Chapter 10 Exercise Solutions
Note: MINITAB’s Tsquared functionality does not use summary statistics, so many of
these exercises have been solved in Excel.
10-1.
Phase 2 T 2 control charts with m = 50 preliminary samples, n = 25 sample size, p = 2
characteristics. Let  = 0.001.
p(m  1)(n  1)
UCL 
F , p ,mn m p 1
mn  m  p  1
2(50  1)(25  1)

F0.001,2,1199
50(25)  50  2  1
  2448 1199  (6.948)  14.186
LCL = 0
Excel : workbook Chap10.xls : worksheet Ex10-1
Sample No.
xbar1
xbar2
1
58
32
2
60
33
3
50
27
4
54
31
5
63
38
6
53
30
7
42
20
8
55
31
9
46
25
3
2
5
3
-5
-3
-1
1
8
8
-2
0
-13
-10
0
1
-9
-5
0.0451
1.1268
0.1268
3.1690
0.1268
3.1690
0.0817
2.0423
0.5408
13.5211
0.0676
1.6901
0.9127
22.8169
0.0282
0.7042
0.4254
10.6338
14.1850
0
14.1850
0
14.1850
0
14.1850
0
14.1850
0
14.1850
0
14.1850
0
14.1850
0
14.1850
0
diff1
diff2
matrix calc
t2 = n * calc
UCL =
LCL =
OOC?
In control
In control
In control
In control
In control
In control
Above UCL
In control
In control
…
T^2 Control Chart for Quality Characteristics
60.00
50.00
T^2
40.00
30.00
20.00
10.00
0.00
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Sample Number
Process is out of control at samples 7 and 14.
10-1
Chapter 10 Exercise Solutions
10-2.
Phase 2 T 2 control limits with m = 30 preliminary samples, n = 10 sample size, p = 3
characteristics. Let  = 0.001.
p (m  1)(n  1)
UCL 
F , p ,mn m  p 1
mn  m  p  1
3(30  1)(10  1)

