Chapter 7 Exercise Solutions
Note: Several exercises in this chapter differ from those in the 4th edition. An “*”
indicates that the description has changed. A second exercise number in parentheses
indicates that the exercise number has changed. New exercises are denoted with an “”.
7-1.
ˆ  x  74.001; R  0.023; ˆ  R d 2  0.023 2.326  0.010
SL  74.000  0.035  [73.965, 74.035]
USL  LSL 74.035  73.965
Cˆ p

 1.17
6ˆ
6(0.010)
ˆ  LSL 74.001  73.965
Cˆ pl 

 1.20
3ˆ
3(0.010)
USL  ˆ 74.035  74.001
Cˆ pu 

 1.13
3ˆ
3(0.010)


Cˆ pk  min Cˆ pl , Cˆ pu  1.13
7-2.
In Exercise 5-1, samples 12 and 15 are out of control, and the new process parameters are
used in the process capability analysis.
n  5; ˆ  x  33.65; R  4.5; ˆ  R d 2  1.93
USL  40; LSL  20
USL  LSL 40  20
Cˆ p

 1.73
6ˆ
6(1.93)
ˆ  LSL 33.65  20
Cˆ pl 

 2.36
3ˆ
3(1.93)
USL  ˆ 40  33.65
Cˆ pu 

 1.10
3ˆ
3(1.93)


Cˆ pk  min Cˆ pl , Cˆ pu  1.10
7-1
Chapter 7 Exercise Solutions
7-3.
ˆ  x  10.375; Rx  6.25; ˆ x  R d 2  6.25 2.059  3.04
USL x  [(350  5)  350]  10  50; LSL x  [(350  5)  350]  10  50
xi  (obsi  350)  10
USL x  LSL x 50  (50)
Cˆ p 

 5.48
6ˆ x
6(3.04)
The process produces product that uses approximately 18% of the total specification
band.
USL x  ˆ 50  10.375
Cˆ pu 

 4.34
3ˆ x
3(3.04)
ˆ  LSL x 10.375  (50)
Cˆ pl 

 6.62
3ˆ x
3(3.04)
Cˆ pk  min(Cˆ pu , Cˆ pl )  4.34
This is an extremely capable process, with an estimated percent defective much less than
1 ppb. Note that the Cpk is less than Cp, indicating that the process is not centered and is
not achieving potential capability. However, this PCR does not tell where the mean is
located within the specification band.
T  x 0  10.375

 3.4128
S
3.04
Cˆ p
5.48
Cˆ pm 

 1.54
1V 2
1  (3.4128) 2
Since Cpm is greater than 4/3, the mean  lies within approximately the middle fourth of
the specification band.
V
ˆ 
ˆ  T 10.375  0

 3.41
ˆ
3.04
Cˆ pkm 
Cˆ
1.54
pk

 0.43
1  3.412
1  ˆ 2
7-2
Chapter 7 Exercise Solutions
7-4.
n  5; x  0.00109; R  0.00635; ˆ x  0.00273 ; tolerances: 0  0.01
USL  LSL 0.01  0.01
Cˆ p 

 1.22
6ˆ
6(0.00273)
The process produces product that uses approximately 82% of the total specification
band.
USL  ˆ 0.01  0.00109
Cˆ pu 

 1.09
3ˆ
3(0.00273)
ˆ  LSL 0.00109  (0.01)
Cˆ pl 

 1.35
3ˆ
3(0.00273)
Cˆ  min(Cˆ , Cˆ )  1.09
pk
pl
pu
This process is not considered capable, failing to meet the minimally acceptable
definition of capable Cpk  1.33
T  x 0  0.00109

 0.399
S
0.00273
Cˆ
1.22
p
ˆ
C pm 

 1.13
1V 2
1  (0.399) 2
Since Cpm is greater than 1, the mean  lies within approximately the middle third of the
specification band.
V
ˆ 
ˆ  T 0.00109  0

 0.399
ˆ
0.00273
Cˆ pkm 
Cˆ
1.09
pk

 1.01
2
2
ˆ
1

0.399
1 
7-3
Chapter 7 Exercise Solutions
7-5.
ˆ  x  100; s  1.05; ˆ x  s c4  1.05 0.9400  1.117
(a)
USL  LSL (95  10)  (95  10)
Potential: Cˆ p 

 2.98
6ˆ
6(1.117)
(b)
ˆ  LSL x 100  (95  10)
Cˆ pl 

 4.48
3ˆ x
3(1.117)
USL x  ˆ (95  10)  100
Actual: Cˆ pu 

 1.49
3ˆ x
3(1.117)
Cˆ  min(Cˆ , Cˆ )  1.49
pk
pl
pu
(c)
pˆ Actual  Pr{x  LSL}  Pr{x  USL}
 Pr{x  LSL}  1  Pr{x  USL}
LSL  ˆ  
USL  ˆ  


