All DMAIC Paper for BB New

7/31/2019 All DMAIC Paper for BB New

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Q1.

Q2. Answer the questions about the following process. (To the second decimal place)85

100

6 3 4

Disposal Disposal Disposal Disposal15 5 3 6

5 13. The following is the road map to calculate the yield value of discrete data with Z value. What is1) Yft > YNA > e -DPU > Zlt2) Yft > YNA > Yrt > Zlt3) e -DPU > Yrt > (Yrt) 1/OPP > Zlt4) e -DPU > Yrt > (Yft) 1/OPP > Zlt

7 16. The following is the process map for the order/request process by mail order. Answer the following questions. Ans

Order Entry Credit Review Shipping 75.22%Yrt = " A " 75% Yrt = " C " 93.13%YNA = " B " YNA = " D " 9429%

61.64%Contact Order Review Order Matching 88.61%

93% 75% 78%

Availability Scheduleing86% 85%

Order Build W/house Loading99% 98.60%

Pricing Shipping95% " E "

Ship'g truck Load Shipping direct90.90% 97.80%

P2100 90 70 35

1015

5

5

waste waste

P1 P3

a ) Calculating a RTY on the PROCESS

b) Calculating a Normalized Yield on the PROCESS .

3510 w

P4

40%

60%

P 1 P 2 P 3 P 4


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1) " A " 75.22% 94.29%2) " B " 93.13%3) " C " 61.64%4) " D " 88.61%


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0.9 0.777778 0.875 0.777778

47.6%

83.08%

100 85 80 7785 74 74 67

0.85 0.870588 0.925 0.87013 59.56%

0.85 0.870588 0.925 0.87013

59.56%

orrect ? ( 3 )

aste

aste


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Q7. The following are the characteristics of central composite design. What is incorrecta. When the centre point is away from the mean connecting lines then, it can be curvedb. The experiment of 2 level is carried out under the assumption that the effect of factors are non-linear.c. If repeats the cube point in 2 level DOE, it is not possible to find curved regressiond. The experiment of 3 level can be said to be central composite design.

Q8. When your number of factors are 7, then which type of Fractional Factorial canbe useda) 1/16 Factorial Design

b) 1/8 Factorial Designc) 1/4 Factorial Designd) 1/2 Factorial Design

Q9. Which of the following i s incorrect explanation of experimental Pure error/Lack-of-Fit ?(more than one can be the answer)a. To analyze Lack-of-Fit, the independent variables must be repeated more than twice at the same levelb. For the regression eq to be significant, p value of Lack-of-Fit should be less than Alpha value.c. The smaller p value of Lack of fit, the larger p value of regression.d. To make the regression equation to be calculated significant, P value of Lack-of-Fit must be significant.e. Pure error value shows the reproducibility of response in a fixed independent variable.f. None of these

Q10. Which among the following has the highest correlation coefficient

LSL T


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Q1. From the following Normality Effect Plot, Answer the below questions

7 Term Tot ABC

AC ABBC

ABC

a. Write down all the terms which are insignificant C,AC,BC,ABCb. Identify which variable has the maximum influence on the response Ac. Choice the wrong statement (more than one can be the answer)

1. AC is not the significant term2. Screening designs is used to identify the "vital" few factors or key variables that

influence the response3. Normal probability plot identifies important effects using a = 0.10, by default4. Points that do not fall near the line usually signal insignificant terms

Q2. From the following Gage Linearity and Bias Study, Answer the below questions


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c. MS Error 3.700d. R-Sqr 34.39


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6222

alSigSigIn singIn singSigIn singIn sing

61212


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624.3 24.4 0.125.1 24.4 -0.724.4 24.4 0.024.8 24.4 -0.424.6 24.4 -0.2

0.320936 24.4 0.2

riation 1.92561712.4635

61.51.51.51.5

61212

USL

USL


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56.89 57.22 57.22 57.25

56.62 56.62 56.62 56.60

57.65 57.65 57.70 57.65

56.83 56.84 56.00 56.83

53.54 53.55 53.54 53.00

54.12 54.10 54.12 54.12

54.50 54.50 54.52 54.5256.52 56.52 56.50 56.50

a. What is the Gage R&R % SV value 9.11b. Write down the Repeatibility % T Value 12.23c. What is the ttl variation study variance value 8.06179d. Find out which operator has minimum difficulty in measuring the readings Operator 1


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103 2.3863 18.6514

102

233

103


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cts 34

NightScratch,Sumadgand peel

Incorrect

CorrectIncorrect

1022213

102233


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Q1. a. C, AC, BC, ABCb. Ac. 3, 4

Q2.

-0.016-0.016-0.016-0.0120.004

a. 0.0112b. 0.007976c. 1.23%

d. 3

Q3.

% Bias Value 40%

Reference Value

B i a s

302826242220

0.8

0.6

0.4

0.2

0.0 0

Data Avg Bias

Gage Bias

24.4 0.24 40.0 0.17

Reference Bias %Bias Average 0.24 40.0 0.17

Gage name:Date of study :

Reported by :Tolerance:Misc:

Gage Linearity and Bias Study for C3


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Q4. 1 C2 B3 A4 D

Q5. One-way ANOVA: C2 versus Subscripts

Source DF SS MS F PSubscripts 2 **.** 20.38 *.** 0.012Error 21 77.75 *.**Total 23 118.50

S = 1.924 R-Sq = **.**% R-Sq(adj) = 28.14%

a. SS factor 40.760b. F factor 5.505c. MS Error 3.702d. R-Sqr 34.39%

Q1.

a. 2.378b. 18.649c. 135.67 21.6

6.7835 1.087.40666

Q2. a. 4b. R-Sqr = 75.4%, R-Sqr (Adj) = 61.0%c. A= 5.688, B=0.59, C=3.9087d. Y = 12.6358 + 0.8898A - 1.7662B - 2.0904C - 0.2576A*A + 1.1793B*B - 0.6001A*B + 0.6951

Response Surface Regression: Y versus A, B, CThe analysis was done using coded units.Estimated Regression Coefficients for Y

Sample

S a m p l e M e a n

7654321

20.50

20.25

20.00

19.75

19.50

_ _ X=20.011

UC L=20.660

LCL=19.363

Sample

S a m p l e R a n g e

7654321

2.4

1.8

1.2

0.6

0.0

_ R=1.125

UC L=2.378

LCL=0

Xbar-R Chart of C2

Sample

S a m p l e M e a n

7654321

20

19

18

17

_ _ X=18

UC L=18.649

LCL=17.351

Sample


7654321

2.4

1.8

1.2

0.6

0.0

_ R=1.125

UC L=2.378

LCL=0

111

11

11

Xbar-R Chart of C2

n A2 A31 2.660 3.760

2 1.880 2.659

3 1.023 1.954

4 0.729 1.628

5 0.577 1.427

6 0.483 1.2877 0.419 1.182 0.

