑ =1
=
200
Assume that the quality characteristic is normally distributed. (a) Compu Compute te contro controll limits for the x and R control charts. (b) All points points on both both control control charts charts fall between between the control limits computed in part (a). What are the natural tolerance limits of the process? (c) If the specif specification ication limits are 41 ± 5.0 5.0,, wha whatt are are your conclusions regarding the ability of the process to produce items within these specifications? (d) Assumin Assuming g that if an item exceeds exceeds the the upper specspec-
6.13.
ification limit ification limit it can be reworked reworked,, and if it is below the lower specification limit it must be scrapped, scrapp ed, what percent percent scrap and rework rework is the process producing? (e) Make suggestio suggestions ns as to how the the process perforperformance could be improved. Samples of n = 4 items are taken from a process at regular intervals. A normally distributed quality characteristic is measured and x and s values are calculated for each sample. After 50 subgroups have been analyzed, analyze d, we have 50
∑ xi i =1
6.14.
and ∑ si
=
x3
x4
1
1 38 38 .1 .1 1 10 10 .8 .8
2
1 49 49 .3 .3 14 14 2. 2. 1 1 05 05 .0 .0 1 34 34 .0 .0
3 4
x5
1 38 38 .7 .7 1 3 7. 7. 4 1 25 25 .4 .4
x¯
R
1 30 30 .1 .1
2 7. 7. 9
9 2. 2. 3
1 24 24 .5 .5
5 7. 7. 0
1 15 15 .9 .9 1 35 35 .6 .6
1 24 24 .2 .2 1 5 5. 5. 0 1 17 17 .4 .4
1 29 29 .6 .6
3 9. 9. 1
1 18 18 .5 .5 1 16 16 .5 .5
1 30 30 .2 .2 1 2 2. 2. 6 1 00 00 .2 .2
1 17 17 .6 .6
3 0. 0. 0
5
1 08 08 .2 .2 1 23 23 .8 .8
1 17 17 .1 .1 1 4 2. 2. 4 1 50 50 .9 .9
1 28 28 .5 .5
4 2. 2. 7
6
1 02 02 .8 .8 1 12 12 .0 .0
1 35 35 .0 .0 1 3 5. 5. 0 1 45 45 .8 .8
1 26 26 .1 .1
4 3. 3. 0
7
1 20 20 .4 .4
8 4. 4. 3 1 12 12 .8 .8 1 18 18 .5 .5 1 19 19 .3 .3
1 11 11 .0 .0
3 6. 6. 1
8
1 32 32 .7 .7 1 51 51 .1 .1
1 24 24 .0 .0 1 2 3. 3. 9 1 05 05 .1 .1
1 27 27 .4 .4
4 6. 6. 0
9
1 36 36 .4 .4 1 26 26 .2 .2
1 54 54 .7 .7 1 2 7. 7. 1 1 73 73 .2 .2
1 43 43 .5 .5
4 6. 6. 9
10
1 35 35 .0 .0 1 15 15 .4 .4
1 49 49 .1 .1 1 38 38 .3 .3 1 30 30 .4 .4
1 33 33 .6 .6
3 3. 3. 7
11
1 39 39 .6 .6 1 27 27 .9 .9
1 51 51 .1 .1 1 43 43 .7 .7 1 10 10 .5 .5
1 34 34 .6 .6
4 0. 0. 6
12
1 25 25 .3 .3 1 60 60 .2 .2
1 30 30 .4 .4 1 52 52 .4 .4 1 65 65 .1 .1
1 46 46 .7 .7
3 9. 9. 8
13
1 45 45 .7 .7 1 01 01 .8 .8
1 49 49 .5 .5 1 13 13 .3 .3 1 51 51 .8 .8
1 32 32 .4 .4
5 0. 0. 0
14
1 38 38 .6 .6 13 13 9. 9. 0 1 31 31 .9 .9 1 40 40 .2 .2 1 41 41 .1 .1
1 38 38 .1 .1
9. 2
15
1 10 10 .1 .1 1 14 14 .6 .6
1 65 65 .1 .1 1 13 13 .8 .8 1 39 39 .6 .6
1 28 28 .7 .7
5 4. 4. 8
16
1 45 45 .2 .2 1 01 01 .0 .0
1 54 54 .6 .6 1 20 20 .2 .2 1 17 17 .3 .3
1 27 27 .6 .6
5 3. 3. 3
17
1 25 25 .9 .9 1 35 35 .3 .3
1 21 21 .5 .5 1 47 47 .9 .9 1 05 05 .0 .0
1 27 27 .1 .1
4 2. 2. 9
18
1 29 29 .7 .7
9 7. 7. 3 1 30 30 .5 .5 1 09 09 .0 .0 1 50 50 .5 .5
1 23 23 .4 .4
5 3. 3. 2
19
1 23 23 .4 .4 1 50 50 .0 .0
1 61 61 .6 .6 1 48 48 .4 .4 1 54 54 .2 .2
1 47 47 .5 .5
3 8. 8. 3
20
1 44 44 .8 .8 1 38 38 .3 .3
1 19 19 .6 .6 1 51 51 .8 .8 1 42 42 .7 .7
1 39 39 .4 .4
32 .2 .2
Table 6E.8 presents 20 subgroups of five measurements on the critical dimension of a part produced by a machining process. (a) Se Sett up x an and d R control charts on this process. Verify that the process is in statistical control. (b) Foll Followin owing g the establishment establishment of control control charts charts in part (a) above, above, 10 new samples samples in Table Table 6E.9
50 = 1000
TA B L E 6 E 8
72
i =1
(a) Compute Compute the contro controll limit for the x and s control charts. (b) Assume that that all points points on both both charts plot plot within within the control limits. What are the natural tolerance limits of the process? (c) If the specif specification ication limits are 19 ± 4.0 4.0,, wha whatt are are your conclusions regarding the ability of the process to produce items conforming to specifications? (d) Assumin Assuming g that if an item exceeds exceeds the the upper specspecification ifica tion limit limit it can be reworked reworked,, and if it is below the lower specification limit it must be scrapped, scrapp ed, what percent percent scrap and rework rework is the process now producing? (e) If the proce process ss were center centered ed at m = 19 19.0 .0,, wh what at would be the effect on percent scrap scra p and rework?
TA B L E 6 E 9
Additional Data Data for Exercise Exercise 6.14, part (b) Sample Number x1 x2
x3
x4
x5
x¯
R
1
1 31 31 .0 .0 1 84 84 .8 .8
1 82 82 .2 .2 1 4 3. 3. 3 2 12 12 .8 .8
1 70 70 .8 .8
8 1. 1. 8
2
1 81 81 .3 .3 1 93 93 .2 .2
1 80 80 .7 .7 1 6 9. 9. 1 1 74 74 .3 .3
1 79 79 .7 .7
2 4. 4. 0
3
1 54 54 .8 .8 1 70 70 .2 .2
1 68 68 .4 .4 2 0 2. 2. 7 1 74 74 .4 .4
1 74 74 .1 .1
4 8. 8. 0
4
1 57 57 .5 .5 1 54 54 .2 .2
1 69 69 .1 .1 1 4 2. 2. 2 1 61 61 .9 .9
1 57 57 .0 .0
2 6. 6. 9
5
2 16 16 .3 .3 1 74 74 .3 .3
1 66 66 .2 .2 1 5 5. 5. 5 1 84 84 .3 .3
1 79 79 .3 .3
6 0. 0. 8
6
1 86 86 .9 .9 1 80 80 .2 .2
1 49 49 .2 .2 1 7 5. 5. 2 1 85 85 .0 .0
1 75 75 .3 .3
3 7. 7. 8
7
1 67 67 .8 .8 1 43 43 .9 .9
1 57 57 .5 .5 1 7 1. 1. 8 1 94 94 .9 .9
1 67 67 .2 .2
5 1. 1. 0
8
1 78 78 .2 .2 1 86 86 .7 .7
1 42 42 .4 .4 1 5 9. 9. 4 1 67 67 .6 .6
1 66 66 .9 .9
4 4. 4. 2
9 10
1 62 62 .6 .6 1 43 43 .6 .6 1 72 72 .1 .1 1 9 91 1 .7 .7
1 32 32 .8 .8 1 6 8. 8. 9 1 77 77 .2 .2 2 03 03 .4 .4 1 5 50 0 .4 .4 1 9 96 6 .3 .3
1 57 57 .0 .0 1 82 82 .8 .8
4 4. 4. 5 53 .0 .0
276
Variables Chapter 6 Control Charts for Variables
TABLE 6E 10
Strength Data for Exercise 6.15
Sample Number x1 x2
Sample Number x1 x2
6.17.
