A particular auto manufacturer ships their finished cars by rail. Before loading
ID: 3301761 • Letter: A
Question
A particular auto manufacturer ships their finished cars by rail. Before loading cars on any particular railcar, it is inspected for certain problems. At the auto manufacturer's discretion, they can reject a railcar and not use it. The auto manufacturer claims that the number of railcars rejected is not only unacceptably high, but is very erratic. Below are the results of 15 consecutive days of operation Each day, the total number of railcars inspected and the total number rejected is recorded (notice all 15 values of "number inspected" are different) # of Railcars # of Railcars InspectedRe iected Day 115 221 209 227 215 256 443 365 253 299 281 330 340 226 293 4073 2 4 6 18 25 21 15 46 43 7 35 30 37 42 12 15 Totals 412Explanation / Answer
Solution
Let ni = number of railcars inspected and xi = number of railcars rejected on day i, i = 1,2,….,15
Back-up Theory
Central Line for p-chart = pbar = {[1,15]xi}/{[1,15]ni}
Control Limits for the ith day:
LCL = pbar - 3{pbar(1 – pbar)/ni}
UCL = pbar + 3{pbar(1 – pbar)/ni}
Part (b)
The central line = pbar = {[1,15]xi}/{[1,15]ni} = 412/4073 = 0.1012 ANSWER 1
Control Limits for the first day:
LCL = pbar - 3{pbar(1 – pbar)/n1}
= 0.1012 – 3{(0.1012 x 0.8988)/115}
= 0.1012 – 0.0281
= 0.0731
UCL = pbar + 3{pbar(1 – pbar)/n1}
= 0.1012 + 0.0281
= 0.1293
p1 = 16/115 = 0.1391> 0.1293 => out of control situation ANSWER 2
Part (c)
The LCL, UCL and control status for each day are tabulated below:
pbar =
0.1012
pbar(1 - pbar) =
0.090959
day (i)
xi
ni
pi
LCL
UCL
pi status
1
16
115
0.13913
0.073076
0.129324
pi > UCL
2
18
221
0.081448
0.080913
0.121487
3
25
209
0.119617
0.080338
0.122062
4
21
227
0.092511
0.081183
0.121217
5
19
215
0.088372
0.080632
0.121768
6
15
256
0.058594
0.08235
0.12005
pi < LCL
7
46
443
0.103837
0.086871
0.115529
8
43
365
0.117808
0.085414
0.116986
pi > UCL
9
13
253
0.051383
0.082239
0.120161
pi < LCL
10
35
299
0.117057
0.083758
0.118642
11
30
281
0.106762
0.083208
0.119192
12
37
330
0.112121
0.084598
0.117802
13
42
340
0.123529
0.084844
0.117556
pi > UCL
14
19
226
0.084071
0.081138
0.121262
15
33
293
0.112628
0.083581
0.118819
As is clear from the above table, 5 out of 15 pi’s are out of limits, this is very high, 33%.
Further, these out-of-limits pi’s also do not follow any pattern, suggesting enough evidence to conclude that the railcars’ behavior is quite erratic. ANSWER
Part (d)
The valuable information contained in the above table would be the basis for initiating corrective actions.
As a first step, it would be worthwhile to eliminate the out-of-control pi’s and revise the limits. Using these as preliminary limits, the process can be brought under some sort of stability and then final limits can be drawn up for perpetual control.
DONE
pbar =
0.1012
pbar(1 - pbar) =
0.090959
day (i)
xi
ni
pi
LCL
UCL
pi status
1
16
115
0.13913
0.073076
0.129324
pi > UCL
2
18
221
0.081448
0.080913
0.121487
3
25
209
0.119617
0.080338
0.122062
4
21
227
0.092511
0.081183
0.121217
5
19
215
0.088372
0.080632
0.121768
6
15
256
0.058594
0.08235
0.12005
pi < LCL
7
46
443
0.103837
0.086871
0.115529
8
43
365
0.117808
0.085414
0.116986
pi > UCL
9
13
253
0.051383
0.082239
0.120161
pi < LCL
10
35
299
0.117057
0.083758
0.118642
11
30
281
0.106762
0.083208
0.119192
12
37
330
0.112121
0.084598
0.117802
13
42
340
0.123529
0.084844
0.117556
pi > UCL
14
19
226
0.084071
0.081138
0.121262
15
33
293
0.112628
0.083581
0.118819
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