A filing machine for an infant milk powder manufacturer produces approximately 9
ID: 3143125 • Letter: A
Question
A filing machine for an infant milk powder manufacturer produces approximately 90 milk powder cans per minute. The process manager monitors the weight of potassium that is placed into each 200g can A subgroup of taken every 12 minutes for 10 consecutive time periods. The results are given in Table Q1 Weight in Grams 4.48 429 4.56 4.38 4.50 4.39 4.34 4.40 4.24 4.40 4.27 4.47 4.544.23 4.43 428 4.53 4.35 4.25 4.56 4.30 4.45 4.32 4.59 4.444.39 4.534.34 4.42 4.38 4.27 4.51 4.34 4.44 4.3 4.51 4.52 4.31 4.59 4.24 4.52 431 426 4.56 4.31 4.55 4.24444 4.42 4.38 4.50 425 445 4.30 4.23 4.53 4.37 4.42 4.30 4.60 10 Table Q1-Weight of Potassium in Milk Powder Cans a) Calculate the upper and lower control limits of a Chart for the data (Constants for Upper and Lower Control limits forX Charts are given in the appendox) Worth 10marks) 01S SEPT 30 Page 3 afs b) Plot the and Range charts and hence determine the process is in control. Worth 10marks) c) Explain why it is not necessary to know the underlying distrbton of the process Worth 5 marks) d) New research finds that potassium levels greater than 21% by weight have an impact on infant health. your customer requires a defect rate of less than 3.4 per million cans, and the lowest setting on the machine is 3.6g of potassium per can, is your company capable of producing to the new suggested maximum levels of potassium with existing equipment? Explain your answer Worth 5 marks) EmailSave m SaveExplanation / Answer
Solution
Back-up Theory
Let xij be the jth observation in the ith sub-group; i = 1,2,…..,k; j = 1,2,….n
Average of the ith sub-group, xiBar = (sum of xij over j)/ni
Range of ith sub-group, Ri = Max observation - min onservation of the ith sub-group
Grand Average, xdouble bar = (sum of xi bar over i)/k
Average Range, R bar = (sum of Ri over i)/k
LCL (Lower Control Limit) for x bar chart: x double bar - (A2).(R bar)
UCL (Upper Control Limit) for x bar chart: x double bar + (A2).(R bar)
LCL (Lower Control Limit for R chart): (D3).(R bar)
UCL (Upper Control Limit) for R chart: (D4).(R bar)
Estimate of Standard Deviation, = R bar/d2
Values of A2, D3, D4 and d2 are taken from the Appendix of the given Question.
Now, to work out the problem,
Part (a)
We have n = 6 and k = 10
A2 =
0.483
D3 =
0
D4 =
2.004
d2 =
2.5
From the calculations given in the table below,
x double bar =
4.399
R bar =
0.287
LCL for x bar chart:
4.261
UCL for x bar chart:
4.538
LCL for R chart:
0
UCL for R chart:
0.575
Excel Worksheet
i
xi1
xi2
xi3
xi4
xi5
xi6
xibar
Ri
1
4.45
4.29
4.56
4.38
4.5
4.34
4.42
0.27
2
4.34
4.4
4.24
4.4
4.27
4.47
4.35
0.23
3
4.54
4.23
4.43
4.28
4.53
4.35
4.39
0.31
4
4.25
4.56
4.3
4.45
4.32
4.59
4.41
0.34
5
4.44
4.39
4.53
4.34
4.42
4.38
4.42
0.19
6
4.27
4.51
4.34
4.44
4.31
4.51
4.40
0.24
7
4.52
4.31
4.59
4.24
4.52
4.31
4.42
0.35
8
4.26
4.56
4.31
4.55
4.24
4.44
4.39
0.32
9
4.42
4.38
4.5
4.25
4.45
4.3
4.38
0.25
10
4.23
4.53
4.37
4.42
4.3
4.6
4.41
0.37
Part (b)
Comparing the control limits and the entries in the last two columns of the above table, it is seen that all xibar values and Ri values are within the respective control limits.
=> the process is under control.