x3
x4
x5
x¯
R
x3
x4
x5
x¯
R
1
1 31 31 .5 .5 1 43 43 .1 .1
1 18 18 .5 .5
1 03 03.2
1 21 21 .6 .6
1 23 23.6
3 9. 9.8
1
83.0
81.2
7 8 .7
75.7
7 7 .0
79.1
7 .3
2
111.0 1 27 27.3
1 1 0 .4
91.0
1 4 3 .9
116.7
5 2 .8
2
88.6
78.3
7 8 .8
71.0
8 4 .2
80.2
1 7 .6
3
129.8
98.3
1 3 4 .0
105.1
1 3 3 .1
120.1
3 5 .7
3
85.7
75.8
8 4 .3
75.2
8 1 .0
80.4
1 0 .4
4
145.2 1 32 32.8
1 0 6 .1
131.0
9 9 .2
122.8
4 6 .0
4
80.8
74.4
8 2 .5
74.1
7 5 .7
77.5
8 .4
5
1 14 14 .6 .6 1 11 11 .0 .0
1 08 08 .8 .8
1 77 77.5
1 21 21 .6 .6
1 26 26.7
6 8. 8.7
5
83.4
78.4
8 2 .6
78.2
7 8 .9
80.3
5 .2
6
125.2
86.4
6 4 .4
1 37 37.1
1 17 17.5
106.1
7 2 .6
6
75.3
79.9
8 7 .3
89.7
8 1 .8
82.8
1 4 .5
7
145.9 1 09 09.5
8 4 .9
129.8
1 1 0 .6
116.1
6 1 .0
7
74.5
78.0
8 0 .8
73.4
7 9 .7
77.3
7 .4
8
123.6 1 14 14.0
1 3 5 .4
83.2
1 0 7 .6
112.8
5 2 .2
8
79.2
84.4
8 1 .5
86.0
7 4 .5
81.1
1 1 .4
156.3 15
1 1 9 .7
96.2
1 53 53.0
122.2
7 0 .6
9
80.5
86.2
7 6 .2
64.1
8 0 .2
81.4
9 .9
1 07 07 .4 .4 1 48 48 .7 .7
1 27 27 .4 .4
1 25 25.0
1 27 27 .5 .5
1 27 27.2
4 1. 1.3
10
75.7
75.2
7 1 .1
82.1
7 4 .3
75.7
1 0 .9
11
80.0
81.5
7 8 .4
73.8
7 8 .1
78.4
7 .7
12
80.6
81.8
7 9 .3
73.8
8 1 .7
79.4
8 .0
13
82.7
81.3
7 9 .1
82.0
7 9 .5
80.9
3 .6
14
79.2
74.9
7 8 .6
77.7
7 5 .3
77.1
4 .3
15
85.5
82.1
8 2 .8
73.4
7 1 .7
79.1
1 3 .8
16
78.8
79.6
8 0 .2
79.1
8 0 .8
79.7
2 .0
17
82.1
78.2
7 5 .5
78.2
8 2 .1
79.2
6 .6
18
84.5
76.9
8 3 .5
81.2
7 9 .2
81.1
7 .6
19
79.0
77.8
8 1 .2
84.4
8 1 .6
80.8
6 .6
20
84.5
73.1
7 8 .6
78.7
8 0 .6
79.1
1 1 .4
10
6.16.
TABLE 6E 11
New Data for Exercise Exercise 6.14, 6.14, part (c)
9
6.15.
85.8
were collected. Plot the x and R values on the control chart you established in part (a) and draw conclusions. (c) Suppo Suppose se that the the assignable assignable cause cause responsible responsible for the action signals generated in part (b) has been identified and adjustments made to the process to correct its performance. Plot the x and R values from the new subgroups shown in Table 6E.10 which were taken following the adjustment, against the control control chart limits established in part (a). What are your conclusions? Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples of fiv fivee parts each are collected, and the compressive strengths (in psi) are shown in Table 6E.11. (a) Es Estab tabli lish sh x and R control charts for compressive strength using these data. Is the process in statistical control? (b) After establishin establishing g the control control charts in part part (a), 15 new subgroups were collected and the compressive strengths are shown in Table 6E.12. Plot the x and R values against the control units from part (a) and draw conclusions. Reconsider the data presented in Exercise 6.15. (a) Rewo Rework rk both parts parts (a) and (b) of of Exercise Exercise 6.15 using the x and s charts. (b)) Do (b Does es the the s chart detect the shift in process variability more quickly than the R chart did originally in part (b) of Exercise 6.15? Consider the x and R charts you established in Exercise 6.1 using n = 5. (a) Suppose Supp ose thatcharacteristic you wished you wished using to continue con chcharts arting this quality x tinue and Rcharting based on a sample size of n = 3. What limits would be used on the x and R charts?
TABLE 6E 12
New Data for Exercise Exercise 6.15, 6.15, part (b) Sample Number x1 x2
x3
x4
x5
x¯
R
1
68.9
81.5
78.2
80.8
81.5
78.2
12.6
2
69.8
68.6
80.4
84.3
83.9
77.4
15.7
3
78.5
85.2
78.4
80.3
81.7
80.8
6.8
4
76.9
86.1
86.9
94.4
83.9
85.6
17.5
5
93.6
81.6
87.8
79.6
71.0
82.7
22.5
6
65.5
86.8
72.4
82.6
71.4
75.9
21.3
7
78.1
65.7
83.7
93.7
93.4
82.9
27.9
8
74.9
72.6
81.6
87.2
72.7
77.8
14.6
9
78.1
77.1
67.0
75.7
76.8
74.9
11.0
10
78.7
85.4
77.7
90.7
76.7
81.9
14.0
11
85.0
60.2
68.5
71.1
82.4
73.4
24.9
12
86.4
79.2
79.8
86.0
75.4
81.3
10.9
13
78.5
99.0
78.3
71.4
81.8
81.7
27.6
14
68.8
62.0
82.0
77.5
76.1
73.3
19.9
15
83.0
83.7
73.1
82.2
95.3
83.5
22.2
(b) What would would be be the impact impact of the decision decision you made in part (a) on the ability of the x chart to detect a 2s shift in the mean?
Exercises
6.18.
6.19.
(c) Suppose Suppose you wished wished to continue continue charting charting this this quality characteristic using x and R charts based on a sample size of n = 8. What limits would be used on the x and R charts? (d) Wha Whatt is the impact impact of using using n = 8 on the ability of the x chart to detect a 2s shift in the mean? Consider the x and R chart that you established in Exercise 6.9 for the piston ring process. process . Suppose that you want to continue control charting piston ring diameter using n = 3. What limits would be used on the x and R chart? Control charts for x and R are maintained for an important quality characteristic. The sample size is n = 7; x and R are computed for each sample. After 35 samples, samples, we have found found that 35
35
∑ xi i =1
=
7805 and ∑ Ri
= 1200
i =1
(a) Set Set up x an and d R charts using these data. (b) Assumi Assuming ng that both charts exhi exhibit bit contro control, l, estimate the process mean and standard deviation.
6.20.
(c) If the quality quality characteris characteristic tic is normally normally distribdistributed and if the specifications are 220 ± 35 35,, ca can n the process meet the specifications? Estimate the fraction nonconforming. (d) Assumi Assuming ng the varia variance nce to remain consta constant, nt, state where the process mean should be located to minimize the fraction nonconforming. What would be the value of the fraction nonconforming under these conditions? Samples of size n = 5 are taken from a manufacturing process every hour. A quality characteristic is measure meas ured, d, and x and R are computed for each sample. After After 25 samples have been analyzed, analyzed, we have 25
∑ xi i =1
6.21.
=
25
662.50 and ∑ Ri
=
9.00
i =1
The quality characteristic is normally distributed. (a) Find the contro controll limits for the x and R charts. (b) Assume that that both charts charts exhibit exhibit control. control. If the specifications are 26.40 ± 0.50, estimate the the fraction nonconforming. (c) If the the mean of the the process process were were 26.40, 26.40, what fracfraction nonconforming would result? Samples of size n = 5 are collected from a process every half hour. After 50 samples have been collected, we calculate calculate x = 20.0 and s = 1.5. Assume that both charts exhibit control and that the quality characteristic is normally distributed. (a) Estimat Estimatee the process process standard standard deviat deviation. ion. (b) Find the contro controll limits on the x and s charts.
6.22.