 Pr  z 
  1  Pr  z 

ˆ  
ˆ



85  100  

 105  100 
 Pr  z 
  1  Pr  z 

1.117  
1.117 


  (13.429)  1   (4.476) 
 0.0000  1  0.999996
 0.000004
85  95  

 105  95 
pˆ Potential  Pr  z 
  1  Pr  z 

1.117  
1.117 


 (8.953)  1  (8.953) 
 0.000000  1  1.000000
 0.000000
7-4
Chapter 7 Exercise Solutions
7-6.
n  4; ˆ  x  199; R  3.5; ˆ x  R d 2  3.5 2.059  1.70
USL = 200 + 8 = 208; LSL = 200 – 8 = 192
(a)
USL  LSL 208  192
Potential: Cˆ p 

 1.57
6ˆ
6(1.70)
The process produces product that uses approximately 64% of the total specification
band.
(b)
USL  ˆ 208  199
Cˆ pu 

 1.76
3ˆ
3(1.70)
ˆ  LSL 199  192

 1.37
Actual: Cˆ pl 
3ˆ
3(1.70)
Cˆ  min(Cˆ , Cˆ )  1.37
pk
pl
pu
(c)
The current fraction nonconforming is:
pˆ Actual  Pr{x  LSL}  Pr{x  USL}
 Pr{x  LSL}  1  Pr{x  USL}
LSL  ˆ  
USL  ˆ 


 Pr  z 
  1  Pr  z 

ˆ  
ˆ



208  199  
 192  199  

 Pr  z 
  1  Pr  z 

1.70  
1.70  


  (4.1176)  1   (5.2941) 
 0.0000191  1  1
 0.0000191
If the process mean could be centered at the specification target, the fraction
nonconforming would be:
 192  200 
pˆ Potential  2  Pr  z 

1.70 

 2  0.0000013
 0.0000026
7-5
Chapter 7 Exercise Solutions
7-7.
n  2; ˆ  x  39.7; R  2.5; ˆ x  R d 2  2.5 1.128  2.216
USL = 40 + 5 = 45; LSL = 40 – 5 = 35
(a)
USL  LSL 45  35
Potential: Cˆ p 

 0.75
6ˆ
6(2.216)
(b)
USL  ˆ 45  39.7
Cˆ pu 

 0.80
3ˆ
3(2.216)
ˆ  LSL 39.7  35

 0.71
Actual: Cˆ pl 
3ˆ
3(2.216)
Cˆ  min(Cˆ , Cˆ )  0.71
pk
pl
pu
(c)
V
Cˆ pm
x  T 39.7  40

 0.135
s
2.216
Cˆ p
0.75


 0.74
2
1V
1  (0.135) 2
Cˆ pkm 
Cˆ pk

0.71
 0.70
1V
1  (0.135) 2
The closeness of estimates for Cp, Cpk, Cpm, and Cpkm indicate that the process mean is
very close to the specification target.
2
(d)
The current fraction nonconforming is:
pˆ Actual  Pr{x  LSL}  Pr{x  USL}
 Pr{x  LSL}  1  Pr{x  USL}
LSL  ˆ  
USL  ˆ 


 Pr  z 
  1  Pr  z 

ˆ  
ˆ



35  39.7  
45  39.7 


 Pr  z 
  1  Pr  z 

2.216  
2.216 


  (2.12094)  1   (2.39170) 
 0.0169634  1  0.991615
 0.025348
7-6
Chapter 7 Exercise Solutions
7-7 (d) continued
If the process mean could be centered at the specification target, the fraction
nonconforming would be:
35  40 

pˆ Potential  2  Pr  z 

2.216 

 2  Pr{z  2.26}
 2  0.01191
 0.02382
7-8 (7-6).
ˆ  75; S  2; ˆ  Sˆ c4  2 0.9400  2.13
(a)
USL  LSL
2(8)
Potential: Cˆ p 

 1.25
6ˆ
6(2.13)
(b)
ˆ  LSL 75  (80  8)
Cˆ pl 

 0.47
3ˆ
3(2.13)
USL  ˆ 80  8  75

 2.03
Actual: Cˆ pu 
3ˆ
3(2.13)
Cˆ  min(Cˆ , Cˆ )  0.47
pk
pl
pu
(c) Let ˆ  80
pˆ Potential  Pr{x  LSL}  Pr{x  USL}
LSL  ˆ 
USL  ˆ 