F0.001,3,268
30(10)  30  3  1
 837 

 (5.579)
 268 
 17.425
LCL = 0
Excel : workbook Chap10.xls : worksheet Ex10-2
Sample No.
xbar1
xbar2
xbar3
1
3.1
3.7
3
2
3.3
3.9
3.1
3
2.6
3
2.4
4
2.8
3
2.5
5
3
3.3
2.8
6
4
4.6
3.5
7
3.8
4.2
3
8
3
3.3
2.7
9
2.4
3
2.2
10
2
2.6
1.8
11
3.2
3.9
3
12
3.7
4
3
13
4.1
4.7
3.2
14
3.8
4
2.9
15
3.2
3.6
2.8
diff1
diff2
diff3
0.1
0.2
0.2
0.3
0.4
0.3
-0.4
-0.5
-0.4
-0.2
-0.5
-0.3
0
-0.2
0
1
1.1
0.7
0.8
0.7
0.2
0
-0.2
-0.1
-0.6
-0.5
-0.6
-1
-0.9
-1
0.2
0.4
0.2
0.7
0.5
0.2
1.1
1.2
0.4
0.8
0.5
0.1
0.2
0.1
0
0.0528
0.5279
0.1189
1.1887
0.1880
1.8800
0.2372
2.3719
0.0808
0.8084
1.0397
10.3966
1.0593
10.5932
0.0684
0.6844
0.3122
3.1216
0.8692
8.6922
0.1399
1.3990
0.6574
6.5741
2.0793
20.7927
1.1271
11.2706
0.0852
0.8525
matrix calc
t2 = n * calc
UCL =
LCL =
OOC?
17.4249 17.4249 17.4249 17.4249 17.4249 17.4249
17.4249 17.4249 17.4249 17.4249 17.4249 17.4249
17.4249 17.4249 17.4249
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
In control In control In control In control In control In control In control In control In control In control In control In control Above UCL
In control In control
25.00
T^2 Control Chart for Quality Characteristics
20.00
25.00
15.00
10.00
20.00
5.00
0.00
15.00
2
3
4
5
6
7
8
9
T^2
1
10.00
5.00
0.00
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Sample Number
Process is out of control at sample 13.
10-2
10
11
12
13
14
15
Chapter 10 Exercise Solutions
10-3.
Phase 2 T 2 control limits with p = 2 characteristics. Let  = 0.001.
Since population parameters are known, the chi-square formula will be used for the upper
2
control limit: UCL  2, p  0.001,2
 13.816
Excel : workbook Chap10.xls : worksheet Ex10-3
Sample No.
xbar1
xbar2
1
2
UCL =
LCL =
OOC?
4
5
6
7
8
9
10
11
12
13
14
15
60
33
50
27
54
31
63
38
53
30
42
20
55
31
46
25
50
29
49
27
57
30
58
33
75
45
55
27
3
2
5
3
-5
-3
-1
1
8
8
-2
0
-13
-10
0
1
-9
-5
-5
-1
-6
-3
2
0
3
3
20
15
0
-3
0.0451
1.1268
0.1268
3.1690
0.1268
3.1690
0.0817
2.0423
0.5408
13.5211
0.0676
1.6901
0.9127
22.8169
0.0282
0.7042
0.4254
10.6338
0.2676
6.6901
0.2028
5.0704
0.0676
1.6901
0.0761
1.9014
2.1127
52.8169
0.2535
6.3380
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
13.8150
0
diff1
diff2
matrix calc
t2 = n * calc
3
58
32
In control In control In control In control In control In control Above UCL
In control In control In control In control In control In control Above UCL
In control
Phase II T^2 Control Chart
60.0000
50.0000
T^2
40.0000
30.0000
20.0000
10.0000
0.0000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Sample Number
Process is out of control at samples 7 and 14. Same results as for parameters estimated
from samples.
10-3
Chapter 10 Exercise Solutions
10-4.
Phase 2 T 2 control limits with p = 3 characteristics. Let  = 0.001.
Since population parameters are known, the chi-square formula will be used for the upper
2
control limit: UCL  2, p  0.001,3
 16.266
Excel : workbook Chap10.xls : worksheet Ex10-4
Sample No.
xbar1
xbar2
xbar3
1
3.1
3.7
3
2
3.3
3.9
3.1
3
2.6
3
2.4
4
2.8
3
2.5
5
3
3.3
2.8
6
4
4.6
3.5
7
3.8
4.2
3
8
3
3.3
2.7
9
2.4
3
2.2
10
2
2.6
1.8
11
3.2
3.9
3
12
3.7
4
3
13
4.1
4.7
3.2
14
3.8
4
2.9
15
3.2
3.6
2.8
diff1
diff2
diff3
0.1
0.2
0.2
0.3
0.4
0.3
-0.4
-0.5
-0.4
-0.2
-0.5
-0.3
0
-0.2
0
1
1.1
0.7
0.8
0.7
0.2
0
-0.2
-0.1
-0.6
-0.5
-0.6
-1
-0.9
-1
0.2
0.4
0.2
0.7
0.5
0.2
1.1
1.2
0.4
0.8
0.5
0.1
0.2
0.1
0
0.0528
0.5279
0.1189
1.1887
0.1880
1.8800
0.2372
2.3719
0.0808
0.8084
1.0397
10.3966
1.0593
10.5932
0.0684
0.6844
0.3122
3.1216
0.8692
8.6922
0.1399
1.3990
0.6574
6.5741
2.0793
20.7927
1.1271
11.2706
0.0852
0.8525
matrix calc
t2 = n * calc
UCL =
LCL =
OOC?
16.2660 16.2660 16.2660 16.2660 16.2660 16.2660
16.2660 16.2660 16.2660 16.2660 16.2660 16.2660
16.2660 16.2660 16.2660
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
In control In control In control In control In control In control In control In control In control In control In control In control Above UCL
In control In control
Phase II T^2 Control Chart
25.00
Phase II T^2 Control Chart
20.00
25.00
T^2
15.00
10.00
20.00
5.00
0.00
15.00
1
2
3
4
5
6
7
8
9
T^2
Sample Number
10.00
5.00
0.00
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Sample Number
Process is out of control at sample 13. Same as results for parameters estimated from
samples.
10-4
10
11
12
13
14
15
Chapter 10 Exercise Solutions
10-5.
m = 30 preliminary samples, n = 3 sample size, p = 6 characteristics,  = 0.005
(a)
Phase II limits:
p (m  1)(n  1)
UCL 
F , p ,mn  m  p 1
mn  m  p  1
6(30  1)(3  1)

F0.005,6,55
30(3)  30  6  1
 372 

 (3.531)
 55 
 23.882
LCL = 0
(b)
2
chi-square limit: UCL  2, p  0.005,6
 18.548
The Phase II UCL is almost 30% larger than the chi-square limit.
(c)
Quality characteristics, p = 6. Samples size, n = 3.  = 0.005. Find "m" such that exact
Phase II limit is within 1% of chi-square limit, 1.01(18.548) = 18.733.
Excel : workbook Chap10.xls : worksheet Ex10-5
m
30
40
50
60
70
80
90
100
num denom
372
55
492
75
612
95
732
115
852
135
972
155
1092
175
1212
195
F
3.531
3.407
3.338
3.294
3.263
3.240
3.223
3.209
UCL
23.8820
22.3527
21.5042
20.9650
20.5920
20.3184
20.1095
19.9447
717
718
719
720
721
722
8616
8628
8640
8652
8664
8676
3.107
3.107
3.107
3.107
3.107
3.107
18.7337
18.7332
18.7331
18.7328
18.7325
18.7324
…
1429
1431
1433
1435
1437
1439
720 preliminary samples must be taken to ensure that the exact Phase II limit is within
1% of the chi-square limit.
10-5
Chapter 10 Exercise Solutions
10-6.
m = 30 preliminary samples, n = 5 sample size, p = 6 characteristics,  = 0.005
(a)
Phase II UCL:
p (m  1)(n  1)
UCL 
F , p ,mn  m  p 1
mn  m  p  1
6(30  1)(5  1)

