All DMAIC Paper for B B

7/31/2019 All DMAIC Paper for B B

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Hypothesis testing

Level-3 1.The following are the hypothesis testing procedures.Connect the stage

Level-5 2.The following figure explains the concept of hypothesis testing.Have a look at the example below and choose a correct answer.

[ Example ]

Acceptance Rejection region Right decision

Consumer'sProducer's risk Significant level

1) Write a correct term in the region A.

Consumer's risk

2) Write a correct term in the region B.

Right decision

3) Write a correct term in the region C.

Right decision

4) Write a correct term in the region D.

Producer risk

Level-5 3. The following is the data examined on defective types (A,B,C) by the fac

Fill the blank.

A B C Total

1 20 40 33 9328.83 40.92 23.25

2 35 49 11 95

29.45 41.8 23.753 32 43 31 106

Chi-Square Test: Defect A, B and C

1

2

3

4

5

Set up H1 and Ho.

Select the significant level. - Normally

Select the testing method. (Select the menu from

By using the data observed, calculate p v

By comparing P value and the significant level, dra

A

C

Ha(false)

Ho

Ha

Fact

Decisionmaking

1


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32.86 46.64 26.5

Total 93 132 75 300

Chi-Sq= 2.704 0.021 4.089

1.046 1.240 6.845

0.023 0.284 0.764 17.015

DF= 4 P-Value = 0.007

1) Calculate the number 1 value in the .

41.8


1.24


4

Level-7 4. According to the data, the size of pipes coming out of one cutting machiThe following are experimental results to improe it is tested under two c

4-1.What is an incorrect explanation? ( 2 )1) For the significant difference test for the average of two conditions, t- test is desi

2) Before carrying out the average value test, it is desirable to perform the homoge

3) Before carrying out the average value test, the normality test must be carried ou

4) The null hypothesis under these experimental conditions is that there is no impro

4-2. Calculate the pooled StDev value.0

Level-5 5. Answer the following questions.

T-test of the Mean

Test of mu 5.39 vs mu not = 5.39

variable N Mean StDev SE Mean T

10 5.39 0 (1) (2)

5.1) Calculate the standard error value of (1).

0.00035

CTQ size

+

+

+

+

+

+ =

2

3

Testing results:

Condition 1 vs Condition 2N Mean StDev SE Mean

Condition 1 10 5.3899 0.0011 0.00035

Condition 2 10 5.3947 0.00116 0.00037

95% CI for mu condition 1 - mu condition 2: (-0.000586 , -0.000374)

T-Test mu condition 1 = mu condition 2 (vs not =): T = (-9.5) DF = 18


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5.2) Calculate the T statistic value of (2).

-11.79 FORMULA ,=(D99-C97)/B102 ,=,(SAMPLE M

-11.79


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th the process.

ory.

41.8

%

the minitab.)

ue.

a conclusion.

D

B

Ho(true)


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e is different by time zone in a factory mass-producing pipes.nditions. Answer the questions.

able.

eity test of variance.

ement effect for two experimental condition.

P

0


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AN-TEST MEAN)/SEM


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Question Text

1. Define (1-1)

"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 data is followings and the population is known as

normal distribution.

----------------------------------------------------------------------------------

Sample data : , 52.3, 50.6, 72.3, 49.3, 76.8, 58.7, 62.6, 56.8, 5

----------------------------------------------------------------------------------

(Round off up to 1 decimal place) (5 points)

Upper =64.9 & lower =54.6

5. Control

The following table is to plot control chart of certain continuous data.

-------------------------------------------------------------------

Sample No Average Width,X bar Range, R (Sample Size=6)

1 17 25

2 35 30

3 -1 2

4 -3 18

5 10 30

6 19 24

7 13 40

8 13 24

9 15 36

10 11 16

--------------------------------------------------------------------

3. Analyze

We summarized the defective types of refrigerator by shift for 3 months, and are

going to check if there is a relation between the shift and the defective type.

Answer the following questions.

(1-1) Calculate the upper limit of 90% confidential intervals of the population.

(5-1) Calculate UCL of X bar (Round off up to 3 decimal place) 24.734

(5-2) Calculate UCL of R bar (Round off up to 3 decimal place) 49.098

(5-3) How many points are located outside of the UCL and LCL of X bar Chart? 3

(5-4) How many points are located outside of the UCL and LCL of R bar Chart? 0


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(Round off up to 2 decimal place)3.45

the second shift.(Round off up to 2 decimal place)

11.81

We summarized the defective types of refrigerator by shift for 3 months,

and are going to check if there is a relation between the shift and the defective type.


Calculate the expected value for poor adhesion of the workers in the second shift.(4 Marks, Round off up to 2 decimal place)

11.81

Many samples have been made while developing washing machines of new design.

We are going to have an experiment on the effect of washing time and amount of water to clean clothes.


The following are explanations of the interaction analysis results.

When amount of water is (-) level and washing time is at (+) level,

The development department uses certain adhesive material to stick two products.

Now, it is going to develop an adhesive material with high adhesive strength.

If the adhesive strength is more than 20, the condition will selected and

be put under process control. Answer the following questions.

(All data are normally distributed.)

Test the homogeneity of variance and calculate P-value. (5% significant difference test)






Calculate the upper confidence interal limit value of standard deviation for the adhesive material with level 2.

There was a newspaper report that the children in a smoking family went to

hospital more often than the children in a nonsmoking family.

To support the report objectively, we sampled 300 smoking families and 200 nonsmoking

families at random, and examined the families with the children going to hospital more than

two times for a month on the average. The result was 90 of smoking families

(3-1)Calculate the P-value.(Round off up to 3 decimal place)

(3-2)Calculate the chi-square value for leakage of the workders in the third shift.

(3-3)Calculate the expected value for poor adhesion of the workers in

calculate the response value(washing ability)? (5 Mark , round off upto 2 decimal place) =14.9

( 3 Marks, round off upto 3 decimal place ) =.452

(95% of confidence interval)(3 Marks, round off upto 3 decimal place ) =6.324


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and 50 of nonsmoking families. With the significant level of 5%,

we tested that it could be said that the children of smoking family went to hospital more often than

the children of nonsmoking family.

Answer the questions. (Use the minitab)

By using the following data, calculate the upper control limit value of p control chart.(Use the Minitab, round off upto 3 decimal place)(5 Marks)

0.1

By using the following data, Find out how many samples are over the upper control limit?

Using the below given data.

Y=A+B , Target of Y = 15 and both sided tolerance would be necessary.

Mean of A = 10, Mean of B = 5Short term Standard deviation of A = 0.1 , Short term Standard deviation of = 0.3

A and B are independent.

Calculate Upper Spec Limit of Y for meeting 6 level.

(Round off upto 2 decimal place) (5 Marks)

using data below.

Y=A+B , Target of Y = 15 and both sided tolerance would be necessary.

Mean of A = 10, Mean of B = 5

Short term Standard deviation of A = 0.1 , Short term Standard deviation of = 0.3

A and B are dependent and correlation is 0.7.

Calculate Upper Spec Limit of Y for meeting 6 level.

(Round off upto 2 decimal place) (5 Marks)






What is P-value ? (5 Marks, round off upto 3 decimal place) =.217

(Use the Minitab)(5 Marks) =0

What is the adhesive material with the adhesive strength of more than 20 ? (4 Marks) (level 3)


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52.3

50.672.3

49.3

76.8

58.7

62.6

56.8

58.3

59.744

64.9

54.6

3.16

17 25

35

-1 X bar Range

-3

6

17 25

10 35 30

19 -1 2

13 -3 18

13 10 30

15 19 24

11 13 40

13 24

15 36

11 16

12.9

24.5

24.734

1.067

Upper 49.098

0


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Level3 3

2119

21

16

23

No. Washing time(mi

1 10

2 10

3 10

4 10

5 20

6 20

7 20

8 20

9 10


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11 10

12 10

13 20

14 20

15 20

16 20


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1 17 25

2 35 30

3 -1 2

4 -3 18

5 10 30

6 19 24

7 13 40

8 13 24

9 15 36

10 11 16


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Cut Leakage Switch def poor adhesion

15 21 45 13

26 31 34 533 17 49 20

in.) Amount of water(gal.) Washing ability1

4 20.4

4 19.3

4 17.6

4 16.3

8 17.4

8 17.7

8 23.2

8 20.4

8 9.7


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.