 Pr  z 
  1  Pr  z 

ˆ 
ˆ



72  80 
88  80 


 Pr  z 
  1  Pr  z 

2.13 
2.13 


  (3.756)  1   (3.756)
 0.000086  1  0.999914
 0.000172
7-7
Chapter 7 Exercise Solutions
7-9 (7-7).
Assume n = 5
Process A
ˆ  xA  100; sA  3; ˆ A  sA c4  3 0.9400  3.191
USL  LSL (100  10)  (100  10)
Cˆ p 

 1.045
6ˆ
6(3.191)
USL x  ˆ (100  10)  100
Cˆ pu 

 1.045
3ˆ x
3(3.191)
ˆ  LSL x 100  (100  10)
Cˆ pl 

 1.045
3ˆ x
3(3.191)
Cˆ  min(Cˆ , Cˆ )  1.045
pk
pl
pu
x  T 100  100

0
s
3.191
Cˆ p
1.045
Cˆ pm 

 1.045
2
1V
1  (0) 2
pˆ  Pr{x  LSL}  Pr{x  USL}
 Pr{x  LSL}  1  Pr{x  USL}
V
LSL  ˆ 
USL  ˆ 


 Pr  z 
  1  Pr  z 

ˆ 
ˆ



90  100 

 110  100 
 Pr  z 
  1  Pr  z 

3.191 
3.191 


  (3.13)  1   (3.13)
 0.00087  1  0.99913
 0.00174
Process B
ˆ  xB  105; sB  1; ˆ B  sB c4  1 0.9400  1.064
Cˆ
p

USL  LSL
6ˆ

(100  10)  (100  10)
 3.133
6(1.064)
ˆ  LSL x 105  (100  10)
Cˆ pl  x

 4.699
3ˆ x
3(1.064)
USL x  ˆ x (100  10)  105
Cˆ pu 

 1.566
3ˆ x
3(1.064)
Cˆ pk  min(Cˆ pl , Cˆ pu )  1.566
7-8
Chapter 7 Exercise Solutions
7-9 continued
x  T 100  105
V

 4.699
s
1.064
Cˆ p
3.133
Cˆ pm 

 0.652
1V 2
1  (4.699) 2
90  105 

 110  105 
pˆ  Pr  z 
  1  Pr  z 

1.064 
1.064 


 (14.098)  1  (4.699)
 0.000000  1  0.999999
 0.000001
Prefer to use Process B with estimated process fallout of 0.000001 instead of Process A
with estimated fallout 0.001726.
7-10 (7-8).
Process A: ˆ A  20(100)  2000; ˆ A  20ˆ 2  20(3.191) 2  14.271
Process B: ˆ B  20(105)  2100; ˆ B  20ˆ 2  20(1.064)2  4.758
Process B will result in fewer defective assemblies. For the parts
indicates that more parts from Process B are within
Cˆ
 1.045  1.566  Cˆ

pk , A
 
pk , B

specification than from Process A.
7-9
Chapter 7 Exercise Solutions
7-11 (7-9).
MTB > Stat > Basic Statistics > Normality Test
Probability Plot of 1-kg Containers (Ex7-9Wt)
Normal
99
Mean
StDev
N
AD
P-Value
95
90
0.9968
0.02167
15
0.323
0.492
Percent
80
70
60
50
40
30
20
10
5
1
0.950
0.975
1.000
Ex7-9Wt
1.025
1.050
A normal probability plot of the 1-kg container weights shows the distribution is close to
normal.
x  p50  0.9975; p84  1.0200
ˆ  p84  p50  1.0200  0.9975  0.0225
6ˆ  6(0.0225)  0.1350
7-12.
LSL = 0.985 kg
ˆ  LSL 0.9975  0.985
C pl 

