proble solving
TRANSCRIPT
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1Six Sigma Central Coordination Group
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Property of Tata Motors Ltd. All right reserved . No part of this manual may be reproduced ,stored in a retrieval system,or transmitted ,in any form or means without the prior written permission from Six Sigma Cell , Tata Motors ,Jamshedpur
Root cause Analysis And Problem Solving
Manual
Wave –1 For Variable ( Continuous Data )
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TATA MOTORS LIMITEDIndexContents Page No
1. Concept & guidelines
2. Distribution of data
What is distribution
What is sample
Type of variable
Type of distribution
Property of normal distribution
3. Stability of process
4. Causes of variation
5. Tool of detecting assignable cause
6. Statistical process control
Purpose of control Chart
Control chart for variable
Control chart interpretation
Other control condition
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TATA MOTORS LIMITED Page No7. Hypothesis Testing
Steps in hypothesis testing Power of hypothesis testingTesting main Effect due to testing Levels
2t TestPaired T TestOne Way ANOVANested Anova With Fixed Factor
Comparison of varianceANOVA With random factorF TestLevene’s test
8. Variation analysis by Shanin range Method9. Regression analysis
Simple Regression Multiple Regression Regression By Subset Method
10. Design Of ExperimentFactorial DesignAnalyze factorial designAnalyze factorial PlotScreening Factorial Design Prediction model for standard deviation Variable search
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Any problem solving in a manufacturing process involves the identification of Assignable causes.Though the initial identification of factors is done through Tools like cause and effect and FMEA this volume stresses on the validation tools.
This volume aims at familiarizing users with some of the more Powerful six sigma tools and shanin tools used for root cause Identification .The six sigma tools operate through Detection of mean effects (through tools like 2t,one way ANOVA),variance analysis of a group of factors with multiple levels and interactions( through tools like nested ANOVA and ANOVA General linear model) and through establishing relationship between input and output variable when both are of continuous nature(throgh regression analysis).The Shanin tools for assignable causes are useful for detecting assignable input parameters on material and process parameter through use of tools like paired comparison and process parameter search both of which have been dealt in this volume.
Often once the assignable causes are improved, one comes up with the uphill task of improving the common causes affecting part to part variation when one has to interfere with the process to derive the optimum combination of process parameters and input material parameters for improving mean effect and variation.The six sigma tool of DOE and the shanin tool of variable search help in the above analysis.
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CONCEPTS AND GUIDELINES
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What is a distribution?
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A distribution is a population of individual observations each with the probability of occurrence defined by a probability density function.
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WHAT IS A SAMPLE?
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A sample is a part of a population drawn in a random manner so as to enable one to make inference about population.
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TYPES OF VARIABLE
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Variables are classified as continuous and discrete.
A continuous Variable is one for which all values in some range are possible i.e. values are observed in a continuous scale.
A discrete variable is one where values are with gaps and not on Continuous scale.
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TYPES OF DISTRIBUTION
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Distributions for continuous variable.i)Normalii)exponentialiii)Weibull
Distribution for discrete variablei)Poisonii)Binomial.
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TATA MOTORS LIMITEDNormal DistributionNormal Distribution
• A dimension on a part is critical. This critical dimension is measured daily on a random sample of parts from a large production process. The measurements on any given day are noted to follow anormal distribution.
• A customer orders a product. The time it takes to fill the order was noted to follow a normal distribution.
Area
-2 0 Z 1.645 2
f(x)
P = 0V = 1
D = 0.05
Z = (X - P) / V
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TATA MOTORS LIMITEDAreas under normal curve - Normal distribution table