8 14.8

8 12.3

4 15.0

4 14.5

4 15.2

4 15.6


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No.

1

2

3

4


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Question Text

By using the following data, calculate the upper and lower control limit value of Xbar & R control chart.

(Use the Minitab)(2 Marks)

1) xbar ucl =2.52

2) xbar lcl =.627

3) R ucl =3.464) R lcl = 0

write down the two decimal point.

By using the following data, calculate the upper and lower control limit

of P Chart. (Use the Minitab, write down the 4 decimal point)(2 Marks)

1)P UCL =.1312

2)P LCL =0

By using the following data, calcuate the upper and lower control limit

value of c control chart. (Use the Minitab)

----------------------------------------------

month 1 2 3 4 5 6 7 8 9 10

No. of 2 4 1 1 4 5 2 1 2 4

Defects

-----------------------------------------------

Ucl=7.437 ,LCL=0

The followings are explanation of continous control chart. Find out 2 wrong answers. ( 1 Mark )

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.


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1.66 1.7

1.88 1.6

1.02 1.9

1.73 1.61.58 1.4


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Gage R&R

Level-2 1. When the same product is measured with a measuring instrument, whatmeasured values and the base value of characteristic values called? ( 4

Level-2 2. What is the evaluation of measurement coherence within the range of

Level-2 3. When measuring the same part with a measuring system, what is the va

Level-3 4. What is the following graph about?( 1 )

Level-3 5. The following are matters to be generally considered before evaluating

Level-3 6. The following are matters to be considered when the bias is larger than

Level-5 7. The following are matters to be considered when the linearity of a meas

5) Check the discrimination of the measuring instrument used by the measuring per

Level-7 8. Connect related ones to each other.

Level-3 9. The following are explanations of statistical characteristics that the me

Level-5 10. The following are to ascertain the error of a measuring system. What i

1) Repeatability 2) Reproducibility 3) Precision 4) Inaccuracy

1) Test and rectification 2) Linearity 3) Accuracy 4) Orthogonali

1) Process variation amount 2) Within group variation 3) Between group v

1) Bias 2) Linearity 4) Stability 4) Repeatability 5) Re

1) The number of evaluators, samples and repeatition must be fixed in advance.

2) Specially for main parts, the size of samples and the frequency of repeatition mu

3) Evaluators must be composed of the personnel participating in the project.4) Samples must be taken during the process, and they must represent the conditio

5) The scale of a measuring instrument must have the discrimination possible to re

1) Check if the base value is accuate.

2) Check if the measuring instrument is worn out.

3) Check if the measured location is accurate.

4) Check if the measuring instrument is properly rectified.

5) Check if the measuring person used the measuring instrument properly.

6) Check if the stability of the measuring instrument was evaluated.

1) Examine if the scale of upper or lower part of the measuring range is proper.

2) Check if the base value is accurate.

3) Examine if there is any problem in the design of the measuring instrument itself.

4) Check if the measuring instrument is worn out.

1) The measuring system must be under the statistical control condition.

2) The dispersion by the measuring system must samller than the dispersion by the

3) The dispersion must be smaller than the product specification.

4) The measuring instrument must be able to read up to the scale one larger than t

1) Stability 2) Bias 3) Straightness 4) Repeatability

Linearity Bias Stability Reproducibility


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2) The measuring person 2 generally has tendency to measure the second measure

4) For the part number 7, when considering that the measurement values of 3 meas

Level-9 12. The grapth in the lower right shows the results ascertained the linearit

gage R&R analysis. What is incorrect? ( 1 )

Level-7 13. 3 measuring persons measured 10 samples twice repeatedly. As a resu

What is a correct explanation? ( 1 )

Level-7 14. 3 operators measured 10 samples twice repeatedly. As a result, the folWhat is a correct explanation ? ( 1 )

Level-7 15. 3 measuring persons measured 10 samples twice repeatedly. As a resuWhat is a correct explanation ? ( 4 )

L l 7 16 3 i d 10 l t i t dl A

measurement value.

3) For the part number 10, every operator shows differenct measurement values so

examine the cause closely.

it seems that they did not carry out the blind test.

1) R-Squared value is 97.8%, and it can be said that the linearity is good.

2) A larger dot ( ) indicates the average value of bias by part.

3) A small dot indicates the bias value of individual data by part.

4) When measuring a sample with a small value, it provides a smaller value than th

When measuring a sample with a large value, it provides a larger value than the

Therefore, it can be said that the linearity is not good.

1) It indicates that the selected samples reflect the dispersion of process properly.2) If the dispersion of measurement data values is uniform,

it means that the samples reflect the dispersion of process properly.

3) It indicates that if each operator is measuring in a different way according to the

4) It indicates that the weight of R&R out of the whole dispersion is small enough.

5) It indicates that the repeated measurement values by operator are stabilized.

1) It indicates that if there is significant difference between operators.




5) It indicates that if the capability to discriminate each part different from the othe





5) If the dispersion of measurement data values is uniform,



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Level-7 17. 3 measuring persons measured 10 samples twice repeatedly. As a resu

What is a correct explanation ? ( 1 )

Level-7 18. 3 measuring persons measured 10 samples twice repeatedly. As a resuWhat is a correct explanation ? ( 2 )

Level-9 19. The following are analysis results of variance of gage R&R. Answer the(3 measuring persons measured 10 samples twice repeatedly.)

Two-Way ANOVA Table With Interaction

Source DF SS MS F P

Part 9 2.06 0.22875 39.72 0.00000

Operator 2 0.05 0.02400 4.17 0.03256Operator*Part 18 0.1 0.00576 4.46 0.00016

Repeatability 30 0.04 0.00129

Total 59 2.25

1) Calculate the variance value (variation amount) 0.001292 .Number 10

.Variation amount of repea

2) Calculate the variance value (variation amount) .Variation amount of repro0.003146 ^2 op =

^2 op*prt =

3) Calculate the total variation amount of gage R& .Variation amount of repro0.004438 ^2 RPD= ^2o

.Variation amount of ^2



















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1 2 2 0

2 2 2 0.01

3 2.01 2 0

4 2 2 0.01

5 2 2.01 0 Sum of range verage of range

0.02 0

0.01

49.05%

1. Spec= 2.000.015 , 2. For d value, see the teaching materials.

Part uring per uring per Range

1) Calcuate the gage error.

2) Calculate the gage error for the allowable error.


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is the difference of the average of)

easurement of the measuring instrument called? ( 2 )

riation degree of measured results in course of time called? ( 4 )

age R&R. Which one is incorrect? ( 3 )

expected. Which one is incorrect? ( 6 )

uring instrument was bad. Which is incorrect? ( 5 )

son.

suring system must have. What is incorrect? ( 4 )

different from the others? ( 4 )

5)Stability

y 5)Stability

riation 4) Stability 5) Straightness

roducibility

t be increased to raise the reliability.

n of whole process.

d up to 1/10 of the specification at least.

manufacturing process.

e specifiation.


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ment value smaller than the first

uring persons are similar,

y of a measuring instrument after

t, the following graph is obtained.

owing graph is obtained.


t th f ll i h i bt i d

t is necessary to

practical value.

ractical value.

sample.

sample.

s is enough.

sample.

s is enough.


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following questions.

Number of repeatition= 2 .Number of measuring persons= 3

tability(variance)= ^2RPT= ^2e=MSe= 0.001292

ucibility(variance)= ^2op + ^2op*prt

0.000912

0.002234

ucibility(variance)= ^2op + ^2op*prt

p + ^2op*prt = 0.003146

&R= ^2RPT + ^2 RPD 0.004438

s is enough.

sample.

s is enough.

sample.

s is enough.


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Allowance

0.03


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< Logic Tree/Pareto/Brainstorming

3 1. The following are explanstions of when to use the brainstor

2 5. The following are explanations of four principles of brainsto

3 6. The following are precautions at the brainstorming. What is

5 7. The following are explanations of pareto chart. Which is inc

5 8. The following are explanations of a cause-and-effect diagra

3 9. Have a look at the following example, and connect related o

1) When it is necessary to find out all fundamental causes possible for

2) When it is necessary to decide what kind of improvement activities

3) When it is necessary to find improvement methods for a process, pr

4) When it is necessary to promote the creativity of a team

5) When it is necessary to make a decision

1) It is strictly prohibited to criticize as good or bad.