8 0.373 1.099 0.

9 0.337 1.032 0.

10 0.308 0.975 0.

Hi

Lo1.0000D

Optimal

Cur 0.

2

0.950

6.290 A

[5.6880] [0.


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Term Coef SE Coef T PConstant 10.2373 0.3378 30.303 0.000

A -0.9646 0.4434 -2.175 0.050B 0.3090 0.4429 0.698 0.499C 0.7662 0.4439 1.726 0.110

A*A -1.8363 0.7225 -2.541 0.026B*B 1.6700 0.7220 2.313 0.039

A*B -1.9068 0.9704 -1.965 0.073 A*C 3.3404 0.9740 3.430 0.005

S = 0.9752 R-Sq = 75.4% R-Sq(adj) = 61.0%

Analysis of Variance for Y

Source DF Seq SS Adj SS Adj MS F PRegression 7 34.972 34.972 4.9960 5.25 0.006

Linear 3 7.797 7.797 2.5990 2.73 0.090Square 2 12.316 12.316 6.1582 6.47 0.012Interaction 2 14.859 14.859 7.4293 7.81 0.007

Residual Error 12 11.413 11.413 0.9511Lack-of-Fit 7 8.874 8.874 1.2677 2.50 0.166Pure Error 5 2.540 2.540 0.5079

Total 19 46.385

Estimated Regression Coefficients for Y using data in uncoded unitsTerm Coef Constant 12.6358

A 0.8898B -1.7662C -2.0904

A*A -0.2576B*B 1.1793

A*B -0.6001

A*C 0.6951

Q3.

a. Nightb. All --> Smudge, Peel & Scratchc. Mark the incorrect Statement

1. In Evening 90% of the defects are contributed by Smudges & Scratch

d = 1.0000

Targ: 14.0Y

y = 14.0000

Defects Type

C o u

n t

PeelSmudgeScratchNIL

10.0

7.5

5.0

2.5

0.0


10.0

7.5

5.0

2.5

0.0

Period = Day Period = Evening

Period = Night

Defects Typ eNILScratchSmudgePeel

Pareto Chart of Defects Type by Period

C o u

n t

Defects

CuPe


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2. At night, maximum number of defects have occurred3. In evening shift, Smudges frequency is minimum4. Relationship exists between the type of defects and the work shift producing the Sheets.

Q4. a. 0.212b. 0.599c. 3.70d. Tick the wrong explaination

1. Means of all models are same2. Means of all models are different3. Ho means means are equal4. Ha means means are different

e. Average 4.375Tinv 2.079614S 1.924n 85.78963

Q5 Gage R&R Study - XBar/R Method

%ContributionSource VarComp (of VarComp)Total Gage R&R 0.01497 0.83

Repeatability 0.01497 0.83Reproducibility 0.00000 0.00

Part-To-Part 1.79038 99.17Total Variation 1.80535 100.00

Study Var %Study Var %ToleranceSource StdDev (SD) (3.9199 * SD) (%SV) (SV/Toler)Total Gage R&R 0.12234 0.47956 9.11 7.99

Repeatability 0.12234 0.47956 9.11 7.99Reproducibility 0.00000 0.00000 0.00 0.00

Part-To-Part 1.33805 5.24502 99.58 87.42Total Variation 1.34363 5.26690 100.00 87.78

Number of Distinct Categories = 15

a. 9.11b. 7.99c. 5.2669d. Operator 1

S u b s c r i p t s

95%Bonferroni Confidence

LEVEL 3

LEVEL 2

LEVEL 1

321

Test for Equal V

C2

P e r c e n t

6420

99

95

90

80

70

60504030

20

10

5

1

ProbabilityNor


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5

P5


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P e r c

e n t

Typeount

20.0m % 40.0 60.0 80.0 100.0

4 2 2 2rcent 40.0 20.0 20.0

ScratchPeelNILSmudge

10

8

6

4

2

0

100

80

60

40

20

0

Pareto Chart of Defects Type by PeriodPeriod = Night


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One-way ANOVA: C2 versus Subscripts

Source DF SS MS F PSubscripts 2 40.75 20.38 5.50 0.012Error 21 77.75 3.70Total 23 118.50

S = 1.924 R-Sq = 34.39% R-Sq(adj) = 28.14%

Intervals for StDevs654

Bartlett's Test

0.453

Test Statistic 1.03P -Valu e 0.599

Levene's Test

Test Statistic 0.82P-Value

ariances for C2

12108

Mean

0.212

5.75StDev 2.270N 24

AD 0.480P-Value

Plot of C2al


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o its mean variance are not equal .So we will stop here

84.7pooling c is the main factor

actor AA96.8895.69

.2 4.54.03.53.0

99

90

50

10

1

15MIN 10MIN

P-Value 0.071

15MINMean 3.7StDev 0.1886N 10

AD

Mean

0.462P-Value 0.201

20MINM ean 3.16StDev 0.8422N 10

3.53

AD 0.411P-Value 0.275

StDev 0.1829N 10

AD 0.629

t of 10MIN, 15MIN, 20MINNormal - 95% CI


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DEFECT Nos DEFECTS Nos 5 A 23 H 543 TTl Defect 3690B 34 I 671 Other Defect' 176 95.23%C 45 J 234 Remaning 3514D 32 K 564 Accptable 95.23E 678 L 8F 34 M 90G 678 N 56

Q2 ) FIND OUT THE VALUE OF THE FOLLOWING 5

: DEFECT 10unit 1Opportunity 36DPU = 10DPO = 0.27777778 (1-dpo)^opportuPPM = 999991.831 yield 8.16865E-06ZLT = -4.3098395ZST = -2.8098395

Regression Analysis: RES versus PRESSURE, TEMP, TIME 5

The regression equation is******************************************* 858-30.38 pressure+1.74 temp-38.81timePredictor Coef SE Coef T PConstant 858 1560 0.55 0.595PRESSURE -30.38 92.89 -0.33 0.750TEMP 1.74 54.40 0.03 0.975TIME -38.81 25.67 -1.51 0.161

S = 291.999 R-Sq = **.**% R-Sq(adj) = 0.0% 21.1985263.1

Analysis of Variance 83224.6923Source DF SS MS F P -0.0244928Regression 3 229290 76430 0.90 0.476 -2.4492823Residual Error 10 852631 85263Total 13 1081921 0.788071403