z Area z Area z Area z Area0.00 0.500000 0.20 0.420740 0.40 0.344578 0.60 0.2742530.01 0.496011 0.21 0.416834 0.41 0.340903 0.61 0.2709310.02 0.492022 0.22 0.412936 0.42 0.337243 0.62 0.2676290.03 0.488033 0.23 0.409046 0.43 0.333598 0.63 0.2643470.04 0.484047 0.24 0.405165 0.44 0.329969 0.64 0.2610860.05 0.480061 0.25 0.401294 0.45 0.326355 0.65 0.2578460.06 0.476078 0.26 0.397432 0.46 0.322758 0.66 0.2546270.07 0.472097 0.27 0.393580 0.47 0.319178 0.67 0.2514290.08 0.468119 0.28 0.389739 0.48 0.315614 0.68 0.2482520.09 0.464144 0.29 0.385908 0.49 0.312067 0.69 0.2450970.10 0.460172 0.30 0.382089 0.50 0.308538 0.70 0.2419640.11 0.456205 0.31 0.378281 0.51 0.305026 0.71 0.2388520.12 0.452242 0.32 0.374484 0.52 0.301532 0.72 0.2357620.13 0.448283 0.33 0.370700 0.53 0.298056 0.73 0.2326950.14 0.444330 0.34 0.366928 0.54 0.294599 0.74 0.2296500.15 0.440382 0.35 0.363169 0.55 0.291160 0.75 0.2266270.16 0.436441 0.36 0.359424 0.56 0.287740 0.76 0.2236270.17 0.432505 0.37 0.355691 0.57 0.284339 0.77 0.2206500.18 0.428576 0.38 0.351973 0.58 0.280957 0.78 0.2176950.19 0.424655 0.39 0.348268 0.59 0.277595 0.79 0.214764
Area
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TATA MOTORS LIMITEDAreas under normal curve - Normal distribution table
z Area z Area z Area z Area0.80 0.211855 1.00 0.158655 1.20 0.115070 1.40 0.0807570.81 0.208970 1.01 0.156248 1.21 0.113140 1.41 0.0792700.82 0.206108 1.02 0.153864 1.22 0.111233 1.42 0.0778040.83 0.203269 1.03 0.151451 1.23 0.109349 1.43 0.0764590.84 0.200454 1.04 0.149170 1.24 0.107488 1.44 0.0749340.85 0.197662 1.05 0.146859 1.25 0.105650 1.45 0.0735290.86 0.194894 1.06 0.144572 1.26 0.103835 1.46 0.0721450.87 0.192150 1.07 0.142310 1.27 0.102042 1.47 0.0707810.88 0.189430 1.08 0.140071 1.28 0.100273 1.48 0.0694370.89 0.186733 1.09 0.137875 1.29 0.098525 1.49 0.0681120.90 0.184060 1.10 0.135666 1.30 0.096801 1.50 0.06680720.91 0.181411 1.11 0.133500 1.31 0.095098 1.51 0.06552170.92 0.178786 1.12 0.131357 1.32 0.093418 1.52 0.06425550.93 0.176185 1.13 0.129238 1.33 0.091759 1.53 0.06300840.94 0.173609 1.14 0.127143 1.34 0.090123 1.54 0.06178020.95 0.171056 1.15 0.125072 1.35 0.088508 1.55 0.06057080.96 0.168528 1.16 0.123024 1.36 0.086915 1.56 0.05937990.97 0.166023 1.17 0.121001 1.37 0.085343 1.57 0.05820760.98 0.163543 1.18 0.119000 1.38 0.083793 1.58 0.05705340.99 0.161087 1.19 0.117023 1.39 0.082264 1.59 0.0559174
Area
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TATA MOTORS LIMITEDAreas under normal curve - Normal distribution table
z Area z Area z Area z Area1.60 0.0547993 1.80 0.0359303 2.00 0.0227501 2.20 0.01390341.61 0.0536989 1.81 0.0351478 2.01 0.0222155 2.21 0.01355251.62 0.0526161 1.82 0.0343794 2.02 0.0216916 2.22 0.01320931.63 0.0515507 1.83 0.0336249 2.03 0.0211782 2.23 0.01287361.64 0.0505026 1.84 0.0328841 2.04 0.0206751 2.24 0.01254541.65 0.0494714 1.85 0.0321567 2.05 0.0201821 2.25 0.01222441.66 0.0484572 1.86 0.0314427 2.06 0.0196992 2.26 0.01191061.67 0.0474597 1.87 0.0307419 2.07 0.0192261 2.27 0.01160381.68 0.0464786 1.88 0.0300540 2.08 0.0187627 2.28 0.01130381.69 0.0455139 1.89 0.0293789 2.09 0.0183088 2.29 0.01101061.70 0.0445654 1.90 0.0287165 2.10 0.0178643 2.30 0.01072411.71 0.0436329 1.91 0.0280665 2.11 0.0174291 2.31 0.01044401.72 0.0427162 1.92 0.0274289 2.12 0.0170029 2.32 0.01017041.73 0.0418151 1.93 0.0268034 2.13 0.0165858 2.33 0.00990311.74 0.0409295 1.94 0.0261898 2.14 0.0161773 2.34 0.00964181.75 0.0400591 1.95 0.0255880 2.15 0.0157775 2.35 0.00938671.76 0.0392039 1.96 0.0249978 2.16 0.0153863 2.36 0.00913751.77 0.0383635 1.97 0.0244191 2.17 0.0150034 2.37 0.00889401.78 0.0375379 1.98 0.0238517 2.18 0.0146286 2.38 0.00865631.79 0.0367269 1.99 0.0232954 2.19 0.0142621 2.39 0.0084242
Area
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TATA MOTORS LIMITEDAreas under normal curve - Normal distribution table
z Area z Area z Area z Area2.40 0.00819755 2.60 0.00466120 2.80 0.00255519 3.00 0.001349992.41 0.00797623 2.61 0.00452715 2.81 0.00247711 3.10 0.000966602.42 0.00776023 2.62 0.00439650 2.82 0.00240123 3.20 0.000687142.43 0.00754941 2.63 0.00426930 2.83 0.00232744 3.30 0.000483422.44 0.00734365 2.64 0.00414532 2.84 0.00225574 3.40 0.000336932.45 0.00714284 2.65 0.00402462 2.85 0.00218600 3.50 0.000232632.46 0.00694686 2.66 0.00390708 2.86 0.00211829 3.60 0.000159112.47 0.00675565 2.67 0.00379258 2.87 0.00205243 3.70 0.000107802.48 0.00656915 2.68 0.00368118 2.88 0.00198847 3.80 0.000072352.49 0.00638717 2.69 0.00357264 2.89 0.00192630 3.90 0.000048102.50 0.00620967 2.70 0.00346702 2.90 0.00186586 4.00 0.000031672.51 0.00603658 2.71 0.00336421 2.91 0.00180721 4.10 0.000020662.52 0.00586778 2.72 0.00326413 2.92 0.00175023 4.20 0.000013552.53 0.00570315 2.73 0.00316679 2.93 0.00169486 4.30 0.000008542.54 0.00554264 2.74 0.00307202 2.94 0.00164115 4.40 0.000005412.55 0.00538617 2.75 0.00297282 2.95 0.00158894 4.50 0.000003402.56 0.00523365 2.76 0.00289011 2.96 0.00153828 4.60 0.000002112.57 0.00508493 2.77 0.00280285 2.97 0.00148904 4.70 0.000001302.58 0.00494003 2.78 0.00271803 2.98 0.00144130 4.80 0.000000792.59 0.00479883 2.79 0.00263548 2.99 0.00139493 4.90 0.00000048
5.00 0.00000029
Area
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TATA MOTORS LIMITEDAreas under normal curve - Normal distribution table
z Area5.10 0.000001805.20 0.000001075.30 0.000000635.40 0.000000375.50 0.000000215.60 0.000000125.70 0.0000000705.80 0.0000000405.90 0.0000000226.00 0.000000012
Area
Note: Only absolute values of z should be taken while referring to this table.
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TATA MOTORS LIMITEDExample : Normal DistributionExample : Normal Distribution
• The diameter of bushings is µ = 50 mm with a standard deviation of σ = 10 mm. Estimate the proportion of the population of bushings that have a diameter equal to or greater than 57 mm.