2) Free spirited atmosphere is to be assured.

3) Quality of idea is to be pursued rather than quantity of idea.

4) Add oneself's idea to another person's idea and develop it.

1) It is unsuitable for a wide or complex problem.

2) It is unsuitable for conditions that have to go through trials and erro

3) It has nothing to do with the leadership.

4) During the brainstoring, walking around or having a discussion sepa

1) A full line is drawn in the upper part of the bar graph to indicate tha

2) Important causes of 20% occur 80% of the whole problem and it is

3) It is a kind of bar graph, and bars are arranged from larger ones to

4) The pareto chart is easy to make out and has a great effect. Theref

1) It is a diagramatic representation of arranging the results of a work

2) The results of a work is called a characteristic. A characteristic indic

3) A factor is the larger one among causes. One out of those causes is

recorded in the cause-effect diagram as a factor.

4) When making out a cause-effect diagram, experts in improving the


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7 10. What is a correct order to make out a pareto chart? ( 2

2) Stat > Quality tools > Pare

4) Graph > Character Graph

9 11. What is a correct order to make out a pareto chart? ( 4

1) Stat > Quality tools > Paret

3) Graph > Character Graph

. To calculate the percentage and cumulative percentage for each it

. To measure related elemements

. To make a ranking list in order from the category (items) with the h. To draw bars representing each category by using the data values i

. To make out a chart

. To indicate cumulative percentage in the pareto chart

. To make out a list of all elements

1) ,,,,, 2) ,,,,, 3) ,,,,, 4) ,,


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ing. What is incorrect? ( 5 )

ming. What is incorrect? ( 3 )

incorrect? ( 3 )

rrect ? ( 1 )

. What is incorrect ? ( 4 )

es to a cause-effect diamgram with each other.

the problems

hould be done

oduct or service

rs.

rately should be avoided .

which factor has a strong influence.

articulary called "20:80 law".

maller ones and from left to right.

re, it is used as a technique to improve the quality.

and the causes having influence on those results.

ates the length, velocity, defective proportion, etc., and it is an abbreviation of q

to be thought as it has influence on the characteristic and

uality must participate and make progress.

Middle bone

Back bone

Small bone

Grandchild bone


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)

o chart > Chart defects table > ok

Set Options > Chart defects table > ok

)

o chart > Chart defects data in > ok

Set Options > Chart defects in > ok

m

ighest value to the category with the lowest valuesn the list

,,,,


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uality characteristic.


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< RTY >

3 1. The following are explanations of discrete data. What is incorrect1) Discrete data is the number of pass or fail.

2) Discrete data is measured with the frequency of occurrence, and can be e

3) Yield is a general type of discrete data.

4) It is meaningful when discrete data is divided more in detail.

7 2. The following are definition of terms. What is incorrect ? ( 4 )1) All kinds of unfitness in a specific product are called a defect.

2) Opportunity means everything that is examined or measured.

3) Unit means a part, assembly or system to be tested.

4) The meaning of defect in a dictionary is what is already found, confirmed

5 3. There are 1100 defects among the 1000 washing machines currenWhat is a correct explanation ? ( 4 )

1) DPU is 0.91.

2) It means that all units of manufactured product has statistical significant

3) It means that all units of manufactured product probabilistically contain o

4) It means that all units of manufactured product contain one defect on the

5 4. In 1000 manufactured washing machines, there are 10 opportunit800 defects in total were detected. What is a correct explanation ?

1) DPO is 0.8.

2) It means that the defect by unit region of opportunity is about 0.1 on the

3) It means that the defective probability by unit region of opportunity is abo4) DPO is a ground to decide Z value with using discrete data.

3 5. What is correct ? ( 4 )1) To convert the results of discrete data into Z value, it must be converted i

2) It is possible to calculate Z value with using DPO and Z-table. At this time,

3) If DPO is 0.08, Z value is 1.91.

4) It takes a lot of samples and time to collect discrete process data and so

it is impossible to calcuate Zst in general.

5 6. The followng are explanations of Yft. What is incorrect ? ( 3

1) It is the yield at each stage of process.2) It is the yield after examination or test.

3) Reworking and disposal are not included.

4) It always has a larger value than Yrt.

5) It provides a probability to pass a stage of process without defect.

5 7. The following are explanations of Yrt. What is incorrect ? ( 2 )1) It is the yield at the final stage of process.

2) It is the yield before examination or test.

3) By-products of reworking and disposal are evaluated.

4) It always has a smaller value than Yft.

5) It provides a probability to pass the whole stage of process without defect


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7 8. Calculate Yrt. (To the second decimal place)

42% Roll Form

97%

Stamping Painting Painting

90.70% 96% 88%

Injection&Molding Assembly

91% 83%

Supplier System

91%

5 9. When the process to manufacture a product is divided into 200 stof 6 , what is Yrt(%)? (To the second decimal place)

Stage

% RTY

7 10. Calculate Yrt and normalized average yield. (To the second deci

Step 1 Step 2 Yrt99% 99% ? 98.01%

Yrt 0.9801 99.00%99%

5 11. When the yield (Yrt) of stage by process in the factory manufactby stage is 20, calculate the value of normalized average yield. (

99.80%

9 12. When the yield Yrt of stage by process in the factory manufactu

by stage is 9, calculate Zlt value.

YNA 0.99

Zlt 2.46

5 13. The following is the road map to calculate the yield value of disc

9 14. Calculate Z value for the following process. (To the second deci

Process 1 Process 2 Process 3

YNA

YNA

1) Yft > YNA > e-DPU > Zlt

2) Yft > YNA > Yrt > Zlt

3) e-DPU > Yrt > (Yrt)1/OPP > Zlt

4) e-DPU > Yrt > (Yft)1/OPP > Zlt


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97% 96%

Sub 1

99%

Process 4

Sub 2 89%97%

Process 5

Sub 3 78%

96%

Yrt 59.59%

0.93735

P(defectes) 0.06265

Zlt 1.53

9 15. Answer the questions about the following process. (To the secon

100

6 3 4

Disposal Disposal Disposal Disposal

15 5 3 0

p1 0p2 #DIV/0! p1p3 0 p2

p4 0 p3

Yrt #DIV/0! p4

yrt

1) Calculate Yrt.

#DIV/0!

2) Calculate the value of normalized average yield.

#DIV/0!

7 16. The following is the process map for the order/request process bAnswer the following questions.

Order Entry Credit Review Shipping

Yrt = " A " 75% Yrt = " C "YNA = " B " YNA = " D "

Contact Order Review Order Matching

93% 75% 78%

Availability Scheduleing

86% 85%

YNA

P 1 P 2 P 3 P 4


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Order Build W/house Loading

99% 98.60%

Pricing Shipping

95% E 94.287%

Ship'g truck Load Shipping direct

90.90% 97.80%

1) " A " 75.22%

2) " B " 93.13%

3) " C " 61.64%

4) " D " 88.61%


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? ( 4 )

pressed as a level.

nd examined.

ly manufactured.

1.1

fference.

e defect.

verage.

es for defect and( 4 )

1000

verage. 10

ut 0.1. 8000.8

0.08

0.45

to DPU first.

Z value is Zst. 0.92

1.41

41.851143691410900%


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0

Final Assembly

83%

Yrt 1 10.51

50.54

1 0.93

ges, and all processes have the process capability

200

0.0000

0.99999660 99.932%

al place)

ring a product is 96%, and the number of processo the second decimal place)

99.80%

ng a product is 94%, and the number of process

0.992.46

ete data with Z value. What is correct ? ( 3 )

al place)


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d decimal place)

RTY

Input repair scrap out putp1 100 0 15 85p2 85 6 5 74

p3 80 3 3 74

p4 77 4 0 73

P1 0.85

P2 0.87

P3 0.93

P4 0.95

y mail order. 64.89%

75.2211900%


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93.13%

61.637045387273600%

88.605%


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[ Performance evaluatoin of MFG

1. The following are the number of defective cases for examined items.

Analize the pareto chart and write all defective items having the cumulative p

Defective No. of cases

Dent 23

Poor Seal 34

O-Ring 48

Finish 12

Finish 45

Dent 33

Screw 9 1) What are the contents of defective items hav

Connecto 2 Whicg are in red color

O-Ring 26

Scratch 52

3. Which is not a method for brainstorming ? ( 4 )

1) Free Wheeling

2) Round Robin

3) Card Method

4) Association method

5. When % tolerance is 6.78 from above GRR analysis, what is the tolerance?

% tolerance 6.78 % toleran

Total GRR 0.27 Tolerance

Answere= 4.00 Tolerance

6. In A company, QA group carried out a data analysis for "B" cooperative compa

the population standard deviation is 10. The sample size is 10 selected at ran

the population is supposed to have a normal distribution.