78.80714026Q4) FIND OUT THE %STUDY VARIANCE, % STUDY TOLERANCE & NUMBER OF DISTINCT CATEGORIES

Study Var %Study Var %Tolerance 5Source StdDev (SD) (6 * SD) (%SV) (SV/Toler) St var %Total Gage R&R 0.083661 0.50197 ** 42.41940254 8.3662

Repeatability 0.061128 0.36677 30.99 3.06

Reproducibility 0 .057118 0.34271 28.96 2 .86 5.0197Operator 0.024438 0.14663 12.39 1.22Operator*Part 0.051626 0.30976 26.18 2.58

Part-To-Part 0.178602 1.07161 90.56 8.93Total Variation 0.197226 1.18335 100.00 9.86Number of Distinct Categories = *** 3.010110087 0.42419403

3.010

ALSO FIND OUT THE SV/Toler VAL

10

Q1 ) FIND OUT WHAT WILL BE THE COMBINED EFFECT AFTER THE FIRST % WHEN TTL OF 6 DEFCTS ARE LYING IN OTHER

Q3 ) FIND OUT THE EQUATION FOR THE FOLLOWING MINITAB OUTPUT & R-Sq VALUE AND FIND OUT THE VALUE OF REGR12, TEMP = 23 & TIME = 6

28


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Q5) AT 92% CONFIENCE INTERVAL WHAT CAN BE THE VALUE OF SEM & CI5

One-Sample Z: C2Test of mu = 40 vs not = 40The assumed standard deviation = 1.3

.Variable N Mean StDev SE Mean 92% CI Z PC2 67 44.3955 3.3588 ****** (******, ******) 27.68 0.000

0.15882 44.1175 44.67 0.15882Q6 ) FILL IN THE BLANKS 5

Chi-Square Test: DAY1, DAY2, DAY3Expected counts are printed below observed countsChi-Square contributions are printed below expected counts

DAY1 DAY2 DAY3 Total Day 11 **** 521 452 1537 564

559.54 344.39 633.08 564**** 90.573 51.792 0.04 0.036

2 452 157 498 1107403.00 248.04 455.965.958 33.415 3.876

3 521 268 789 1578574.46 353.57 649.964.976 20.711 29.742

Total **** 946 1739 4222 1537Chi-Sq = 241.078, DF = 4, P-Value = 0.00

Test for Equal Variances: C1, C2 589% Bonferroni confidence intervals for standard deviations

N Lower StDev Upper C1 10 ****** 3.35345 6.03758 2.3234 11.2456269C2 10 1.43548 2.07136 ******* 3.72929 4.29053225

F-Test (normal distribution)Test statistic = *.**, p-value = 0.167 2.62

Levene's Test (any continuous distribution)Test statistic = 2.03, p-value = 0.172

Q8 ) CALCULATE THE UPPER & LOWER VALUES OF THE X-BAR R CHART GIVEN THAT

Avg RANGE ALSO FIND OUT HOW MANY VALUES ARE CROSSING THE CONTROL LIMITS13 2 519 3 x bar upper 20.424 4 lower 16.59 8 value are crossing the control limit21 625 1 R bar Upper 6.9727 2 lower 011 315 418 612 2

S a m p

l e S i z e =

5

Q7 ) 2 - VARIANCE TEST WAS CONDUCTED TO CHECK THE VARIANCE OF TWO SAMPLES.FIND OUT THE CONFIDENCE INTERVAL FOR SD OF THE ****** AREAS

29


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ity

E AT TTL TOLERANCE =

S CATGY.

ESSION AT PRESSURE =

30


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31


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GIVEN THAT PROCESS VARIES FROM 30 + 5.1) 1.372 0.832) 0.954 0.95423) 2.7444) 1.892

Q8 ) CHOICE THE WRONG STATEMENT (MORE THAN ONE CAN BE THE ANSWER)1) IN CASE OF 1SAMPLE Z TEST, TEST MEAN SHOULD LIE BETWEEN THE CONFIDENCE INTERVAL.

2) RTYparallel = YFT1 x YFT2 x YFT3.3) YOUR R-Sqr (adj) VALUE WILL BE ALWAYS GREATER THAN R-Sqr VALUE.4) C-CHART SHOULD BE USED WHEN SUBGROUP SIZE IS SAME.

Q9 ) IF THE NO. OF DISTINCT CATEGORIES ARE LESS THAN 5 WHAT DOES IT MEAN (MORE THAN ONE CAN BE THE ANSWER1) RATIONAL SUB GROUPING IS NOT PROPER2) GAGE SELECTION IS NOT PROPER3) ENTIRE RANGE IS NOT COVERED4) GAGE VARIATION HAS GOT MAX CONTRIBUTION IN TTL VARIATION OF THE MEASUREMENT SYSTEM

Q10 ) FOLLOWING ARE THE THREE TYPES OF REGRESSION MODES. AMONG THESE FIND OUT WHICH ONE IS FOR QUADRA

Q12 ) Gage R&R studies are conducted to determine all of the following except:1) Reproducibility of measurements between operators2) The resolution of the gage

3) The ability of the measurement system to detect process changes4) Repeatability of the measurement system

Q13 ) The method that attempts to delineate all possible failures and their effect on a system is calleda. PDCAb. FMEAc. Cause-and-effect diagramsd. Pareto analysis

Q14 ) FROM THE FOLLOWING GRAPH WHAT DOES X-BAR CHART INDICATES

1) NO. OF DISTINCT CATEGORIES ARE LESS THAN 5

3) SELECTED GAGE IS OK TO MEASURE THE PARTS4) HIGH VARAIATION IN OPERATOR 2 AS COMPARED T

2) VERY LOW VARIATION OBSERVED BY THE MEASUDUE TO THE OPERATOR CHANGE

Q11 ) On an X-bar and R control chart for a gage R&R study the control chart for the Ranges of repeated readings is in statistical control.However, nearly all of the averages on the X-bar chart are outside of the control limits. This indicates:

2

(1) (2) (3)

Gage name:Date of study :

Reported by :Tolerance:M isc:

Gage R&R (A NOVA ) for C3


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P e r c e n t

Part-to-PartReprodRepeatGage R&R

200

100

0

% Contribution% Study Var% Tolerance


10

5

0

_ R=3.9

UCL=12.74

LCL=0

1 2

S a m p l e M e a n 60

55

50

_ _ X=56

UCL=63.33

LCL=48.67

1 2

PART7654321

64

56

48

OPER 1

64

56

48

PART

A v e r a g e

87654321

60

55

50

Components of Variation

R Chart by OPER

Xbar Chart by OPER

C3 by PART

C3 by OPER

OPER * PART Interaction


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3020


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)