Part (c)
It is necessary to know the exact distribution of the variable involved [weight of potassium] because by Central Limit Theorem, average has an asymptotic Normal distribution, irrespective of the distribution of the variable involved.
Part (d)
Given condition that potassium must not be more than 2.1% => X = weight of potassium must not be more than 4.2 gm per can since content in a can is 200g.
Estimate of Standard Deviation, = R bar/d2 = 0.113
At the possible minimum setting of 3.6 gm and a standard deviation of 0.113,
P(X < 4.2) = P[Z < {(4.2 – 3.6)/0.115}], where Z ~ N(0. 1).
= P(Z < 5.0847) = 0.999999816
=> proportion not meeting the limit on potassium is: 0.000000816 which is equivalent to 0.8 per million which is less than the stipulated limit of 3.4 per million.
=> company is capable of producing to the new suggested potassium levels ANSWER
Solution
Back-up Theory
Let xij be the jth observation in the ith sub-group; i = 1,2,…..,k; j = 1,2,….n
Average of the ith sub-group, xiBar = (sum of xij over j)/ni
Range of ith sub-group, Ri = Max observation - min onservation of the ith sub-group
Grand Average, xdouble bar = (sum of xi bar over i)/k
Average Range, R bar = (sum of Ri over i)/k
LCL (Lower Control Limit) for x bar chart: x double bar - (A2).(R bar)
UCL (Upper Control Limit) for x bar chart: x double bar + (A2).(R bar)
LCL (Lower Control Limit for R chart): (D3).(R bar)
UCL (Upper Control Limit) for R chart: (D4).(R bar)
Estimate of Standard Deviation, = R bar/d2
Values of A2, D3, D4 and d2 are taken from the Appendix of the given Question.
Now, to work out the problem,
Part (a)
We have n = 6 and k = 10
A2 =
0.483
D3 =
0
D4 =
2.004
d2 =
2.5
From the calculations given in the table below,
x double bar =
4.399
R bar =
0.287
LCL for x bar chart:
4.261
UCL for x bar chart:
4.538
LCL for R chart:
0
UCL for R chart:
0.575
Excel Worksheet
i
xi1
xi2
xi3
xi4
xi5
xi6
xibar
Ri
1
4.45
4.29
4.56
4.38
4.5
4.34
4.42
0.27
2
4.34
4.4
4.24
4.4
4.27
4.47
4.35
0.23
3
4.54
4.23
4.43
4.28
4.53
4.35
4.39
0.31
4
4.25
4.56
4.3
4.45
4.32
4.59
4.41
0.34
5
4.44
4.39
4.53
4.34
4.42
4.38
4.42
0.19
6
4.27
4.51
4.34
4.44
4.31
4.51
4.40
0.24
7
4.52
4.31
4.59
4.24
4.52
4.31
4.42
0.35
8
4.26
4.56
4.31
4.55
4.24
4.44
4.39
0.32
9
4.42
4.38
4.5
4.25
4.45
4.3
4.38
0.25
10
4.23
4.53
4.37
4.42
4.3
4.6
4.41
0.37
Part (b)
Comparing the control limits and the entries in the last two columns of the above table, it is seen that all xibar values and Ri values are within the respective control limits.
=> the process is under control.
Part (c)
It is necessary to know the exact distribution of the variable involved [weight of potassium] because by Central Limit Theorem, average has an asymptotic Normal distribution, irrespective of the distribution of the variable involved.
Part (d)
Given condition that potassium must not be more than 2.1% => X = weight of potassium must not be more than 4.2 gm per can since content in a can is 200g.
Estimate of Standard Deviation, = R bar/d2 = 0.113
At the possible minimum setting of 3.6 gm and a standard deviation of 0.113,
P(X < 4.2) = P[Z < {(4.2 – 3.6)/0.115}], where Z ~ N(0. 1).
= P(Z < 5.0847) = 0.999999816
=> proportion not meeting the limit on potassium is: 0.000000816 which is equivalent to 0.8 per million which is less than the stipulated limit of 3.4 per million.
=> company is capable of producing to the new suggested potassium levels ANSWER
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