44.1 52.3 50.6 72.3 49.3 76.8 58.7

1) Calculate the confidence interval of population average 90%. 63.4

53.0

2) Calculate the confidence interaval of population average 95%. 64.4

52.0

7. There was a newspaper report that the children in a smoking family went to hin a nonsmoking family. To support the report objectively, we sampled 300 sfamilies at random, and examined the families with the children going to hosp

on the average. The result was 90 of smoking families and 50 of nonsmoking f5%, we tested that it could be said that the children of smoking family went tof nonsmoking family. Answer the questions. (Use the minitab)

1) What is P-value ?

0.22

6 level. As the result of the data measuring CTQ of " " part supplied from B

Defective itemCount

20.1 19.7 18.3 12.0 3.9

Cum% 26.1 46.1 65.8 84.2

74

96.1 100.0

57 56 52 34 11

Percent 26.1

OtherPoor SealScratchDentFinishO-Ring

300

250

200

150

100

50

0

100

80

60

40

20

0


42/125

2) What is the result of hypothesis testing ? (Choose between Ho accept and Ha accept.)

Accept ho

8. We summarized the defective types of refrigerator by shift for 3 months, andif there is a relation between the shift and the defective type. Answer the foll

Shift Cut Leakage Switch defect Poor adhesion

1 15 21 45 13

2 26 31 34 5

3 33 17 49 20

1) Calculate the expected value for poor adhesion of the workers in the second shift.

11.81

2) Calculate the chi-square value for leakage of the workders in the third shift.

3.45

3) Calculate the P-value.

0

9. The development department uses certain adhesive material to stick two proNow, it is going to develop an adhesive material with high adhesive strength.is more than 20, the condition will selected and be put under process control.

Adhesive material with le esive material with lev ive material with level 3

9 18 21

12 15 19

14 14 21

13 17 16

18 15 23

1) Test the homogeneity of variance and calculate P-value. (5% significant difference test)

0.45

2) Calculate the upper confidence interal limit value of standard deviation for the adhesiv

6.32

3) What is the adhesive material with the adhesive strength of more than 20 ? ( 3 )

Adhesive material with level 1



10. Many samples have been made while developing washing machines of new dWe are going to have an experiment on the effect of washing time and amou


Washing timAmount of water(g Washing ability1

10 4 20.410 4 19.3

10 4 17.6

10 4 16.3

20 8 17.420 8 17.7

20

19

18


43/125

20 8 23.2

20 8 20.4

10 8 9.7

10 8 16.4

10 8 14.8

10 8 12.320 4 15.0

20 4 14.5

20 4 15.2

20 4 15.6

1) The following are explanations of the interaction analysis results. What is correct? ( 4

There is no interaction.

There is some interaction.

When amount of water increases from (-) level to (+) level and washing time is at (-) lev

As amount of water goes from (+) level to (-) level, washing ability appears high.

2) When amount of water is (-) level and washing time is at (+) level, calculate the respon

15.08

4

17

16

15

14

13


44/125

project ]

ercentage of 84.2%.

ng the cumulative percentage of 84.2%?

44.1

52.3

50.6

72.3

49.3

76.8

Formula 58.7

e = .=Gage R&R*100/tolerance 62.6

,=(Gage R&R*100)/%Tolerance 56.8

.=(.27109*100)/6.78 Mean 58.17

ny to receive QTA parts of n 10

st dev 10

om,data is given bellow. AT 90% confidence i

62.6 56.8 At 95 % confidence

ospital more often than the childrenoking families and 200 nonsmokingital more than two times for a month

amilies. With the significant level ofhospital more often than the children

cooperative company in last year,


45/125

are going to checkwing questions.

ucts.If the adhesive strengthAnswer the following questions. (All data are normally distributed.)

95% Bonferroni confidence intervals for standard deviations

C4 N Lower StDev Upper

Adhesive material with level 1 5 1.76780 3.27109 12.5901



Bartlett's Test (normal distribution)

Test statistic = 1.59, p-value = 0.452

material with level 2.(95% of confidence interval)

Levene's Test (any continuous distribution)


esign.t of water to clean clothes.

Washing

time(min.)

10

20




20


46/125

)

el, washing ability appears high.

e value(washing ability).

8

8

4

2010

Amount of water(gal.)

Washing time(min.)

19.675

15.07518.400

13.300


47/125

nterval 63.4

53.0

nterval 64.4

52.0


48/125

141210864

Bartlett's Test

0.706

Test Statistic 1.59

P-Value 0.452

Levene's Test

Test Statistic 0.36

P-Value


49/125


50/125

Q1. For finding the relationship between Y and two independent variables X1 & X2,

you collected total of 15 readings and found the regression model eqn as

Y = a + bX1 + cX2, Find out what would be the degree freedom of residual error

gree of freedom

12

Q2. The following are the explanations of FMEA, Mark the inncorrect statement

a. It checks the potential failure mode

b. It analyzes the influence of failure mode on the achievement of system mission

c. It is to be carried out to prevent disadvantages of reliability in advance

d. It is to be carried out to bring out CTQ from the requirements of customers.

Q3. The following are the explainations of the factorial design. Find out the incorrect explaination

1. 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.

Q4. In analysis of variance, when the data by factor does not satisfy the normality, what is

the decision making criterion that can be used in the homogeneity 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

Q5. 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 perform

d. The response surface experiment are generally used for the case of continous factors.

Q6. When actual mean is less than the Target Mean & Zlt = 0, then from the followinggraphs identify which type of process will it have

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 curved

b. 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 regression

d. 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 can

be used

a) 1/16 Factorial Design

b) 1/8 Factorial Design

c) 1/4 Factorial Design

d) 1/2 Factorial Design

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

LSL T USL


51/125

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 more than twice at the same level

b. 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

1


52/125

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

a. Write down all the terms which are insignificant c,ca.cb,cc

b. Identify which variable has the maximum influence on the response A

c. Choice the wrong statement (more than one can be the answer)

1. AC is not the significant term

2. Screening designs is used to identify the "vital" few factors or key variables that

influence the response

3. Normal probability plot identifies important effects using a = 0.10, by default

4. 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


53/125

a. What can be the Average Bias Value -0.01

b. What can be the historical Standard deviation Value

c. Find out the %Linearity 0.88

d. Choice the wrong statement

1. For Acceptable Gage Linearity & Bias, both should be less than equal to 5%

2. %Linearity will increases, if Slope(S) will increase3. With increase in R-Sqr value, %Linearity will decrease

4. For good linearity, Average Bias value should be same for all the parts

Q3. For measuring the measurement system accuracy, same part was measured with actual gage

repeated 5 times. The reading found were as

24.3 24.3

25.1 25.1

24.4 24.4

24.8 24.8

24.6 24.6

While measuring with the Reference gage, the part was measured to be 24.4

Find out what would be the % Bias value. -40

Q4. Match the position (A,B,C,D) of 4-Block diagram with the appropriate data distribution

curves (1,2,3,4)

C

4- Block Diagram

Accuracy

Pression A

Q5. Find out the missing values

One-way ANOVA: C2 versus Subscripts

Source DF SS MS F P

Subscripts 2 **.** 20.38 *.** 0.012 44.76 5.5Error 21 77.75 *.** 3.7

Total 23 118.50

R sq

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

R sq adj

a. SS factor

b. F factor

c. MS Error

For accepting the part, the tolerance range should be 24 + 0.3

A

D

B

C

1 2 3 4 5 6

1.51.0

0.5

2.0

2.5

3)

LSL USL

LSL USL

1)

LSL

2)

4)