F0.005,6,115
30(5)  30  6  1
 744 

 (3.294)
 115 
 21.309
(b)
2
chi-square UCL: UCL  2, p  0.005,6
 18.548
The Phase II UCL is almost 15% larger than the chi-square limit.
(c)
Quality characteristics, p = 6. Samples size, n = 5.  = 0.005. Find "m" such that exact
Phase II limit is within 1% of chi-square limit, 1.01(18.548) = 18.733.
Excel : workbook Chap10.xls : worksheet Ex10-6
m
30
40
50
60
70
80
90
100
num
744
984
1224
1464
1704
1944
2184
2424
denom
115
155
195
235
275
315
355
395
F
3.294
3.240
3.209
3.189
3.174
3.164
3.155
3.149
UCL
21.3087
20.5692
20.1422
19.8641
19.6685
19.5237
19.4119
19.3232
390
400
410
411
412
9384
9624
9864
9888
9912
1555
1595
1635
1639
1643
3.106
3.105
3.105
3.105
3.105
18.7424
18.7376
18.7330
18.7324
18.7318
…
411 preliminary samples must be taken to ensure that the exact Phase II limit is within
1% of the chi-square limit.
10-6
Chapter 10 Exercise Solutions
10-7.
m = 25 preliminary samples, n = 3 sample size, p = 10 characteristics,  = 0.005
(a)
Phase II UCL:
p (m  1)(n  1)
UCL 
F , p ,mn m  p 1
mn  m  p  1
10(25  1)(3  1)

F0.005,10,41
25(3)  25  10  1
 520 

 (3.101)
 41 
 39.326
(b)
2
chi-square UCL: UCL  2, p  0.005,10
 25.188
The Phase II UCL is more than 55% larger than the chi-square limit.
(c)
Quality characteristics, p = 10. Samples size, n = 3.  = 0.005. Find "m" such that exact
Phase II limit is within 1% of chi-square limit, 1.01(25.188) = 25.440.
Excel : workbook Chap10.xls : worksheet Ex10-7
m
25
35
45
55
65
75
85
95
105
num
520
720
920
1120
1320
1520
1720
1920
2120
denom
41
61
81
101
121
141
161
181
201
F
3.101
2.897
2.799
2.742
2.704
2.677
2.657
2.641
2.629
UCL
39.3259
34.1991
31.7953
30.4024
29.4940
28.8549
28.3808
28.0154
27.7246
986
987
988
989
990
19740
19760
19780
19800
19820
1963
1965
1967
1969
1971
2.530
2.530
2.530
2.530
2.530
25.4405
25.4401
25.4399
25.4398
25.4394
…
988 preliminary samples must be taken to ensure that the exact Phase II limit is within
1% of the chi-square limit.
10-7
Chapter 10 Exercise Solutions
10-8.
m = 25 preliminary samples, n = 5 sample size, p = 10 characteristics,  = 0.005
(a)
Phase II UCL:
p (m  1)(n  1)
UCL 
F , p ,mn m  p 1
mn  m  p  1
10(25  1)(5  1)

F0.005,10,91
25(5)  25  10  1
 1040 

 (2.767)
 91 
 31.625
(b)
2
chi-square UCL: UCL  2, p  0.005,10
 25.188
The Phase II UCL is more than 25% larger than the chi-square limit.
(c)
Quality characteristics, p = 10. Samples size, n = 5.  = 0.005. Find "m" such that exact
Phase II limit is within 1% of chi-square limit, 1.01(25.188) = 25.440.
Excel : workbook Chap10.xls : worksheet Ex10-8
m
25
35
45
55
65
75
85
95
105
num denom
F
UCL
1040
91 2.767 31.6251
1440
131 2.689 29.5595
1840
171 2.648 28.4967
2240
211 2.623 27.8495
2640
251 2.606 27.4141
3040
291 2.594 27.1011
3440
331 2.585 26.8651
3840
371 2.578 26.6812
4240
411 2.572 26.5335
540
541
542
543
544
545
21640
21680
21720
21760
21800
21840
2151
2155
2159
2163
2167
2171
2.529
2.529
2.529
2.529
2.529
2.529
25.4419
25.4413
25.4408
25.4405
25.4399
25.4394
544 preliminary samples must be taken to ensure that the exact Phase II limit is within
1% of the chi-square limit.
10-8
Chapter 10 Exercise Solutions
10-9.
p = 10 quality characteristics, n = 3 sample size, m = 25 preliminary samples. Assume 
= 0.01.
Phase I UCL:
p (m  1)(n  1)
UCL 
F , p ,mn  m  p 1
mn  m  p  1
10(25  1)(3  1)

F0.01,10,41
25(3)  25  10  1
 480 

 (2.788)
 41 
 32.638
Phase II UCL:
p (m  1)(n  1)
UCL 
F , p ,mn  m  p 1
mn  m  p  1
10(25  1)(3  1)

F0.01,10,41
25(3)  25  10  1
 520 

 (2.788)
 41 
 35.360
10-10.
Excel : workbook Chap10.xls : worksheet Ex10-10
(a)
 1 0.7 0.7 0.7 
0.7 1 0.7 0.7 

