4) (2 points) A machining process has a required dimension on a part of 0.515 ±

Question

4)   (2 points) A machining process has a required dimension on a part of 0.515 ± 0.007 inches. Twenty-five parts were measured.

a)   How far is the process mean from the target, measured in standard deviations?

b)   What is the sigma-level of the process?

c)   Is the process capable of producing within acceptable limits?

d)   If the process were centered at the target, would the process be capable?

C1 1 Problem 4. 2 Machining Process -I 5 0.497 0.519 0.517 0.504 0.512 6 0.506 0.518 0.514 0.513 0.504 7 0.517 0.517 0.513 0.519 0.520 8 0.508 0.522 0.515 0.513 0.518 9 0.505 0.516 0.516 0.512 0.5 14 T Problem 3 Problem 4 Problem 5 Chapter 4 Datasets Excel AT A Arial

Explanation / Answer

From the given data of a sample of 25 measurements,

sample average = 0.5124 and sample standard deviation = 0.0049825.

This is standard deviation for the individual measurement. So, standard deviation for the sample average = 0.0049825/5 = 0.0009785 [5 being the square root of sample size]

The given requirement (also known as specification limits) is 0.515 ± 0.007.

This means: target is 0.515 and tolerance is ± 0.007, i.e., total tolerance = 2 x 0.007 = 0.014.   

a)   How far is the process mean from the target, measured in standard deviations?

Process mean = sample average = 0.5124 => it is 0.0026 far from the target. 0.0026 is 2.66 times the standard deviation for the sample average.

So, the answer is 2.66.

b)   What is the sigma-level of the process?

Sigma-level of the process = total tolerance/sample standard deviation = ± 1.43 sigma level.

c)   Is the process capable of producing within acceptable limits?

Process is capable of producing within acceptable limits if

6 times the sample standard deviation < total tolerance.

In the given case, 6 times the sample standard deviation = 0.0294 > 0.014.

So, process is not capable of producing within acceptable limits

d)   If the process were centered at the target, would the process be capable?

           So far as 6 times the sample standard deviation > total tolerance, by centering the process at the target will not alter the process capability. In this case, by centering the process at the target will bring down the proportion of units above the upper limit, but then the proportion of units below the lower limit will increase.

DONE

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