Solution of Control chart

1. Following data are diameters (in mm) of a type of shaft manufactured by a new machine are obtained from an engineering workshop.

Sample No.
No. of Observation

1
2
3
4
5
1
5.4
5.5
6.0
5.3
5.5
2
5.6
6.0
5.4
5.4
5.5
3
5.9
5.6
5.2
5.2
5.6
4
5.5
5.5
5.9
5.3
5.3
5
5.5
5.4
5.8
5.3
5.2
6
5.4
6.0
5.9
5.3
5.6
7
5.6
5.2
5.3
5.6
5.9
8
5.8
5.2
5.3
5.6
5.5

Construct control charts for the shaft diameter for 3-sigma limit of confidence. Find whether the process is in control or not?

Test with following data and find whether it meets specifications or not.

Sample No.
No. of Observation

Sample No.
No. of Observation

1
2
3
4
5

1
2
3
4
5
Data after 25 months of Installation

Data after 26 months of Installation
1
5.6
5.7
5.4
5.4
5.5

1
5.3
5.6
5.2
5.4
5.6
2
5.5
5.5
5.7
5.3
5.3

2
5.4
5.5
5.3
5.2
5.5
3
5.4
5.4
5.8
5.5
5.6

3
5.5
5.4
5.6
5.3
5.4
Data after 27 months of Installation

Data after 28 months of Installation
1
5.6
5.5
5.4
5.6
5.5

1
5.6
5.5
5.5
5.3
5.3
2
5.4
5.3
5.6
5.3
5.6

2
5.5
5.2
5.3
5.6
5.4
3
5.6
5.2
5.3
5.6
5.3

3
5.8
5.6
5.5
5.4
5.5
Data after 29 months of Installation

1
5.6
5.4
5.4
5.7
5.3

2
5.6
5.3
5.6
5.4
5.5

3
5.8
5.4
5.5
5.6
5.4

Solution:

SN

Mean
Range
1
5.4
5.5
6
5.3
5.5
5.54
0.7
2
5.6
6
5.4
5.4
5.5
5.58
0.6
3
5.9
5.6
5.2
5.2
5.6
5.5
0.7
4
5.5
5.5
5.9
5.3
5.3
5.5
0.6
5
5.5
5.4
5.8
5.3
5.2
5.44
0.6
6
5.4
6
5.9
5.3
5.6
5.64
0.7
7
5.6
5.2
5.3
5.6
5.9
5.52
0.7
8
5.8
5.2
5.3
5.6
5.5
5.48
0.6

5.525
0.65

µ = 5.525, Mean Range, R̅ = 0.65, A2 = 0.577 (from factor table)

UCL = µ + A2 R̅ = 5.525 + 0.577 *0.65 = 5.90

LCL = µ - A2 R̅ = 5.525 - 0.577 *0.65 = 5.15

Mean Range Control Chart is