Question

Samples of n5 units are taken from a process every hour. The and R values for a particular quality char acteristic are determined. After 25 samples have been collected, we calculate 20 and R 4.56 (a) What are the three-sigma control limits for x and R? (b) Both charts exhibit control. Estimate the process standard deviation (c) Assume that the process output is normally dis- tributed. If the specifications are 19 t 5, what are your conclusions regarding the process capability? (d) If the process mean shifts to 24, what is the prob- ability of not detecting this shift on the first sub- sequent sample

Answer 1

Given Data :

sample sapce = 5 and

after 25 samples have been collected. we calculate x bar =20 and R bar 4.56

a) the three control limits for X bar and R bar are:

X-bar n is the number of observations k is the number of subgroups Upper control limit: Lower control limit CL2-R Range Range xmin k is the number of subgroups. Lower control limit Upper control limit

Tabular values for X-bar and range charts Subgroup Size 3.268 2.574 2.282 2.114 2.004 1.924 1.864 1.816 1.880 1.128 1.023 1.693 0.729 2.059 0.577 2.326 0483 2.534 419 2.704 0.076 0.373 .847 0.136 0.337 2.970 0.184 0.308 3078 0.223 0.285 3.173 0.256 0.266 3.258 0.283 0.249 3.336 0.307 0.235 3.4070.328 0.223 3.472 0.347 0.212 3.532 0.363 0.203 3.588 0.378 0.194 3.640 0.391 0.187 3.689 0.403 0.180 3.735 0.415 0.1733.778 0.425 0.167 3.819 0.434 0.162 3.858 0.443 0.157 3.895 0,451 0.153 3.931 0,459 10 1.744 12 13 1.693 1.672 1.653 1.637 1.622 1.608 1.597 1.585 1.575 1.566 1.557 1.548 1.541 15 16 17 19 20 21 23 24 25

So,
The three-sigma control limits for xbar is

lower control limit :
20-0.153*4.56 =19.30232

upper control limit:
20+0.153*4.56 = 20.69768

the three-sigma control limits for R is
lower control limit :

0.459*4.56 =2.09304


upper control limit:
1.514*4.56 = 6.90384

  (b)
We use Average of Subgroup Ranges to estimate the process standard deviation:

ts d2

= 4.56/3.931
=1.1600

(c)

the conclusion regarding the process capability by assuming that the process output is normally distributed with the specifications are 19(+ or -)5

because the specificantions are within the range

(d)

The probability of not detecting this shift on the first subsequent sample is
P(X<24) = P((X-mean)/s <(24-20)/1.16001)
=P(Z<3.45)
=0.9997 (from standard normal table)

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