277
(c) If the the process process mean shifts to 22, 22, what is is the probability of concluding that the process is still in control? An x chart is used to control the the mean of a normally distributed quality characteristic. It is known that s = 6.0 and n = 4. The center line = 200 200,, UCL = 209, and LCL = 191. If the process mean shifts to 188, find the probability that this shift is detected on the
first subsequent sample. A critical dimension of a machined part has specifications 100 ± 10. Control chart analysis indicates = 104 and that the process is in control with x = R = 9.30. The control charts use samples of size n = 5. If we assume that the characteristic is normally distrib distributed, uted, can the the mean be located located (by (by adjusting the tool position) so that all output meets specifications? What is the present capability of the process? 6.24. A process is to be monitored with standard values = 2.5. The sample size is n = 2. m = 10 and s = (a) Find the the center line and contro controll limits for the x chart. (b) Find the the center line line and control control limits for for the R chart. (c) Find the the center line line and control control limits limits for the s chart. 6.25. Samples of n = 5 units are taken from a process every hour. The x and R values for a particular quality characteristic are determined. After 25 samples have been – collected, collect ed, we calculate calculate x = 20 and R = 4.56. (a) What are are the three-sigm three-sigmaa control control limits limits for x and R? (b) Both charts charts exhibit exhibit control. control. Estimate the the process standard deviation. (c) Assu Assume me that the process process output output is normally normally disdistributed. If the specifications are 19 ± 5, wh what at are your conclusions regarding the process capability? (d) If the process mean shifts shifts to 24, what is is the probprobability of not detecting this shift on the first subsequent sample? 6.26. A TiW layer is deposited on a substrate using a sputtering tool. Table 6E.13 contains layer thickness measurements (in Angstroms) on 20 subgroups of four substrates. (a) Se Sett up x an and d R control charts on this process. Is the process in control? Revise the control limits as necessary. (b) Estimat Estimatee the mean and standard standard deviatio deviation n of the process. (c) Is the layer layer thickness thickness normally normally distribut distributed? ed? (d) If the the specificati specifications ons are are at 450 450 ± 30, esti estimate mate the process capability capa bility.. 6.23.
278
Variables Chapter 6 Control Charts for Variables
TABLE 6E 13
Layer Thickness Data for Exercise 6.26 Subgroup
x1
x2
x3
TABLE 6E 15
Additional Thickness Data for Exercise 6.28.
x4
Subgroup
6.27.
6.28.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
459 443 457 469 443 444 445 446 444 432 445 456 459 441 460 453 451
449 440 444 463 457 456 449 455 452 463 452 457 445 465 453 444 460
435 442 449 453 445 456 450 449 457 463 453 436 441 438 457 451 450
450 442 444 438 454 457 445 452 440 443 438 457 447 450 438 435 457
18 19 20
422 444 450
431 446 450
437 448 454
429 467 454
21 22 23 24 25 26 27 28 29 30
6.29. 6.30.
Continuation of Exercise 6.26. Table 6E.14 contains 10 new subgroups of thickness data. Plot this data on the control charts constructed in Exercise 6.26 (a). Is the process in statistical control? Continuation of Exercise 6.26. Suppose that following the construction of the x and R control charts in Exercise Exercise 6.26, the process process engineers engineers decided to change the subgroup size to n = 2. Table 6E.15 contains 10 new subgroups of thickness data. Plot this
Subgroup
x1
x2
x3
21 22 23 24 25 26 27
454 449 442 443 446 454 458
449 441 442 452 459 448 449
443 444 442 438 457 445 453
461 455 450 430 457 462 438
28 29 30
450 443 457
449 440 450
445 443 452
451 451 437
6.32.
x2
449 441 442 452 459 448 449 449 440 450
trol charts. Control charts for x and R are to be established to control the tensile strength of a metal part. Assume that tensile strength is normally distributed. Thirty samples of size n = 6 parts are collected over a period of time with the following results:
i =1
x4
data on the control charts from Exercise 6.26 (a) based on the new subgroup size. Is the process in statistical control? Rework Exercises 6.26 and 6.27 using x and s con-
∑ xi
Additional Thickness Data for Exercise 6.27.
x1
454 449 442 443 446 454 458 450 443 457
30
6.31.
TA B L E 6 E 1 4
30 =
6000 and ∑ Ri
= 150
i =1
(a) Calculate Calculate contro controll limits for x and R. (b) Both charts charts exhibit exhibit control. control. The specification specificationss on tensile strength are 200 ± 5. What are your conclusions regarding process capability? (c) Fo Forr the abo above ve x cha chart, rt, fin find d the b-risk when the true process mean is 199. An x chart has a center center line of 100, uses three-si three-sigma gma control limits, and is based on a sample size of four. four. The process standard deviation is known to be six. If the process process mean shifts from from 100 to 92, 92, what is the probability of detecting this shift on the first sample following the shift? The data in Table 6E.16 were collected from a process manufacturing power supplies. The variable of interest is output output voltage, voltage, and n = 5. (a) Compu Compute te center lines lines and control control limits suitable suitable for controlling future production. (b) Assume that that the quality quality character characteristic istic is nornormally distributed. Estimate the process standard deviation. (c) What are the the apparent apparent three-sigma three-sigma natural natural tolertolerance limits of the process?
279
Exercises
TA B L E 6 E 1 6
Voltage Data for Exercise 6.32. Sample Number
¯ x
1 2 3 4 5 6 7 8 9 10
103 102 104 105 104 106 102 105 106 104
Data for Exercise 6.34.
R
Sample Number
¯ x
R
4 5 2 11 4 3 7 2 4 3
11 12 13 14 15 16 17 18 19 20
105 103 102 105 104 105 106 102 105 103
4 2 3 4 5 3 5 2 4 2
Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(d) What would would be your your estimate estimate of the process process fraction nonconforming if the specifications on the characteristic were 103 ± 4? (e) What approaches approaches to to reducing reducing the fraction fraction nonnon6.33.
conforming can you suggest? Control charts on x and R for samples of size n = 5 are to be maintained on the tensile strength in pounds of a yarn. yarn. To To start the the charts, charts, 30 samples samples were were selected, and the mean and range of of each computed. computed. This yields 30
∑ xi i =1
6.34.
6.35.
607.8 and ∑ Ri
1 9 0 1 −3 −7 −3 0 2 0 −3 −16 −6 −3 −1
x2
9 4 9 1 0 2 −1 −2 0 2 −2 2 −3 −5 −1
x3
6 3 0 0 −1 0 −1 −3 −1 −1 −1 0 0 5 −1
x4
9 0 3 2 0 0 0 −3 −3 −1 −1 −4 0 0 −2
x5
6 3 2 1 −4 2 −2 −2 −1 2 2 −1 −8 5 −1
TA B L E 6 E 1 8
Sample Number 16 17 18 19 20 21 22 23 24 25
= 144
i =1
(a) Compute Compute the center center line and and control control limits for for the x and R control charts. (b) Suppo Suppose se both charts charts exhibit exhibit control. control. There is a sinsingle lower specification limit of 16 lb. If strength is normally normal ly distributed, distributed, what fraction of yarn would fail to meet specifications? 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 s hown in Table Table 6E.17 (measured as deviations from 0.3210 in 0.0001 in.). (a)) Se (a Sett up up an an R chart and examine the process for statistical control. (b) What parameters parameters would would you you recommend recommend for an an R chart for on-line control? (c) Estimat Estimatee the standard standard deviatio deviation n of the process. process. (d) What is the the process process capability capability?? Continuation of Exercise 6.34. Reconsider the data from Exercise 6.34 and establish x and R charts with appropriate trial control limits. Revise these trial limits as necessary to produce a set of control charts for monitoring future production. Suppose that the new data in Table 6E.18 are observed.
x1
New Data for Exercise 6.35
30 =
TA B L E 6 E 1 7
6.36.
x1
x2
x3
2 10 1 0 9 1 9 5 0 9 8 −3 0 5 2 10 10 9 −5 4 0 0 2 −5 10 0 3 −1 2 5 0 −1 2
x4
6 9 2 1 3 6 4 1 6 5
x5
5 4 5 4 1 −1 6 5 −3 −2
(a) Plot these these new observat observations ions on the the control chart. chart. What conclusions can you draw about process stability? (b) Use all 25 observa observations tions to revise revise the the control control limits for the x and R charts. What conclusions can you draw now about the process? Two parts are assembled as shown in Figure 6.28. Assume that the dimensions x and y are normally dis y and standard deviatributed withs means m x and. mThe tions s x and respectively. parts are produced produced y, respectively on different machines and are assembled at random. Control charts are maintained on each dimension for
280
Variables Chapter 6 Control Charts for Variables
6.40. x
y
An x chart with three-sigma limits has parameters as follows:
UCL = 104 UCL FIGURE 6 28
Center line
Parts for Exercise
6.36.
=
100
LCL = 96 LCL n = 5
the range of each sample ( n = 5). Both range charts are in control. (a) Giv Given en that for for 20 samples samples on the range range chart concontrolling x and 10 samples on the range chart controlling y, we hav havee 20
∑ R xi
=
i =1
6.37.