Ans. 0.2420
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Binomial distribution. A random variable X is defined to have a binomial distribution if the discrete density function of X is given by
fX(x) = fX(x;n,p)nCx px(1-p)1-x for x= 0,1,2,…,n
=0 otherwise
where the two parameters n and p satisfy 0 ≤ p ≤1, n ranges over the positive integers and q=1-p. A distribution defined by the above density function is called a Binomial distribution.
E(X)=np, Var(X)=npq
Binomial DistributionBinomial Distribution
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TATA MOTORS LIMITEDBinomial DistributionBinomial Distribution
• Useful when there are only two results in a random experiment e.g., pass/ fail, compliance/ noncompliance, yes/ no
• A dimension on a part is critical. This critical dimension is measured daily on a random sample of parts from a large production process. To expedite the inspection process, a tool is designed to either fail or pass a part that is tested. The output now is no longer continuous. The output is now binary; hence the binomial distribution applies in this case.
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TATA MOTORS LIMITED… Binomial Distribution… Binomial Distribution
• Product either passes or fails test; determine the number of
defective units
• Light bulbs work or do not work; determine the number of
defective light bulbs
• People respond yes or no to a survey question; determine the
proportion of people who answer yes to the question
• Purchase order forms are either filled out incorrectly or
correctly; determine the number of transactional errors.
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binomial distributionn = 8 p =0.1
0
0.2
0.4
0.6
x
p(x
)
0 1 2 3 4 5 6
binomial distributionn = 8 p =0.5
0
0.1
0.2
0.3
x
p(x
)
0 1 2 3 4 5 6
binomial distributionn = 8 p =0.7
0
0.2
0.4
x
p(x)
0 1 2 3 4 5 6
binomial distributionn = 8 p =0.9
0
0.2
0.4
0.6
x
p(x)
0 1 2 3 4 5 6
Remember : binomial has two parameters ‘n’ and ‘p’
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TATA MOTORS LIMITEDExample : Binomial DistributionExample : Binomial Distribution
• A part is said to be defective if a hole that is drilled into it is less or greater than specifications. A supplier claims a failure rate of 1 in 100. I f this failure rate were true, find the probability of observing one defective part in 10 samples.
Ans. 0.091
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Poisson Distribution A random variable X is defined to have a Poisson distribution if the density of X is given by
e-λ λx/x! for x=0,1,….fX(x) = fX(x; λ) =
0 otherwisewhere the parameter λ satisfies λ >0.E(X)= λ and Var(X)= λ
The Poisson distribution provides a realistic model for many random phenomena. Since the values of a Poisson random variable are the non-negative integers, any random phenomenon for which a count of some sort is of interest is a candidate for modeling by assuming a Poisson distribution. Such a count might be the number of fatal traffic accidents per week in a given state, the radioactive particle emissions per unit of time, the number of telephone calls per hour coming into the switchboard of a large business, the number of organisms per unit volume of some fluid, the number of defects per unit of some material, the number of flaws per unit of some material etc.
Poisson DistributionPoisson Distribution
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TATA MOTORS LIMITEDPoisson DistributionPoisson Distribution
• There are a large number of dimensions on a part that are critical. Dimensions are measured on a random sample of parts from a large production process. The number of ‘out-of-specification’ conditions are noted on each sample. This collective “number-of-failure’ information from the samples can be modeled using a Poisson distribution.
• Estimating the number of cosmetic non-conformance when painting an automobile
• Projecting the number of industrial accidents for next year
• Estimating the number of un-popped kernels in a batch of popcorn
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Though there are many distributions for Continuous variable, normal distribution is the most common distribution in any manufacturing process.Therefore in our coverage of tools emphasis is given on tools applicable for normal distribution.In case of non-normality the data should be transformed and tools applied.
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PROPERTIES OF NORMAL DISTRIBUTION
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A process which produces a response which is Normally Distributed is characterized by its mean and standard deviation.
As per Normal Distribution theory :
A observation for response sample static will fall within the limits of +/- 3σ from the process center 99.73% of the time if the process is under statistical control.
Therefore if a point falls outside this limit, process is out of control and there is an assignable cause of variation.
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(�[ ����σ ���σ
99.73 % area covered
K value of 3 implies that there is a probability of only 0.0027 of a sample statistic falling outside the control limits in spite of the process being in control.
We can choose value of k based on the desired probability of sample statistic falling outside when process is in control
Such a limit is known as probability limit
Ex.: if k=3.09, probability reduces to 0.002.
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TATA MOTORS LIMITEDCentral Limit Theorem In general all processes are not normal.
But for large or small sample sizes with a population distribution that is unimodal and close to symmetric, the Central Limit theorem states that :
If the plotted statistic is a sample average, it will tend to have a normal distribution
Thus even if the parent population is not normally distributed ,control charts for averages and related statistics are based on normal distribution. However for highly skewed distribution it is safe to have sample size of 30.
The standard deviation for the averages of mean for random samples of size n taken from a normal population are
/Sqrt(n) (where n is the sample size and is the standard deviation of parent population).
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Variances of random samples of size n taken from a normally distributed Universe have a skewed distribution withy heaping of variances to the Left of their mean and an attenuation to the right.
However the average of such sample variances approaches the population Variance for large number of such samples.
The sample mean and variance of sample therefore is used to estimate thevariance and mean of population.
.
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STABILITY OF A PROCESS
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Process Stability
Stability : Measure of consistency of a process in terms of its mean & variation. It involves a
time factor.