Σ
0.7 0.7 1 0.7 


0.7 0.7 0.7 1 
(b)
2
UCL  2, p  0.01,4
 13.277
10-9
Chapter 10 Exercise Solutions
10-10 continued
(c)
T 2  n  y -    1  y -  
 3.5 0    1 0.7 0.7 0.7  1  3.5 0  
    


3.5 0    0.7 1 0.7 0.7   3.5 0  


1


 3.5 0    0.7 0.7 1 0.7   3.5 0  
      
      
 3.5 0    0.7 0.7 0.7 1   3.5 0  
 15.806
Yes. Since T 2  15.806    UCL  13.277  , an out-of-control signal is generated.
(d)
1
T(1)2  n  y (1) -  (1)   (1)
 y (1) - (1) 
 3.5 0    1 0.7 0.7  1  3.5  0  




 1 3.5  0   0.7 1 0.7   3.5  0  
 3.5 0   0.7 0.7 1   3.5 0  
    
    
 15.313
2
2
2
2
T(1)  T(2)
 T(3)
 T(4)
 15.313
di  T 2  T(2i )
d1  d2  d3  d4  15.806  15.313  0.493
2
0.01,1
 6.635
2
No. First, since all di are smaller than 0.01,1
, no variable is identified as a relatively large
contributor. Second, since the standardized observations are equal (that is, all variables
had the same shift), this information does not assist in identifying which a process
variable shifted.
(e)
Since (T 2  28.280)  (UCL  13.277) , an out-of-control signal is generated.
(f)
2
0.01,1
 6.635
T(1)2  15.694; d1  12.585
2
T(2)
 21.979; d 2  6.300
2
T(3)
 14.479; d3  13.800
2
T(4)
 25.590; d 4  2.689
Investigate variables 1 and 3.
10-10
Chapter 10 Exercise Solutions
10-11.
Excel : workbook Chap10.xls : worksheet Ex10-11
(a)
 1 0.8 0.8
  0.8 1 0.8
0.8 0.8 1 
(b)
2
UCL  2, p  0.05,3
 7.815
(c)
T 2  11.154
Yes. Since T 2  11.154    UCL  7.815  , an out-of-control signal is generated.
(d)
2
0.05,1
 3.841
T(1)2  11.111; d1  0.043
2
T(2)
 2.778; d 2  8.376
2
T(3)
 5.000; d3  6.154
Variables 2 and 3 should be investigated.
(e)
Since (T 2  6.538)  (UCL  7.815) , an out-of-control signal is not generated.
(f)
2
0.05,1
 3.841
T(1)2  5.000; d1  1.538
2
T(2)
 5.000; d 2  1.538
T(3)2  4.444; d3  2.094
Since an out-of-control signal was not generated in (e), it is not necessary to calculate the
diagnostic quantities. This is confirmed since none of the di’s exceeds the UCL.
10-11
Chapter 10 Exercise Solutions
10-12.
Excel : workbook Chap10.xls : worksheet Ex10-12
m  40
 4.440 0.016 
x  15.339 0.104 ; S1  

 0.016 0.001 
121.101 0.256 
 1.553 0.003
VV  
; S2  


 0.256 0.071 
 0.003 0.001 
10-13.
Excel : workbook Chap10.xls : worksheet Ex10-13
m  40
 4.440 0.016 5.395 
x  15.339 0.104 88.125 ; S1   0.016 0.001 0.014 
 5.395 0.014 27.599 
121.101 0.256 43.720 
 1.553 0.003 0.561


