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Name:
Six Sigma Green Belt Certification Test.
Time: 2 hours Marks: 50
The Question paper is given to you in word document as a softcopy. Please
enter your name on the Top left hand Corner.
The Question paper has 3 sets of Questions. All Questions are Compulsory.
Set 1 is Multiple Choice. You need to change the color of the right answer
to green and save the file. It has 22 Questions each of 1 mark. There is no
negative marking.Set 2 is Short Answers. It has 2 Questions of 6 marks each. You need to
insert your answers after each question and save the file.
Set 3 is analysis and has 1 Question of 16 marks. You need to insert your
answers after the question and save the file.
Note please copy analysis / graphs and any other thing that you deem fit
from your Minitab analysis, onto this word document to support your
answers. Also pen the same in your answers sheets hard copy. Set one needs
to be answered on the Question paper itself.
Set 1 Multiple Choice
1. A team has been asked to reduce the cycle time for a process. The team
decides to collect baseline data. It will do this by
A. seeking ideas for improvement from all stakeholders
B. researching cycle times for similar processes within the organizationC. obtaining accurate cycle times for the process as it currently runs
D. benchmarking similar processes outside the organization
2. A Six Sigma project designed to solve a particular problem needs a
definition/scope statement to help avoid:A. going beyond the problem into other problems
B. failing to cover the entire problemC. misunderstanding and disagreement between team members regarding
problem boundaries
D. all of the above
E. none of the above
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3. There are 14 different defects that can occur on a completed time card. The
payroll department collects 328 cards and finds a total of 87 defects. Then DPU
=_____
A. 87/328 B. 87/(328x14) C. 87/14 D. (87x1,000,000)/(14x328)
4. There are 14 different defects that can occur on a completed time card. Thepayroll department collects 328 cards and finds a total of 87 defects. Then
DPMO =____
A. 87/328 B. 87/(328x14) C. 87/14 D. (87x1,000,000)/(14x328)
5. For a manufacturer making tetrapacks, the customer is willing to accept the
product only if it weighs between 490 gms and 510 gms. The machine is able to
produce packs with a mean of 500 gms and a standard deviation of 1gm. Which
of the following is true
A. USL = 506 gms and LSL = 494 gms B. USL = 510 gms and LSL = 490 gmsC. USL = 490 gms and LSL = 510 gms D. USL = 497 gms and LSL = 503 gms
6. In Box plot, the inside horizontal line describes the
A. mean B. median C. mode D. none of the above
7. The mean, median and mode of a distribution have the same value. What can
be said about the distribution:
A. it is exponential B. it is normal C. it is poison D. none of the
above(unimodal)
8. For a normal distribution, two standard deviations on each side of the mean
would include approx what percentage of the total population?
A. 47% B. 68% C. 95% D. 99% E. None of the above
9. In regression analysis. R = 0.88 signifies
A. 88% of the variation in X is explained by variation in YB. 12% of the variation in Y is explained by variation in X
C. 12% of the variation in X is explained by variation in Y
D. 88% of the variation in Y is explained by variation in X
10. Which is true?A. Cp >= Cpk B. Cp < Cpk C. Cp, Cpk are not comparable D. none of
the above
11. Which one of the following is true ?
A. ZLT = ZST + 1.5 B. ZST = ZLT C. ZST = ZLT + 1.5 D. none of the above
12. Pareto diagram is used:A. To identify vital few from trivial many B.To calculate mean and s.d.
C. To identify cause and effect relationship D. none of the above
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13. Measurement uncertainty is known and suitable if
A. %R&R = 50% B. %R&R 30% D. None of the above
14. Variation in a process exists due to
A. Chance Causes B. assignable causesC. Chance & assignable causes D. It happens without any causes
15. FMEA stands forA. Failure Mode and Effect Analysis B. Failure Mode and Efficient Analysis
C. Failure Model for Effectiveness Analogy D. Fatal modes for Electrical
Assemblies.