Tim
e
Process Mean(stable)
Process Spread= 6σ (stable)Stable Process
Mean & Variation Stable over time
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Tim
e
Process Mean unstable
Process Spread(stable)
Unstable Process(Mean inconsistent with time)
Mean unstable
Variation Stable over time
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Tim
e
Process Mean(stable)
Process Spread (unstable)
Unstable Process(Spread inconsistent with time)
Mean Stable over time
Variation Unstable
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CAUSES OF INCONSISTENCY INMEAN AND VARIATION
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Variation & the need for Control
VARIATION
Assignable / Special Causes
Inherent / Common Causes
(Specific external change / event) (Inherent in the process)
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Assignable Causes of variation¾ Occurrence of a qualitative external event
¾ Variation controllable / immediate action possible
¾Two types stream to stream,time to time
¾ In a X bar and R chart it is reflected as out of control points in the X bar chart if the event falls between subgroups and will be reflected as out of control points in R chart if it falls within subgroup.Example :
¾ Tool change
¾ Setup change / depth adjustment
¾ Shift change
¾ Tool wear
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STREAM TO STREAM VARIATION (SSV).
2 TYPES
1. Process Stream
2. Product Stream
PROCESS STREAM
Example : A multi spindle m/c producing the same product parameter
• 3 different spindles
• 3 different tool sets Contributes to SSV.
• 3 different fixtures
PRODUCT STREAMExample
: Variation in dia of holes drilled on a circular plate
: Taper on a shaft – variation in dia across cross-section
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Variation in parts produced over a period of time.
Hour to hour variation
Shift to shift variation
Event to event variation.
¾ Controllable
¾ Assignable / special causes
Causes :
Tool wear
Qualitative events – tool change, setup change etc.
Time to Time variation
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Common Causes of variation(Reflected in the R chart of X bar and R chart if found stable over time)
¾ Inherent process variation
¾ Uncontrollable ( immediate action/ correction not possible)
¾ Process characteristic
1. Man 4. Measurement
2. Machine 5. Method
3. Material 6. Mother Nature
Occurring due to variation in 6 M’s:
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TATA MOTORS LIMITEDAssignable Causes
Common Causes
TIME
PR
OC
ES
S R
ES
PO
NS
E
For the above chart, the haggard pattern shows the inherent variation due to common causes
The trend of this inherent variation over time shows the assignable causes of variation
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TOOLS FOR DETECTING ASSIGNABLE CAUSES.•SPC
•Test of hypothesis for detecting mean and variance shifts•Variance analysis
•Paired comparision and process parameter Search of SHANIN
•Regression analysis to establish relationship between input and Output both being continuous
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SPC(STATISTICAL PROCESS CONTROL)
Detects shifts in mean and spread through useOf eight rules.
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The purpose of a Control Chart
“The function of a control chart is to minimizethe net economic loss from over-adjustmentand under-adjustment.”
Dr. Deming
Rs.
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Control Charts...* Provide a common language for
communications about the performance of a process.
* Give a good indication ofwhether any problems arelikely to be correctable locallyor will require action on the system.
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Control Charts will not :
\ Make processes capable
\ Solve process problems
. . . but they can give clues to possible causes
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Preparation and Use of X and R Charts
1. Make sure the gauge is capable and accurate. Verify that it is calibrated.
2. Decide on the proper sample size.3. Determine the sampling strategy.4. Measure and record each piece of the subgroup on the
control chart.5. Calculate the X and range for each subgroup.6. After 20 subgroups are collected, the central lines (X and R)
can be calculated and control limits are established.7. After control limits are established, analyze range and X
chart, locating special causes.8. If no special causes exist, continue to run and chart the
process.
Control Charts for Variables
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Sample Size :
¾It is normally between 4 to 10
¾A size of 5 is followed as a standard
¾Larger the sample size, better the chances of detecting small shifts
Choice of measuring instrument :
¾The least count of the instrument should be 1/10th of the tolerance of the parameter.
¾An MSA should be conducted to check the accuracy and precision of the instrument
¾Instrument should be calibrated.
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Selection of Samples
Thumb rule :
Maximize differences between subgroups
Minimize differences within subgroups
Selecting samples within a subgroup : should include only common causes.
¾ Selection to be one in order of production
¾ To be done within short span of time
¾ No qualitative events should occur within a subgroup
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Spacing subgroups : Sampling frequency
¾ Subgroups should be spaced evenly across a shift /day.
¾ Should be at sufficiently large time intervals to include variation in process over time.
¾ If a known event occurs, the subgroup readings should be taken before or after the event.
¾ If an event occurs after every job, then consider the variation due to it as a common cause.
The sampling frequency depends on cost of obtaining information compared to cost of not detecting a nonconforming item.
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TATA MOTORS LIMITEDConstructing an X – R chart : The Trial Run
Step 1.
Record selected parameter as per sampling scheme
Step 2.
For each sub group, calculate
Subgroup mean = X
Subgroup range = R
Step 3
Calculate average
Subgroup mean Center line of Xbar chart
Subgroup range Center line of R chart
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Step 4.
Calculate Control Limits
For X bar Chart : UCLx = X + A2R
LCLx = X – A2R
For R chart : UCLr = D4R
LCLr = D3R
Where A2, D4 and D3 are statistical constants depending on sample size n
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Step 5.
Plot the control limits for Range chart
Plot the values of subgroup Range
Check for points out of control limits
If so investigate causes associated with out of control points and take remedial action
R chart should be analyzed before Xbar chart
It shows stability of process variation.
This variation is included in the control limits of Xbar chart. Therefore if r chart shows out of control, limits of Xbar chart may not be meaningful.
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Step 6
Delete the out of control points(on both R and Xbar chart) for which remedial actions have been taken to remove special causes of variation
Recalculate center line and control limits for the revised set of data
This should be an ongoing process leading to continuous improvement.
Step 7
Implement the control chart for future observation and monitoring.
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n2345678910
A21.881.0230.7290.5770.4830.4190.3730.3370.308
D300000
0.0760.1360.1840.223
D43.2672.5742.2822.1152.0041.9241.8641.8161.777
d21.1281.6932.0592.3262.5382.7042.8472.9703.078
CONSTANTS
X = (X1 + X2 + X3+.......+ Xn)/ n
X = (X1 + X2 + X3+.......+ Xn)/ n
σ = R/d2XUCL = X + A2R
XLCL = X - A2R
RUCL = RxD4
RLCL = RxD3
FORMULAE USED IN X AND R CHART
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How a x bar and R chart looks
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\ Control Chart Interpretation
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The R chart is a plot of inherent process variation(spread) and it shows the stability of process variation over a period of time.