54/125

d. R-Sqr


55/125


56/125

Linearity .=Slope*process variation

% Linearity .=Slope*100

Process variation .=average bias/%Bias*100

% Bias .=Bias/Process variation*100

Process variation .=Linearity/slope

4.79 0.82.19

6

24.4 0.1 Linearity .=Slope*process variation

24.4 -0.7 % Linearity .=Slope*100

24.4 0.0 Process va.=average bias/%Bias*100

24.4 -0.4% Bias .=Bias/Process variation*100

24.4 -0.2 Process va .=Linearity/slope

Average -0.2

61.5

1.5

1.5

1.5B

d

61

2

1

2

0.34

34.39

0.28

28.14

USL

LSL USL


57/125


58/125

Q1. Conduct the X-Bar R Chart for the following data & answer the below

a) From the R Chart, find out the UCL R Value 2.38

b) At CL 18, what would be the UCL X-Bar Value 18.65

c) At X-Bar Ttl = 135.67 & R Ttl = 21.6, What would be the value of X-Bar UCL Given that

Ttl Subgroups are 20 & Subgroup size is 5 7.41

1 2 3 4 5 6 7

20.1 20.1 20.1 19.9 19.8 20.3 19.6

19.8 19.8 20.4 21.2 19.7 19.5 21.4

19.5 20.3 20.5 20.3 19.7 19.8 20.3

20.4 19.4 19.5 20.4 20.4 20.2 19.8

20.1 19.5 20.3 19.7 19.5 19.6 19.5

Q2. a) Find out the Wrong explaination

1. More the Insignificant regression equation, Lesser the P-value of Lack-Of-Fit

2. Interaction terms are much more significant than the Linear Terms3. The interaction term B * C is not significant

4. Using 95% CI & removing all possible insignificant terms, the maximum

upper CL value would be 14.67

b) Find out the R-Sqr & R-Sqr (Adj) value (by completely removing all insignificant terms)

c) a=5.680 b=5.900

d) After screening the insignificant terms, What can be the regression model equation

.=102373-.9646A+.3090B+7662C-1.8363AA+1.6700BB-1.9068AB+3.3404AC

A B C Y

2.03 2.49 3.49 11.85

3.62 2.97 2.42 11.06

3.62 0.59 2.42 12.08

3.62 1.78 2.42 11.02

3.62 1.78 2.42 10.145.21 2.49 3.49 11.03

0.95 1.78 2.42 8.26

3.62 1.78 2.42 9.50

3.62 1.78 4.22 10.43

5.21 1.07 3.49 10.90

3.62 1.78 2.42 10.22

3.62 1.78 2.42 11.53

3.62 1.78 2.42 10.53

5.21 2.49 1.35 7.71

6.29 1.78 2.42 7.87

2.03 2.49 1.35 13.19

2.03 1.07 3.49 8.945.21 1.07 1.35 8.44

3.62 1.78 0.62 7.98

2.03 1.07 1.35 11.28

Q3. An ABC company (manufactures Sheet) found that increased number of sheets are being rejected

at 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 defe

At 14 + 5, what would be the optimum values of your factors


59/125

Conduct the Pareto Chart of the following and answer the below mentioned question

Defects Typ Period Defects Typ Period Defects Typ Period

NIL Day NIL Evening Scratch Night

Scratch Day NIL Evening NIL Night

NIL Day Smudge Evening Smudge Night

NIL Day NIL Evening Peel NightSmudge Day NIL Evening NIL Night

NIL Day Peel Evening Smudge Night

NIL Day Scratch Evening Peel Night

NIL Day NIL Evening Scratch Night

Peel Day NIL Evening Smudge Night

Scratch Day Scratch Evening Smudge Night

a. In which Work Shift, peel frequency is maximum Night

b. At Night, 90% of the defects are contributed by? All defects

c. Mark the incorrect Statement

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

2. At night, maximum number of defects have occurred

3. In evening shift, Smudges frequency is minimum

4. Relationship exists between the type of defects and the work shift producing the Sheets.

Q4. a. Find out the P-value of Normality Plot (Take entire group as one) 0.21

b. What would be the Test Statistics value while doing Test for equal variance 1.03

c. What would be the MSW Value 20.38

d. Tick the wrong explaination

1. Means of all models are same

2. Means of all models are different

3. Ho means means are equal

4. Ha means means are different

e. What would be the upper confidence interval value of LEVEL 25.79

LEVEL 1 LEVEL 2 LEVEL 3

3 4 6

5 4 7

7 3 8

4 6 6

8 7 5

4 4 9

3 2 10

9 5 9

Q5. Calculate the Gage R&R for the following using X-Bar and R Method at 95% Confidence Interval

Operator 1 Operator 2READING1 READING2 READING1 READING2

54.29 54.29 54.29 55.10

55.42 55.42 55.42 55.53

56.89 57.22 57.22 57.25

given that tolerance = 54+ 3 and readings are as follows


60/125

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 54.52

56.52 56.52 56.50 56.50

a. What is the Gage R&R % SV value 9.11

b. Write down the Repeatibility % T Value 7.99

c. What is the ttl variation study variance value 5.27

d. Find out which operator has minimum difficulty in measuring the readings Operator 2


61/125

103

3

4

102

23

3

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

c=3.9087

103

3


62/125

4

102

2

2

1

3

102

2

3

3


63/125


64/125

Q1. a. C, AC, BC, ABC

b. A

c. 3, 4

Q2.

a. 0.01

b. 1.07

c. 1.23%

d. 3

Q3.

% Bias Value 40%

Q4. 1 C

2 B

3 A

4 D

302826242220

0.8

0.6

0.4

0.2

0.0 0

Data

Avg Bias

Gage Bias

24.4 0.24 40.0 0.175

Reference Bias %Bias P

Average 0.24 40.0 0.175

Gage name:

Date of study :

Reported by:

Tolerance:

Misc:


65/125

Q5. One-way ANOVA: C2 versus Subscripts

Source DF SS MS F P

Subscripts 2 **.** 20.38 *.** 0.012

Error 21 77.75 *.**

Total 23 118.50

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

a. SS factor 40.760

b. F factor 5.505

c. MS Error 3.702

d. R-Sqr 34.39%

Q1.

a. 2.38

b. 18.65

c. 135.67 21.6

6.78 1.08

7.41

Q2. a. 4

b. 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, C

The analysis was done using coded units.

Estimated Regression Coefficients for Y

Term Coef SE Coef T P

Constant 10.2373 0.3378 30.303 0.000

A -0.9646 0.4434 -2.175 0.050

B 0.3090 0.4429 0.698 0.499

C 0.7662 0.4439 1.726 0.110

7654321

20.50

20.25

20.00

19.75

19.50

__X=20.011

UCL=20.660

LCL=19.363

7654321

2.4

1.8

1.2

0.6

0.0

_R=1.125

UCL=2.378

LCL=0

7654321

20

19

18

17

__X=18

UCL=18.649

LCL=17.351

7654321

2.4

1.8

1.2

0.6

0.0

_R=1.125

UCL=2.378

LCL=0

1

11

11

11

n A2 A3

1 2 . 6 6 0 3 . 7 6 0

2 1 . 8 8 0 2 . 6 5 9

3 1 . 0 2 3 1 . 9 5 4

4 0 . 7 2 9 1 . 6 2 8

5 0 . 5 7 7 1 . 4 2 7

6 0 . 4 8 3 1 . 2 8 7

7 0 . 4 1 9 1 . 1 8 2 0

8 0 . 3 7 3 1 . 0 9 9 0

9 0 . 3 3 7 1 . 0 3 2 0

1 0 0 . 3 0 8 0 . 9 7 5 0

Hi

Lo1.0000D

Optimal

Cur

d = 1.0000

Targ: 14.0

Y

y = 14.0000

0.5900

2.970

0.950

6.290B A

[5.6880] [0.5900]


66/125

A*A -1.8363 0.7225 -2.541 0.026

B*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 P

Regression 7 34.972 34.972 4.9960 5.25 0.006

Linear 3 7.797 7.797 2.5990 2.73 0.090

Square 2 12.316 12.316 6.1582 6.47 0.012

Interaction 2 14.859 14.859 7.4293 7.81 0.007

Residual Error 12 11.413 11.413 0.9511

Lack-of-Fit 7 8.874 8.874 1.2677 2.50 0.166

Pure Error 5 2.540 2.540 0.5079

Total 19 46.385

Estimated Regression Coefficients for Y using data in uncoded units

Term Coef

Constant 12.6358A 0.8898

B -1.7662

C -2.0904

A*A -0.2576

B*B 1.1793

A*B -0.6001

A*C 0.6951

Q3.

a. Night

b. All --> Smudge, Peel & Scratch

c. Mark the incorrect Statement

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

2. At night, maximum number of defects have occurred

3. In evening shift, Smudges frequency is minimum

4. Relationship exists between the type of defects and the work shift producing the Sheets.

d = 1.0000

y = .