18.608 and ∑ Ryi
30
6.978 6.42.
and ∑ Ri
30 i =1
Chart
UCL = 201.88
s
Chart
UCL
Center line = 200.00 LCL = 198.12
= 1350
i =1
(a) Compute Compute control control limits on the the R chart. (b) Ass Assumin uming g that that the the R chart exhibits exhibits control, control, estimate the parameters m and s . (c) If the proce process ss output output is normal normally ly distrib distributed uted,, and if the specifications are 440 ± 40, can the the process process meet the specifications? Estimate the fraction nonconforming. (d) If the varia variance nce remain remainss constan constant, t, whe where re should should the mean be located to minimize the fraction nonconforming? Control charts for x and s are maintained on a quality characteristic. The sample size is n = 4. After 30 samples, samp les, we obtain obtain
∑ xi
x
=
2.266
Center line = 1.000 LCL = 0
30 = 12, 870
i =1
6.39.
=
i =1
Estimate s x and s y. (b) If it is desired desired that that the probabili probability ty of a smaller smaller − y) than 0.09 should be 0.006, clearance (i.e., x − what distance between the average dimensions (i.e., m x – m y) should be specified? Control charts for x and R are maintained on the tensile strength of a metal fastener. After 30 samples of size n = 6 are analyzed, analyzed, we find that
∑ xi
6.38.
6.41.
10
Suppose the process quality characteristic being controlled is normally distributed with a true mean of 98 and a standard deviation of 8. What is the probability that the control chart would exhibit lack of control by at least the third point plotted? Consider the x chart defined in Exercise 6.40. Find the ARL1 for the chart. Control charts for x and s with n = 4 are maintained on a quality characteristic. The parameters of these charts are as follows:
30 = 12, 870
and ∑ si
=
6.44.
chart such that the probability of a type I error is 0.05. Specifications on a normally distributed dimension are 600 ± 20. x and R charts are maintained on this dimension and have been in control over a long period of time. The parameters of these control charts are as follows (n = 9).
410
i =1
(a) Find the three-sig three-sigma ma limits limits for the s chart. (b) Assumin Assuming g that that both charts exhib exhibit it contro control, l, estimate the parameters m and s . An x chart on a normally distributed quality characteristic is s to = withthe thefollowing: standard values = 100, =be m 8, established and an d n = 4. Find (a) The two-s two-sigma igma contro controll limits. limits. (b) The 0 0.005 .005 probab probability ility limits.
6.43.
Both charts exhibit control. Specifications on the quality characteristic are 197.50 and 202.50. What can be said about the ability of the process to produce product that conforms to specifications? Statistical monitoring of a quality characteristic uses both an x and an s chart. The charts are to be = 10, based on the standard values m = 200 and s = with n = 4. (a) Find three-si three-sigma gma control control limits limits for the s chart. (b) Find a center center line line and control control limits for for the x
x
Chart
UCL = 616 Center line = 610 LCL = 604
s
Chart
UCL = 32.36 Center line = 17.82 LCL = 3.28
(a) What are your your conclusio conclusions ns regarding regarding the the capability of the process to produce items within specifications?
Exercises
6.45.
(b) Construct Construct an OC curve for the x chart assumin assuming g that s is constant. Thirty samples each of size 7 have been collected to establish control over a process. The following data were collected: 30
∑ xi
6.50.
30 =
2700 and ∑ Ri
i =1
Chart
UCL = 815 Center line = 800 LCL = 785
6.48. 6.49.
(d) What would would be the appropriat appropriatee control control limits for the x chart if the type I error probability were to be 0.01? A normally distributed quality characteristic is monitored through use of an x and an R chart. These charts have the following parameters ( n = 4):
= 120
i =1
x
(a) Calculate Calculate trial control control limits limits for the the two charts. charts. (b) On the aassumpt ssumption ion that the R chart is in control, estimate the process standard deviation. (c) Su Supp ppos osee an an s chart were desired. What would be the appropriate control limits and center line? 6.46. An x chart is to be established based on the standard = 12 values m = 600 and s = 12,, wi with th n = 9. The control limits are to be based on an a -risk -risk of 0.01. What are the appropriate control limits? 6.47. x and R charts with n = 4 are used to monitor a normally distributed quality characteristic. The control chart parameters are x
R
UCL = 46.98 Center line = 20.59 LCL = 0
UCL = 363.0 Center line = 360.0 LCL = 357.0
R
R
Chart
UCL = 18.795 Center line = 8.236 LCL = 0
Both charts exhibit control. (a) What is the the estimated standar standard d deviation deviation of of the process? (b)) Su (b Supp ppos osee an s chart were to be substituted for the R chart. What would be the appropriate parameters of the s chart? (c) If specificatio specifications ns on the produc productt were 610 ± 15, what would be your estimate of the process frac-
6.51.
tion nonconforming? (d) What could could be done to to reduce this this fraction fraction nonconforming? (e) What is the the probability probability of of detecting detecting a shift in the the process mean to 610 on the first sample following the shift (s remains constant)? (f) What is the probabi probability lity of detecting detecting the the shift in part (e) by at least the third sample after the shift occurs? Control charts for x and s have been maintained on a process and have exhibited statistical control. The sample size is n = 6. The control chart parameters are as follows: x
Chart
Chart
UCL = 626.0 Center line = 620.0 LCL = 614.0
Chart
Both charts exhibit control. What is the probability that a shift in the process mean to 790 will be detected on the first sample following the shift? Consider the x chart in Exercise 6.47. Find the average run length for the chart. Control charts for x and R are in use with the following parameters: x
281
Chart
s
Chart
Chart
UCL = 16.18 Center line = 8.91 LCL = 1.64
UCL = 708.20 Center line = 706.00 LCL = 703.80
UCL = 3.420 Center line = 1.738 LCL = 0.052
The sample size is n = 9. Both charts exhibit control. The quality characteristic is normally distributed. (a) Wh What at is th thee a -risk -risk associated with the x chart? (b) Specif Specification icationss on this quality quality characteristic characteristic are 358 ± 6. What are your conclusions regarding the ability of the process to produce items within
(a) Estimate Estimate the mean and standar standard d deviation deviation of the the process. (b) Estimate the the natural natural tolerance tolerance limits for for the process. (c) Assume that that the process process output output is well well modeled modeled by a normal distribution. If specifications are 703 and 709, estimate the fraction nonconform nonconforming. ing.
specifications? (c) Supp Suppose ose the mean mean shifts shifts to 357. 357. What What is the probability that the shift will not be detected on the first sample following the shift?
(d) Suppose Supp ose the deviation process mean process shifts shifts to 702.00 702.00 whilise while the standard remains constant. What the probability of an out-of-control signal occurring on the first sample following the shift?
282
6.52.
Variables Chapter 6 Control Charts for Variables
(e) For the shift shift in part (d), (d), what is is the probabi probability lity of of detecting the shift by at least the third subsequent sample? The following x and s charts based on n = 4 have shown statistical control: x
Chart
UCL = 710 Center line = 700 LCL = 690
s
6.54.
6.55.
6.56.
TABLE 6E 19
Can Weight Data for Exercise 6.53 Can Number
Heat
Hardness (coded)
1 2 3 4 5 6 7 8
52 51 54 55 50 52 50 51
UCL = 18.08 Center line = 7.979 LCL = 0
x chart on the first subsequent sample. (e) For For the shift shift of part part (d), (d), fin find d the aver average age run run length. One-pound coffee cans are filled by a machine, sealed, and then weighed automatically. After adjusting for the weight weight of the the can, any package package that that weighs less than 16 oz is cut out of the conveyor. conveyor. The weights of 25 successive cans are shown in Table 6E.19. Set up a moving range control chart and a control chart for individuals. Estimate the mean and standard deviation of the amount of coffee packed in
Weight
Can Number
Weight
1 2 3 4 5 6 7 8 9 10
16.11 16.08 16.12 16.10 16.10 16.11 16.12 16.09 16.12 16.10
14 15 16 17 18 19 20 21 22 23
16.12 16.10 16.08 16.13 16.15 16.12 16.10 16.08 16.07 16.11
11 12 13
16.09 16.07 16.13
24 25
16.13 16.10
TABLE 6E 20
Hardness Data for Exercise 6.54
Chart
(a) Estimate Estimate the the process process paramet parameters ers m and s . (b) If the specif specification icationss are at 705 ± 15 15,, an and d the the processs output is normally distribute proces distributed, d, estimate the fraction nonconforming. (c) Fo Forr the x cha chart, rt, fin find d the pro probab babili ility ty of a typ typee I error,, assumin error assuming g s is constant. (d) Supp Suppose ose the process process mean shifts shifts to 693 and and the standard deviation simultaneously shifts to 12. Find the probability of detecting this shift on the
6.53.