VV   0.256 0.071
0.950  ; S 2   0.003 0.001 0.012 
 43.720 0.950 587.000 
 0.561 0.012 7.526 
10-14.
Excel : workbook Chap10.xls : worksheet Ex10-14
xbar
xbar1
xbar2
10.607
21.207
S1
3.282
3.305
3.305
5.641
V'V
133.780 80.740
80.740 67.150
S2
2.307 1.392
1.392 1.158
10-12
Chapter 10 Exercise Solutions
10-15.
Excel : workbook Chap10.xls : worksheet Ex10-15
p=
mu' =
Sigma =
y' =
4
0
0
0
0
1
0.75
0.75
0.75
0.75
1
0.75
0.75
0.75
0.75
1
0.75
0.75
0.75
0.75
1
1
1
1
1
y' Sigma-1 =
0.308 0.308 0.308 0.308
y' Sigma-1 y =
1.231
delta =
1.109
ARL0 =
200
Sigma-1 =
3.0769
-0.9231
-0.9231
-0.9231
y=
-0.9231
3.0769
-0.9231
-0.9231
-0.9231
-0.9231
3.0769
-0.9231
-0.9231
-0.9231
-0.9231
3.0769
1
1
1
1
From Table 10-3, select (lambda, H) pair that closely minimizes ARL1
1.5
1
delta =
0.2
0.1
lambda =
12.73 13.87
UCL = H =
12.17 6.53
ARL1 =
Select  = 0.1 with an UCL = H = 12.73. This gives an ARL1 between 7.22 and 12.17.
10-13
Chapter 10 Exercise Solutions
10-16.
Excel : workbook Chap10.xls : worksheet Ex10-16
p=
mu' =
Sigma =
y' =
4
0
0
0
0
1
0.9
0.9
0.9
0.9
1
0.9
0.9
0.9
0.9
1
0.9
0.9
0.9
0.9
1
1
1
1
1
y' Sigma-1 =
0.270 0.270 0.270 0.270
y' Sigma-1 y =
1.081
delta =
1.040
ARL0 =
500
Sigma-1 =
y=
7.568
-2.432
-2.432
-2.432
-2.432
7.568
-2.432
-2.432
-2.432
-2.432
7.568
-2.432
-2.432
-2.432
-2.432
7.568
1
1
1
1
From Table 10-4, select (lambda, H) pair
delta =
1
1.5
lambda =
0.105 0.18
UCL = H =
15.26 16.03
ARLmin =
14.60 7.65
Select  = 0.105 with an UCL = H = 15.26. This gives an ARLmin near 14.60.
10-14
Chapter 10 Exercise Solutions
10-17.
Excel : workbook Chap10.xls : worksheet Ex10-17
p=
mu' =
Sigma =
y' =
2
0
0
1
0.8
0.8
1
1
1
y' Sigma-1 =
0.556 0.556
y' Sigma-1 y =
1.111
delta =
1.054
ARL0 =
200
Sigma-1 =
2.7778 -2.2222
-2.2222 2.7778
y=
1
1
From Table 10-3, select (lambda, H) pair that closely minimizes ARL1
delta =
1
1 1.5
1.5
lambda =
0.1
0.2 0.2
0.3
UCL = H =
8.64 9.65 9.65 10.08
ARL1 =
10.15 10.20 5.49 5.48
Select  = 0.2 with an UCL = H = 9.65. This gives an ARL1 between 5.49 and 10.20.
10-15
Chapter 10 Exercise Solutions
10-18.
(a)
Note: In the textbook Table 10-5, the y2 values for Observations 8, 9, and 10 should be
100, 103, and 107.
Stat > Control Charts > Variables Charts for Individuals > Individuals
I Chart of y2 Output Variable (Tab10-5y2)
110
1
Individual Value
105
1
5
100
UCL=100.12
5
6
95
_
X=91.25
90
6
85
6
6
5
5
5
5
LCL=82.38
80
4
8
12
16
20
24
Observation
28
32
36
40
Test Results for I Chart of Tab10-5y2
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 9, 10
TEST 5. 2 out of 3 points more than 2 standard deviations from center line (on
one side of CL).
Test Failed at points: 8, 9, 10, 11, 35, 37, 39, 40
TEST 6. 4 out of 5 points more than 1 standard deviation from center line (on
one side of CL).
Test Failed at points: 5, 10, 11, 12, 36, 37, 38, 39, 40
TEST 8. 8 points in a row more than 1 standard deviation from center line
(above and below CL).
Test Failed at points: 40
10-16
Chapter 10 Exercise Solutions
10-18 continued
(b)
Stat > Regression > Regression
Regression Analysis: Tab10-5y2 versus Tab10-5x1, Tab10-5x2, ...
The regression equation is
Tab10-5y2 = 215 - 0.666 Tab10-5x1 - 11.6 Tab10-5x2 + 0.435 Tab10-5x3
+ 0.192 Tab10-5x4 - 3.2 Tab10-5x5 + 0.73 Tab10-5x6 + 6.1 Tab10-5x7
+ 10.9 Tab10-5x8 - 215 Tab10-5x9
Stat > Control Charts > Variables Charts for Individuals > Individuals
I Chart of Regression Model Residuals (Ex10-18Res)
7.5
1
1
UCL=6.57
5
5
Individual Value
5.0
2.5
_
X=-0.00
0.0
-2.5
-5.0
5
LCL=-6.57
4
8
12
16
20
24
Observation
28
32
36
40
Test Results for I Chart of Ex10-18Res
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 7, 18
TEST 5. 2 out of 3 points more than 2 standard deviations from center line (on
one side of CL).
Test Failed at points: 19, 21, 25