16. Which of the following are not included in the project charter
A. Project scope B. Project goal C. Project title D. Tools to be used
17. CTQs are derived from
A. Histogram B. VOC C. Run chart D. Pareto diagram
18. There is a group of 10 students in a class. Their weights in kgs are as
follows: 20, 22, 30, 30, 30, 45, 50, 56, 56, 60. Which of the following is true?
A. Mean is 39.9 kg, Median is 37.5 kg & Mode is 30 kg
B. Mean is 37.5 kg, Median is 39.9 kg & Mode is 30 kg
C. Mean is 30 kg, Median is 37.5 kg & Mode is 37.5 kg
D. Mean is 39.9 kg, Median is 60 kg & Mode is 20 kg
19. The weight of the pouches of a particular product is normally distributed
with a mean of 200 kg. We know that 68.26% of the pouches weigh between
196 kg and 204 kg. Which of the following is likely to be the standard deviation
of the distribution?
A. 8 B. 2 C. 4 D. 5
20. The process capability is defined by the formula
A. (Mean-LSL)/(6*) B.(USL-LSL)/(6*) C. (USL-LSL)/()D. (USL-LSL)/(3*)
21. Which of the following is true?
A. All the non- value adding activities can be eliminated
B. All the processes in the TO BE map should be value-added activities
C. All the activities in the TO BE map need not necessarily be value -addingactivities
D. Delays can always be eliminated
22. Which of the following hold true for Normal Distribution?
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A. A normally distributed process has only 0.0000002% parts outside the +/- 6
limits
B. Taken along with a long term mean shift of +/- 1.5 it denotes 3.4 ppm or
virtually Zero Defect
C. It is a Bell shaped Curve
D. All of the above
Set 2 Short Answers
1 . The Performance output of 4 operators was measured and the same is givenbelow. Check using ANOVA if performance of operators is equal or Not.
Operator 1 Operator 2 Operator 3 Operator 4
21.4 22.2 20.2 23.223.2 23.1 22.5 24.1
21.5 21.5 20.7 23.1
22.2 22.7 23.2 23.0
24.8
01-09-2013 14:55:16
Welcome to Minitab, press F1 for help.
Executing from file: C:\Program Files\Minitab\Minitab 16\English\Macros\Startup.mac
This Software was purchased for academic use only.
Commercial use of the Software is prohibited.
Gage R&R Study - ANOVA Method
* ERROR * Design is not balanced; execution aborted.
One-way ANOVA: Operator1, Operator 2, Operator3, Operator 4
Source DF SS MS F P
Factor 3 7.15 2.38 2.05 0.156Error 13 15.10 1.16
Total 16 22.25
S = 1.078 R-Sq = 32.14% R-Sq(adj) = 16.47%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ---------+---------+---------+---------+
Operator1 4 22.075 0.830 (---------*---------)
Operator 2 5 22.860 1.238 (--------*-------)
Operator3 4 21.650 1.429 (--------*---------)
Operator 4 4 23.350 0.507 (---------*--------)
---------+---------+---------+---------+
21.6 22.8 24.0 25.2
Pooled StDev = 1.078
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Individual Value Plot of Operator1, Operator 2, Operator3, Operator 4
Boxplot of Operator1, Operator 2, Operator3, Operator 4
Normplot of Residuals for Operator1, Operator 2, Operator3, Operator 4
Residual Histogram for Operator1, Operator 2, Operator3, Operator 4
One-way ANOVA: Operator1, Operator 2, Operator3, Operator 4
Source DF SS MS F P
Factor 3 7.15 2.38 2.05 0.156
Error 13 15.10 1.16
Total 16 22.25
S = 1.078 R-Sq = 32.14% R-Sq(adj) = 16.47%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ---------+---------+---------+---------+
Operator1 4 22.075 0.830 (---------*---------)
Operator 2 5 22.860 1.238 (--------*-------)
Operator3 4 21.650 1.429 (--------*---------)