 0.19
3ˆ
3(0.0225)
LSL  ˆ 
0.985  0.9975 


pˆ  Pr  z 
  Pr  z 
  (0.556)  0.289105
ˆ 
0.0225



7-10
Chapter 7 Exercise Solutions
7-13.
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate  and .)
Probability Plot of Disk Height (Ex7-13Ht)
Normal
99
95
90
84
Percent
80
Mean
StDev
N
AD
P-Value
20.00
0.009242
25
0.515
0.174
70
60
50
40
30
50
19.99986
10
5
1
19.98
19.99
20.00
Disk Height, mm
20.00905
20
20.01
20.02
A normal probability plot of computer disk heights shows the distribution is close to
normal.
x  p50  19.99986
p84  20.00905
ˆ  p84  p50  20.00905  19.99986  0.00919
6ˆ  6(0.00919)  0.05514
7-11
Chapter 7 Exercise Solutions
7-14.
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate  and .)
Probability Plot of Cycle Time (Ex7-14CT)
Normal
99
95
90
84
Percent
80
Mean
StDev
N
AD
P-Value
13.2
4.097
30
0.401
0.340
70
60
50
40
30
50
20
1
13.2
5
5
17.27
10
10
15
20
Reimbursement Cycle Time, Days
25
A normal probability plot of reimbursement cycle times shows the distribution is close to
normal.
x  p50  13.2
p84  17.27
ˆ  p84  p50  17.27  13.2  4.07
6ˆ  6(4.07)  24.42
7-12
Chapter 7 Exercise Solutions
7-15.
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate  and .)
Probability Plot of Response Time (Ex7-15Resp)
Normal
99
95
90
84
Percent
80
Mean
StDev
N
AD
P-Value
98.78
12.27
40
0.463
0.243
70
60
50
40
30
50
20
1
98.78
5
70
80
110.98
10
90
100
110
Response Time, minutes
120
130
A normal probability plot of response times shows the distribution is close to normal.
(a)
x  p50  98.78
p84  110.98
ˆ  p84  p50  110.98  98.78  12.2
6ˆ  6(12.2)  73.2
(b)
USL = 2 hrs = 120 mins
USL  ˆ 120  98.78
C pu 

 0.58
3ˆ
3(12.2)
USL  ˆ 
USL  ˆ 


 120  98.78 
pˆ  Pr  z 
  1  Pr  z 
  1  Pr  z 

ˆ
ˆ
12.2






 1   (1.739)  1  0.958983  0.041017
7-13
Chapter 7 Exercise Solutions
7-16 (7-10).
MTB > Stat > Basic Statistics > Normality Test
(Add percentile lines at Y values 50 and 84 to estimate  and .)
Probability Plot of Hardness Data (Ex5-59Har)
Normal
99
95
90
84
Percent
80
Mean
StDev
N
AD
P-Value
53.27
2.712
15
0.465
0.217
70
60
50
40
30
50
20
1
53.27
5
46
48
50
52
54
Hardness
55.96
10
56
58
60
A normal probability plot of hardness data shows the distribution is close to normal.
x  p50  53.27
p84  55.96
ˆ  p84  p50  55.96  53.27  2.69
6ˆ  6(2.69)  16.14
7-14
Chapter 7 Exercise Solutions
7-17 (7-11).
MTB > Stat > Basic Statistics > Normality Test
Probability Plot of Failure Times (Ex7-17FT)
Normal
99
Mean
StDev
N
AD
P-Value
95
90
1919
507.1
10
0.272
0.587
Percent
80
70
60
50
40
30
20
10
5
1
1000
1500
2000
Ex7-17FT
2500
3000
The plot shows that the data is not normally distributed; so it is not appropriate to
estimate capability.
7-15
Chapter 7 Exercise Solutions
7-18 (7-12).
LSL = 75; USL = 85; n = 25; S = 1.5
(a)
USL  LSL 85  75
Cˆ p 

 1.11
6ˆ
6(1.5)
(b)
  0.05
2
12 / 2,n 1   0.975,24
 12.40
2
2/ 2,n 1   0.025,24
 39.36
12 / 2,n 1
2/ 2, n 1
ˆ
ˆ
Cp
 Cp  Cp
n 1
n 1
1.11
12.40
39.36
 C p  1.11
25  1
25  1
0.80  C p  1.42
This confidence interval is wide enough that the process may either be capable
(ppm = 27) or far from it (ppm  16,395).
7-19 (7-13).
n  50
Cˆ  1.52
p
1    0.95
2
12 ,n 1   0.95,49
 33.9303
2
Cˆ p 1 ,n 1  C p
n 1
1.52
33.9303
 1.26  C p
49
The company cannot demonstrate that the PCR exceeds 1.33 at a 95% confidence level.
1.52
12 ,49
49
 1.33
2
 1.33 

 49 
  37.52
 1.52 
1    0.88
2
1 ,49
  0.12
7-16
Chapter 7 Exercise Solutions
7-20 (7-14).
n  30; x  97; S  1.6; USL  100; LSL  90
(a)
USL x  ˆ x 100  97
Cˆ pu 

 0.63
3ˆ x
3(1.6)
ˆ  LSL x 97  90
Cˆ pl  x

 1.46
3ˆ x
3(1.6)
Cˆ pk  min(Cˆ pl , Cˆ pu )  0.63
(b)
  0.05
z / 2  z0.025  1.960

1
1 
  C pk  Cˆ pk
Cˆ pk 1  z / 2

2
2(n  1) 
9nCˆ pk



1
1 
1  z / 2


2
2(n  1) 
9nCˆ pk






1
1
1
1
0.63 1  1.96


  C pk  0.63 1  1.96

2
2
9(30)(0.63)
2(30  1) 
9(30)(0.63)
2(30  1) 


