STAT3365 – Exercise 15
1. The data that follow give the number of nonconforming bearing and seal assemblies in
samples of size 100.
Sample
Number of
Sample
Number of
Sample
Number of
Sample
Number of
Number
Nonconforming
Number
Nonconforming
Number
Nonconforming
Number
Nonconforming
Assemblies
Assemblies
Assemblies
Assemblies
1
7
6
8
11
6
16
1
2
4
7
10
12
15
17
4
3
1
8
5
13
0
18
5
4
3
9
2
14
9
19
7
5
6
10
7
15
5
20
12
Construct a fraction nonconforming control chart for these data. If any points plot out
of control, assume that assignable causes can be found and determine the revised control
limits.
Answer.
Summary statistics are first calculated: n = 100, m = 20,
UCL
=
0.0585 + 3 ×
Central Line
=
0.0585
LCL
=
0.0585 − 3 ×
Pm
i =1
D i = 117, p̄ = 117/(100 × 20) = 0.0585.
q
0.0585(1−0.0585)
= 0.1289059
100
q
assign
0.0585(1−0.0585)
= −0.01190595 = 0
100
0.15
0.20
The lower control limit as calculated is a negative value, therefore, zero is chosen for the lower limit.
●
0.10
●
●
●
●
●
●
0.05
●
●
●
●
●
●
●
●
●
●
0.00
●
●
●
5
10
15
20
Sample 12 is out of control, so remove from control limit calculation: n = 100, m = 19,
Pm
i =1
D i = 102, p̄ = 102/(100 ×
19) = 0.05368421.
UCL
=
0.05368421 + 3 ×
Central Line
=
0.05368421
LCL
=
0.05368421 − 3 ×
q
0.05368421(1−0.05368421)
= 0.1213023
100
q
assign
0.05368421(1−0.05368421)
= −0.01393383 = 0
100
0.15
0.20
Again, zero is chosen for the lower control limit.
●
0.10
●
●
●
●
0.05
●
●
●
●
●
●
●
●
●
●
●
0.00
●
●
●
●
5
10
1
15
20
2. The following data represent the results of inspecting all units of a personal computer
produced for the last 10 days.
Day
Units
Nonconforming
Fraction
Inspected
Number
Nonconforming
Day
Units
Nonconforming
Fraction
Inspected
Number
Nonconforming
1
80
4
0.050
2
110
7
0.064
6
120
6
0.050
7
70
4
3
90
5
0.057
0.056
8
125
5
4
75
0.040
8
0.107
9
105
8
5
130
0.076
6
0.038
10
95
7
0.074
Does the process appear to be in control?
Answer.
The summary statistics are m = 10,
1
3
4
5
6
7
8
9
10
i =1
n = 1000,
Pm
i =1
D i = 60, p̄ = 60/1000 = 0.06.
LCL
Sample Number
2
Pm
UCL
Central Line
q
−0.06)
0.06 − 3 × 0.06(180
q
0.06(1−0.06)
0.06 − 3 ×
110
q
−0.06)
0.06 − 3 × 0.06(190
q
0.06(1−0.06)
0.06 − 3 ×
75
q
−0.06)
0.06 − 3 × 0.06(1
130
q
0.06(1−0.06)
0.06 − 3 ×
120
q
−0.06)
0.06 − 3 × 0.06(170
q
−0.06)
0.06 − 3 × 0.06(1
125
q
0.06(1−0.06)
0.06 − 3 ×
105
q
−0.06)
0.06 − 3 × 0.06(195
= 0.1396555
0.06
= 0.1279304
0.06
= 0.1350999
0.06
= 0.1422679
0.06
= 0.1224869
0.06
= 0.1250385
0.06
= 0.1451553
0.06
= 0.1237244
0.06
= 0.1295290
0.06
= 0.1330969
0.06
q
−0.06)
0.06 + 3 × 0.06(180
q
0.06(1−0.06)
0.06 + 3 ×
110
q
−0.06)
0.06 + 3 × 0.06(190
q
0.06(1−0.06)
0.06 + 3 ×
75
q
−0.06)
0.06 + 3 × 0.06(1
130
q
0.06(1−0.06)
0.06 + 3 ×
120
q
−0.06)
0.06 + 3 × 0.06(170
q
−0.06)
0.06 + 3 × 0.06(1
125
q
0.06(1−0.06)
0.06 + 3 ×
105
q
−0.06)
0.06 + 3 × 0.06(195
= −0.019655508 ⇒ 0
= −0.007930446 ⇒ 0
= −0.015099933 ⇒ 0
= −0.022267855 ⇒ 0
= −0.002486922 ⇒ 0
= −0.005038450 ⇒ 0
= −0.025155320 ⇒ 0
= −0.003724407 ⇒ 0
= −0.009529028 ⇒ 0
= −0.013096908 ⇒ 0
0.15
Here is the p-chart
0.10
●
●
●
●
●
●
●
●
●
0.00
0.05
●
2
4
6
8
10
The process appears to be in statistical control.