Heat
Hardness (coded)
9 10 11 12 13 14 15
58 51 54 59 53 54 55
each can. Is it reasonable to assume that can weight is normally distributed? If the process remains in controll at this level, what percentage contro percentage of cans will be underfilled? Fifteen successive heats of a steel alloy are tested for hardness. The resulting data are shown in Table 6E.20. Set up a control chart for the moving range and a control chart for individual hardness measurements. Is it reasonable to assume that hardness is normally distributed? The viscosity of a polymer is measured hourly. Measurements for the last twenty hours are shown in Table 6E.21. (a) Does viscosity viscosity follow follow a normal distribut distribution? ion? (b) Set up a control control chart on on viscosity viscosity and a moving moving range chart. Does the process exhibit statistical control? (c) Estimate the the process process mean and standard standard deviat deviation. ion. Continuation of Exercise 6.55. The next five measuremen sur ements ts on viscosi viscosity ty are 3163, 3163, 319 3199, 9, 305 3054, 4, 314 3147, 7, and 3156. Do these measurements indicate that the process is in statistical control?
TABLE 6E 21
Viscosity Data for Exercise 6.55 Test
Viscosity
Test
Viscosity
1 2 3 4 5 6 7
2838 2785 3058 3064 2996 2882 2878
11 12 13 14 15 16 17
3174 3102 2762 2975 2719 2861 2797
8 9 10
2920 3050 2870
18 19 20
3078 2964 2805
Exercises
TABLE 6E 22
Additional Data for Exercise 6.57, part (c)
Oxide Oxide Wafer Thickness Wafer Thickness 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
6.57.
45.4 48.6 49.5 44.0 50.9 55.2 45.5 52.8 45.3 46.3 53.9 49.8 46.9 49.8 45.1
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Oxide Oxide Wafer Thickness Wafer Thickness
58.4 51.0 41.2 47.1 45.7 60.6 51.0 53.0 56.0 47.2 48.0 55.9 50.0 47.9 53.4
1 2 3 4 5 6 7 8 9 10
43.4 46.7 44.8 51.3 49.2 46.5 48.4 50.1 53.7 45.6
11 12 13 14 15 16 17 18 19 20
50.0 61.2 46.9 44.9 46.2 53.3 44.1 47.4 51.3 42.5
continuous chemical process are shown in Table 6E.24. (a) A normal probabi probability lity plot plot of the concentratio concentration n data is by shown inpass Figure 6.29. The straight was fit eye to approximately throughline the twentieth and eightieth percentiles. Does the normality assumption seem reasonable here? (b) Set up an individ individuals uals and moving moving range range control control chart for the concentration data. Interpret the charts. (c) Const Construct ruct a normal probabi probability lity plot plot for the natural log of concentration. Is the transformed tr ansformed variable normally distributed?
(a) Thirty observ observations ations on the oxide oxide thickness thickness of individual silicon wafers are shown in Table 6E.22. Use these data to set up a control chart on oxide thickness and a moving range chart. Does the process exhibit statistical control? Does oxide thickness follow a normal distribution? (b) Foll Followin owing g the establishment establishment of of the control control charts in part (a), ten new wafers wafers were observed. observed. The oxide thickness measurements are as follows:
Wafer
Oxide Thickness
Wafer
Oxide Thickness
1 2 3 4 5
54.3 57.5 64.8 62.1 59.6
6 7 8 9 10
51.5 58.4 67.5 61.1 63.3
Plot these observations against the control limits determined in part (a). Is the process in control? (c) Supp Suppose ose the assignabl assignablee cause responsible responsible for for the out-of-control signal in part (b) is discovered and removed from the process. Twenty additional wafers are subsequently sampled. Plot the oxide thickness against the part (a) control limits. limi ts. What
6.58.
TABLE 6E 23
Data for Exercise 6.57
283
conclusions can6E.23. you draw? The new data are shown in Table Thirty observations on concentration (in g/l) of the active ingredient in a liquid cleaner produced in a
TA B L E 6 E 2 4
Data for Exercise 6E.58 Observa Obse rvation tion
Concentr Conc entratio ation n
Observa Obse rvation tion Conc Concentr entratio ation n
1
60.4
16
99.9
2
69.5
17
59.3
3
78.4
18
60.0
4
72.8
19
74.7
5
78.2
20
75.8
6
78.7
21
76.6
7
56.9
22
68.4
8
78.4
23
83.1
9
79.6
24
61.1
10
100.8
25
54.9
11
99.6
26
69.1
12
64.9
27
67.5
13 14
75.5 70.4
28 29
69.2 87.2
15
68.1
30
73.0
284
Variables Chapter 6 Control Charts for Variables 99.9 t n e c r e p e v i t a l u m u C
TABLE 6E 26
Additional Velocity Velocity of Light Data for Exercise 6.60.
99 95 80
Measurement Measur ement Veloci elocity ty Measur Measurement ement Veloc elocity ity
50 20
21 22 23 24 25 26 27 28 29 30
5 1 0.1 54
64
74 84 Concentration
94
104
FIGURE 6 29
Normal Probability Plot of the Concentration Data for Exercise 6.58
(d) Repeat part (b), (b), using the natural natural log of of concenconcentration as the charted variable. Comment on any differences in the charts you note in comparison to those constructed in part (b). In 1879, A. A. Michelson Michelson measured the velocity velocity of light in air using a modification of a method proposed by the French physicist Foucauld. Twenty of these measurements are in Table 6E.25 (the value reported is in kilometers per second and has 299,000 subtracted from it). Use these data to set up an individuals and a moving range control chart. Is there some evidence that the measurements of the velocity of light are normally distributed? Do the measurements exhibit statistical control? Revise the control limits if necessary. Continuation of Exercise 6.59. 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.26. (a) Plot these these new new measurements measurements on the the control control charts constructed in Exercise 6.59. Are these new measurements in statistical control? Give a practical interpretation of the control charts.
6.59.
6.60.
6.61.
Uniformity Data for Exercise 6.61 Wafer
Uniformity
Wafer
Uniformity
11 16 22 14 34 22 13 11 6 11 11 23
16 17 18 19 20 21 22 23 24 25 26 27
15 16 12 11 18 14 13 18 12 13 12 15
14 12 7
28 29 30
21 21 14
850 1000 740 980 900 930 1070
11 12 13 14 15 16 17
850 810 950 1000 980 1000 980
8 9 10
650 930 760
18 19 20
960 880 960
13 14 15
Measurement Measur ement Veloci elocity ty Measur Measurement ement Veloc elocity ity
800 830 850 800 880 790 900 760 840 800
TABLE 6E 27
1 2 3 4 5 6 7
Velocity of Light Data for Exercise 6E.59
31 32 33 34 35 36 37 38 39 40
(b) Is there evidence that the variability in the measurements has decreased between trial 1 and trial 2? The uniformity of a silicon wafer following an etching process is determined by measuring the layer thickness at several locations and expressing uniformity as the range of the thicknesses. Table 6E.27 presents uniformity determinations for 30 consecutive wafers processed through the etching tool. (a) Is there evidence evidence that that uniformity uniformity is normally normally distributed? If not, find a suitable transformation for the data. (b) Cons Construct truct a control control chart chart for individu individuals als and a moving range control chart for uniformity for the etching process. Is the process in statistical control?
1 2 3 4 5 6 7 8 9 10 11 12
TABLE 6E 25
960 830 940 790 960 810 940 880 880 880
Exercises
TA B L E 6 E 2 8
Purity Data for Exercise 6.62.
6.62.
6.63.
6.64.
6.65.
6.66.
6.67.