IC

3

Yes

O OPERATOR 1

RING SYSTEM


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1098

2

109

OPER 12


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65040


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Q1. For regression equation to be significant, Residuals should be normally & randomly distributedFrom the following graphs identify which of them are randomly distributed

Q2. Identify among the following which is th ewrong explaination of Anovaa. Anova is a tool which can compare several meansb. Is used to identify significant factor X which has influence on the response Y

c. Is used to determine whether MSB is greater than the MSW or notd. We can screen data to identify Vitel Fewe. To determine the average of each response when selected from same population

Q3. For finding the relationship between Y and two independent variables X1 & X2, Df regressioyou collected total of 19 readings and found the regression model eqn as Df error Y = a + bX1 + cX2, Find out what would be the degree freedom of residual error Df total

Q4. Which among the following has the highest correlation coefficient

Q5. Conduct Simple regression of the following data and answer the below questionsa. Write down the Regression equation The regression equb. Using regression eq find out what could be the Fitted value at 630, 730 & 960c. Write down the R-Sqr & R-sqr (adj) valued. Using 90% Confidence interval, What will be the maximum upper CL value of responsee. For X at 520, what would be the upper & lower Confidence limit value (use 95% CI) Lower

F What will the residual P valueX Y

500 0.18 Y = - 1.12 + 0.00268 X540 0.37 0.5684 92.5580 0.35 0.8364620 0.50 1.4528

Fi

Fit Fit Fit

Residual Residual ResidualResidual

(a) (b) (c) (d)

Fit


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660 0.56700 0.75740 1.02780 1.18820 1.05860 0.94

900 1.50940 1.56980 1.65

1000 1.401020 1.501060 1.63

Q6. In analysis of variance, when the data by factor does not satisfy the normality, what isthe decision making criterion that can be used in the homogeneity of variance analysis?1. P value of Bartletts Test2. P value of Levenes Test

3. Whether the dispersion for variation by factor is confounded or not.4. P value of normality analysis

Q7. "ABC" division want to analyze the process capability of supplier "DEF" to improve CTQ to 6 sigma level.The historical standard deviation of the population of CTQ "X" is known as 10.0. And randomly sampled 10 dis followings and the population is known as normal distribution.

Sample data : 40, 52.3, 50.6, 72.3, 49.3, 76.8, 58.7, 62.6, 56.8, 58.3

(1) Calculate the upper limit of 90% confidential intervals of the population. 63.07(2) Calculate the upper limit of 95% confidential intervals of the population. 64.07

Q8. Development dept. has used specified adhesives to stick 2 parts together. They want to develop one with higadhesive strength. If the adhesive strength is more than 20, they will select that one and control the process.

Answer to the questions. (All of the data are normally distributed.)

(1) Conduct the homogeneity of Variation, and get the P-Value. 2.39(2) Calculate the upper confidence interval limit value of Standard deviation for the adhesive strength for

adhesived in level 2. (90% of confidence interval) 3.898(3) Calculate the Lower confidence interval l imit of Mean for adhesive level 3 23.042

Adhesive level 1 Adhesive level 2 hesive level39 18 21

12 15 1914 14 2113 17 2218 15 23

(3) Which adhesives have adhestive strength more than 20? (A) Adhesives in level 1 (B) Adhesives in level 2 (C) Adhesives in level 3


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(D) No adhesives

Q9. a) Find out the P-value of Normality Plot (Take entire group as one) 0.212b) What would be the Test Statistics value while doing Test for equal variance 1.03c) What would be the MSW Value 3.7

d) Tick the wrong explaination1) The Upper SD limit for Gr2 at 90% CI will be 3.487212) Means of all models are different3) The Lower CI limit of Mean for Gr3 at 95% CI would be 8.9154) The Ferror Value is 0.428

e) When total number of observations are 30, and number of factor levels is 3, then what wouldbe the DFerror value

Gr1 Gr2 Gr33 4 65 4 77 3 8

4 6 68 7 54 4 93 2 109 5 9

Q10. From the following Minitab Anova output, Answer the below mentioned questions

Individual 95% CIs For Mean Based onPooled StDev N

Level N Mean StDev -------+---------+---------+---------+-- 8LEVEL 1 8 4.625 2.560 (------*-------) 8LEVEL 2 8 4.125 1.246 (-------*------) 8LEVEL 3 8 9.125 2.031 (-------*------)

-------+---------+---------+---------+--4.0 6.0 8.0 10.0

a) What is the Pooled Standard Deviation Value 2.018b) What is the Upper Mean value of LEVEL 1 at 95% CI 7.984c) At 95% Confidence Interval, what could be the Fcritical Value 3.4668d) Tick the wrong explaination

1. Means of all models are same2. Means of all models are different3. Ho means means are equal

4. Ha means means are different

Q11. An ABC Company wants to identify whether vendor change can be a factor or not.They selcted 2 vendors, Vendor 1 and Vendor 2 and tested around 100 samples of Vendor 1 and 150 of Vendor 2. The test results are as followsVendor 1 --> OK material : 87 & Vendor 2 --> NG material : 28

a. Find out which vendor is better Vendor 1 is better


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Analysis of VarianceSource DF SS MS F PRegression 1 3.5816 3.5816 171.62 0.000Residual Error ** 0.2922 *.**** 14 0.020871Total 15 3.8738

a. What would be the F Critical Value 4.60011b. Write down the R-Sqr & R-Sqr (Adj) Value 92.46% 91.92

AlphaDF numer DF denom

F critical Answere


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Answer D

2 Total no of fctor 16 DF total-DF regression18 Total no of observation -1

1

ation is Y = - 1.12 + 0.00268 X0.5684 0.8364 1.4528

Rsq=92.5 Rsq adj=92%1.83650.1333 Upper 0.4171

0.21 DO normality test of residual value

90

Fit

Residual(e)


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40

ta

data

40.=xbar+1.64*sem 62.95614 52.3.=xbar+1.96*sem 63.96806 50.6

73.349.3

her 76.858.762.656.858.3


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27

Mean St dev Sem4.625 2.56 1.190376 45.87524.125 1.245 0.612996 10.850189.125 2.031 0.672347 28.87473

4.076195Pooled sd 2.018959


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0.15

No improvementSampling is ok

= 2, P-Value = 0.005, DF = 2, P-Value = 0.005


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0.05 Given1 Given

14 Given

.=FINV(ALPHA,DFn,DFd)4.60011


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Q1. The following are the explanations of control limit determination in the control chart.What is incorrect.1. The control limit uses + 3 s traditionally2. + 3 s is the area of 99.73%3. If increase the number of samples, the control limit widens4. If it is the variation by a chance cause, the data is mostly within the control limit.

