$3.$ Is the manufacturing process Cpk capable?

As the Bright Star Stamping Company of Yantai, China, is pursuing an expansion drive, it stands at a juncture where one essential quality management$-$related problem would lurk down. The main aim of this company is to expand its markets from primarily a local market it currently has into an international platform, even with all the peculiarities that come along with the Chinese business environment. It does that through its synergy with a local company engaged in trading, which is significant due to high$-$quality control $($Seidelson$,2022).$ Recently, the challenge of this task was emphasized by a massive buy order for $8$ million parts with stringent quality management standards. For this order assurance, the company has to prepare a PPAP that will imply needful awareness of measurement system analysis regarding process stability capacity and quality sampling requirements, as well as the probability of making the wrong product regarding strength competence reactions $($Seidelson$,2022).$ The PO states a Cpk of $1\mathrm{.}67$ or higher looks for the process to be controlled and stable in the Quality Plan.

Calculations

To Calculate $\backslash$sigma :

The $70$ samples are divided into $10$ groups of $7$ samples each $($from the $10$ hours of data$)$

To calculate the R$-$bar, use data from Table $2($for Operator $1)$ and Table $3($for Operator $2).$

For group $1,$ the ranges are $0\mathrm{.}6,0\mathrm{.}5,0\mathrm{.}4,0\mathrm{.}3,0\mathrm{.}7$ and $0\mathrm{.}6$

The average range for group $1$ is $(0\mathrm{.}6+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}3+0\mathrm{.}7+0\mathrm{.}6)/6=0\mathrm{.}5$

For $10$ groups, the average ranges are

$$ Group $1$: $0\mathrm{.}5$

$$ Group $2$: $0\mathrm{.}5$

$$ Group $3$: $0\mathrm{.}4$

$$ Group $4$: $0\mathrm{.}6$

$$ Group $5$: $0\mathrm{.}7$

$$ Group $6$: $0\mathrm{.}5$

$$ Group $7$: $0\mathrm{.}4$

$$ Group $8$: $0\mathrm{.}5$

$$ Group $9$: $0\mathrm{.}5$

$$ Group $10$: $0\mathrm{.}5$

The average of these $10$ averages is R$-$bar $=(0\mathrm{.}5+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}6+0\mathrm{.}7+0\mathrm{.}5+0\mathrm{.}4+0\mathrm{.}5+0\mathrm{.}5+0\mathrm{.}5)/10=0\mathrm{.}51$ mm

The sample size, n$,$ is $7$

From Table $5,$ the d$2$ constant for n$=7$ is $2\mathrm{.}704$ Table $2.$ Operator # $1$ Process Study $($mm$)$

Therefore, $\backslash$sigma $=$ R$-$bar$/$d$2$ Sample $1$ Sample $2$ Sample $3$ Sample $4$ Sample $5$ Sample $6$ Sample $7$

$=0\mathrm{.}51$ mm $/2\mathrm{.}704$ Hour $13030\mathrm{.}230\mathrm{.}33030\mathrm{.}230\mathrm{.}630$

$=0\mathrm{.}189$ mm Hour $230\mathrm{.}530\mathrm{.}43030\mathrm{.}33030\mathrm{.}330$

$=0\mathrm{.}19$ mm Hour $330\mathrm{.}230\mathrm{.}230\mathrm{.}43030\mathrm{.}13030$

To calculate the x$-$bar: Hour $430\mathrm{.}23030\mathrm{.}4303030\mathrm{.}630$

The sum of all $70$ measurements is: Hour $53030\mathrm{.}730\mathrm{.}63030\mathrm{.}33030$

Table $2$: $1056\mathrm{.}5$ mm Average $30\mathrm{.}1830\mathrm{.}3030\mathrm{.}3430\mathrm{.}0630\mathrm{.}1230\mathrm{.}3030\mathrm{.}00$

Table $3$: $1055\mathrm{.}6$ mm

Total $=1056\mathrm{.}5+1055\mathrm{.}6$ Table $3.$ Operator # $2$ Process Study $($mm$)$

$=2112\mathrm{.}1$ mm Sample $1$ Sample $2$ Sample $3$ Sample $4$ Sample $5$ Sample $6$ Sample $7$

The number of samples, n$,$ is $70$ Hour $130\mathrm{.}0030\mathrm{.}1030\mathrm{.}3030\mathrm{.}0030\mathrm{.}2030\mathrm{.}6030\mathrm{.}10$

Therefore, x$-$bar $=$ total$/$n $=2112\mathrm{.}1/70$ Hour $230\mathrm{.}5030\mathrm{.}4030\mathrm{.}0030\mathrm{.}4030\mathrm{.}0030\mathrm{.}3030\mathrm{.}00$

$=30\mathrm{.}17$ mm Hour $330\mathrm{.}1030\mathrm{.}0030\mathrm{.}0030\mathrm{.}0030\mathrm{.}1030\mathrm{.}1030\mathrm{.}00$

The specification limits given are: Hour $430\mathrm{.}2030\mathrm{.}0030\mathrm{.}4030\mathrm{.}0030\mathrm{.}0030\mathrm{.}2030\mathrm{.}00$

LSL $=29\mathrm{.}5$ mm Hour $530\mathrm{.}0030\mathrm{.}6030\mathrm{.}6030\mathrm{.}1030\mathrm{.}2030\mathrm{.}0030\mathrm{.}10$

USL $=30\mathrm{.}5$ mm Mean $30\mathrm{.}160030\mathrm{.}220030\mathrm{.}260030\mathrm{.}100030\mathrm{.}100030\mathrm{.}240030\mathrm{.}0400$

For Cpk calculation

Cpk $=$ min $[($USL $-$ Mean$)/(3$Sigma$),($Mean $-$ LSL$)/(3$Sigma$)]$

USL $-$ Mean $=30\mathrm{.}5-30\mathrm{.}17$ mm

$=0\mathrm{.}33$ mm

Mean $-$ LSL $=30\mathrm{.}17-29\mathrm{.}5$

$=0\mathrm{.}67$ mm

$3*$Sigma $=3*0\mathrm{.}165$ Taking average SD $(0\mathrm{.}16+0\mathrm{.}17)/2$

$=0\mathrm{.}495$ mm

Therefore:

$($USL $-$ Mean$)/(3*$Sigma$)$

$=0\mathrm{.}33/0\mathrm{.}495$

$=0\mathrm{.}667$

$($Mean $-$ LSL$)/(3*$Sigma$)$

$=0\mathrm{.}67/0\mathrm{.}495$

$=1\mathrm{.}3535$

Taking the minimum, Cpk $=0\mathrm{.}667$

$4.$ Based on the calculated Cpk$,$ on a six sigma scale, what is the sigma of this process?

Give me proper calculation on how it is calculated