Chapter 7 TN

2. a.

= .889

b. The process is potentially capable of producing the desired quality, but at present it is not capable due to the process center. The process center should be adjusted (back to target of 1).

3.  a. Ten defectives were found in 10 samples of size 15.

= .067

= .0645

UCL = + 1.96 = .067 + 1.96(.0645) = .194

UCL = - 1.96 = .067 - 1.96(.0645) = -.060 zero

b.  Stop the process and look for the special cause, it is out of statistical control.

6.

Sample / 1 / 2 / 3 / 4 / mean / range
1 / 1010 / 991 / 985 / 986 / 993.00 / 25
2 / 995 / 996 / 1009 / 994 / 998.50 / 15
3 / 990 / 1003 / 1015 / 1008 / 1004.00 / 25
4 / 1015 / 1020 / 1009 / 998 / 1010.50 / 22
5 / 1013 / 1019 / 1005 / 993 / 1007.50 / 26
6 / 994 / 1001 / 994 / 1005 / 998.50 / 11
7 / 989 / 992 / 982 / 1020 / 995.75 / 38
8 / 1001 / 986 / 996 / 996 / 994.75 / 15
9 / 1006 / 989 / 1005 / 1007 / 1001.75 / 18
10 / 992 / 1007 / 1006 / 979 / 996.00 / 28
11 / 996 / 1006 / 997 / 989 / 997.00 / 17
12 / 1019 / 996 / 991 / 1010 / 1004.00 / 28
13 / 981 / 991 / 989 / 1003 / 991.00 / 22
14 / 999 / 993 / 988 / 984 / 991.00 / 15
15 / 1013 / 1002 / 1005 / 992 / 1003.00 / 21

999.1, = 21.733

Control limits for X-bar chart:

UCL, LCL = , = 999.1 + .73(21.733) = 1014.965, 983.235

Control limits for R chart:

UCL = = 2.28(21.7333) = 49.551

LCL = = 0(21.7333) = 0.00

The process is in statistical control.

8. a. = .06

= .0075

UCL = + 2 = .06 + 2(.0075) = .075

UCL = - 2 = .06 - 2(.0075) = .045

b. The chart indicates that the process is out of control. The administrator should investigate the quality of the patient meals.

14. a. = .8333

b. = .1667

c. Many defects will be produced. Assuming a normal distribution, the left tail is z = (92 - 90)/4 = .50, which corresponds to a probability of .1915. The right tail is z = (110-92)/4 = 4.5, which approximately corresponds to a probability of .5000. Therefore, .6915 (.1915 + .5000) are inside the specifications, while 1- .6915 = .3085 or approximately 31% are outside the specification limits.

Chapter 9

1.

Plastic / Year 1 / Year 2 / Year 3 / Year 4
Demand for plastic sprinklers / 97 / 115 / 136 / 141
Percentage of capacity used / 48.5% / 57.5% / 68.0% / 70.5%
Machine requirements / .485 / .575 / .680 / .705
Labor requirements / 1.94 / 2.30 / 2.72 / 2.82
Bronze / Year 1 / Year 2 / Year 3 / Year 4
Demand for bronze sprinklers / 21 / 24 / 29 / 34
Percentage of capacity used / 58.3% / 66.7% / 80.6% / 94.4%
Machine requirements / 1.75 / 2.00 / 2.42 / 2.83
Labor requirements / 3.50 / 4.00 / 4.84 / 5.66


2. Requirements for plastic remain unchanged.

Bronze / Year 1 / Year 2 / Year 3 / Year 4
Demand for bronze sprinklers / 32 / 36 / 41 / 52
Percentage of capacity used / 88.9% / 100.0% / 113.9% / 144.4%
Machine requirements / 2.67 / 3 / 3.42 / 4.33
Labor requirements / 5.33 / 6 / 6.84 / 8.67

It is obvious that not enough capacity is available after year two to meet the increased demand. AlwaysRain will have to consider purchasing additional machines for the bronze operations.

Chapter 11

16.  a. FSeptember = (170 + 180 + 140)/3 = 163.3

b. FSeptember = .50(170) + .30(180) + .20(140) = 167.0

a.  FJuly = FJune + a(AJune – FJune) = 130 + .3(140 - 130) = 133.00

FAugust = FJuly + a(AJuly – FJuly) = 133.00 + .3(180 – 133.00) = 147.10

FSeptember = FAugust + a(AAugust – FAugust) = 147.10 + .3(170 – 147.10) = 153.97

19.

Period / Forecast / Actual / Deviation / RSFE / Absolute deviation / Sum of Absolute deviations / MAD / TS
1 / 1500 / 1550 / 50 / 50 / 50 / 50 / 50.0 / 1.00
2 / 1400 / 1500 / 100 / 150 / 100 / 150 / 75.0 / 2.00
3 / 1700 / 1600 / -100 / 50 / 100 / 250 / 83.3 / 0.60
4 / 1750 / 1650 / -100 / -50 / 100 / 350 / 87.5 / -0.57
5 / 1800 / 1700 / -100 / -150 / 100 / 450 / 90.0 / -1.67


a.  For period 5, the MAD = 90.00, and the TS = -1.67

b.  The model seems acceptable since the tracking signal is 1.67 off the mean and is reasonable. However, the downward trend in the graph does present a concern.

20.

Month (t) / Demand / 3-mo. MA / Absolute deviation / 3-mo WMA / Absolute deviation / Ft / Absolute deviation / Tt / Ft / FITt / Absolute deviation
1 / 62 / 61.00 / 1.80 / 60.00 / 61.80
2 / 65 / 61.30 / 1.82 / 61.86 / 63.68
3 / 67 / 62.41 / 1.94 / 64.07 / 66.02
4 / 68 / 64.67 / 3.33 / 65.40 / 2.60 / 63.79 / 4.21 / 2.03 / 66.31 / 68.33 / 0.33
5 / 71 / 66.67 / 4.33 / 67.10 / 3.90 / 65.05 / 5.95 / 2.00 / 68.23 / 70.23 / 0.77
6 / 73 / 68.67 / 4.33 / 69.30 / 3.70 / 66.84 / 6.16 / 2.07 / 70.46 / 72.53 / 0.47
7 / 76 / 70.67 / 5.33 / 71.40 / 4.60 / 68.68 / 7.32 / 2.11 / 72.67 / 74.78 / 1.22
8 / 78 / 73.33 / 4.67 / 74.10 / 3.90 / 70.88 / 7.12 / 2.22 / 75.14 / 77.36 / 0.64
9 / 78 / 75.67 / 2.33 / 76.40 / 1.60 / 73.02 / 4.98 / 2.28 / 77.55 / 79.83 / 1.83
10 / 80 / 77.33 / 2.67 / 77.60 / 2.40 / 74.51 / 5.49 / 2.11 / 79.28 / 81.39 / 1.39
11 / 84 / 78.67 / 5.33 / 79.00 / 5.00 / 76.16 / 7.84 / 1.99 / 80.98 / 82.96 / 1.04
12 / 85 / 80.67 / 4.33 / 81.60 / 3.40 / 78.51 / 6.49 / 2.08 / 83.27 / 85.35 / 0.35
MAD / 4.07 / 3.46 / 6.17 / 0.89

Based upon MAD, the exponential smoothing with trend component appears to be the best method.

Solutions to Part II 7 7