Operator 4 4 23.350 0.507 (---------*--------)
---------+---------+---------+---------+
21.6 22.8 24.0 25.2
Pooled StDev = 1.078
Individual Value Plot of Operator1, Operator 2, Operator3, Operator 4
Boxplot of Operator1, Operator 2, Operator3, Operator 4
Residual Plots for Operator1, Operator 2, Operator3, Operator 4
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210-1-2
99
95
90
80
70
60
50
40
30
20
10
5
1
Residual
Percent
Normal Probability Plot(responses are Operator1, Operator 2, Operator3, Operator 4)
Operator 4Operator3Operator 2Operator1
25
24
23
22
21
20
Data
Individual Value Plot of Operator1, Operator 2, Operator3, Operator 4
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Operator 4Operator3Operator 2Operator1
25
24
23
22
21
20
Data
Boxplot of Operator1, Operator 2, Operator3, Operator 4
2.01.51.00.50.0-0.5-1.0-1.5
5
4
3
2
1
0
Residual
Frequency
Histogram(responses are Operator1, Operator 2, Operator3, Operator 4)
2. Draw a Cause & Effect Diagram using Mintab for the data given in file
Cause & Effect Data.xlsx
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nt
Environme
Measurements
Methods
Material
Machines
Personnel
Courtesy
Motivation
Experience
Training
Age
Cleanliness
Maintence
Amount
Age
Up to date
Order
Consistency
Procedure
Technique
Scale
Lighting
Humidity
Power
Wrapper
Boxes
Printer
Temperature
Humidity
Cause-and-Effect Diagram
Or
The data of Electricity interruptions in given in the file Electricity Interruptions
Reasons.xlsx. Which interruptions will you address in priority and why. Draw
graphs/ charts where necessary.
Set 3 Analysis
Analyze the data given in the file Oil Bottle Data.xlsx using Minitab and give
your inferences.
Stat >BasicStatistics> Normality Test>Select Oil>Ok
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240235230225220215210
99.9
99
95
90
80
7060504030
20
10
5
1
0.1
Oil
Percent
Mean 220.9
StDev 3.621
N 240
AD 3.322P-Value Main Effects Plot>Response:Oil; Factors:Branch, Shift, Study
Hour, Employee
BA
224
222
220
21
2019181716151413121110987654321
224
222
220
121110987654321
Branch
Mean
Shift
Study hour Employee
Main Effects Plot for OilData Means
Split Worksheet Branchwise
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Branch B
235230225220215210
99.9
99
95
90
80
7060504030
20
10
5
1
0.1
Oil
Percent
Mean 222.6
StDev 3.591
N 120
AD 2.640
P-Value
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21
220
218
216
214
2019181716151413121110987654321
121110987654321
220
218
216
214
Shift
Mean
Study hour
Employee
Main Effects Plot for OilData Means
Looks affected by EMPLOYEE 3
BRANCH B
21
230.0
227.5
225.0
222.5
220.0
2019181716151413121110987654321
121110987654321
230.0
227.5
225.0
222.5
220.0
Shift
Mean
Study hour
Employee
Main Effects Plot for Oil
Data Means
\
Looks affected by EMPLOYEE 6
Subset Worksheet of Branch A excluding employee 3 and only employee 3
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Branch A excluding employee 3
Normality Test
226224222220218216214212
99.9
99
95
90
80
7060504030
20
10
5
1
0.1
Oil
Percent
Mean 219.6
StDev 2.135
N 110
AD 0.178
P-Value 0.917
Probability Plot of OilNormal
Subset worksheet of Branch A only employee 3
225220215210205
99
95
90
80
70
60
50
40
3020
10
5
1
Oil
Percent
Mean 214.5
StDev 3.856
N 10
AD 0.247
P-Value 0.672
Probability Plot of OilNormal
Here p value is > 0.05 in both cases hence we fail to reject null hypothesisand conclude data
is normal in both cases
Process Capability of Branch 1 without Employee 3