3. A control chart indicates that the current process fraction nonconforming in 0.02. If 50
item are inspected each day, what is the probability of detecting a shift in the fraction
nonconforming to 0.04 on the first day after the shift? By the end of the fifth day following
the shift?
Answer.
Given that p̄ = 0.02 and n = 50, we compute the control limits,
UCL
=
0.02 + 3 ×
Central Line
=
0.02
LCL
=
0.02 − 3 ×
q
0.02(1−0.02)
= 0.07939697
50
q
0.02(1−0.02)
= −0.03939697 ⇒ 0
50
Since p new = 0.04 < 0.1 and n = 50 is large, we use the Normal approximation to the binomial with mean p new = 0.04
and variance p new (1 − p new )/ n = 0.000768. P (detect|shift) = P (D / n > UCL) + P (D / n < LCL) = 1 − Φ((0.07939697 −
p
p
0.04)/ 0.000768) + Φ((0 − 0.04)/ 0.000768) ≈ 0.1520263. P (detected by 5th sample) = 1 − (1 − 0.1520263)5 ≈ 0.5615582
2
4. Diodes used on printed circuit boards are produced in lot of size 1000. We wish to control
the process producing there diodes by taking samples of size 64 from each lot. If the nominal value of the fraction is p = 0.10 determine the parameters of the appropriate control
chart. To what level must the fraction nonconforming increase to make the β-risk equal
to 0.50. To what level is the minimum sample size that would give a positive lower level
control limit for this chart?
Answer.
Given that p̄ = 0.01 and n = 64, we compute the control limits,
UCL
=
0.1 + 3 ×
Central Line
=
0.01
LCL
=
0.1 − 3 ×
q
0.1(1−0.1)
= 0.2125
64
q
0.1(1−0.1)
= −0.0125 ⇒ 0
64
The beta risk is determined with the following:
β
=
P (LCL < D / n < UCL| p)
=
P (D / n < UCL| p) − P (D / n < LCL| p)
=
P (D /64 < 0.2125| p) − P (D /64 < 0| p)
=
P (D < 13.6| p) − P (D < 0| p)
D has Bin(64, p) distribution and we compute the probability of P (D ≤ 13) and P (D ≤ 0) with various p. Here we give
0.0
0.2
0.4
0.6
0.8
1.0
the following plot of β against p,
0.0
0.1
0.2
0.3
0.4
From the figure, we could identify that β = 0.5 with 0.21 < p < 0.22. Then we do further computation,
p
P (D ≤ 13)
P (D ≤ 0)
β
0.210
0.5192784
2.806298 × 10−7
0.5192782
0.211
0.5114091
2.587785 × 10−7
0.5114088
0.212
0.5035536
2.386041 × 10−7
0.5035533
0.213
0.4957148
2.199798 × 10−7
0.4957146
0.214
0.4878955
2.027883 × 10−7
0.4878953
0.215
0.4800987
1.869210 × 10−7
0.4800985
0.216
0.4723270
1.722773 × 10−7
0.4723268
0.217
0.4645831
1.587643 × 10−7
0.4645829
0.218
0.4568697
1.462959 × 10−7
0.4568696
0.219
0.4491895
1.347926 × 10−7
0.4491894
0.220
0.4415450
1.241808 × 10−7
0.4415449
The fraction of non-conforming would need to increase to 0.212. Assuming L = 3 sigma control limits, the minimum
sample size to achieve a positive lower control limit is:
n
>
=
=
3
1− p 2
L
p
1 − 0. 1 2
3
0.1
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