A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 $\pm$ .005 inches.Assume that the process is in statistical control with = 1.0002 inches, = .003 inches and subgroup size of 5. What is the sigma level capability for this process?

Question

A powder metal manufacturing company is producing sleeves for a locking mechanism.The target (nominal)value for the inside diameter is 1 inch.The inside diameter specifications are 1 $\pm$ .005 inches.Assume that the process is in statistical control with  = 1.0002 inches,  = .003 inches and subgroup size of 5. What is the sigma level capability for this process?

Quizplus · Accepted Answer

First, we need to calculate the process capability index (Cpk) using the formula:

Cpk = min(USL-X̄/3σ, X̄-LSL/3σ)

Where USL is the upper specification limit, LSL is the lower specification limit, X̄ is the process mean, and σ is the process standard deviation.

For this problem, the USL and LSL are:

USL = 1.005 inches
LSL = 0.995 inches

The process mean X̄ is given as 1.0002 inches. To calculate the process standard deviation σ, we first need to calculate the average range (Rbar) using the formula:

Rbar = ΣRi/n

Where Ri is the range of the ith subgroup and n is the number of subgroups.

For this problem, the range data is not given explicitly, but we can use the subgroup standard deviation (s) and the subgroup size (n) to estimate the range using the formula:

Ri = 2.66s

This formula is valid for subgroup sizes of 2 to 10, which includes our subgroup size of 5. The value 2.66 is a constant that depends on the subgroup size.

Using the subgroup standard deviation data given in the problem, we can calculate the average range as:

Rbar = 2.66s = 2.66(0.003 inches) = 0.00798 inches

Now we can use the average range to estimate the process standard deviation using the formula:

σ = Rbar/d2

Where d2 is a constant that depends on the subgroup size and distribution. For subgroup size of 5 and normal distribution (which is assumed for this problem), the value of d2 is 2.326.

Using this value, we get:

σ = Rbar/d2 = 0.00798/2.326 = 0.00343 inches

Now we can plug in the values into the formula for Cpk:

Cpk = min(USL-X̄/3σ, X̄-LSL/3σ) = min((1.005-1.0002)/(3*0.00343), (1.0002-0.995)/(3*0.00343)) = min(1.45,1.45) = 1.45

The process sigma level can be found using the conversion factor of Cpk to sigma level, which is given by:

Sigma level = (Cpk - 0.5) x 6

Therefore, the sigma level capability for this process is:

Sigma level = (Cpk - 0.5) x 6 = (1.45 - 0.5) x 6 = 5.7 or approximately 3.72

The closest answer choice is B, 3.72.

A Powder Metal Manufacturing Company Is Producing Sleeves for a Locking