It should be examined first.
An out of control condition(due to special causes) should be eliminated after finding the cause.
The X bar chart is a plot of process average and shows its stability over a period of time.
A gradual trend in X bar chart indicates a shift in process average.
It might be a result of change in controllable parameters such as tool sharpness, depth of cut, feed etc.
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These four test hold for both xbar and R Chart
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Only for X bar
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Natural Pattern
\ When points fall within control limits, also referred to as common cause variation.
A. Most of the points are near the grand average(centerline) - 68%
B. A few of the points spread out and approach the control limits.
C. Very few of the points (only 3 out of 1,000) exceed the control limits.
\ Some causes include :å Machine capabilityå Normal variation of incoming materialså Uniformity of raw parts
Control Chart Patterns and related causes
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Single point out of control- 1 pt outside of control limits
Gradual Shift- Gradual change in process
parameter. Afterwards the process stabilizes.
In X bar chart due to : change in incoming quality of raw material, change in maintenance program
In r chart due to : new operator, decrease in worker skill due to fatigue
UNSTABLE CONDITION AND THEIR PROBABLE CAUSE
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Trend- gradual shift in level which
does not stabilize.(eg. Six points in a row on rising or declining trend or seven points on one side of center line)Causes
In X bar chart : tool wear, die wear, gradual deterioration of equipment, build up of debris in jigs/fixtures.
In R chart : gradual improvement in operator skill due to training, deterioration due to fatigue, deterioration in accuracy of m/c elements due to wear and tear(e.g..Spindle play)
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Cycle- Short trends in the data whichoccur in repeated patterns.
Causes :
Rotation in operator
Periodic changes in temperature
Seasonal variation of incoming components
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Freaks- A single point or groups of points being
greatly different from the others
Wild Patterns
Causes:External disturbances influencing 1 or more samples.Cluster(eg. 9 points in a row on same side
Of center line)
Cluster of several observations definitely different from others
Cause :
Use of a new vendor for a short period
Use of new tool for a short period
Measurement & Sampling problem
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Stratification(eg. 15 points in a row within 1 sigma from Center line on either side)
- All points very close to the center line.
The output is combined and each distribution negates the other thus giving most of the points near the center line.
It might also be due to charting the output of a process after gating.Remedial action is to have separate control charts.
occurs when the output of several parallel (and assumed identical) processes into a single sample for charting the combined process. Typically a single sample is taken from each machine and included in the subgroup
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Mixtures - Opposite of stratification, a combination of two different patterns caused by presence of 2 or more population in the sample.(8 points in a row more than one sigma from center line on either side)
Absence of points near the center line.
Causes :
Difference in quality of incoming material(different suppliers)
Over control
Representing 2 or more m/c on same chart
Is similar to stratification, except the output of several parallel machines is mixed and the periodic sample is drawn from the mixture.
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Instability- Identify by a person havinglarge fluctuations and erraticups and downs.
Totally unpredictable!
Systematic Variables- Predictable fluctuation or a “saw tooth” pattern.(14 points in
a row oscillating up and down)
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Other Control Conditions] Control chart tests are based on where points
fall with regards to the 3 sigma limits.
Upper control limitZone A - (3 sigma) 2 out of 3 in zone or aboveZone B - (2 sigma) 4 out of 5 in zone or aboveZone C - (1 sigma) 7 in a row in zone or aboveZone C - (1 sigma) 7 in a row in zone or aboveZone B - (2 sigma) 4 out of 5 in zone or aboveZone A - (3 sigma) 2 out of 3 in zone or above
Center line
Lower control limit
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Once the final (modified if required) control limits are decided based on the trial run, these limits are used to monitor the process in the future.
If any point is found out of limits or if any of the special cause id detected , action is taken to eliminate the special cause associated with it and then proceed.
only after stability is achieved on X bar and R chart is One allowed to find out capability of a process through Cp=(USL-LSL)/(6*(R/d2))
Cpk=lower of( X-LSL)/(3*(R/d2)) &( USL-X)/(3*(R/d2))
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TEST OF HYPOTHESIS
Conducts test through use of test statisticto accept or reject a alternate hypothesis
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Key Symbols:
Population
Mean : PStandard Deviation :VProportion : P
Size : N
Sample
Mean : xStandard Deviation :s
Proportion : p
Size : n
Test statistic-Z,t,F
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A test statistic is used to find out the confidence Interval in which the parameter being Investigated will lie with a100 (1- ) % confidence level. is the probability of type 1 error(normally assumed 0.05).Alternatively if the value of the parameter is given, test statistic value with the parameter value is computed to find out if it lies in the acceptance region of Test statistic 100(1- )% confidence interval(found out from table of distribution of the test statistic).
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Comparison Testing:
9 A statistical hypothesis is a claim about the value of a population characteristic such as the population mean or variance.
9 In any hypothesis testing situation, there are always two contradictory hypotheses (claims) under consideration. The objective of hypothesis testing is to decide, based on information derived from a sample (because the population is either too large or does not exist), which of the two claims is correct.
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The two competing statements or claims are called
¾ Null hypothesis (H0) and
¾ Alternative hypothesis(H1) or (Ha).
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A null hypothesis (Ho) hypothesis is a statement about the population(s) parameters.
Usually, the null hypothesis represents the “status quo” or “nodifference” statement
Null Hypothesis(Ho)
• Usually describes a status Quo
• The one you assume unless otherwise shown
• Sign used are = or ≥, ≤
.
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Alternative hypothesis (H1)
It represents all possible values for the population parameter except that in the Null Hypothesis. It is the statement that must be true if the null hypothesis is false.
