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

Answer:

Step-by-step explanation:

Given that :

sample mean = 4.001 inches

sample standard deviation = 0.002 inch

a.

$C_{pk} = min [\dfrac{(USL- \bar X)}{(3*std \ dev)} \ ; \ \dfrac{ (\bar X- LSL)}{(3*std \ dev)}]$

specification = $4 \pm 0.003$

Upper specification limit USL = 4 + 0.003  = 4.003

Lower specification limit LSL =  4 - 0.003 = 3.997

$\dfrac{ (\bar X- LSL)}{(3*std \ dev)} = \dfrac{ (4.001-3.997)}{(3*0.002)}$  $= 0.667$

$\dfrac{(USL- \bar X)}{(3*std \ dev)} =\dfrac{(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 .