PeelSmudgeScratchNIL

10.0

7.5

5.0

2.5

0.0

PeelSmudgeScratchNIL

10.0

7.5

5.0

2.5

0.0

Period = Day Period = Evening

Period = Night

Defects Type

NIL

Scratch

Smudge

Peel

Defects TypeCount

Cum %

Percent

10

8

6

4

2

0


67/125

Q4. a. 0.21

b. 0.6

c. 3.70

d. Tick the wrong explaination





e. Average 4.38

Tinv 2.08

S 1.92

n 8

5.79

Q5 Gage R&R Study - XBar/R Method

%Contribution

Source VarComp (of VarComp)

Total Gage R&R 0.01497 0.83

Repeatability 0.01497 0.83

Reproducibility 0.00000 0.00

Part-To-Part 1.79038 99.17

Total Variation 1.80535 100.00

Total Gage R&R 0.12234 0.47956 9.11 7.99

Repeatability 0.12234 0.47956 9.11 7.99

Reproducibility 0.00000 0.00000 0.00 0.00

Part-To-Part 1.33805 5.24502 99.58 87.42

Total Variation 1.34363 5.26690 100.00 87.78

Number of Distinct Categories = 15

a. 9.11

b. 7.99

c. 5.27

d. Operator 1

Study Var %Study Var %Tolerance

Source StdDev (SD) (3.9199 * SD) (%SV) (SV/Toler)

LEVEL 3

LEVEL 2

LEVEL 1

321

6420

99

95

90

80

70

60

50

40

30

20

10

5

1

Nor


68/125


69/125

*C

D3 D4 B3 B4 d2 c4

- - - - - -

0 3 . 2 6 7 0 3 . 2 6 7 1 . 1 2 8 0 . 7 9

0 2 . 5 7 5 0 2 . 5 6 8 1 . 6 9 3 0 . 8 8

0 2 . 2 8 2 0 2 . 2 6 6 2 . 0 5 9 0 . 9 2 1

0 2 . 1 1 5 0 2 . 0 8 9 2 . 3 2 6 0 . 9 4

0 2 . 0 0 4 0 . 0 3 1 . 9 7 0 2 . 5 3 4 0 . 9 5 1

. 0 7 6 1 . 9 2 4 0 . 1 1 8 1 . 8 8 2 2 . 7 0 4 0 . 9 5

. 1 3 6 1 . 8 6 4 0 . 1 8 5 1 . 8 1 5 2 . 8 4 7 0 . 9 6

. 1 8 4 1 . 8 1 6 0 . 2 3 9 1 . 7 6 1 2 . 9 7 0 0 . 9 6

. 2 2 3 1 . 7 7 7 0 . 2 8 4 1 . 7 1 6 3 . 0 7 8 0 . 9 7

0.620

4.2200

0.5900

2.970B C

[0.5900] [3.9087]


70/125

20.0

40.0 60.0 80.0 100.0

4 2 2 2

40.0 20.0 20.0

ScratchPeelNILSmudge

100

80

60

40

20

0

Period = Night


71/125

One-way ANOVA: C2 versus Subscripts

Source DF SS MS F P

Subscripts 2 40.75 20.38 5.50 0.012

Error 21 77.75 3.70

Total 23 118.50

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

654

Bartlett's Test

0.453

Test Statistic 1.03

P-Value 0.599

Levene's Test

Test Statistic 0.82

P-Value

12108

Mean

0.212

5.75

StDev 2.270

N 24

AD 0.480

P-Value

al


72/125

INJECTION TIME AT

10MIN 15MIN 20MIN

3.5 3.7 3.9

3.4 3.4 3.6

3.2 3.9 2.8

3.5 3.5 2.0

3.6 3.8 4.13.5 3.6 2.0

3.7 3.9 2.8

3.5 3.9 4.2

3.5 3.8 3.7

3.9 3.5 2.5

VENDOR1 VENDOR2

56 54

45 56

28 8915 57

34 53

89 54

54 56

27 57

26 78

RESP A B C RESP A B C

271.80 40.55 16.66 13.20 267.40 35.96 16.45 14.51

264.00 36.19 16.46 14.11 240.40 37.71 17.37 15.56

238.80 37.31 17.66 15.68 227.20 37.00 18.12 15.83

230.70 32.52 17.50 10.53 196.00 36.76 18.53 16.41

251.60 33.71 16.40 11.00 278.70 34.62 15.54 13.10

257.90 34.14 16.28 11.31 272.30 35.40 15.70 13.63

263.90 34.85 16.06 11.96 267.40 35.96 16.45 14.51

266.50 35.89 15.93 12.58 254.50 36.26 17.62 15.38

229.10 33.53 16.60 10.66 224.70 36.34 18.12 16.10

239.30 33.79 16.41 10.85 181.50 35.90 19.05 16.73

258.00 34.72 16.17 11.41 227.50 31.84 16.51 10.58

257.60 35.22 15.92 11.91 253.60 33.16 16.02 11.28

267.30 36.50 16.04 12.85 263.00 33.83 15.89 11.91

267.00 37.60 16.19 13.58 265.80 34.89 15.83 12.65

259.60 37.89 16.62 14.21 263.80 36.27 16.71 14.06

LEVEL1 LEVEL2

56 54

45 56

28 89

15 57

34 53

89 54

Q1 (i) 4M ANALYSIS WAS DONE TO FIND OUT THE POSSIBLE REASONS FOR WARPAGE OFPANNEL SWITH. USING PROPER TOOL VERIFY WHETHER INJECTION TIME CAN BE ONE OF THEREASONS OR NOT & WHY.

(ii) USING PROPER TOOL FIND OUT WHETHER VENDOR CHANGE CAN BE A FACTOR OR NOTGIVEN THAT THE TTL NO. OF DEFECTS IN CASE OF VENDOR1 & VENDOR2 ARE AS FOLLOWS :

Q2 ) FIND OUT WHETHER THERE IS A RELATIONSHIP BETWEEN RESPONSE AND FACTOR. ALSO FIND OUTWHICH FACTOR SHOULD BE ELIMINATED TO REDUCE THE GAP BETWEEN R-Sq & R-Sq (adj) VALUE

Q3 ) AT 85% CONFIDENCE, FIND OUT THE CONFIDENCE INTERVAL FOR STANDARD DEVIATION OF BOTHLEVEL1 AND LEVEL2 GIVEN THAT THE VALUES AT LEVEL1 & LEVEL2 ARE


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54 56

27 57

26 78


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

DEFECT Nos DEFECTS Nos 5

A 23 H 543

B 34 I 671

C 45 J 234

D 32 K 564

E 678 L 8

F 34 M 90

G 678 N 56

Q2 ) FIND OUT THE VALUE OF THE FOLLOWING 5

yield 1

Regression Analysis: RES versus PRESSURE, TEMP, TIME 5

The regression equation is

*******************************************

Predictor Coef SE Coef T P

Constant 858 1560 0.55 0.595

PRESSURE -30.38 92.89 -0.33 0.750

TEMP 1.74 54.40 0.03 0.975

TIME -38.81 25.67 -1.51 0.161

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

Analysis of Variance

Source DF SS MS F P

Regression 3 229290 76430 0.90 0.476

Residual Error 10 852631 85263

Total 13 ********

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

5

St var %

Total Gage R&R 0.083661 0.50197 * 0

Repeatability 0.061128 0.36677 30.99 3.06

Reproducibility 0.057118 0.34271 28.96 2.86

Operator 0.024438 0.14663 12.39 1.22

Operator*Part 0.051626 0.30976 26.18 2.58

Part-To-Part 0.178602 1.07161 90.56 8.93

Total Variation 0.197226 1.18335 100.00 9.86

Number of Distinct Categories = ***

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

Study Var %Study Var %Tolerance

Source StdDev (SD) (6 * SD) (%SV) (SV/Toler)

ALSO FIND OUT THE SV/Toler VAL10


77/125

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

5

One-Sample Z: C2

Test of mu = 40 vs not = 40

The assumed standard deviation = 1.3

.