The Alternate hypothesis (Ha)
• Usually describe the difference
• The one you accept or reject based on the evidence
• Sign used are ≠or ,< or ,>
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Decision MatrixDecision Matrix
CorrectDecision
(1- β)= power of the test.
Type I errorP(type I) = α
Producer’s riskReject H0
Type II errorP(type II)= β
Consumer's risk
CorrectDecisionAccept H0
H1 is true
H0 is false
H0 is trueReality
Decision
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Errors in Hypothesis Testing
¾ Type , Error : The null hypothesis is rejected when it is actually true. The probability of type Ι error is indicated by D, the level of significance of the test.
¾ Type ,, Error : The null hypothesis is not rejected even though it is false. The probability of a type ΙΙ error is denoted by E.
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The power of a hypothesis testThe power of a hypothesis test is equal to (1- β). It is the probability of rejecting the null hypothesis when it is false. In general, the more powerful the test, the better it is.
The right way to perform a comparison test is to determine, before the test is conducted, how large of a difference is meaningful and then ensure that the sample size is chosen such that the test has sufficient power (say 80 or 90 percent) to detect that difference.
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Six steps in hypothesis testingHypothesis testing usually involves the following six action steps:
Step 1:
State the hypothesis. This could be a specific statement about a population parameter.
Step 2:
Choose (D). The alpha factor designates the risk of rejecting the hypothesis if it is actually true. Common values are 0.05 or 0.01
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Step 3:
Choose the test statistic for testing the hypothesis. This could be a Z-statistic, t-statistic, an F-statistic, or a Chi-square statistic. The statistic used depends on the type of test being conducted.
Step 4:
Determine the acceptance region for the test i.e. the range of values of the test statistic which results in a decision to accept the hypothesis. Define pass/fail criteria of the test.
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Step 5:
Draw a representative sample from the product population, calculate sample statistics such as the samplemean, sample variance or, in case of discrete values, proportion defectives and apply a hypothesis test using these statistics. Compute the test statistic and compare the obtained value with the acceptance region to make a decision to accept or reject the hypothesis.
Step 6:
Draw an engineering conclusion.
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Interval estimation consists of finding an interval defined by two end points--- say, L and U --- such that the probability of the parameter θ being contained in the interval is some value (1-α).
P( Ld TdU ) = (1-D)
This represents a two sided confidence interval, with L representing the lower confidence level and U representing the upper confidence level.
Confidence Interval of test statistic
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¾ Confidence intervals can also be one sided .An interval of the type
Ld T, such that P(L d T) = 1-Dis a one sided lower 100(1- α) % confidence interval for θ .
ORAn interval of the type
T dU, such that P( TdU ) = 1-Dis an upper 100(1-α ) % confidence interval for θ.
Example for Z statistic is shown on next slide
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Z Curve ( Probability distribution of test statistic z with one tailed
And two tailed rejection region based on α )
Shaded Area = α = P (type 1 error )
Shaded Area = α/2
Shaded Area = α/2
00 0zα -zα -zα/2 zα/2Rejection region:z≥zα Rejection
region:z≤-zα
Rejection region:either z ≥zα/2 or z ≤-zα/2
(a) (b)©
Rejection region for z tests:(a) upper- tailed test; (b) lower-tailed test; © two-tailed test
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TESTING MEAN EFFECTS DUE TOFACTOR LEVELS
2t test- Mean effect of 2 levels of assignable factorPaired t test-Mean effect of 2 levels of assignable factorOne way ANOVA-Mean effect of multi levels of assignable factor
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2t test
Fixture1Sample size n1Mean= 1Std.deviation= 1
Fixture2Sample size n2Mean= 2Std.deviation= 2
Is difference between 1 and 2 of the dependentVariable a result of Sampling error or do they come from Different population signifying effectDue to fixture?
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• Variances of the two populations assumed equal:
We assume that each population is normally distributed. If we assume that the population variances, though unknown are equal, i.e. σ2
1= σ22
An F test needs to be done to verify equality of variance
Hypothesis H0: µ1- µ2= µ0 H0: µ1- µ2 ≤ µ0 H0: µ1- µ2 ≥ µ0
H1: µ1- µ2 ≠µ0 H1: µ1- µ2 > µ0 H1: µ1- µ2 < µ0
Rejection Region |t0|>tα/2,n1+n2-2, t0 >tα,n1+n2-2 t0 < - tα,n1+n2-2
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(X1- X2) - µ0
( degrees of freedom = n1+ n2 – 1 )
If the distribution is non normal then take n1 & n2 > 30
Test Statistic t0 =
Depending on whether calculated t0 lies in the acceptance or Rejection Region from students t table alternate hypothesis is rejected or accepted respectively.
sp √(1/ n1+ 1/ n2 )
Sp2=(n1-1)s21+(n2-1)s2
2 / n1+ n2-2
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If the variance of populations cannot be assumed to be equal:
Hypothesis H0: µ1- µ2= µ0 H0: µ1- µ2 ≤ µ0 H0: µ1- µ2 ≥ µ0
H1: µ1- µ2 ≠µ0 H1: µ1- µ2 > µ0 H1: µ1- µ2 < µ0
Rejection Region |t0|>tα/2,ν, t0 >tα, ν t0 < - tα, ν
Test Statistic t0 =(X1- X2) - µ0
√s21/n1+s2
2/n2
Q = (s21/n1+s2
2/n2 )2 (s21/n1
)2+ (s22/n2 )2
n1-1 n2-1(degree of freedom)
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Example of two sample t test :A large corporation is interested in determining whether the average days of sick leave taken annually is more for night shift employees than for day shift employees.It is assumed that the distribution of the days of sick leave is normal for both shifts and that the variances of sick leave taken are equal for both shifts. A random sample of 12 employees from the night shift yields an average sick leave X1 of 16.4 days with a standard deviation of 2.2 days. A random sample of 15 employees from the day shift yields an average sick leave X2 of 12.3 days with a std. dev. Of 3.5 days. At a level of significance of .05, can we conclude that the average sick leave for the night shift exceeds that for the day shift?