0.4287  C pk  0.8313
7-17
Chapter 7 Exercise Solutions
7-21 (7-15).
USL = 2350; LSL = 2100; nominal = 2225; x  2275; s  60; n  50
(a)
USL x  ˆ x 2350  2275
Cˆ pu 

 0.42
3ˆ x
3(60)
ˆ  LSL x 2275  2100
Cˆ pl  x

 0.97
3ˆ x
3(60)
Cˆ pk  min(Cˆ pl , Cˆ pu )  0.42
(b)
  0.05; z / 2  z0.025  1.960


1
1 
1
1 
ˆ 1  z


Cˆ pk 1  z / 2


C

C

pk
pk

/
2
2
ˆ 2 2( n  1) 
2(n  1) 
9nCˆ pk
9
nC


pk







1
1
1
1
0.42 1  1.96


  C pk  0.42 1  1.96

2
2
9(50)(0.42) 2(50  1) 
9(50)(0.42) 2(50  1) 


0.2957  C pk  0.5443
7-22 (7-16).
from Ex. 7-20, Cˆ pk  0.63; z / 2  1.96; n  30


1 
1 
Cˆ pk 1  z / 2
  C pk  Cˆ pk 1  z / 2

2(n  1) 
2(n  1) 






1
1
0.63 1  1.96
  C pk  0.63 1+1.96

2(30  1) 
2(30  1) 


0.47  C pk  0.79
The approximation yields a narrower confidence interval, but it is not too far off.
7-23 (7-17).
 OI  0; ˆ I  3; ˆ Total  5
2
2
2
ˆ Total
 ˆ Meas
 ˆ Process
2
2
ˆ Process  ˆ Total
 ˆ Meas
 52  32  4
7-18
Chapter 7 Exercise Solutions
7-24 (7-18).
(a)
n  2; x  21.8; R  2.8; ˆ Gauge  2.482
MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R
Xbar-R Chart of Part Measurements (Ex7-24All)
30
1
Sample M ean
U C L=27.07
25
_
_
X=21.8
20
15
1
2
4
6
8
1
1
10
Sample
12
14
16
18
LC L=16.53
20
U C L=9.15
Sample Range
8
6
4
_
R=2.8
2
0
LC L=0
2
4
6
8
10
Sample
12
14
16
18
20
Test Results for Xbar Chart of Ex7-24All
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 8, 12, 15, 20
The R chart is in control, and the x chart has a few out-of-control parts. The new gauge
is more repeatable than the old one.
(b) specs: 25  15
6ˆ Gauge
P
6(2.482)

100 
100  49.6%
T USL  LSL
2(15)
7-19
Chapter 7 Exercise Solutions
7-25 (7-19).
MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R
Xbar-R Chart of Part Mesaurements (Ex7-25All)
102.0
1
U C L=100.553
Sample M ean
100.5
99.0
_
_
X=98.2
97.5
96.0
LC L=95.847
1
1
2
3
4
6
5
7
8
9
10
Sample
Sample Range
6.0
U C L=5.921
4.5
3.0
_
R=2.3
1.5
LC L=0
0.0
1
2
3
4
6
5
7
8
9
10
Sample
Test Results for Xbar Chart of Ex7-25All
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 2, 3
The x chart has a couple out-of-control points, and the R chart is in control. This
indicates that the operator is not having difficulty making consistent measurements.
(b)
x  98.2; R  2.3; ˆ Gauge  R d 2  2.3 1.693  1.359
2
ˆ Total
 4.717
2
2
2
ˆ Product
 ˆ Total
 ˆ Gauge
 4.717  1.3592  2.872
ˆ Product  1.695
(c)
ˆ Gauge
ˆ Total
100 
1.359
100  62.5%
4.717
(d)
USL = 100 + 15 = 115; LSL = 100 – 15 = 85
6ˆ Gauge
P
6(1.359)


 0.272
T USL  LSL 115  85
7-20
Chapter 7 Exercise Solutions
7-26 (7-20).
(a)
Excel : workbook Chap07.xls : worksheet Ex7-26
x1  50.03; R1  1.70; x2  49.87; R2  2.30
R  2.00
n  3 repeat measurements
d 2  1.693
ˆ Repeatability  R d 2  2.00 1.693  1.181
Rx  0.17
n  2 operators
d 2  1.128
ˆ Reproducibility  Rx d 2  0.17 1.128  0.151
(b)
2
2
2
ˆ2
ˆ2
ˆ Measurement
Error   Repeatability   Reproducibility  1.181  0.151  1.418
ˆ Measurement Error  1.191
(c) specs: 50  10
6ˆ Gauge
P
6(1.191)