TEST 6. 4 out of 5 points more than 1 standard deviation from center line (on
one side of CL).
Test Failed at points: 21
Plot points on the residuals control chart are spread between the control limits and do not
exhibit the downward trend of the response y2 control chart.
10-17
Chapter 10 Exercise Solutions
10-18 continued
(c)
Stat > Time Series > Autocorrelation
Autocorrelation Function for y2 Output Variable (Tab10-5y2)
(with 5% significance limits for the autocorrelations)
1.0
0.8
Autocorrelation
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
1
2
3
4
5
6
7
8
9
10
Lag
Autocorrelation Function for Regression Model Residuals (Ex10-18Res)
(with 5% significance limits for the autocorrelations)
1.0
0.8
Autocorrelation
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
1
2
3
4
5
6
7
8
9
10
Lag
The decaying sine wave of ACFs for Response y2 suggests an autoregressive process,
while the ACF for the residuals suggests a random process.
10-18
Chapter 10 Exercise Solutions
10-19.
Different approaches can be used to identify insignificant variables and reduce the
number of variables in a regression model. This solution uses MINITAB’s “Best
Subsets” functionality to identify the best-fitting model with as few variables as possible.
Stat > Regression > Best Subsets
Best Subsets Regression: Tab10-5y1 versus Tab10-5x1, Tab10-5x2, ...
Response is Tab10-5y1
T T T T T T T T
a a a a a a a a
b b b b b b b b
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0
- - - - - - - 5 5 5 5 5 5 5 5
Mallows
x x x x x x x x
Vars R-Sq R-Sq(adj)
C-p
S 1 2 3 4 5 6 7 8
1 43.1
41.6
52.9
1.3087
1 31.3
29.5
71.3
1.4378
X
2 62.6
60.5
24.5
1.0760 X
2 55.0
52.5
36.4
1.1799
X
3 67.5
64.7
18.9
1.0171 X
X
3 66.8
64.0
19.9
1.0273 X
X
4 72.3
69.1
13.3 0.95201 X
X X
4 72.1
68.9
13.6 0.95522 X
X
X
5 79.5
76.5
4.0 0.83020 X
X X
X
5 73.8
69.9
13.0 0.93966 X
X
X
X
6 79.9
76.2
5.5 0.83550 X
X X
X
X
6 79.8
76.1
5.6 0.83693 X X X X
X
7 80.3
76.0
6.8 0.83914 X
X X X
X X
7 80.1
75.8
7.1 0.84292 X
X X X X
X
…
Best Subsets Regression: Tab10-5y2 versus Tab10-5x1, Tab10-5x2, ...
Response is Tab10-5y2
T T T T T T T T
a a a a a a a a
b b b b b b b b
1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0
- - - - - - - 5 5 5 5 5 5 5 5
Mallows
x x x x x x x x
Vars R-Sq R-Sq(adj)
C-p
S 1 2 3 4 5 6 7 8
1 36.1
34.4
24.0 4.6816
X
1 35.8
34.1
24.2 4.6921
2 55.1
52.7
8.1 3.9751
X
2 50.7
48.1
12.2 4.1665
X
X
3 61.6
58.4
4.0 3.7288
X X
X
3 59.8
56.4
5.7 3.8160
X
X
4 64.9
60.9
2.9 3.6147 X
X
X
4 64.4
60.4
3.4 3.6387
X X
X
5 67.7
62.9
2.3 3.5208 X
X X
X
5 65.2
60.1
4.7 3.6526 X
X
X
X
6 67.8
62.0
4.2 3.5660 X
X X
X X
6 67.8
61.9
4.3 3.5684 X
X X X
X
7 67.9
60.9
6.1 3.6149 X X X X
X X
7 67.8
60.8
6.2 3.6200 X
X X X
X X
…
T
a
b
1
0
5
x
9
X
X
X
X
X
X
X
X
X
X
X
X
X
******
T
a
b
1
0
5
x
9
X
X
X
X
X
X
X
X
X
X
X
******
For output variables y1 and y2, a regression model of input variables x1, x3, x4, x8, and
x9 maximize adjusted R2 (minimize S) and minimize Mallow’s C-p.
10-19
Chapter 10 Exercise Solutions
10-19 continued
Stat > Regression > Regression
Regression Analysis: Tab10-5y1 versus Tab10-5x1, Tab10-5x3, ...
The regression equation is
Tab10-5y1 = 819 + 0.431 Tab10-5x1 - 0.124 Tab10-5x3 - 0.0915 Tab10-5x4
+ 2.64 Tab10-5x8 + 115 Tab10-5x9
Predictor
Constant
Tab10-5x1
Tab10-5x3
Tab10-5x4
Tab10-5x8
Tab10-5x9
Coef
818.80
0.43080
-0.12396
-0.09146
2.6367
114.81
S = 0.830201
SE Coef
29.14
0.08113
0.03530
0.02438
0.7604
23.65
R-Sq = 79.5%
Analysis of Variance
Source
DF
SS
Regression
5
90.990
Residual Error 34
23.434
Total
39 114.424
T
28.10
5.31
-3.51
-3.75
3.47
4.85
P
0.000
0.000
0.001
0.001
0.001
0.000
R-Sq(adj) = 76.5%
MS
18.198
0.689
F
26.40
P
0.000
Regression Analysis: Tab10-5y2 versus Tab10-5x1, Tab10-5x3, ...
The regression equation is
Tab10-5y2 = 244 - 0.633 Tab10-5x1 + 0.454 Tab10-5x3 + 0.176 Tab10-5x4
+ 11.2 Tab10-5x8 - 236 Tab10-5x9