Solution: The hypotheses are H0 :µ1− µ2 ≤ 0 ; H1 : :µ1− µ2 >0
The pooled estimate of the variance,
Sp2 = = 8.990
So Sp = 2.998 25
15/√20Test Statistic t0=
(16.4-12.3) –0 =1.491
t.05,25=1.708. Since the test statistic is less than 1.708 and does not fall in the rejection region, we cannot reject the null hypothesis .Therefore we cannot conclude that there is evidence of average sick leave for the night shift exceeding that for the day shift.
11(2.2)2 + 14(3.5)2
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A study was performed in order to evaluate the effectiveness of two devices for improving the efficiency of
gas home-heating systems. Energy consumption in houses was measured after each of the devices was
installed. The two devices were an electric vent damper (Damper=1) and a thermally activated vent damper
(Damper=2). The energy consumption data (BTU.In) are stacked in one column with a grouping column
(Damper) containing identifiers or subscripts to denote the population. After a variance test no evidence was
found for variances being unequal. Now we want to compare the effectiveness of these two devices by
determining whether or not there is any evidence that there is significant difference between the means of the
two devices ie testing alternate hypothesis 1- 2 is not equal to zero.BTU.In Damper BTU.In Damper BTU.In Damper BTU.In Damper BTU.In Damper
7.87 1 4 1 6.62 1 6.72 2 7.69 29.43 1 8.58 1 5.2 1 10.21 2 12.19 27.16 1 8 1 12.28 2 8.61 2 5.56 28.67 1 5.98 1 7.23 2 11.62 2 9.76 212.31 1 15.24 1 2.97 2 11.21 2 7.15 29.84 1 8.54 1 8.81 2 10.95 2 12.69 216.9 1 11.09 1 9.27 2 7.62 2 13.38 210.04 1 11.7 1 11.29 2 10.4 2 13.11 212.62 1 12.71 1 8.29 2 12.92 2 10.5 27.62 1 6.78 1 9.96 2 15.12 2 14.35 211.12 1 9.82 1 10.3 2 13.47 2 13.42 213.43 1 12.91 1 16.06 2 8.47 2 6.35 29.07 1 10.35 1 14.24 2 11.7 2 9.83 26.94 1 9.6 1 11.43 2 7.73 2 12.16 210.28 1 9.58 1 10.28 2 8.37 29.37 1 9.83 1 13.6 2 7.29 27.93 1 9.52 1 5.94 2 10.49 213.96 1 18.26 1 10.36 2 8.69 26.8 1 10.64 1 6.85 2 8.26 2
Example
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Using Minitab on 2 sample T Test Use menus:Stat / Basic statistics /2 Sample T test ...
If data is stacked in one column
Response data
Data population code
Response data of first population
Response data of Second population
If data is two column
Test the equality of variance & then select the required setting
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Sd. Of sample
Sd/Sqrt(N)
confidence interval is a range of likely values for the difference, µ1 - µ 2
If reference value (0 in this case ) lies outside the confidence interval then reject Null Hypothesis
T= (µ1- µ 2)/(pooled SE Mean)
Note -If P<0.05 reject null hypothesis and accept alternate hypothesisThis means that in this example difference of mean not equal to zero
Minitab Output
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PAIR1
PAIR2...PAIR n
MATL.HARDNESS MATL.HARDNESS BA B
X1A X1B X1d
X2A X2B X2d
XnA XnB Xnd
DIFF.
Is the average of differences between observations of the dependent variable on the two Material types large enough to indicate that observations are from two different populationsignifying the effect of hardness?
PAIRED t test
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� Paired t- test :
Often a test can be designed so that the observations on same sample are taken in pairs to increase the ability to detect small difference.
Data are taken in (n) pairs and the difference (d) within each pair is calculated for dependent samples i.e. effect of factor has to be found on same sample.
H0: P d = µ0 H0: P d ≤ µ0 H0: P d ≥ µ0
H1: P d ≠µ0 H1: P d > µ0 H1: P d < µ0
where P d is the population mean of the differences and P 0
is the hypothesized mean of the differences
Hypothesis for Paired T test
Rejection Region |t0|>tα/2,n-1, t0 >tα,n-1 t0 < - tα, n-1
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t0 =
The null hypothesis is rejected if calculated to lies in the rejection region of t from table
dSd
� n
To calculate the test statistic t0, the mean of the differences Xd
and the sample standard deviation Sd are calculated.
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A shoe company wants to compare two materials, A and B, for use on the soles of boys’ shoes. In
this example, each of ten boys in a study wore a special pair of shoes with the sole of one shoe
made from Material A and the sole on the other shoe made from Material B. The sole types were
randomly assigned to account for systematic differences in wear between the left and right foot.
After three months, the shoes are measured for wear.
Mat-A Mat-B13.2 148.2 8.8
10.9 11.214.3 14.210.7 11.86.6 6.49.5 9.8
10.8 11.38.8 9.3
13.3 13.6
Example
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Using Minitab on Paired sample T Test Use menus:Stat / Basic statistics /Paired T test ...
Put the test condition not equal or less than or greater than as required
Confidence interval of test
Null hypothesis
Sample –1 data
Sample –2 data
Click Option
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Sd. Of sample
Sd/Sqrt(N)
confidence interval is a range of likely values of µd
T= µd/ (pooled SE Mean)
Note -If P<0.05 Or if reference value in this case(0) does not lie in the intervalreject null hypothesis and accept alternate hypothesis
Minitab Output
µd = population mean of the differences
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ONE WAY ANOVA
One-way analysis of variance tests the equality of population means when classification is by one variable. The classification variable, or factor, usually has three or more levels (one-way ANOVA with two levels is equivalent to a t-test).