100 
100  35.7%
T USL  LSL
60  40
7-21
Chapter 7 Exercise Solutions
7-27 (7-21).
(a)
ˆ Gauge  R d2  1.533 1.128  1.359
Gauge capability: 6ˆ  8.154
(b)
MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R
Xbar-R Chart of Part Measurements (Ex7-27All)
1
1
Sample M ean
25.0
U C L=23.58
22.5
_
_
X=20.7
20.0
LC L=17.82
17.5
1
15.0
1
1
2
3
4
5
6
7
8
Sample
9
10
Sample Range
6.0
11
12
1
1
13
14
15
U C L=5.010
4.5
3.0
_
R=1.533
1.5
0.0
LC L=0
1
2
3
4
5
6
7
8
Sample
9
10
11
12
13
14
15
Test Results for R Chart of Ex7-27All
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 11, 12
Out-of-control points on R chart indicate operator difficulty with using gage.
7-22
Chapter 7 Exercise Solutions
7-28.
MTB > Stat > ANOVA > Balanced ANOVA
In Results, select “Display expected mean squares and variance components”
ANOVA: Ex7-28Reading versus Ex7-28Part, Ex7-28Op
Factor
Ex7-28Part
Ex7-28Op
Type
random
random
Factor
Ex7-28Part
18, 19, 20
Ex7-28Op
Values
1, 2,
Levels
20
3
3,
4,
5,
6,
7,
8,
1, 2, 3
Analysis of Variance for Ex7-28Reading
Source
DF
SS
MS
Ex7-28Part
19 1185.425 62.391
Ex7-28Op
2
2.617
1.308
Ex7-28Part*Ex7-28Op
38
27.050
0.712
Error
60
59.500
0.992
Total
119 1274.592
S = 0.995825
1
2
3
4
9, 10, 11, 12, 13, 14, 15, 16, 17,
R-Sq = 95.33%
Source
Ex7-28Part
Ex7-28Op
Ex7-28Part*Ex7-28Op
Error
F
87.65
1.84
0.72
P
0.000
0.173
0.861
R-Sq(adj) = 90.74%
Variance
component
10.2798
0.0149
-0.1399
0.9917
Error
term
3
3
4
Expected Mean Square
for Each Term (using
unrestricted model)
(4) + 2 (3) + 6 (1)
(4) + 2 (3) + 40 (2)
(4) + 2 (3)
(4)
2
ˆ Repeatability
 MSError  0.992
MSP×O  MS E 0.712  0.992

 0.1400  0
n
2
MSO  MSP×O 1.308  0.712
2
ˆ Operator

=
 0.0149
pn
20(2)
MSP  MSP×O 62.391  0.712
2
ˆ Part

=
 10.2798
on
3(2)
2
ˆ Part×Operator

The manual calculations match the MINITAB results. Note the Part  Operator variance
component is negative. Since the Part  Operator term is not significant ( = 0.10), we
can fit a reduced model without that term. For the reduced model:
ANOVA: Ex7-28Reading versus Ex7-28Part, Ex7-28Op
…
1
2
3
Source
Ex7-28Part
Ex7-28Op
Error
Variance
component
10.2513
0.0106
0.8832
Error
term
3
3
Expected
Mean Square
for Each
Term (using
unrestricted
model)
(3) + 6 (1)
(3) + 40 (2)
(3)
7-23
Chapter 7 Exercise Solutions
(a)
2
2
ˆ Reproducibility
 ˆ Operator
 0.0106
2
2
ˆ Repeatability
 ˆ Error
 0.8832
(b)
2
2
2
ˆ Gauge
 ˆ Reproducibility
 ˆ Repeatability
 0.0106  0.8832  0.8938
ˆ Gauge  0.9454
(c)
6  ˆ Gauge
6  0.9454
 0.1050
USL-LSL
60  6
This gauge is borderline capable since the estimate of P/T ratio just exceeds 0.10.
P /T 

