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 ± .005 inches.Assume that the process is in statistical control with = 1.0002 inches, = .003 inches,and subgroup size of 5.What are the natural tolerance limits for this process? Is the process capable?

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 ± .005 inches.Assume that the process is in statistical control with  = 1.0002 inches,  = .003 inches,and subgroup size of 5.What are the natural tolerance limits for this process? Is the process capable?

Quizplus · Accepted Answer

The natural tolerance limits (NTLs) can be calculated as:
Upper NTL = X̄ + 3σ
Lower NTL = X̄ - 3σ

where X̄ is the mean of the subgroup, and σ is the standard deviation of the subgroup.

Given data:
Target value (μ) = 1 inch
Specifications (USL - LSL) = 1 + 0.005 - 1 = 0.005 inches
Subgroup size (n) = 5
Sample mean (x̄) = 1.0002 inches
Sample range (R) = 0.003 inches

The process is assumed to be in statistical control, so we will use the estimate of sigma based on the range (Rbar/d2) to calculate the natural tolerance limits.

d2 factor for n=5 is 2.326.

Standard deviation (σ) = Rbar/d2 = 0.003 / 2.326 = 0.001288 inches

Upper NTL = 1.0002 + 3(0.001288) = 1.004 theta inches
Lower NTL = 1.0002 - 3(0.001288) = 0.9963 inches

The natural tolerance interval is (0.9963, 1.004) inches.

The entire tolerance specification (0.995, 1.005) is within the natural tolerance limits, so the process is capable. Therefore, the answer is C.

A Powder Metal Manufacturing Company Is Producing Sleeves for a Locking