Determine the upper and lower control limits

Assignment Help Business Management

Reference no: EM1378423

As the production manager for an electronic circuit company you have encountered the following situation.

Part 1

A procedure for manufacturing electronic circuits has achieved very high yield levels. An average of only 10 defective parts per million is currently manufactured.

1.What are the upper and lower control limits for a sample size of 100?

2.Recompute the upper and lower control limits for a sample size of 10,000?

3.Which of these two sample sizes would you recommend? Explain.

In part one I am suppose to address the attribute control, the formulae are the UCL and the LCL. The formulas I have are:
UCL =pbar + 3Sqrt(pbar(1-pbar)/n
LCL = PBAR - 3Sqrt(Pbar(1-pbar)/n
I also noted that the pbar gets a certain percentage. I prefer to use the decimal numbers in this calculation.
So far I have Pbar is 10 defects per million, the past defect rate.
10/1,000,000=00001
I need to make sure that I carry 7 decimal spaces for the attribute control responses.

This was an example from another problem given to me to use as a format, I dont understand it, but this is the way I am suppose to answer my problems.
Suppose samples of 200 records are taken from a data entry operations at 2 hr intervals to control the data entry process. The percentage of records in error for past 11 samples is found to be 5, 1.0,1.5,2.0,1.5,1.0,1.5 and 2.0 percent. The average of these sample % yields a p=1.27%, which is the center line of the control chart.
Attribute Control Chart (example of a solution I am to follow for part 1 )
(.5 + 1.0 + 1.5 +2.0 +1.5 + 1.0 +1.5 +.5 +1.0 + 1.5 + 2.0)11=1.27%.
1.27% = .0127 decimal, average of all samples.
UCL = pbar + 3Sqrt(pbar(1-pbar)/n)
UCL = .0127 + 3Sqrt(.0127(.9873)/200
UCL = .0127 + 3Sqrt(.00006269355)
UCL = .0127 + 2.3753775943E-2
UCL = .0364
LCL = pbar - 3Sqrt(pbar(1-pbar)/N)
LCL= .0127 - 23753775943e-2
CLC = -.0110
I really need some assistance trying to answer part 1 questions.

Part 2
(This is base on variable controls)
Management has reconsidered the method of quality control and has decided to use process control by variables instead of attributes. For variables control a circuit voltage will be measured based on a sample of only five circuits. The past average voltage for samples of size 5 has been 3.1 volts, and the range has been 1.2 volts.

1.What would the upper and lower control limits be for the resulting control charts (average and range)?

2.Five samples of voltage are taken with the results in the table below. What action should be taken if any?

3.Discuss the pros and cons of using this variables control chart versus the control chart discussed in the first part of the assignment. Which do you prefer?

Sample 1 2 3 4 5
x 3.6 3.3 2.6 3.9 3.4
R 2.0 2.6 0.7 2.1 2.3

I also have an example for the variable control that I am trying to use as a guide.
It starts with midwest that control bolts produced by its automatic screw machine. each machine produces 100 bolts and control by separate control charts. Every hour a random sample of six bolts is selected and the diameter of each sample bolt is measured. From each six diameters, an average and range are computed. One sample produced the following six measurements .536, .507, .530, .525, .530, and .520. The average of these measurements is x=.525 and the range is R =.029. All samples have average x=.513 and the grand average range is R =.020.

Average Chart Formulas
UCL = xdoublebar + A2Rbar
LCL = xdoublebar - A2Rbar
Range Chart Formulas
UCL = D4Rbar
LCL = D3Rbar

Variable control Range Chart
2.004 is D4 for a sample size of 6.
0.000 is D3 FOR A SAMPLE SIZE OF 6.
.020 is grand average Range.
UCL = D4Rbar
UCL = 2.004(.020) =.040
LCL = D3Rbar
LCL = 0(.020)=0

Varible Control Average Chart
.513 is grand average of all past samples.
.438 is A2 for a sample size of 6.
.020 is grand average Range
UCL =xdoublebar + A2Rbar
LCL = xdoublebar - A2Rbar
LCL = .513 - .483(.020) = .503

This is the format I am suppose to use to part 2, I really need help with this.

I also need an conclusion are recommendation suggestion.

Reference no: EM1378423

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