Q2. Conduct the X-Bar R Chart for the following data & answer the belowa) From the R Chart, find out the UCL R Value 2.378b) At CL 18, what would be the UCL X-Bar Valu 18.649c) At X-Bar Ttl = 135.67 & R Ttl = 21.6, What would be the value of X-Bar UCL Given that 7.40666 6.7835

Ttl Subgroups are 20 & Subgroup size is 5 1.08

1 2 3 4 5 6 7 A2 020.1 20.1 20.1 19.9 19.8 20.3 19.6 D3 019.8 19.8 20.4 21.2 19.7 19.5 21.4 D4 2.11519.5 20.3 20.5 20.3 19.7 19.8 20.320.4 19.4 19.5 20.4 20.4 20.2 19.820.1 19.5 20.3 19.7 19.5 19.6 19.5

Q3. The following table is to plot a control chart of the continous data. Given Sample Size = 6S.N. Average Range

1 23 1.12 25 1.33 27 1.94 21 3.05 24 1.26 22 1.47 20 1.48 32 0.29 21 3.5

10 16 1.6

a) Find out what would be the UCL of X-Bar Chart 23.90178b) What would be the LCL of R Chart 0c) How many points are located outside the UCL of R Chart 1d) If Central Line value of mean is 22, then what could be the UCL X-Bar Value\ 22.8e) How many points are located outside the UCL of X-Bar Chart 5 With consider UCL 22.8


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Q4. The followings are explanation of continous control chart. F ind out 2 wrong answers.1) If the number of samples are more than 10, Xbar & S control chart will be better than Xbar & R control chart.2) The more number of samples, the narrower line of control limit.3) If 2 points among 3 consecutive ones are between 1 and 2, it can be considered as unstable condition.4) To stabilize the condition, all the points in control chart shall be within 1.

Q5. Find out which is the wrong explaination for the concept of p rocess stablea) We can conclude the process is stable if the process output is consistent with common causeb) Stability means the average and range are located within control limitsc) If there are any pt outside the control limits then we must accept that the process is not stabled) To estimate the process, it is not necessay that process should be stable

Q6. Find out which control chart should be used when given no. of defectives & sample size is constanta) p Chartb) C Chartc) np Chartd) U Chart

Q7. Which type of the following charts can be used for attribute dataa) n Chartb) X-Bar R chartc) U Chartd) X-Bar S Chart

Q8. Which is the wrong explaination of the followinga) U Chart charts the no. of defects per unit sampled in each subgroup. Use U Chart when the subgroup size variesb) P Chart charts the proportion of defectives in each subgroupc) NP Chart charts the number of defectives in each subgroupd) C Chart charts the no. of defects in each subgroup. Use C Chart when the subgroup size variese) None of these


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Q1. Which of the following is the wrong explainations for the Blockinga. If two different suppliers materials were used during the experiment, then the material may be a blockb. It is used to minimize the effect of confounding & lucking variablesc. Used for higher precision of the experimentd. If the test was conducted over 2 days, then the date can be a block factor e. None of these

Q2. Which has the following graph shows highest interaction

Answere c

Q3. Followings are explanation of experiment plan. Find out a wrong answer.a. If experiments are performed on different days, they can be blocked to accomplish highly precise experiments.

b. If there is repetition in 2 factors experiment, F-value of interaction will not be detected.c. Available design helps to select an appropriate fractional design, based on the no. of runs you can performd. The response surface experiment are generally used for the case of continous factors.

Q4. When your number of factors are 7, then which type of Fractional Factorial can be useda) 1/16 Factorial Designb) 1/8 Factorial Designc) 1/4 Factorial Designd) 1/2 Factorial Design

Q5. The following are explanations of response surface experiment. What is incorrect?1. It is possible to determine the spec of each independent variable to meet the spec of dependent variables. correct2. When an independent variable has variation, it is possible to determine the control amount of the other variables. correct

3. It is used to determine the spec of optimal factor after screening experiment. correct4. When the interaction between independent variables is large, it cannot be used.

(a) (b) (c) (d)


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d. Pure error value shows the reproducibility of response in a fixed independent variable. correct


a. Write down all the terms which are insignificant ca,cb,abcb. Identify which variable has the maximum influence on the response Ac. Choice the wrong statement (more than one can be the answer)



Q11. The following are the explainations of the factorial design. Find out the incorrect explaination1. Randomization is to be carried out to increase the reproducibility.2. Confounding is to be used to raise experimental pricision.3. When there is no block effect, it is possible to join together and analyze as a whole.4. If central value is repeated, it is impossible to measure P value of the factor.

Q12. The following are the explainations of the fractional factorial design. Find out which one in incorrect.


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a) For full factorial what would be the frequency of all experiments 24b) For 1/2 factorial design, what would be the ttl number of runs 12

Q15. While developing the washing machines with new design, several samples were manufactured. Now we make anexperiment on cleaness of cloth to washing hour and volume of water.Washing hour : time (min), Volume of water: water(gal), Washing effect (Response variable) : Y

(1) When the volume of water is in (-)level and washing time is in (+) level, get the response value (washing effect)(2) The following are the explainations of the interaction analysis results. What is correct.

a) There is no interaction. incorrectb) There is some interaction. correctc) When amount of water increases from (-) level to (+) level & washing time is at (-) Level,

washing ability appears to be high incorrectd) As amount of water goes from (+) to (-) Level, Washing ability appears high correct

(3) The following is the effect plot from the following data. Identify which bar represents the factor Time b

Time Water Y10 4 20.4

10 4 19.310 4 17.610 4 16.320 8 17.420 8 17.720 8 23.220 8 20.4

T e r m

Standardized Effect

B

A

AB

2.52.01.51.00.50.0

2.179

(a)

(b)

(c) M e a n o f Y

18.5

18.0

17.5

17.0

16.5

16.0


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Q17. Identify which is the wrong explaination for fractional factorial designa) Display available design is used to identify the appropriate Factorial design correctb) If we increase from 1/2 to 1/8 factorial, condounding effect will increase correctc) We select that factorial design which has maximum condounding effect wrongd) 2 3-2 mean 2 level, 3 factors & 1/4 factorial design correct

Q18. a) Using Pooling, find out which factors are insignificant Speed & vibrationb) Find out the R-Sqr & R-Sqr (Adj) value (by completely removing all insignificant termsl) R-Sq = 88.5% R-Sq(adj) = 56.9%c) At 60 + 10, what would be the optimum values of your significant factors Speed 15d) After screening, what could be the maximum Upper 95% Confidence interval value Catal -0.0993

Vib 119.8Speed catal vibration temp density response

10 -1 100 180 3 6910 1 120 140 6 6715 -1 100 140 3 5310 -1 100 140 6 5615 1 100 140 6 6510 1 100 140 3 6310 -1 120 140 3 5315 -1 100 180 6 45

