Question

C-Spec, Inc., is attempting to determine whether an existing machine is capable of milling an engine part that has a key specification of {eq}4 \pm 0.003 {/eq} inches. After a trial run on this machine, C-Spec has determined that the machine has a sample mean of 4.001 inches with a standard deviation of 0.002 inch.

a. Calculate the {eq}C_{pk} {/eq} for this machine
b. Should C-Spec use this machine to produce this part? Why?

Answer 1

Explanation:

Given that :

sample mean = 4.001 inches

sample standard deviation = 0.002 inch

a.

`C_(pk) = min [((USL- \bar X))/((3*std \ dev)) \ ; \ ( (\bar X- LSL))/((3*std \ dev))]`

specification =
`4 \pm 0.003`

Upper specification limit USL = 4 + 0.003 = 4.003

Lower specification limit LSL = 4 - 0.003 = 3.997

`( (\bar X- LSL))/((3*std \ dev)) = ( (4.001-3.997))/((3*0.002))`
`= 0.667`

`((USL- \bar X))/((3*std \ dev)) =((4.003-4.001))/((3*0.002))`
`= 0.333`

Thus ;

`C_(pk) =`
`min (0.333 , 0.667)` = 0.333

`C_(pk)` is a measure of closeness to one's target and the consistency around the average performance.

b) No, C - spec should not use this machine to produce this part because
`C_(pk)` < 1.33 which typical means that the part is not fully capable of hitting the target specification on a consistent basis .