Estimates of variance components, reproducibility, repeatability, and total gauge
variability may also be found using:
MTB > Stat > Quality Tools > Gage Study > Gage R&R Study (Crossed)
Gage R&R Study - ANOVA Method
Two-Way ANOVA Table With Interaction
Source
Ex7-28Part
Ex7-28Op
Ex7-28Part * Ex7-28Op
Repeatability
Total
DF
19
2
38
60
119
SS
1185.43
2.62
27.05
59.50
1274.59
MS
62.3908
1.3083
0.7118
0.9917
F
87.6470
1.8380
0.7178
P
0.000
0.173
0.861
Two-Way ANOVA Table Without Interaction
Source
Ex7-28Part
Ex7-28Op
Repeatability
Total
DF
19
2
98
119
SS
1185.43
2.62
86.55
1274.59
MS
62.3908
1.3083
0.8832
F
70.6447
1.4814
P
0.000
0.232
Gage R&R
Source
Total Gage R&R
Repeatability
Reproducibility
Ex7-28Op
Part-To-Part
Total Variation
VarComp
0.8938
0.8832
0.0106
0.0106
10.2513
11.1451
%Contribution
(of VarComp)
8.02
7.92
0.10
0.10
91.98
100.00
Study Var
Source
StdDev (SD)
(6 * SD)
Total Gage R&R
0.94541
5.6724
Repeatability
0.93977
5.6386
Reproducibility
0.10310
0.6186
Ex7-28Op
0.10310
0.6186
Part-To-Part
3.20176
19.2106
Total Variation
3.33842
20.0305
Number of Distinct Categories = 4
%Study Var
(%SV)
28.32
28.15
3.09
3.09
95.91
100.00
7-24
Chapter 7 Exercise Solutions
7-28 continued
Visual representations of variability and stability are also provided:
Gage R&R (ANOVA) for Ex7-28Reading
Reported by :
Tolerance:
M isc:
G age name:
D ate of study :
Components of Variation
Ex7-28Reading by Ex7-28Part
100
% Contribution
30
Percent
% Study Var
25
50
20
0
Gage R&R
Repeat
Reprod
1
Part-to-Part
2
3
4
5
R Chart by Ex7-28Op
Sample Range
4
1
2
6
7
8
Ex7-28Reading by Ex7-28Op
3
UCL=3.757
30
25
2
_
R=1.15
0
20
LCL=0
1
2
Ex7-28Op
Xbar Chart by Ex7-28Op
1
2
3
Ex7-28Op * Ex7-28Part Interaction
3
30
30
25
UCL=24.55
_
_
X=22.39
20
LCL=20.23
Average
Sample Mean
9 10 11 12 13 14 15 16 17 18 19 20
Ex7-28Part
Ex7-28Op
1
2
3
25
20
1
2
3
4
5
6
7
8
9 10 11 1 2 13 14 1 5 16 17 18 1 9 20
Ex7-28Part
7-25
Chapter 7 Exercise Solutions
7-29.
2
2
ˆ Part
 10.2513; ˆ Total
 11.1451
ˆ P 
2
ˆ Part
10.2513

 0.9198
2
ˆ Total 11.1451
SNR 
DR 
2 ˆ P
2(0.9198)

 4.79
1  ˆ P
1  0.9198
1  ˆ P 1  0.9198

 23.94
1  ˆ P 1  0.9198
SNR = 4.79 indicates that fewer than five distinct levels can be reliably obtained from the
measurements. This is near the AIAG-recommended value of five levels or more, but
larger than a value of two (or less) that indicates inadequate gauge capability. (Also note
that the MINITAB Gage R&R output indicates “Number of Distinct Categories = 4”;
this is also the number of distinct categories of parts that the gauge is able to distinguish)
DR = 23.94, exceeding the minimum recommendation of four. By this measure, the
gauge is capable.
7-30 (7-22).
  1  2  3  100  75  75  250
   12   22   33  42  42  22  6
Pr{x  262}  1  Pr{x  262}
262   

 1  Pr  z 

 

262  250 

 1  Pr  z 

6


 1   (2.000)
 1  0.9772
 0.0228
7-26
Chapter 7 Exercise Solutions
7-31 (7-23).
x1 ~ N (20, 0.32 ); x2 ~ N (19.6, 0.42 )
Nonconformities will occur if y  x1  x2  0.1 or y  x1  x2  0.9
 y  1  2  20  19.6  0.4
 y2   12   22  0.32  0.42  0.25
 y  0.50
Pr{Nonconformities}  Pr{ y  LSL}  Pr{ y  USL}
 Pr{ y  0.1}  Pr{ y  0.9}
 Pr{ y  0.1}  1  Pr{ y  0.9}
 0.1  0.4 
 0.9  0.4 
 
 1  

 0.25 
 0.25 
 (0.6)  1   (1.00)
 0.2743  1  0.8413
 0.4330
7-32 (7-24).
Volume  L  H  W
  L  H W  ( L   L )  H W  ( H   H )  L W  (W  W )  L  H
ˆ Volume  L H W  6.0(3.0)(4.0)  72.0
2
 Volume
  L2 H2  W2   H2  L2 W2  W2  L2 H2
 6.02 (0.01)(0.01)  3.02 (0.01)(0.01)  4.02 (0.01)(0.01)
 0.0061
7-33 (7-25).
Weight  d  W  L  T
 d  W  L T  (W  W )  L T  ( L   L ) W T  (T  T ) W  L 
ˆ Weight  d [ W  L T ]  0.08(10)(20)(3)  48
2
ˆ Weight
 d 2  ˆW2 ˆ L2ˆT2  ˆ L2ˆW2 ˆT2  ˆT2ˆW2 ˆ L2 
 0.082 102 (0.32 )(0.12 )  202 (0.22 )(0.12 )  32 (0.2 2 )(0.32 )   0.00181
ˆ Weight  0.04252
7-27
Chapter 7 Exercise Solutions
7-34 (7-26).
1
(5 x  2); 2  x  4
26
4
4
4
1 5 3
1