Predictor
Constant
Tab10-5x1
Tab10-5x3
Tab10-5x4
Tab10-5x8
Tab10-5x9
Coef
244.4
-0.6329
0.4540
0.1758
11.175
-235.7
S = 3.52081
SE Coef
123.6
0.3441
0.1497
0.1034
3.225
100.3
R-Sq = 67.7%
Analysis of Variance
Source
DF
SS
Regression
5
882.03
Residual Error 34
421.47
Total
39 1303.50
T
1.98
-1.84
3.03
1.70
3.47
-2.35
P
0.056
0.075
0.005
0.098
0.001
0.025
R-Sq(adj) = 62.9%
MS
176.41
12.40
F
14.23
P
0.000
10-20
Chapter 10 Exercise Solutions
10-19 continued
Stat > Control Charts > Variables Charts for Individuals > Individuals
I-MR Chart of y1 Regression Model Residuals (Ex10-19Res1)
1
Individual V alue
2
U C L=2.105
1
_
X=-0.000
0
2
-1
2
-2
LC L=-2.105
4
8
12
16
20
24
O bser vation
28
32
36
40
1
M oving Range
3
U C L=2.586
2
__
M R=0.791
1
0
LC L=0
2
4
8
12
16
20
24
O bser vation
28
32
36
40
Test Results for I Chart of Ex10-19Res1
TEST
Test
TEST
Test
1. One point more than 3.00 standard deviations from center line.
Failed at points: 25
2. 9 points in a row on same side of center line.
Failed at points: 10, 11
Test Results for MR Chart of Ex10-19Res1
TEST
Test
TEST
Test
1. One point more than 3.00 standard deviations from center line.
Failed at points: 26
2. 9 points in a row on same side of center line.
Failed at points: 11
10-21
Chapter 10 Exercise Solutions
10-19 continued
I-MR Chart of y2 Regression Model Residuals (Ex10-19Res2)
Individual V alue
8
1
1
5
U C L=6.52
5
4
_
X=-0.00
0
-4
5
LC L=-6.52
-8
4
8
12
16
20
24
O bser vation
28
32
36
40
1
M oving Range
8
U C L=8.02
6
4
__
M R=2.45
2
0
LC L=0
4
8
12
16
20
24
O bser vation
28
32
36
40
Test Results for I Chart of Ex10-19Res2
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 7, 18
TEST 5. 2 out of 3 points more than 2 standard deviations from center line (on
one side of CL).
Test Failed at points: 19, 21, 25
TEST 6. 4 out of 5 points more than 1 standard deviation from center line (on
one side of CL).
Test Failed at points: 7, 21
Test Results for MR Chart of Ex10-19Res2
TEST 1. One point more than 3.00 standard deviations from center line.
Test Failed at points: 26
For response y1, there is not a significant difference between control charts for residuals
from either the full regression model (Figure 10-10, no out-of-control observations) and
the subset regression model (observation 25 is OOC).
For response y2, there is not a significant difference between control charts for residuals
from either the full regression model (Exercise 10-18, observations 7 and 18 are OOC)
and the subset regression model (observations 7 and 18 are OOC).
10-22
Chapter 10 Exercise Solutions
10-20.
Use  = 0.1 and L = 2.7.
Stat > Control Charts > Time-Weighted Charts > EWMA
EWMA Chart of y1 Regression Model Residuals (Ex10-19Res1)
0.5
+2.7SL=0.435
0.4
0.3
EWMA
0.2
0.1
_
_
X=-0.000
0.0
-0.1
-0.2
-0.3
-0.4
-2.7SL=-0.435
4
8
12
16
20
24
Sample
28
32
36
40
EWMA Chart of y2 Regression Model Residuals (Ex10-19Res2)
2.0
1.5
+2.7SL=1.347
EWMA
1.0
0.5
_
_
X=-0.000
0.0
-0.5
-1.0
-2.7SL=-1.347
-1.5
4
8
12
16
20
24
Sample
28
32
36
40
Test Results for EWMA Chart of Ex10-19Res2
TEST. One point beyond control limits.
Test Failed at points: 21, 22
The EWMA control chart for residuals from the response y1 subset model has no out-ofcontrol signals. However the chart for y2 residuals still indicates a problem beginning
near observation 20. A potential advantage to using the EWMA control chart for
residuals from a regression model is the quicker detection of small shifts in the process.
10-23
Chapter 10 Exercise Solutions
10-21.
(a)
Stat > Multivariate > Principal Components
Note: To work in standardized variables in MINITAB, select Correlation Matrix.
Note: To obtain principal component scores, select Storage and enter columns for
Scores.
Principal Component Analysis: Ex10-21X1, Ex10-21X2, Ex10-21X3, Ex10-21X4