285
TA B L E 6 E 2 9
Data for Exercise 6.67
Batch
Purity
Batch
Purity
1 2 3
0.81 0.82 0.81
11 12 13
0.81 0.83 0.81
1 2 3
10.07 10.47 9.45
14 15 16
9.58 8.80 12.94
4 5 6 7 8 9 10
0 0..8 82 2 0.83 0.81 0.80 0.81 0.82
1 14 5 16 17 18 19 20
0 0..8 82 1 0.85 0.83 0.87 0.86 0.84
4 5 6 7 8 9 10 11 12 13
9 8..4 94 9 7.74 10.63 9.78 9.37 9.95 12.04 10.93 11.54
1 17 8 19 20 21 22 23 24 25
1 10 1..7 28 6 9.48 11.28 12.54 11.48 13.26 11.10 10.82
The purity of a chemical product is measured on each batch. Purity determinations for 20 successive batches are shown in Table 6E.28. (a) Is purity purity normall normally y distrib distributed? uted? (b) Is the process in statistical statistical contro control? l? (c) Estimate the the process process mean and standard standard deviat deviation. ion. Reconsider the situation in Exercise 6.53. Construct an individuals control chart using the median of the span-two moving ranges to estimate variability. Compare this control chart to the one constructed in Exercise 6.53 and discuss. Reconsider the hardness measurements in Exercise 6.54. Construct an individuals control chart using the median of the span-two moving ranges to estimate variability.. Compare this control variability c ontrol chart to the one constructed in Exercise 6.54 and discuss. Reconsider the polymer viscosity data in Exercise 6.55. Use the median of the span-two moving ranges to estimate s and set up the individuals control chart. Compare this chart to the one originally constructed using the average moving range method to estimate s . Continuation of Exercise 6.57. Use all 60 observations on oxide thickness. (a) Set up an indi individual vidualss control control chart with with s estimated by the average moving range method. (b) Set up an indiv individuals iduals control control chart chart with s estimated by the median moving range method. (c) Compar Comparee and discuss discuss the two two control control charts. Consider the individuals measurement data shown in Table 6E.29. (a)) Es (a Esti tima mate te s using the average of the moving of span two. (b) ranges (b) Esti Es timat mate e s using s/c4. (c)) Es (c Esti tima mate te s using the median of the span-two moving ranges.
Observation
x
Observation
x
(d)) Esti (d Estimat matee s using the average of the moving rang ra nges es of of span span 3, 3, 4, . . . , 20 20.. (e) Discuss the results results you you have have obtained. obtained. The vane heights for 20 of the castings from Fig. 6.25 are shown in Table 6E.30. Construct the “between/withi “between /within” n” contro controll charts for these these process
6.68.
TA B L E 6 E 3 0
Vane Heights for Exercise 6.68. Cast Ca stin ing g
Van anee 1
Van anee 2
Van anee 3
Van anee 4
Van anee 5
1
5.77 5. 7779 799 9
5.74 5. 7490 907 7
5.76 5. 7667 672 2 5.748 5.74836 36 5.741 5.74122 22
2
5.79 5. 7909 090 0
5.78 5. 7804 043 3
5.79 5. 7916 163 3 5.793 5.79393 93 5.811 5.81158 58
3
5.77 5. 7731 314 4
5.71 5. 7121 216 6
5.74 5. 7481 810 0 5.772 5.77292 92 5.755 5.75591 91
4
5.77 5. 7703 030 0
5.75 5. 7590 903 3
5.77 5. 7715 157 7 5.796 5.79687 87 5.780 5.78063 63
5
5.72 5. 7204 047 7
5.68 5. 6858 587 7
5.73 5. 7330 302 2 5.704 5.70472 72 5.681 5.68116 16
6
5.77 5. 7726 265 5
5.76 5. 7642 426 6
5.74 5. 7437 373 3 5.713 5.71338 38 5.747 5.74765 65
7
5.70 5. 7058 581 1
5.70 5. 7083 835 5
5.71 5. 7186 866 6 5.712 5.71252 52 5.720 5.72089 89
8
5.76 5. 7646 466 6
5.78 5. 7876 766 6
5.76 5. 7611 115 5 5.775 5.77523 23 5.755 5.75590 90
9
5.79 5. 7939 397 7
5.83 5. 8330 308 8
5.77 5. 7790 902 2 5.811 5.81122 22 5.823 5.82335 35
10
5.78 5. 7867 671 1
5.76 5. 7641 411 1
5.75 5. 7594 941 1 5.756 5.75619 19
5.71 5. 7178 787 7
11
5.75 5. 7535 352 2
5.74 5. 7414 144 4
5.74 5. 7410 109 9 5.768 5.76817 17
5.75 5. 7501 019 9
12
5.72 5. 7278 787 7
5.70 5. 7071 716 6
5.75 5. 7534 349 9 5.723 5.72389 89
5.73 5. 7348 488 8
13
5.79 5. 7970 707 7
5.79 5. 7923 231 1
5.79 5. 7902 022 2 5.796 5.79694 94
5.79 5. 7980 805 5
14
5.73 5. 7376 765 5
5.73 5. 7361 615 5
5.73 5. 7324 249 9 5.740 5.74006 06
5.73 5. 7326 265 5
15
5.72 5. 7247 477 7
5.76 5. 7656 565 5
5.76 5. 7696 963 3 5.749 5.74993 93
5.75 5. 7519 196 6
16
5.73 5. 7319 199 9
5.72 5. 7292 926 6
5.72 5. 7296 963 3 5.722 5.72259 59
5.73 5. 7351 513 3
17
5.79 5. 7916 166 6
5.79 5. 7951 516 6
5.79 5. 7990 903 3 5.785 5.78548 48
5.79 5. 7982 826 6
18
5.74 5. 7497 973 3
5.74 5. 7486 863 3
5.73 5. 7399 994 4 5.744 5.74405 05
5.74 5. 7468 682 2
19
5.76 5. 7644 449 9
5.75 5. 7563 632 2
5.76 5. 7619 197 7 5.766 5.76684 84
5.75 5. 7547 474 4
20
5.75168
5.7557 9 5.
5.7 39 39 79 79 5 .7 .77 96 963 5.76 93 933
286
Variables Chapter 6 Control Charts for Variables
6.69.
6.70.
data using a range chart to monitor the within-castings vane height. Compare these to the control charts shown in Fig. 6.27. The diameter of the casting in Fig. 6.25 is also an important quality characteristic. A coordinate measuring machine is used to measure the diameter of each casting at five different locations. Data for 20 castings are shown in the Table 6E.31.
and 48 wafers. Some processing steps treat each wafer separately, so that the batch batch size for that step is one wafer. It is usually necessary to estimate several compone comp onents nts of var variati iation: on: with withinin-wafe wafer, r, betw betweeneenwafer,, between wafer between-lot, -lot, and the total variatio variation. n. (a) Supp Suppose ose that one one wafer is is randomly randomly selected selected from each lot and that a single measurement on a critical dimension of interest is taken. Which
(a) Set Set up x an and d R charts for for this process, process, assumin assuming g the measurements on each casting form a rational subgroup. (b) Discuss the the charts you you have have constructed constructed in part part (a). (c)) Con (c Constru struct ct “betwe “between/ en/wit within hin”” char charts ts for for this this process. (d) Do you believ believee that the charts charts in part (c) (c) are more informative informative than those in part (a)? Discuss why. (e) Prov Provide ide a practical practical interpretati interpretation on of the “within” “within” chart. In the semiconducto semiconductorr industry industry,, the production production of of microcircuits involves many steps. The wafer fabrication process typically builds these microcircuits on silicon wafers, wafers, and there are many microcircuits microcircuits per wafer. Each production lot consists of between 16
components of variation could be estimated with these data? What type of control charts would you recommend? (b) Suppo Suppose se that each wafer wafer is tested at five five fixed fixed locations locatio ns (say, (say, the center and four four points at the circumference). The average and range of these within-wafer measurements are x ww and Rww, respectively. What components of variability are estimated using control charts based on these data? (c) Supp Suppose ose that one one measurement measurement point point on each each wafer is selected and that this measurement is recorded for five consecutive wafers. The average and range of these between-wafer measurements are x BW and R BW , respecti respectively vely.. What components of variability are estimated using control charts based on these data? Would it be necessary to run separate x and R charts for all five locations on the wafer? (d) Consi Consider der the question question in part part (c). How How would your answer change if the test sites on each wafer were randomly selected and varied from wafer to wafer? (e) What type type of control control charts charts and rational rational subsubgroup scheme would you recommend to control the batch-to-batch variability? Consider the situation described in Exercise 6.70. A critical dimension (measured in mm) is of interest to the process engineer. Suppose that five fixed positions are used on each wafer (position 1 is the center) and that two consecutive wafers are selected from each batch. The data that result from several batches are shown in Table 6E.32. (a) What can you you say about about overall overall process process capability? (b) Can you construct construct control control charts charts that allow allow withinwafer variability to be evaluated? (c) What control control charts charts would you you establish establish to evalevaluate variability between wafers? Set up these charts and use them to draw conclusions about the process.
