You want to develop 3 sigma level process control charts to check the weight of

Question

You want to develop 3 sigma level process control charts to check the weight of boxes that contain trash. Due to different content of trash, weights are varying. You performed 4 samplings. Each time, they measured 3 units (n = 3). Draw control charts and make conclusion whether things are under control.   

Sampling 1

Sampling 2

Sampling 3

Sampling 4

20

28

24

28

24

30

28

20

26

24

26

26

The number of weekly customer complaint is monitored in a small hotel using a c-chart. Develop three sigma level control charts using the data below. Calculate and show values for UCL, CL, and LCL. Is everything under control? Discuss your conclusion.

week

1

2

3

4

5

# of complaint

3

2

3

1

3

A small town is concerned about public reaction to its new high voltage electric tower project that is currently in progress. In order to measure resident’s perception, a weekly survey is conducted. Each time, a sample of 100 residents is surveyed. The results to date are shown below. The P-chart can be used. Draw the graph and plot data. Is everything under control? Analyze data using 5% risk of Type I error (z=1.96).

ABC sells no-spill syrup in bottles. Management is evaluating four different machines with standard deviations of .006, .01, .022, and .04 ounces for Machines A, B, C, and D, respectively. The specification limit is s 11.95 and 12.05 ounces. Which machines should receive more detailed evaluation (which machines are statistically capable)?

Sampling 1

Sampling 2

Sampling 3

Sampling 4

20

28

24

28

24

30

28

20

26

24

26

26

Explanation / Answer

Answer-3

The capability of the machine can be found by using this formula:

Capability = Cp = (UCL-LCL)/6*sigma

where UCL is upper control limit

LCL is lower control limit

sigma is standard deviation

Here for machine A

UCL = 12.05 ; LCL = 11.95 ; sigma = 0.006

Cp = 2.78

For machine B

Cp = 1.67

For machine C

Cp = 0.758

For machine D

Cp = 0.417

As the capability score of machine D is least it requires more detailed eveluation.

Answer-1 Sampling-1 Sampling-2 Sampling-3 Sampling-4 20 28 24 28 24 30 28 20 26 24 26 26 Mean 23.3333333 27.3333333 26 24.66666667 Range 6 6 4 8 Overall mean = 25.3333333 n = 3 overall range = 6 Therefore UCL = 43.3333333 and LCL = 7.33333333 Answer-2 Mean complaints = 2.4 at the z value of 1.96 we have probability of complaints = 0.9750021 and standard deviation = 0.89442719 Therefore control limit = 3.29442719 Lower control limit 1.50557281

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