Eigenanalysis of the Correlation Matrix
Eigenvalue 2.3181 1.0118 0.6088 0.0613
Proportion
0.580
0.253
0.152
0.015
Cumulative
0.580
0.832
0.985
1.000
Variable
Ex10-21X1
Ex10-21X2
Ex10-21X3
Ex10-21X4
PC1
0.594
0.607
0.286
0.444
PC2
-0.334
-0.330
0.794
0.387
PC3
0.257
0.083
0.534
-0.801
PC4
0.685
-0.718
-0.061
0.104
Principal Component Scores
Ex10-21z1
0.29168
0.29428
0.19734
0.83902
3.20488
0.20327
-0.99211
-1.70241
-0.14246
-0.99498
0.94470
-1.21950
2.60867
-0.12378
-1.10423
-0.27825
-2.65608
2.36528
0.41131
-2.14662
Ex10-21z2
-0.60340
0.49153
0.64094
1.46958
0.87917
-2.29514
1.67046
-0.36089
0.56081
-0.31493
0.50471
-0.09129
-0.42176
-0.08767
1.47259
-0.94763
0.13529
-1.30494
-0.21893
-1.17849
Ex10-21z3
0.02496
1.23823
-0.20787
0.03929
0.12420
0.62545
-0.58815
1.82157
0.23100
0.33164
0.17976
-1.11787
-1.19166
-0.19592
0.01299
-1.31445
-0.11243
0.32286
0.64480
-0.86838
10-24
Chapter 10 Exercise Solutions
10-21 continued
(b)
Graph > Matrix Plot > Simple Matrix of Plots
Matrix Plot of Ex10-21z1, Ex10-21z2, Ex10-21z3
Principal Component Scores
-2
0
2
2
0
Ex10-21z1
-2
2
0
Ex10-21z2
-2
1.9
0.7
Ex10-21z3
-0.5
-2
0
2
-0.5
0.7
1.9
(c) Note: Principal component scores for new observations were calculated in Excel.
See Excel : workbook Chap10.xls : worksheet Ex10-21.
Graph > Matrix Plot > Matrix of Plots with Groups
Matrix Plot of Ex10-21z1all, Ex10-21z2all, Ex10-21z3all
Principal Component Scores
-3.0
-0.5
2.0
Ex10-21Obs
New
Original
4
Ex1 0 -2 1 z1 all
0
-4
2.0
-0.5
Ex1 0 -2 1 z2 all
-3.0
2
Ex1 0 -2 1 z3 all
0
-2
-4
0
4
-2
0
2
Although a few new points are within area defined by the original points, the majority of
new observations are clearly different from the original observations.
10-25
Chapter 10 Exercise Solutions
10-22.
(a)
Stat > Multivariate > Principal Components
Note: To work in standardized variables in MINITAB, select Correlation Matrix.
Note: To obtain principal component scores, select Storage and enter columns for
Scores.
Principal Component Analysis: Ex10-22x1, Ex10-22x2, Ex10-22x3, …, Ex10-22x9
Eigenanalysis of the Correlation Matrix
Eigenvalue 3.1407 2.0730 1.3292 1.0520
Proportion
0.349
0.230
0.148
0.117
Cumulative
0.349
0.579
0.727
0.844
Variable
Ex10-22x1
Ex10-22x2
Ex10-22x3
Ex10-22x4
Ex10-22x5
Ex10-22x6
Ex10-22x7
Ex10-22x8
Ex10-22x9
PC1
-0.406
0.074
-0.465
0.022
-0.436
-0.163
-0.425
-0.120
0.448
PC2
0.204
-0.267
0.050
0.409
-0.372
0.579
-0.407
0.145
-0.238
PC3
-0.357
0.662
-0.000
0.575
0.089
0.108
0.175
0.202
-0.115
PC4
-0.261
-0.199
0.156
-0.200
0.048
0.032
-0.014
0.874
0.247
0.6129
0.068
0.912
PC5
0.068
0.508
0.525
-0.431
-0.277
0.332
-0.127
-0.123
0.240
0.3121
0.035
0.947
PC6
-0.513
-0.380
0.232
0.135
0.262
0.419
0.193
-0.368
0.323
0.2542
0.028
0.975
PC7
0.322
0.166
-0.602
-0.162
0.262
0.529
0.188
0.089
0.297
0.1973
0.022
0.997
PC8
0.467
-0.006
0.256
0.471
0.152
-0.244
-0.105
0.021
0.632
0.0287
0.003
1.000
PC9
0.090
-0.124
-0.018
0.099
-0.651
-0.022
0.723
0.035
0.133
(b)
72.7% of the variability is explained by the first 3 principal components.
(c)
Graph > Matrix Plot > Simple Matrix of Plots
Matrix Plot of Ex10-22z1, Ex10-22z2, Ex10-22z3
Principal Component Scores
-3
0
3
3
0
Ex10-22z1
-3
3
Ex10-22z2
0
-3
2
0
Ex10-22z3
-2
-3
0
3
-2
0
2
10-26
Chapter 10 Exercise Solutions
10-22 continued
(d) Note: Principal component scores for new observations were calculated in Excel.
See Excel : workbook Chap10.xls : worksheet Ex10-22.
Graph > Matrix Plot > Matrix of Plots with Groups
Matrix Plot of Ex10-22z1all, Ex10-22z2all, Ex10-22z3all
All Principal Component Scores
-4
0
4
3
Ex10-22Obs
First
Last
0
Ex1 0 -2 2 z1 all
-3
4
0
Ex1 0 -2 2 z2 all
-4
2
0
Ex1 0 -2 2 z3 all
-2
-3
0
3
-2
0
2
Several points lie outside the area defined by the first 30 observations, indicating that the
process is not in control.
10-27