TABLE 6E 31
Diameter Data for Exercise 6.69. Diameter C a st i n g
1
2
3
4
5
1
11.7 11 .762 629 9
11.7 11 .740 403 3
11.7 11 .751 511 1 11.7 11.747 474 4
11.7 11 .737 374 4
2
11.8 11 .812 122 2
11.7 11 .750 506 6
11.7 11 .778 787 7 11.7 11.773 736 6
11.8 11 .841 412 2
3
11.7 11 .774 742 2
11.7 11 .711 114 4
11.7 11 .753 530 0 11.7 11.753 532 2
11.7 11 .777 773 3
4
11.7 11 .783 833 3
11.7 11 .731 311 1
11.7 11 .777 777 7 11.8 11.810 108 8
11.7 11 .780 804 4
5
11.7 11 .713 134 4
11.6 11 .687 870 0
11.7 11 .730 305 5 11.7 11.741 419 9
11.6 11 .664 642 2
6
11.7 11 .792 925 5
11.7 11 .761 611 1
11.7 11 .758 588 8 11.7 11.701 012 2
11.7 11 .761 611 1
7
11.6 11 .691 916 6
11.7 11 .720 205 5
11.6 11 .695 958 8 11.7 11.744 440 0
11.7 11 .706 062 2
8
11.7 11 .710 109 9
11.7 11 .783 832 2
11.7 11 .749 496 6 11.7 11.749 496 6
11.7 11 .731 318 8
9
11.7 11 .798 984 4
11.8 11 .888 887 7
11.7 11 .772 729 9 11.8 11.848 485 5
11.8 11 .841 416 6
10
11.7 11 .791 914 4
11.7 11 .761 613 3
11.7 11 .735 356 6 11.7 11.762 628 8
11.7 11 .707 070 0
11
11.7 11 .726 260 0
11.7 11 .732 329 9
11.7 11 .742 424 4 11.7 11.764 645 5
11.7 11 .757 571 1
12
11.7 11 .720 202 2
11.7 11 .753 537 7
11.7 11 .732 328 8 11.7 11.758 582 2
11.7 11 .726 265 5
13
11.8 11 .835 356 6
11.7 11 .797 971 1
11.8 11 .802 023 3 11.7 11.780 802 2
11.7 11 .790 903 3
14
11.7 11 .706 069 9
11.7 11 .711 112 2
11.7 11 .749 492 2 11.7 11.732 329 9
11.7 11 .728 289 9
15
11.7 11 .711 116 6
11.7 11 .797 978 8
11.7 11 .798 982 2 11.7 11.742 429 9
11.7 11 .715 154 4
16
11.7 11 .716 165 5
11.7 11 .728 284 4
11.7 11 .757 571 1 11.7 11.759 597 7
11.7 11 .731 317 7
17
11.8 11 .802 022 2
11.8 11 .812 127 7
11.7 11 .786 864 4 11.7 11.791 917 7
11.8 11 .816 167 7
18
11.7 11 .777 775 5
11.7 11 .737 372 2
11.7 11 .724 241 1 11.7 11.777 773 3
11.7 11 .754 543 3
19
11.7 11 .775 753 3
11.7 11 .787 870 0
11.7 11 .757 574 4 11.7 11.762 620 0
11.7 11 .767 673 3
20
11.7 11 .757 572 2
11.7 11 .762 626 6
11.7 11 .752 523 3 11.7 11.739 395 5
11.7 11 .788 884 4
6.71.
(d) What control charts control wup ould youcharts use toand evaluate evaluate lotto-lot variability? Setwould these use them to draw conclusions about lot-to-lot variability.
Exercises
TA B L E 6 E 3 2
Data for Exercise 6.71
Lot Wafer Num be r Num be r 1
Position 1
2
3
4
5
Lo t Wafer N u m b er N u m b e r
1
2
3
4
5
1
2. 15
2 .1 3 2.
2.14
2. 09
2 . 08
2
2. 11
2. 1 3 2.
2.10
2. 14
2 . 10
1
2. 03
2 .0 6 2.
2.05
2. 01
2 . 00
2
2. 04
2. 0 8 2.
2.03
2. 10
2 . 07
1
2. 05
2 .0 3 2.
2.05
2. 09
2 . 08
2
2. 08
2. 0 1 2.
2.03
2. 04
2 . 10
1
2. 08
2 .0 4 2.
2.05
2. 01
2 . 08
2
2. 09
2. 1 1 2.
2.06
2. 04
2 . 05
1
2. 14
2 .1 3 2.
2.10
2. 10
2 . 08
2
2. 13
2. 1 0 2.
2.09
2. 13
2 . 15
1
2. 06
2 .0 8 2.
2.05
2. 03
2 . 09
2
2. 03
2. 0 1 2.
1.99
2. 06
2 . 05
1
2. 05
2 .0 3 2.
2.08
2. 01
2 . 04
2
2. 06
2. 0 5 2.
2.03
2. 05
2 . 00
1
2. 15 2. 13
2 . 08
2 . 12
2 . 10
2
2. 13 2. 10
2 . 04
2 . 08
2 . 05
1
2. 02 2. 01
2 . 06
2 . 05
2 . 08
2
2. 03 2. 09
2 . 07
2 . 06
2 . 04
1
2. 13 2. 12
2 . 10
2 . 11
2 . 08
2
2. 03 2. 08
2 . 03
2 . 09
2 . 07
1
2. 04 2. 01
2 . 10
2 . 11
2 . 09
2
2. 07 2. 14
2 . 12
2 . 08
2 . 09
1
2. 16 2. 17
2 . 13
2 . 18
2 . 10
2
2. 17 2. 13
2 . 10
2 . 09
2 . 13
1
2. 04 2. 06
1 . 97
2 . 10
2 . 08
2
2. 03 2. 10
2 . 05
2 . 07
2 . 04
1
2. 04 2. 02
2 . 01
2 . 00
2 . 05
2
2. 06 2. 04
2 . 03
2 . 08
2 . 10
8
1
2. 13 2. 10
2 . 10
2 . 15
2 . 13
18
1
2. 03
2 .0 8 2.
2.04
2. 00
2 . 03
9
2 1
2. 10 2. 09 1. 95 2. 03
2 . 13 2 . 08
2 . 14 2 . 07
2 . 11 2 . 08
19
2 1
2. 04 2. 16
2. 0 3 2. 2..13 2
2.05 2.10
2. 01 2. 13
2 . 04 2 . 12
2
2. 01 2. 03
2 . 06
2 . 05
2 . 04
2
2. 13
2. 1 5 2.
2.18
2. 19
2 . 13
1
2. 04 2. 08
2 . 09
2 . 10
2 . 01
1
2. 06
2 .0 3 2.
2.04
2. 09
2 . 10
2
2. 06 2. 04
2 . 07
2 . 04
2 . 01
2
2. 01
1. 9 8 1.
2.05
2. 08
2 . 06
2
3
4
5
6
7
10
11
Position
12
13
14
15
16
17
20
287
722 4.17.
4.19.
4.21.
4.23.
Answers to Selected Exercises
(a) F 0 = 1.0987. Do not reject H 0. (b) t 0 = 1.461. Do not reject H 0.
6.13.
t 0 = –1.10. There is no difference between mean measurements. 2
(a) χ 0 = 42.75. Do not reject H 0. (b) 1.14 ≤ s ≤ 2.19. n = 3 ( Z a /2 + Z b ) s /d 4
(d) p ˆrework = 0.0275, p ˆscrap
2
Z 0 = 0.3162. Do not reject H 0. 4.29. (a) F 0 = 3.59, P = 0.053.
(a) F 0 = 1.87, P = 0.214.
4.33.
(a) F 0 = 1.45, P = 0.258.
4.35.
(a) F 0 = 30.85, P = 0.000.
(e) p ˆrework = 0.00523, p ˆscrap
Pattern is random.
5.19.
There is a nonrandom, nonrandom, cycl cyclic ic pattern. pattern.
5.21.
Points 17, Points 17, 18, 19, and 20 20 are outs outside ide low lower er 1-sigma area.
0.00069,
=
0.00523,
Total = 1.046% 6.15.
(a) x char chart: t: CL = 79. 79.53, 53, UCL = 84.58, LCL = 74.49 char art: t: CL = 8. 8.75 75,, UC UCL L = 18. 18.49, 49, LCL = 0 R ch Process is in statistical control. (b) Several subgroups exceed UCL on R chart.
6.17.
(a) x char chart: t: CL = 34. 34.00, 00, UCL = 37.50, LCL = 30.50 char art: t: CL = 3. 3.42 42,, UC UCL L = 8. 8.81 81,, LC LCL L=0 R ch
CHAPTER 5 5.17.
=
Total = 2.949%
4.27.
4.31.
(a) x char chart: t: CL = 20 20,, UC UCL L = 22.34, LCL = 17.66 char art: t: CL = 1. 1.44 44,, UC UCL L = 3. 3.26, 26, LC LCL L=0 s ch (b) LNTL = 15. 15.3, 3, UNT UNTL L = 24.7 ˆ = 0.85 (c) C P
(b) Detect shift more quickly.
5.23.
(c) x char chart: t: CL = 34. 34.00, 00, UCL = 36.14, LCL = 31.86
Point Points 16,cen 17,terlin and are of 6, 3 beyond b sigma sig mas of center line. e.18Poi Points nts2 5, 7,eyond 8, and2 9 are of 5 at 1 sigma or beyond of centerline. CHAPTER 6 6.1.
