Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
12th Edition
ISBN: 9781259580093
Author: William J Stevenson
Publisher: McGraw-Hill Education
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Chapter 10, Problem 2.2CQ
Summary Introduction

To determine: The difference shown by the second set of sample from the first one.

Introduction

Company TT is a division of company DM. It was about to launch a new product. Ms. MY, the production manager, asked her assistant JM to check the capability of the oven. JM collected 18 random samples from each of the 20 pieces. After the analysis, the samples proved to be incapable based on the specification width of 1.44 cm.

Table 1

Sample Mean Range
1 45.01 0.85
2 44.99 0.89
3 45.02 0.86
4 45 0.91
5 45.04 0.87
6 44.98 0.9
7 44.91 0.86
8 45.04 0.89
9 45 0.85
10 44.97 0.91
11 45.11 0.84
12 44.96 0.87
13 45 0.86
14 44.92 0.89
15 45.06 0.87
16 44.94 0.86
17 45 0.85
18 45.03 0.88

Quiet disappointed with the end results, the manager was figuring out ways to improve the process and free the capital expenditure of $10,000. A former professor suggested going for more samples with less sample sizes. JM conducted the analysis on 27 samples of 5 observations each and the results are tabulated below:

Table 2

Sample Mean Range
1 44.96 0.42
2 44.98 0.39
3 44.96 0.41
4 44.97 0.37
5 45.02 0.39
6 45.03 0.4
7 45.04 0.39
8 45.02 0.42
9 45.08 0.38
10 45.12 0.4
11 45.07 0.41
12 45.02 0.38
13 45.01 0.41
14 44.98 0.4
15 45 0.39
16 44.95 0.41
17 44.94 0.43
18 44.94 0.4
19 44.87 0.38
20 44.95 0.41
21 44.93 0.39
22 44.96 0.41
23 44.99 0.4
24 45 0.44
25 45.03 0.42
26 45.04 0.38
27 45.03 0.4

Expert Solution & Answer
Check Mark

Answer to Problem 2.2CQ

The second sample reveals the changes in the process more clearly than the first set of data.

Explanation of Solution

Given information:

Table 3

Sample Mean Range
1 45.01 0.85
2 44.99 0.89
3 45.02 0.86
4 45 0.91
5 45.04 0.87
6 44.98 0.9
7 44.91 0.86
8 45.04 0.89
9 45 0.85
10 44.97 0.91
11 45.11 0.84
12 44.96 0.87
13 45 0.86
14 44.92 0.89
15 45.06 0.87
16 44.94 0.86
17 45 0.85
18 45.03 0.88

Formula:

Mean Chart:

UCL=X¯¯+A2R¯LCL=X¯¯-A2R¯

Difference shown by the second set of sample from the first one:

Date set:1

Table 4

Sample Mean Range
1 45.01 0.85
2 44.99 0.89
3 45.02 0.86
4 45 0.91
5 45.04 0.87
6 44.98 0.9
7 44.91 0.86
8 45.04 0.89
9 45 0.85
10 44.97 0.91
11 45.11 0.84
12 44.96 0.87
13 45 0.86
14 44.92 0.89
15 45.06 0.87
16 44.94 0.86
17 45 0.85
18 45.03 0.88
45 0.872777778

Excel worksheet:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 2.2CQ , additional homework tip  1

From factors of three-sigma chart, for n=20, A2 = 0.18; D3 = 0.41; D4 = 1.59

Mean control chart:

X¯¯=i=118X¯n=809.9818=45

Range control chart:

R¯=i=118Rn=15.7118=0.873

Upper control limit:

UCL=X¯¯+A2R¯=45+(0.18×0.873)=45.156

The upper control limit is calculated by adding the product of 0.18 and 0.873 with 45, which yields 45.156.

Lower control limit:

UCL=X¯¯-A2R¯=45(0.18×0.873)=44.842

The lower control limit is calculated by subtracting the product of 0.18 and 0.873 with 45, which yields 44.842.

A graph is plotted using the UCL, LCL and samples values.

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 2.2CQ , additional homework tip  2

Diagram 1

Date set: 2

Table 5

Sample Mean Range
1 44.96 0.42
2 44.98 0.39
3 44.96 0.41
4 44.97 0.37
5 45.02 0.39
6 45.03 0.4
7 45.04 0.39
8 45.02 0.42
9 45.08 0.38
10 45.12 0.4
11 45.07 0.41
12 45.02 0.38
13 45.01 0.41
14 44.98 0.4
15 45 0.39
16 44.95 0.41
17 44.94 0.43
18 44.94 0.4
19 44.87 0.38
20 44.95 0.41
21 44.93 0.39
22 44.96 0.41
23 44.99 0.4
24 45 0.44
25 45.03 0.42
26 45.04 0.38
27 45.03 0.4
44.9959 0.40111

Excel worksheet:

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 2.2CQ , additional homework tip  3

From factors of three-sigma chart, for n=20, A2 = 0.58

Mean control chart:

X¯¯=i=127X¯n=1214.8927=44.99

Range control chart:

R¯=i=127Rn=10.8327=0.4011

Upper control limit:

UCL=X¯¯+A2R¯=44.99+(0.58×0.401)=45.229

The upper control limit is calculated by adding the product of 0.58 and 0.401 with 44.99, which yields 45.229.

Lower control limit:

UCL=X¯¯-A2R¯=44.99(0.58×0.401)=44.763

The lower control limit is calculated by subtracting the product of 0.58 and 0.401 with 44.99, which yields 44.763.

Loose-leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 10, Problem 2.2CQ , additional homework tip  4

Diagram 2

On comparing Diagrams 1 and 2, it is evident that the second set of data has closer range of changes while the first set of data is scattered and reveals no information about the changes in the process.

Hence, the second sample reveals the changes in the process changes more clearly than the first set of data.

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Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of x? x= 156.76 mm (round your response to two decimal places). b) What is the value of R? Day 1 2 3 4 5 Mean x (mm) 158.9 155.2 155.6 157.5 156.6 R = 4.40 mm (round your response to two decimal places). c) What are the UCL and LCL using 3-sigma? Upper Control Limit (UCL) = mm (round your response to two decimal places). Range R (mm) 4.2 4.4 4.3 4.8 4.3 Ç
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