Variable N Mean StDev SE Mean 92% CI Z P

C2 67 44.3955 3.3588 ****** (******, ******) 27.68 0.000

Q6 ) FILL IN THE BLANKS 5

Chi-Square Test: DAY1, DAY2, DAY3

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

DAY1 DAY2 DAY3 Total Day 1

1 **** 521 452 ****

559.54 344.39 633.08

**** 90.573 51.792

2 452 157 498 1107

403.00 248.04 455.96

5.958 33.415 3.876

3 521 268 789 1578574.46 353.57 649.96

4.976 20.711 29.742

Total **** 946 1739 4222

Chi-Sq = 241.078, DF = 4, P-Value = 0.00

Test for Equal Variances: C1, C2 5

89% Bonferroni confidence intervals for standard deviations

N Lower StDev Upper

C1 10 ****** 3.35345 6.03758C2 10 1.43548 2.07136 *******

F-Test (normal distribution)

Test statistic = *.**, p-value = 0.167

Levene's Test (any continuous distribution)


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

SampleSize=5

Avg RANGE ALSO FIND OUT HOW MANY VALUES ARE CROSSING THE CONTROL LIMITS

13 2 5

19 324 4

21 6

25 1

27 2

11 3

15 4

18 6

12 2

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


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CATGY.

ESSION AT PRESSURE =

E AT TTL TOLERANCE =


79/125


80/125

Q 1) FIND OUT THE CORRECT ANSWER FROM THE FOLLOWING

1. P CHART IS USED TO CHART THE PROPORTION OF DEFECTS IN EACH SUB GROUP

2. C CHART IS USED TO CHART THE NUMBER OF DEFECTS WHEN SAMPLE SIZE IS CONSTANT

3. NP CHART IS USED TO CHART THE NO. OF DEFECTS IN EACH SUB GROUP

4. U CHART IS USED TO CHART THE NO. OF DEFTECTS / UNIT IN EACH SUB GROUP

Q 2) WHAT SHOULD BE DONE TO MINIMIZE THE GAP BETWEEN THE R-Sq AND R-Sq (Adj) VALUE

1. LOOK FOR MAX P-VALUE & TEST TO BE CONDUCTED AGAIN AFTER REMOVING THAT P-VALUE

2. SAMPLING TO BE DONE AGAIN

3. MINIMIZE THE VARIATION IN THE RESIDUALS

4 LOOK FOR ADDITIONAL FACTORS WHICH ARE NOT INCLUDED

Q3 ) MATCH THE FOLLOWING

1. CORRELATION COMPARES VARIATION BETWEEN & WITHIN THE GROUP

2. ANNOVA TESTS THE EQUALITY OF POPULATION / SAMPLES VARIANCE

3. 2 - VARIANCE MEASURE THE DEGREE OF LINEAR RELATIONSHIP BETWEEN 2 VARIABLES

PERFORMED WHEN THERE IS A NON-LINAER RELATIONSHIP BETWEEN THE VAR

Q4 ) WHAT ARE THE POSSIBLE REASONS OF SUCH NORMALITY GRAPH1) DATA IS RIGHT SCREWED

2) DATA IS LEFT SCREWED

3) P-VALUE IS < 0.05

4) HIGH VARIATION IN THE DATA

GRAPH - 1 GRAPH - 2

1) GRAPH - 1, FACTOR IS NOT EFFECTING THE RESPONSE. SHOULD LOOK FOR ANOTHER FACTOR

2) GRAPH - 2, RESPONSE SURFACE DESIGN SHOULD BE DONE

3) GRAPH - 2, REGRESSION TO BE DONE AGAIN

4) GRAPH - 1, NO. OF REPLICATES TO BE INCREASED

Q6 ) FIND THE RTY VALUE OF THE FOLLOWING PROCESS

4. RESPONSESURFACE DESIGN

Q5 ) FRACTIONAL FACTORIAL WAS CONDUCTED TO FIND OUT THE OPTIMUM CONDITION. FROM THEFOLLOWING GRAPH FIND OUT WHICH ONE IS INCORRECT AND WHY & WHAT SHOULD BE THE NEXT.

6050403020100-10

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.163

25.5

StDev 14.58

N 50

AD 0.535

P-Value

Normal

180175170

50

48

46

44

42

25

765

50

48

46

44

42

TEMP PRESSURE

TIME

Point Type

Corner

Center

180175170

40

30

20

10

25

765

40

30

20

10

TEMP PRESSURE

TIME

Point Type

Corner

Center

180175170

40

30

20

10

25

765

40

30

20

10

TEMP PRESSURE

TIME

PointTy

Corner

Center

180175170

40

30

20

10

25

765

40

30

20

10

TEMP PRESSURE

TIME

PointType

Corner

Center

98.2 % 96.4% PARALELL

SERIES


81/125

Q7 ) AT 83% CONFIDENCE INTERVAL WHAT WOULD BE THE VALUE OF AREA UNDER THE CURVE

GIVEN THAT PROCESS VARIES FROM 30 + 5.

1) 1.372 0.83

2) 0.954 0.9542

3) 2.744

4) 1.892

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 ANSWER

1) RATIONAL SUB GROUPING IS NOT PROPER

2) GAGE SELECTION IS NOT PROPER

3) ENTIRE RANGE IS NOT COVERED

4) 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 operators

2) The resolution of the gage

3) The ability of the measurement system to detect process changes

4) Repeatability of the measurement system

Q13 ) The method that attempts to delineate all possible failures and their effect on a system is called

a. PDCA

b. FMEA

c. Cause-and-effect diagrams

d. Pareto analysis

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

1) NO. OF DISTINCT CATEGORIES ARE LESS THAN

Q8 ) CHOICE THE WRONG STATEMENT (MORE THAN ONE CAN BE THE ANSWER)

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 contrHowever, nearly all of the averages on the X-bar chart are outside of the control limits. This indicates:

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

67.6% 79.4%

2

(1) (2) (3)

Part-to-PartReprodRepeatGageR&R

200

100

0

%Contribution

%StudyVar

%Tolerance

10

UCL=12.741 2

10987654321

64

56

48

64

Gage name:

Date of study:

Reported by:

Tolerance:

Misc:


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84/125

3

1

RING SYSTEM

OPERATOR1


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Q1. For regression equation to be significant, Residuals should be normally & randomly distributed

From the following graphs identify which of them are randomly distributed

Q2. Identify among the following which is th ewrong explaination of Anova

a. Anova is a tool which can compare several means

b. 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 not

d. We can screen data to identify Vitel Few

e. 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 regressio

you 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 questions

a. Write down the Regression equation The regression equ

b. Using regression eq find out what could be the Fitted value at 630, 730 & 960

c. Write down the R-Sqr & R-sqr (adj) value

d. Using 90% Confidence interval, What will be the maximum upper CL value of response

e. For X at 520, what would be the upper & lower Confidence limit value (use 95% CI) Lower

F What will the residual P value

X Y

500 0.18

540 0.37

580 0.35

620 0.50

660 0.56

bbbbbbbbbbbbb

Fit

Fit Fit Fit

ResidualResidual

ResidualResidual

ResidualResidual ResidualResidual

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

Fit


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700 0.75

740 1.02

780 1.18

820 1.05

860 0.94

900 1.50

940 1.56980 1.65

1000 1.40

1020 1.50

1060 1.63

Q6. In analysis of variance, when the data by factor does not satisfy the normality, what is

the decision making criterion that can be used in the homogeneity of variance analysis?

1. P value of Bartletts Test

2. 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 d

is followings and the population is known as normal distribution.

(1) Calculate the upper limit of 90% confidential intervals of the population. 62.9

(2) Calculate the upper limit of 95% confidential intervals of the population. 64

Q8. Development dept. has used specified adhesives to stick 2 parts together. They want to develop one with hig

adhesive 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

17.55

(3) Calculate the Lower confidence interval limit of Mean for adhesive level 3 19.22

Adhesive level 1 Adhesive level 2 hesive level3

9 18 21

12 15 19

14 14 21

13 17 22

18 15 23

(3) Which adhesives have adhestive strength more than 20?