E ( x)   x   xf ( x)dx   x  (5 x  2)  dx   x  x 2   3.1282
2
26  3 2
2  26


4
4
4

1 5
2
1

E x 2   x 2 f ( x)dx   x 2  (5 x  2)  dx   x 4  x3   10.1026
26  4 2 3 2 
2
 26

s  (3  0.05 x) 2 and f ( x) 
 
 x2  E ( x 2 )   E ( x)   10.1026  (3.1282) 2  0.3170
2
s  g ( x)  3  0.05(  x )   3  0.05(3.1282)   9.9629
2
  g ( x) 
 

 x 
2
s
2
2
 x2
x
  (3  0.05 x) 2 


x


2
 x2
x
 2(3  0.05 x )(0.05) x2
 2 3  0.05(3.1282)  (0.05)(0.3170)
 0.1001
7-35 (7-27).
I  E ( R1  R2 )
I  E (R  R )
1
 I2 

2
E
(  R1   R2 )
2

E
 2   R2
2  R
(R  R )
1
1
2

2
7-28
Chapter 7 Exercise Solutions
7-36 (7-28).
x1 ~ N ( 1 , 0.4002 ); x2 ~ N ( 2 , 0.3002 )
 y  1  2
 y   12   22  0.4002  0.3002  0.5
Pr{ y  0.09}  0.006

0.09   y 
1
Pr  z 
   (0.006)
 y 

0.09   y
 2.512
0.5
 y  [0.5(2.512)  0.09]  1.346
7-37 (7-29).
ID ~ N (2.010,0.0022 ) and OD ~ N (2.004,0.0012 )
Interference occurs if y = ID – OD < 0
 y  ID  OD  2.010  2.004  0.006
2
2
 y2   ID
  OD
 0.0022  0.0012  0.000005
 y  0.002236
Pr{positive clearance}  1  Pr{interference}
 1  Pr{ y  0}
 0  0.006 
 1  

 0.000005 
 1  (2.683)
 1  0.0036
 0.9964
7-38 (7-30).
  0.01
  0.80
2
12 ,4   0.20,4
 5.989
1  2    1 ,4 1  2  0.01  5.989

 
 299


2    4
2  0.01  4
2
n
7-29
Chapter 7 Exercise Solutions
7-39 (7-31).
n = 10; x ~ N (300,102 );   0.10;   0.95 ; one-sided
From Appendix VIII: K = 2.355
UTL  x  KS  300  2.355(10)  323.55
7-40 (7-32).
n = 25; x ~ N (85,12 );   0.10;   0.95 ; one-sided
From Appendix VIII: K = 1.838
x  KS  85  1.838(1)  83.162
7-41 (7-33).
n = 20; x ~ N (350,102 );   0.05;   0.90 ; one-sided
From Appendix VIII: K = 2.208
UTL  x  KS  350  2.208(10)  372.08
7-42 (7-34).
  0.05
  0.90
2
12 ,4   0.10,4
 7.779
1  2    1 ,4 1  2  0.05  7.779

 
 77


2    4
2  0.05  4
2
n
After the data are collected, a natural tolerance interval would be the smallest to largest
observations.
7-30
Chapter 7 Exercise Solutions
7-43 (7-35).
x ~ N  0.1264, 0.00032 
(a)
 = 0.05;  = 0.95; and two-sided
From Appendix VII: K = 2.445
TI on x : x  KS  0.1264  2.445(0.0003)  [0.1257, 0.1271]
(b)
  0.05; t / 2,n1  t0.025,39  2.023
CI on x : x  t / 2,n1 S

n  0.1264  2.023 0.0003

40  [0.1263, 0.1265]
Part (a) is a tolerance interval on individual thickness observations; part (b) is a
confidence interval on mean thickness. In part (a), the interval relates to individual
observations (random variables), while in part (b) the interval refers to a parameter of a
distribution (an unknown constant).
7-44 (7-36).
  0.05;   0.95
n
log(1   ) log(1  0.95)

 59
log(1   ) log(1  0.05)
The largest observation would be the nonparametric upper tolerance limit.
7-31