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228225222219216213210
LSL USL
LSL 210
Target *
USL 230
Sample Mean 219.565Sample N 110
StDev(Within) 2.10543
StDev(Overall) 2.1352
Process Data
Cp 1.58CPL 1.51
CPU 1.65
Cpk 1.51
Pp 1.56
PPL 1.49
PPU 1.63
Ppk 1.49
Cpm *
Overall Capability
Potential (Within) Capability
PPM < LSL 0.00
PPM > USL 0.00
PPM Total 0.00
Observed Pe rformance
PPM < LSL 2.77
PPM > USL 0.36
PPM Total 3.13
Exp. Within Performance
PPM < LSL 3.74
PPM > USL 0.51
PPM Total 4.25
Exp. Overall Performance
Within
Overall
Process Capability of Oil
Process Capability only Employee 3 Branch A
228224220216212208
LSL USL
LSL 210
Target *
USL 230
Sample Mean 214.505
Sample N 10
StDev(Within) 2.5858
StDev(Overall) 3.85554
Process Data
Cp 1.29
CPL 0.58
CPU 2.00
Cpk 0.58
Pp 0.86
PPL 0.39
PPU 1.34
Ppk 0.39
Cpm *
Overall Capability
Potential (Within) Capability
PPM < LSL 100000.00
PPM > USL 0.00
PPM Total 100000.00
Observed Pe rformance
PPM < LSL 40749.37
PPM > USL 0.00
PPM Total 40749.37
Exp. Within Performance
PPM < LSL 121333.91
PPM > USL 29.22
PPM Total 121363.14
Exp. Overall Performance
Within
Overall
Process Capability of Oil
WE see that process capability of Branch A excluding employee 3 is far better than that of
Branch A Employee 3 only
Hence Employee 3 must be fixed (Trained)
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121086420
235
230
225
220
215
210
Employee
Oil
Marginal Plot of Oil vs Employee
Branch B excluding employee 6 and only employee 6
Branch B excluding employee 6
Normality Test
230225220215
99.9
99
95
90
80
7060
504030
20
10
5
1
0.1
Oil
Percent
Mean 221.9
StDev 2.703
N 110
AD 0.748
P-Value 0.050
Probability Plot of OilNormal
Only Employee 6
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237.5235.0232.5230.0227.5225.0
99
95
90
80
70
60
50
40
30
20
10
5
1
Oil
Percent
Mean 230.6
StDev 2.383
N 10
AD 0.674P-Value 0.054
Probability Plot of OilNormal
P value is >.05 hence we fail to reject null hypothesis and conclude that data is normal in both
the cases
Process capability index
Without Employee6
231228225222219216213210
LSL USL
LSL 210
Target *
USL 230
Sample Mean 221.919
Sample N 110
StDev(Within) 2.59898
StDev(Overall) 2.70298
Process Data
Cp 1.28
CPL 1.53
CPU 1.04
Cpk 1.04
Pp 1.23
PPL 1.47
PPU 1.00
Ppk 1.00
Cpm *
Overall Capability
Potential (Within) Capability
PPM < LSL 0.00
PPM > USL 9090.91
PPM Total 9090.91
Observed Pe rformance
PPM < LSL 2.26
PPM > USL 937.87
PPM Total 940.13
Exp. Within Performance
PPM < LSL 5.18
PPM > USL 1396.72
PPM Total 1401.89
Exp. Overall Performance
Within
Overall
Process Capability of Oil
Only Employee 6
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236232228224220216212
LSL USL
LSL 210
Target *
USL 230
Sample Mean 230.578Sample N 10
StDev(Within) 1.86072
StDev(Overall) 2.38343
Process Data
Cp 1.79CPL 3.69
CPU -0.10
Cpk -0.10
Pp 1.40
PPL 2.88
PPU -0.08
Ppk -0.08
Cpm *
Overall Capability
Potential (Within) Capability
PPM < LSL 0.00
PPM > USL 400000.00
PPM Total 400000.00
Observed Pe rformance
PPM < LSL 0.00
PPM > USL 621898.86
PPM Total 621898.86
Exp. Within Performance
PPM < LSL 0.00
PPM > USL 595758.02
PPM Total 595758.02
Exp. Overall Performance
Within
Overall
Process Capability of Oil
WE see that process capability of Branch B excluding employee 6 is far better than that of
Branch A Employee 6 only
Hence Employee 6 must be fixed (Trained)
Marginal Plot
12108642
235
230
225
220
215
210
Employee
Oil
Marginal Plot of Oil vs Employee