15 -1 120 140 6 5515 1 100 180 3 9310 -1 120 180 6 4910 1 120 180 3 9515 -1 120 180 3 6015 1 120 180 6 8210 1 100 180 6 7815 1 120 140 3 61

Q19. From the following analysis of variance for response, measure the following valuesa) R- Sqr Valueb) R-Sqr (Adj) Value

Analysis of Variance for response

Source DF Seq SS Adj SS Adj MS F PRegression 5 3260.75 3260.75 652.150 92.83 0.000 Rsq Rsq AdLinear 3 2437.50 2437.50 812.500 115.66 0.000 0.98 0.968365


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Interaction 2 823.25 823.25 411.625 58.59 0.000 97.89102 96.83654Residual Error 10 70.25 70.25 7.025

Lack-of-Fit 2 7.25 7.25 3.625 0.46 0.647Pure Error 8 63.00 63.00 7.875

Total 15 3331.0 3331

Q20. We are intending to open the appended file (Response Optimizer) and analyse the shape : 100 + 5.

Execute the response optimizer for the shape in accordance with the following conditions.

a) What is the optimal shape valueb) When time is 4.3 and pressure is 4.7, what is the new value of shape?c) What should be the spec of independent variable 'time' to meet the spec of dependent variables.


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Q6. Factorial was conducted to find out the optimum condition. From the following graphs indentify which of thesehas a non-linear response & what should be done instead.

GRAPH - 1 GRAPH - 2 is a non linear model

1) GRAPH - 1, FACTOR IS NOT EFFECTING THE RESPONSE. SHOULD LOOK FOR ANOTHER FACTOR2) GRAPH - 2, RESPONSE SURFACE DESIGN SHOULD BE DONE3) GRAPH - 2, REGRESSION TO BE DONE AGAIN4) GRAPH - 1, NO. OF REPLICATES TO BE INCREASED

Q7. The following are the characteristics of central composite design. What is incorrect

a. The experiment of 2 level is carried out under the assumption that the effect of factors are linear.b. When the outcome values measured at the central & factorial point are different, it can be curvedc. The experiment of 3 level can be said to be central composite design.d. If repeats the cube point in 2 level DOE, it is possible to find curved regression

Q8. Which of the following is incorrect explanation of experimental Pure error/Lack-of-Fit ?1. To analize Lack-of-Fit, the independent variables must be repeated more than twice at the same level. correct2. Pure error value shows the reproducibility of response in a fixed independent variable. correct3. To make the regression equation to be calculated significant, P value of Lack-of-Fit must be significant.4. As Pure error value is smaller, the probability of regression equation to be significant is higher. correct

Q9. Which of the following is incorrect explanation of experimental Pure error/Lack-of-Fit ?(more than one can be the answer)

a. To analyze Lack-of-Fit, the independent variables must be repeated at the same levelb. For the regression eq to be significant, p value of Lack-of-Fit should be less than Alpha value.c. The smaller p value of Lack of fit, the larger p value of regression.

180175170

50

48

46

44

42

TEMP

180175170

40

30

20

10

TEMP Point TypeCornerCenter

Point TypeCornerCenter


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d. Pure error value shows the reproducibility of response in a fixed independent variable. correct


a. Write down all the terms which are insignificant ca,cb,abcb. Identify which variable has the maximum influence on the response Ac. Choice the wrong statement (more than one can be the answer)



Q11. The following are the explainations of the factorial design. Find out the incorrect explaination1. Randomization is to be carried out to increase the reproducibility.2. Confounding is to be used to raise experimental pricision.3. When there is no block effect, it is possible to join together and analyze as a whole.4. If central value is repeated, it is impossible to measure P value of the factor.

Q12. The following are the explainations of the fractional factorial design. Find out which one in incorrect.


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a) It is used when it is hard to run the whole experiment for time or economical reasonsb) In case of fractional faxctorial, there is always some probability of confoundingc) It is possible to know the interaction effects on the 2^(3-1) designd) If the experimental increase, the DF of total increases.

Q13. The following are the explainations of the confounding of 2 (5-1) fractional factorial designWhat is incorrect explaination.

a) Factor A & B are confounded with factor C,D & E correctb) The interaction effect of higher than 3rd order does not significantly effect the responsec) It is possible to reduce the experimental runs from 32 to 16 correctd) It is impossible to know the effect of third order interaction effect in correct

A B C D E-1 -1 -1 -1 -11 1 1 1 11 -1 -1 1 -1-1 -1 -1 -1 -1-1 -1 1 1 1-1 1 -1 1 -1-1 1 -1 -1 11 -1 1 1 -11 1 -1 -1 -1

1 1 -1 1 1-1 1 1 1 -11 -1 1 -1 1-1 -1 -1 -1 11 -1 -1 1 1-1 1 -1 -1 11 1 1 1 -1

Q14. We examined the level of each factor to make a experimental plan. When th elevel of each factor is as below,what is the frequency of all experiments.

Experiment ConditionFact Level Repeat : 2 times

A 3 Full factorial

B 2C 2


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a) For full factorial what would be the frequency of all experiments 24b) For 1/2 factorial design, what would be the ttl number of runs 12

Q15. While developing the washing machines with new design, several samples were manufactured. Now we make anexperiment on cleaness of cloth to washing hour and volume of water.Washing hour : time (min), Volume of water: water(gal), Washing effect (Response variable) : Y

(1) When the volume of water is in (-)level and washing time is in (+) level, get the response value (washing effect)(2) The following are the explainations of the interaction analysis results. What is correct.

a) There is no interaction. incorrectb) There is some interaction. correctc) When amount of water increases from (-) level to (+) level & washing time is at (-) Level,

washing ability appears to be high incorrectd) As amount of water goes from (+) to (-) Level, Washing ability appears high correct

(3) The following is the effect plot from the following data. Identify which bar represents the factor Time b

Time Water Y10 4 20.4

10 4 19.310 4 17.610 4 16.320 8 17.420 8 17.720 8 23.220 8 20.4

T e r m

Standardized Effect

B

A

AB

2.52.01.51.00.50.0

2.179

(a)

(b)

(c) M e a n o f Y

18.5

18.0

17.5

17.0

16.5

16.0


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10 8 9.710 8 16.410 8 14.810 8 12.320 4 1520 4 2420 4 15.620 4 15.2

17.45

Q16. a) Find out the Wrong explaination1. More the Insignificant regression equation, Lesser the P-value of Lack-Of-Fit2. Interaction terms are much more significant than the Linear Terms3. The interaction term B * C is not significant4. Using 95% CI & removing all possible insignificant terms, the maximum

upper CL value would be 14.67b) Find out the R-Sqr & R-Sqr (Adj) value (by completely removing all insignificant terms) Rsq 77.7 rsq ad 61.5c) At 14 + 5, what would be the optimum values of your factors A 5.8978 B 0.59 Cd) After screening the insignificant terms, What can be the regression model equation

