Gb Paper Imi - 1 Sep 2013 Manish

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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



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



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Operator 4Operator3Operator 2Operator1

25

24

23

22

21

20

Data


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


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



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


PPM < LSL 100000.00

PPM > USL 0.00

PPM Total 100000.00


PPM < LSL 40749.37

PPM > USL 0.00

PPM Total 40749.37


PPM < LSL 121333.91

PPM > USL 29.22

PPM Total 121363.14


Within

Overall


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


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


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



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


PPM < LSL 0.00

PPM > USL 9090.91

PPM Total 9090.91


PPM < LSL 2.26

PPM > USL 937.87

PPM Total 940.13


PPM < LSL 5.18

PPM > USL 1396.72

PPM Total 1401.89


Within

Overall


Only Employee 6


16/16

236232228224220216212

LSL USL

LSL 210

Target *

USL 230

Sample Mean 230.578Sample N 10



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


PPM < LSL 0.00

PPM > USL 400000.00

PPM Total 400000.00


PPM < LSL 0.00

PPM > USL 621898.86

PPM Total 621898.86


PPM < LSL 0.00

PPM > USL 595758.02

PPM Total 595758.02


Within

Overall


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

Documents

Gb Paper Imi - 1 Sep 2013 Manish