(a) Samples 12 and 15 exceed x UCL. (b) p = 0.00050.
6.3.
(a) x char chart: t: CL = 10. 10.9, 9, UC UCL L = 50.35, LCL = –28.55 char art: t: CL = 54 54.1 .1,, UC UCL L = 123 123.5, .5, LCL = 0 R ch Process is in statistical control. ˆ = 1.43. (b) s x = 23.3. (c) C P
6.5.
6.7.
6.9.
6.11.
(a) x char chart: t: CL = − 0.0 0.003, 03, UCL = 1.037, LCL = −1.043 char art: t: CL = 1.0 1.066, 66, UCL = 1.830, s ch LCL = 0.302 (b) R ch char art: t: CL = 3. 3.2, 2, UC UCL L = 5.686, LCL = 0.714 (c) s2 ch char art: t: CL = 1.2 1.2057, 057, UCL = 3.6297, LCL = 0.1663 char art: t: CL = 10. 10.33, 33, UCL = 14.73, x ch LCL = 5.92 s ch char art: t: CL = 2.7 2.703, 03, UCL = 6.1 6.125, 25, LCL = 0 (a) char chart: t: CL = 74. 74.001 00118, 18, UCL = 74.01458, LCL = 73.98777 char art: t: CL = 0.0 0.0232 2324, 4, UCL = 0.04914, R ch ˆ P = 1.668. LCL = 0 (b) No. (c) C chart rt:: CL = 80 80,, UC UCL L = 89.4 89.49, 9, LCL = 70.51 x cha chart: rt: CL = 9.7 9.727, 27, UCL = 16. 16.69, 69, LCL = 2.76 s cha
char art: t: CL = 5. 5.75 75,, UC UCL L R ch LCL = 0.78 6.19.
10.72,
(a) x char chart: t: CL = 22 223, 3, UC UCL L = 237.37, LCL = 208.63 R cha chart: rt: CL = 34. 34.29, 29, UC UCL L = 65. 65.97, 97, LCL = 2.61 ˆ = 22 3; σ ˆ x = 12.68 (b) µ ˆ (c) C P
6.21.
=
=
0.92. (d) pˆ
=
0.00578
ˆ x = 1.60 (a) σ (b) x chart chart:: UCL = 22. 22.14, 14, LCL = 17.86 chart: rt: UCL = 3. 3.13 13,, LC LCL L=0 s cha
6.23. 6.25.
(c) Pr{in control} = 0.57 5792 926 6 ˆ = 0.83 8338 38 C P (a) x chart chart:: UCL = 22. 22.63, 63, LCL = 17.37 chart: rt: UCL = 9. 9.64, 64, LC LCL L=0 R cha ˆ ˆ x = 1.96 . (c) C = 0.85 . (b) σ P 0593 938 8 (d) Pr{not detect} = 0.05
6.27.
The process continues to be in a state of statistical control.
6.29.
(a) x char chart: t: CL = 449 449.68 .68,, UCL = 462.22, LCL = 437.15 char art: t: CL = 17. 17.44, 44, UCL = 7. 7.70, 70, LC LCL L=0 s ch
6.31.
Pr{detecting shift on 1st sample} = 0.37
6.33.
(a) x = 20.26, UCL x = 23.03, LCL x = 17.49; R = 4.8, UCL R = 10.152, LCL R = 0 (b) pˆ = 0.01 0195 95
Answers to Selected Exercises
6.35.
6.37.
(a) x Recalculating limits without samples 1, 12,, an 12 and d 13: 13: char art: t: CL = 1. 1.45 45,, UC UCL L = 5. 5.46, 46, LC LCL L = (2.57 x ch R ch char art: t: CL = 6. 6.95 95,, UC UCL L = 14. 14.71, 71, LCL = 0 (b)) Sam (b Sampl ples es 1, 12 12,, 13 13,, 16 16,, 17 17,, 18 18,, an and d 20 ar aree out-of-control,, for a total 7 of the out-of-control the 25 samples, with runs of points both above and below the centerline. This This suggests that the process is inherently unstable, and that the the sources of variation need to be identified and removed.
(a) The process is in statistical control. The normality assumption is reasonable. (b) It is clear that the process is out of control during this period of operation. (c) The process has been returned to a state of statistical control.
6.59.
The measurements are approximately normally distributed. The out-of-control signal on the moving range chart indicates a significantly large difference between successive measurements (7 and 8). Consider the process to be in a state of statistical process control.
(b) m ˆ = 429.0, s ˆ x = 17.758 ˆ p = 0.751; p (c) C ˆ = 0.0533
6.39.
6.41.
(a) UCL x
=
108, LCL x
(b) UCL x
=
111.228, LCL x
=
88.772 .
ARL1 = 2.986
6.43.
=
=
(a) CL s 9.213, UCL s 20.88, LCL s (b) UCL x = 209.8, LCL x = 190.2 . 6.45.
(a) x = 90, UCL x
=
91.676, LCL x
=
0.
6.47.
Pr{detect shift on 1st sample} = 0.15 1587 87
6.49.
ˆ = 0.667. 0026 26. (b) C (a) α = 0.00 P (c) Pr{not detect on 1st sample} = 0.50 5000 00. (d) UCL x (a) mˆ
=
=
362.576, LCL x
ˆ x 706.00; s
=
=
= 16 x chart: x 16.1 .11, 1, UC UCL L x = 16 16.1 .17, 7, LC LCL L x = 16.04 MR chart: MR2 = 0.0 0.0236 2365, 5, UCLMR2 = 0.0772 0.0 7726, 6, LCLMR2 = 0
6.65.
= 29 2929 29,, UC UCL L x = 33 3338 38,, LC LCL L x = 2520 x chart: x MR chart: MR2 = 15 153. 3.7, 7, UC UCL LMR2 = 502.2, LCLMR2 = 0
88.324; = 0.304.
=
=
(b) UNTL 711.48, LNTL (c) pˆ = 0.10 1006 06 .
6.67.
ˆ x , span 19 σ 6.69.
=
=
0. 02 021055;
MR2 = 0.02 0237 375 5. Assumption of normally distributed coffee can weights is valid. %underfilled = 0.0003%. 6.55.
(a) Viscosity Viscosity measurements appear to follow a normal distribution. (b) The process appears to be in statistical control, contro l, with no out-of-cont out-of-control rol points, points, runs, trends,, or other patte trends patterns. rns. ˆ = 292 8.9; σ ˆ x = 131.346; (c) µ MR2
=
148 14 8.15 158 8
ˆ x , span 1.210, σ 4
ˆ x , span 1.406, σ 20
=
=
1.262, 1.2 62, ... ...,,
1.435
(a) x char chart: t: CL = 11. 11.76, 76, UCL = 11.79, LCL chart 11.72 (within): (withi n): CL = 0.06109, R UCL = 0.1 0.1292, 292, LCL = 0 (c) I ch char art: t: CL = 11. 11.76, 76, UCL = 11.87, LCL = 11.65 MR2 chart chart (between) (between):: CL = 0.04161, UCL = 0.1 0.1360, 360, LCL = 0
(e) Pr{detect by 3rd sample} ˆ x x = 16.1052; σ
=
=
=
700.52 .
(d) Pr{detect on 1st sample} = 0.99 9920 20
6.53.
ˆ x = 1.157. (b) σ ˆ x = 1.682. (c) σ ˆ x = 1.137 (a) σ ˆ x , span (d) σ 3
357.424 .
1.827.
(a) The data are not normally distributed. The distribution of the natural-log transformed uniformity measurements is approximately normally distributed. (b) x char chart: t: CL = 2.6 2.653, 53, UCL = 3.586, LCL = 1.720 R ch char art: t: CL = 0.3 0.351, 51, UCL = 1.1 1.146, 46, LCL = 0
6.63.
=
4, UCL R = 7.696, LCL R ˆ x = 1.479 . (b) σ (c) s = 1.419, UCLs = 2.671, LCLs = 0.167.
R
6.51.
6.61.
92 . =
723
6.57.
(a) R = 45.0, UCL R = 90.18, LCL R = 0
(d) To minimize fraction nonconforming the mean should be located at the nominal dimension (440) for a constant variance.
6.71.
(b) R chart (withi (within): n): CL = 0.06725, UCL = 0. 0.142 1422, 2, LCL = 0 (c) I ch char art: t: CL = 2.0 2.074, 74, UCL = 2.159, LCL = 1.989 MR2 chart chart (between) (between):: CL = 0.03210, UCL = 0.1 0.1049, 049, LCL = 0 (d) Need lot average, average, moving range between lot averages, averages, and range within within a lot. I ch char art: t: CL = 2.0 2.074, 74, UCL = 2.133, LCL = 2.015
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