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

adhesived in level 2. (90% of confidence interval)

(A) Adhesives in level 1

(B) Adhesives in level 2

(C) Adhesives in level 3

(D) No adhesives


88/125

Q9. a) Find out the P-value of Normality Plot (Take entire group as one) 0.21

b) What would be the Test Statistics value while doing Test for equal variance 1.03

c) What would be the MSW Value 3.7

d) Tick the wrong explaination

1) The Upper SD limit for Gr2 at 90% CI will be 3.48721

2) Means of all models are different

3) 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 would

be the DFerror value

Gr1 Gr2 Gr3

3 4 6

5 4 7

7 3 8

4 6 6

8 7 5

4 4 9

3 2 10

9 5 9

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

Individual 95% CIs For Mean Based on

Pooled StDev N

Level N Mean StDev -------+---------+---------+---------+-- 8

LEVEL 1 8 4.625 2.560 (------*-------) 8

LEVEL 2 8 4.125 1.246 (-------*------) 8

LEVEL 3 8 9.125 2.031 (-------*------)

-------+---------+---------+---------+--

4.0 6.0 8.0 10.0

a) What is the Pooled Standard Deviation Value 2.02

b) What is the Upper Mean value of LEVEL 1 at 95% CI

c) At 95% Confidence Interval, what could be the Fcritical Value 3.47

d) Tick the wrong explaination





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 follows

Vendor 1 --> OK material : 87 & Vendor 2 --> NG material : 28

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

b. Can vendor change be a factor or not Yes vendor change is a factor


89/125

Q12. Comp line registered a six sigma project of reducing the Rejection of line. After 3 months

it was found that the line % Rejection has improved from 23% to 15%.

Inorder to confiorm that the Six Sigma Team randomly selected 120 samples, and found

that the NG material was only 17.

a. Using sutable tool, find out what the Six Sigma team finally concluded from the data

b. Find out whether the samling method done by the team was correct or not

Q13. For the following data find out whether Shift is depended upon Machine or not

a) Tick the correct answer

1) Depended since Pearsons P Value is \0.05

b) What is the Pearson Chi-Square Value 10.61

c) What is the Chi-Square Critical Value 5.99

d) By using the New Machine (0), in which shift you will get the Best Result Shift 1

Shift Machine Defects

1 0 24

1 1 54

1 0 63

1 1 65

2 0 57

2 1 23

2 0 24

2 1 36

3 0 84

3 1 46

3 0 51

3 1 64

Q14. Which method can be used to analyze the condition of the factors as given below

Factor Level : 2 Level

Number of factor : 4

Repeat : 2 times repeat

Kinds of factors : Fixed, Random mixed

Data in the cell : There are no response data on 4th & 10th Run

a) One-Way Anova

b) Balanced Anova

c) GLM

d) Two-Way Anova

Q15. The following is a part of regression analysis results for a certain data. From the following output

Analysis of Variance

Source DF SS MS F P

Regression 1 3.5816 3.5816 171.62 0.000

Residual Error ** 0.2922 *.**** 14 0.02


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Total 15 3.8738

a. What would be the F Critical Value 3.4

b. Write down the R-Sqr & R-Sqr (Adj) Value 0.92 91.92

Alpha

DF numeraDF denom

F critical

Answere


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Answer D

2 Total no of fctor

16 DF total-DF regression

18 Total no of observation -1

1

ation is Y = - 1.12 + 0.00268 X

0.57 0.84 1.45

Rsq=92.5 Rsq adj=92%

1.84

0.13 Upper 0.42

0.21 DO normality test of residual value

Fit

ResidualResidual

(e)


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ta

data

.=xbar+1.64*sem 62.96 52.3

.=xbar+1.96*sem 63.97 50.6

73.3

49.3

er 76.8

58.7

62.6

56.8

58.3


93/125

27

Mean St dev Sem

4.63 2.56 1.19 45.88

4.13 1.25 0.61 10.85

9.13 2.03 0.67 28.87

4.08

Pooled sd 2.02


94/125

0.15

No improvement

Sampling is ok


95/125

0.05Given

1 Given14 Given

.=FINV(ALPHA,DFn,DFd)

4.6


96/125

Q1. Which of the following is the wrong explainations for the Blocking

a. If two different suppliers materials were used during the experiment, then the material may be a

b. It is used to minimize the effect of confounding & lucking variables

c. Used for higher precision of the experiment

d. If the test was conducted over 2 days, then the date can be a block factore. None of these

Q2. Which has the following graph shows highest interaction

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 precis

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 c

d. 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 used

a) 1/16 Factorial Design

b) 1/8 Factorial Design

c) 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 dependen

2. When an independent variable has variation, it is possible to determine the control amount of t

3. 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.

Q6. Factorial was conducted to find out the optimum condition. From the following graphs indentify which of the

has a non-linear response & what should be done instead.

GRAPH - 1 GRAPH - 2 is a non linear model

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

TEMP PRESSURE

TIME

Point Type

Corner

Center

TEMP PRESSURE

TIME

111111

11

11

11

11

11

11

11

11

11

TEMP

TIME

111111111

11

11

11

1111

111

11

11

11

11

TEMP PRESSURE

TIME

PointType

C ornerCenter


97/125

1) GRAPH - 1, FACTOR IS NOT EFFECTING THE RESPONSE. SHOULD LOOK FOR ANOTHER FACTO

2) GRAPH - 2, RESPONSE SURFACE DESIGN SHOULD BE DONE

3) 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 linea

b. When the outcome values measured at the central & factorial point are different, it can be curve

c. 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


1. To analize Lack-of-Fit, the independent variables must be repeated more than twice at the sam

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 sig

4. As Pure error value is smaller, the probability of regression equation to be significant is higher.


(more than one can be the answer)

a. To analyze Lack-of-Fit, the independent variables must be repeated at the same level

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

Q10. From the following Normality Effect Plot, Answer the below questions

765

48

46

44

42

765

30

20

10


98/125

a. Write down all the terms which are insignificant ca,cb,abc

b. Identify which variable has the maximum influence on the response A

c. Choice the wrong statement (more than one can be the answer)

1. AC is not the significant term

2. Screening designs is used to identify the "vital" few factors or key variables that

influence the response

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

Q11. The following are the explainations of the factorial design. Find out the incorrect explaination

1. 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 incorre

a) It is used when it is hard to run the whole experiment for time or economical reasons

b) In case of fractional faxctorial, there is always some probability of confounding

c) It is possible to know the interaction effects on the 2^(3-1) design

d) If the experimental increase, the DF of total increases.

Q13. The following are the explainations of the confounding of 2^(5-1) fractional factorial design

What is incorrect explaination.

a) Factor A & B are confounded with factor C,D & E

b) The interaction effect of higher than 3rd order does not significantly effect the response

c) It is possible to reduce the experimental runs from 32 to 16

d) It is impossible to know the effect of third order interaction effect

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

1 1 1 1 1

1 -1 -1 1 -1

-1 -1 -1 -1 -1

-1 -1 1 1 1

-1 1 -1 1 -1

-1 1 -1 -1 1

1 -1 1 1 -1

1 1 -1 -1 -1

1 1 -1 1 1

-1 1 1 1 -1

1 -1 1 -1 1

-1 -1 -1 -1 1

1 -1 -1 1 1

-1 1 -1 -1 1

1 1 1 1 -1

Q14. We examined the level of each factor to make a experimental plan. When th elevel of each factor i

what is the frequency of all experiments.


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Experiment Condition

Fact Level Repeat : 2 times

A 3 Full factorial

B 2

C 2

a) For full factorial what would be the frequency of all experiments 24

b) 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 ma

experiment 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

(2) The following are the explainations of the interaction analysis results. What is correct.

a) There is no interaction.

b) There is some interaction.

c) When amount of water increases from (-) level to (+) level & washing time is at (-) Level,washing ability appears to be highd) As amount of water goes from (+) to (-) Level, Washing ability appears high

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

Time Water Y

10 4 20.4

10 4 19.3

10 4 17.6

10 4 16.3

20 8 17.4

20 8 17.7

20 8 23.2

20 8 20.4

10 8

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