.9646A+.3090B+.7622C-1.8363A*A+1.6700*B*B-1.9068A*A*B+3.3404*A*C A B C Y

2.03 2.49 3.49 11.853.62 2.97 2.42 11.06

3.62 0.59 2.42 12.083.62 1.78 2.42 11.023.62 1.78 2.42 10.145.21 2.49 3.49 11.030.95 1.78 2.42 8.263.62 1.78 2.42 9.503.62 1.78 4.22 10.435.21 1.07 3.49 10.903.62 1.78 2.42 10.223.62 1.78 2.42 11.533.62 1.78 2.42 10.535.21 2.49 1.35 7.716.29 1.78 2.42 7.872.03 2.49 1.35 13.192.03 1.07 3.49 8.94

5.21 1.07 1.35 8.443.62 1.78 0.62 7.982.03 1.07 1.35 11.28


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Q17. Identify which is the wrong explaination for fractional factorial designa) Display available design is used to identify the appropriate Factorial design correctb) If we increase from 1/2 to 1/8 factorial, condounding effect will increase correctc) We select that factorial design which has maximum condounding effect wrongd) 2 3-2 mean 2 level, 3 factors & 1/4 factorial design correct

Q18. a) Using Pooling, find out which factors are insignificant Speed & vibrationb) Find out the R-Sqr & R-Sqr (Adj) value (by completely removing all insignificant termsl) R-Sq = 88.5% R-Sq(adj) = 56.9%c) At 60 + 10, what would be the optimum values of your significant factors Speed 15d) After screening, what could be the maximum Upper 95% Confidence interval value Catal -0.0993

Vib 119.8Speed catal vibration temp density response

10 -1 100 180 3 6910 1 120 140 6 6715 -1 100 140 3 5310 -1 100 140 6 5615 1 100 140 6 6510 1 100 140 3 6310 -1 120 140 3 5315 -1 100 180 6 45

15 -1 120 140 6 5515 1 100 180 3 9310 -1 120 180 6 4910 1 120 180 3 9515 -1 120 180 3 6015 1 120 180 6 8210 1 100 180 6 7815 1 120 140 3 61

Q19. From the following analysis of variance for response, measure the following valuesa) R- Sqr Valueb) R-Sqr (Adj) Value

Analysis of Variance for response

Source DF Seq SS Adj SS Adj MS F PRegression 5 3260.75 3260.75 652.150 92.83 0.000 Rsq Rsq AdLinear 3 2437.50 2437.50 812.500 115.66 0.000 0.98 0.968365


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Q1. Tick the incorrect explaination for the Gage Linearity & Bias study1. With increase in Slope (S), %Linearity will also increases2. For %Linearity less than 5%, Average Bias value should be same for all the parts3. For Acceptable Gage Linearity & Bias, both should be less than equal to 5%

4. With increase in R-Sqr value, %Linearity will decrease

Q2. The following are explanations of response surface experiment. What is incorrect?1. It is possible to determine the spec of each independent variable to meet the spec of dependent varia2. When an independent variable has variation, it is possible to determine the control amount of the othe3. It is used to determine the spec of optimal factor after screening experiment.4. When the interaction between independent variables is large, it cannot be used.

Q3. When actual mean is greater than Target Mean & Zlt = 0, then from the followinggraphs identify which type of process will it have

Q4. The following are explanations for ANOVA. Which one is incorrect?1. It is a tool to compare averages forcondinuous data.

2. It is a technique used to determine the correlation of the dependent variable (Y) and independentvariables for discrete groups of more than two by the statistic importance.3. It is a method to screen latent Vaital Few Xs.4. It determines if the between group variation value is larger than the rationally expected value.

Q5. Which of the following is incorrect explanation of experimental Pure error/Lack-of-Fit ?1. To analize Lack-of-Fit, the independent variables must be repeated more than twice at the same level.2. Pure error value shows the reproducibility of response in a fixed independent variable.3. To make the regression equation to be calculated significant, P value of Lack-of-Fit must be significant.4. As Pure error value is smaller, the probability of regression equation to be significant is higher.

Q6. Identify which is the wrong explaination of the following Gage R&R Study Graph

(a) (b) (c) (d)

LSL T USL


40

20

0

_ R=11.04

UCL=36.06

LCL=0

1 2 3

1 2 3

R Chart by operator

Xbar Chart by operator


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1. Gage R&R is acceptable since Part to Part variation is more than the Measurement system variation2. Gage R&R is not acceptable since Measurement System Variation is more than the Part to Part Variation3. Reproducibility value is less than the Repeatibility value4. For the above case at % SV = 100%, that means maximum variation is contributed by the Instrument

Q7. The following are to ascertain the error of a measuring system. What is different from the others1. Stability2. Bias3. Straightness4. Repeatability

Q8. If we want to know the trend of population we can divide the population into specific intervals for certain process and get samples from the intervals which we had divided.This type of sampling method is called as1. Systematic Sampling2. Multi - Stage Sampling3. Stratified Sampling4. Random Sampling

Q9. For regression equation to be significant, Residuals should be normally & randomly distributedFrom the following graphs identify which of them are randomly distributed

Q10. The following are the explanations of control limit determination in the control chart.What is incorrect.1. The control limit uses + 3 s traditionally2. + 3 s is the area of 99.73%3. If increase the number of samples, the control limit widens4. If it is the variation by a chance cause, the data is mostly within the control limit.

Q11. Which has the following graph shows highest interaction

S a m p l e M e a n

50

30

10

_ _ X=28.96

UCL=49.71

LCL=8.20

F Fit Fit Fit

Residual Residual Residual Residual(a) (b) (c) (d)

(a) (b) (c) (d)


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Q12. The following are the characteristics of central composite design. What is incorrecta. The experiment of 2 level is carried out under the assumption that the effect of factors are linear.b. When the outcome values measured at the central & factorial point are different, it can be curvedc. The ecperiment of 3 l3vel can be said to be central composite design.d. If repeats the cube point in 2 level DOE, it is possible to find curved regression


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3

3les.variables.

3

3

3

3


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3

3

3


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Q1. An ABC company (manufactures Sheet) found that increased number of sheets are being rejectedat final inspection due to scratches, peels & smudges in their plant.Inorder to check that 10 samples from each work shift were studied with respect to the Types of defectsConduct the Pareto Chart of the following and answer the below mentioned question

Defects Type Period Defects Type Period Defects Type PeriodNIL Day NIL Evening Scratch NightScratch Day NIL Evening Scratch NightNIL Day Smudge Evening Smudge NightNIL Day Peel Evening Peel NightSmudge Day NIL Evening NIL NightNIL Day Peel Evening Smudge NightNIL Day Scratch Evening Peel NightNIL Day NIL Evening Scratch NightNIL Day NIL Evening Smudge NightScratch Day Scratch Evening Smudge Night

a. In which Work Shift, Smudges are maximumb. Which Shift has maximum number of defectsc. Mark the incorrect Statement

1. In evening shift, Smudges frequency is minimum2. Relationship exists between the type of defects and the work shift producing the Sheets.3. At night, maximum number of defects have occurred4. At night 90% of the defects are contributed by Smudges & Scratch

Q2. Conduct the sutable Process Capability of the given data and answer the below questions (Ans in 4 decimala. What is the Minimum & Maximum DPU Valueb. If the opportunity of failing a unit is 100, then what would be the maximum yield valuec. Considering Max Yield as RTY & Processes as 100 then, what would be the normalized yield valued. Also from the same find out the Zlt & Zst value (assume Zshift as 1.2)

Q3. Conduct Simple regression of the following data and answer the below questionsa. Write down the Regression equationb. Using regression eq find out what could be the response value at 600, 720 & 920 (Round off to 2 decimal placc. Write down the R-Sqr & R-sqr (adj) value (Round off to 3 decimal places)d. Using 90% Confidence interval, What will be the maximum CL value (Round off to 1 decimal places)

RPM evaporation500 0.18540 0.37 Response is evaporation

580 0.35620 0.50660 0.56 evaporation = - 1.12 + 0.00268 RPM700 0.75 0.488740 1.02 0.8096780 1.18 1.3456820 1.05 Re Seq 0.925325184860 0.94 Re Ad900 1.50


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940 1.56980 1.65

1000 1.401020 1.501060 1.63


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10

226

l places) 122433

122

s) 244


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Q1.

a. Nightb. Nightc. Option 4th

Q2.

Defects Type

C o u

n t


10.0

7.5

5.0

2.5

0.0


10.0

7.5

5.0

2.5

0.0

Period = Day Period = Evening

Period = Night

Defects TypeNILScratchSmudgePeel

Pareto Chart of Defects Type by Period

Sample

S a m p l e C o u

n t P e r

U n i t

1009080706050403020101

0.075

0.050

0.025

0.000

_ U=0.02652

UC L=0.06904

LCL=0

D P U

0.030

0.025

0.020

0.015

Summary S tats

0.0000M a x D PU : 0. 07 53Targ DPU: 0 .0000

(using 95.0% confidence)

Mean DPU: 0.0265Low e r CI : 0. 02 37U ppe r CI : 0. 02 95Min DPU:

Sample Size

D P U

140120100

0.075

0.050

0.025

0.000

4

16

12

8

4

0

Tar

1

Poisson Capability Analysis of number of defectsU Chart

Tests performed w ith unequal sample sizes

Cumulative DPU

Defect Rate

Dist of DPU


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Maximum DPU = 0.0753Minimum DPU = 0.0000

b. 0.9274

c. 0.9992Zlt = 3.173525Zst = 4.373525

Q3.

a. evaporation = - 1.12 + 0.00268 RPM

600 = 0.49720 = 0.81920 = 1.35

c. Source DF SS MS F PRegression 1 3.5427 3.5427 173.50 0.000Residual Error 14 0.2859 0.0204Total 15 3.8286R-Sqr = 92.533%R-Sqr (Adj) = 92.008%

d. 1.8

b.

a.

d.

Sample

0 . 0

0 . 0

0 . 0

0 . 0

0 . 0

0 . 0

0 . 0

0 . 0


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Lot_No quantity of test number of defects DPU1Lot 132 2 0.0151522Lot 130 4 0.030769

3Lot 120 3 0.0254Lot 124 1 0.0080655Lot 138 2 0.0144936Lot 148 5 0.0337847Lot 101 2 0.0198028Lot 102 5 0.049029Lot 124 4 0.03225810Lot 119 1 0.00840311Lot 120 6 0.0512Lot 123 3 0.0243913Lot 101 3 0.02970314Lot 121 6 0.049587

15Lot 133 1 0.00751916Lot 138 4 0.02898617Lot 113 1 0.0088518Lot 119 8 0.06722719Lot 128 1 0.00781320Lot 103 4 0.03883521Lot 140 4 0.02857122Lot 150 2 0.01333323Lot 121 4 0.03305824Lot 140 2 0.01428625Lot 114 1 0.00877226Lot 140 2 0.01428627Lot 136 2 0.01470628Lot 114 3 0.02631629Lot 149 4 0.02684630Lot 110 4 0.03636431Lot 100 1 0.0132Lot 138 0 033Lot 118 4 0.03389834Lot 116 6 0.05172435Lot 131 5 0.03816836Lot 146 11 0.07534237Lot 147 1 0.00680338Lot 142 4 0.02816939Lot 140 2 0.014286

40Lot 142 4 0.02816941Lot 136 2 0.01470642Lot 139 2 0.01438843Lot 147 3 0.02040844Lot 122 5 0.04098445Lot 149 1 0.00671146Lot 142 1 0.00704247Lot 116 2 0.01724148Lot 146 2 0.013699

Question Number 2


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49Lot 140 5 0.03571450Lot 129 3 0.02325651Lot 100 3 0.0352Lot 124 6 0.04838753Lot 141 5 0.03546154Lot 130 8 0.061538

55Lot 102 5 0.0490256Lot 110 2 0.01818257Lot 134 4 0.02985158Lot 145 4 0.02758659Lot 110 3 0.02727360Lot 105 3 0.028571 061Lot 148 4 0.02702762Lot 144 3 0.02083363Lot 100 4 0.0464Lot 102 2 0.01960865Lot 142 6 0.04225466Lot 105 4 0.038095

67Lot 133 2 0.01503868Lot 129 5 0.0387669Lot 108 3 0.02777870Lot 103 4 0.03883571Lot 132 2 0.01515272Lot 108 2 0.01851973Lot 111 4 0.03603674Lot 107 7 0.06542175Lot 108 5 0.04629676Lot 102 1 0.00980477Lot 111 7 0.06306378Lot 102 2 0.01960879Lot 137 2 0.01459980Lot 128 0 081Lot 120 1 0.00833382Lot 124 5 0.04032383Lot 100 2 0.0284Lot 135 2 0.01481585Lot 148 4 0.02702786Lot 103 6 0.05825287Lot 127 3 0.02362288Lot 104 3 0.028846

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