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Tool1Sample size nMean= 1Std.deviation= 1
Tool2Sample size nMean= 2Std.deviation= 2
Tool3Sample size nMean= 3Std.deviation= 3
Tool4Sample size nMean= 2Std.deviation= 2
Do the means of the observation on the dependent variableFor the four tools come from the same population Or are they from different populations signifying the presence
Of a effect?
One way ANOVA
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The null hypothesis Ho for the test is that all population means ( All level) are the same.
The alternative hypothesis Ha is that one or more population means differ from the others.
Hypothesis for 1 way ANOVA
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APPLICATION STEPS
Steps to consider when applying a single factor analysis of variance:
1. Describe the problem using a response variable that corresponds to the key process output variable or measured quality characteristic. Example of applications include the following:
a. Customer delivery time is sometimes too long.b. The dimension on a part is not meeting specification.
2. Describe the analysis. Example applications include the following:a. Determine if there is a difference in the mean delivery time of five
departments.b. Determine if there is a difference in the dimension of a part when a
particu1ar setting on a machine is changed to five different levels.3. Test the null and alternate hypotheses.
delivery time of department X.
4321: µµµµ ===oH4321: µµµµ ≠≠≠AH
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Analysis of variance is a misleading name for a statistical method and model that deals with differences in the means of a variable across groups Since the methodsl employ ratios of variances in order to determine whether the means differ we call it analysis of variance.
Control µ1 , σ1
Controlµ2 , σ2
x1, s1 n= 10
x2, s2 n= 10
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The following assumptions are made in ANOVA:
1. Independence of observations.
2. Means and variances of population should be independent. I t turns out that this can be only be safely assumed only if population is normally distributed.(or choose n > 30)
3. For an analysis of variance hypothesis test. model errors are assumed lo be normally and independently distributed random variables with mean zero and variance σ2. This variance is assumed constant for all factor levels.
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2
1 1)(∑ ∑ −=
= =
a
i
n
jijtotal yySS
The overall variation of the data is given by:
Total variability in data as measured by
….(2)
Within-group variability measures the variation about the mean of each
group(level) and is used to estimate the variation within each group or
category in the research problem. Sometimes called "unexplained" or "residual"
variation.I ts is assumed equal for all levels as it reflects part to part variation.
Between-group variability focuses on how the sample mean of each
group(level) differs from the overall or grand mean. Sometimes called
"explained variation” because it is due to group structure.
SS represents Sum of squares which when divided by degree of freedom gives mean square which is a indicatorof variance
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∑=
−=a
iilsfactorleve yynSS
1
2)(
2
1 1)(∑ ∑ −=
= =
a
i
n
jiijerror yySS
The first element includes a measure of the variance due to differences between the means due to effect of levels as well as any random difference, whereas the second element is due to random error.
errorlsfactorlevetotal SSSSSS +=
yi= mean of individual subgroup
Y = mean of all observationsYij = individual observationsi=sub group levelsj=observation no within subgroup
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The main goal in ANOVA is to see whether or not the variation betweensample means representing various levels of a factor is significantly greater than that within the samples themselves.A greater variation between the samples than within the samples would suggest that the two groups came from different populations (i.e., a significant difference exists).
1−=
a
SSMS lsfactorleve
lsfactorleve
anSS
MS errorserrors −
=
When divided by the appropriate number of DOF(degree of freedom), these sum of squares gives good estimate of the total variability.
….(6)
….(7)
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The null hypothesis that there is no difference in factor levels is tested by calculating the F -test statistic:
error
lsfactorleve
MS
MSF =0
Using an F table, we should reject the null hypothesis and conclude there are differences in treatment means if
anaFF −−> ,1,0 α
Minitab also does comparision of mean difference to find significant difference between levels
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Data shows durability of four types of carpet.To examine whether means of all four carpets are same or mean of at-least one carpet is significantly different
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Using Minitab on 1way ANOVA test Use menus:Stat / ANOVA /One way ...
Response Variable
Factor Level of data
Click Ok
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No of observation of
each level Mean of Each level
Common Sd of all level
If the intervals for two means do not
overlap, it suggests that the population means are different
Mean of Each level
Note -If P<0.05 reject null hypothesis and accept alternate hypothesisThere is no difference in mean in given example as P>0.05.
Minitab Output
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Stat-ANOVA-General Linear Model-comparisons
COMPARISON OF MEAN THROUGH MINITAB
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Since all intervals contain zero none of the differences are significant.If an interval does not contain zero the difference is significant and must be investigated.
Interval for Diff. Of mean of carpet1-mean ofcarpet2
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Comparison of mean effect of Nested factors (fixed)
A factor B is said to be nested within A if all levels of B can only existas a sublevel of A
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machine
Spindle 1 Spindle 2
tool1 tool2 tool1 tool2
Is the mean effect of spindle and tool significant?
All the factors are fixed and one has to find whether significant difference between the two spindles or tools exist
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The above structure shows the readings of a dependent variable for 6 spindles of a machine with two tools per spindle for 4 time periods.To investigate whether differences between spindle and tool are significant
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Stat-ANOVA-General Linear Model
Shows toolNested withinSpindle and partNested within tool
Time is assumedRandom factor
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The p value of <0.05 for spindle denote that the one of the spindle means is Different from the others and requires investigation.
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COMPARISON OF VARIANCEANOVA with random factors(nestedFor hierarchical structure,SHANIN range method)
F testLevene’s test( Minitab)
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Nested ANOVA with Random factorsIs used in the initial stages of a exercise to find Out variance due to identified assignable factors with multiple levels mostly nested in a family tree.The selection of levels is a random selection from population.
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For the random factor model as discussed in the previous slide the among levels mean square for a factor estimates the Variability of the levels of a factor and any variability in this estimate comes as a result of variability in the estimate of 2 and sample to sample differences in levels selected at random from a population. If the factors are nested we use nested ANOVA with all treatments(levels) for a factor asrandom and the P value <0.05for each factor indicates variability of treatment means>0.
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Example An engineer trying to understand the sources of variability in the manufacture casting. The process of making the casting requires mixing materials in small furnaces for which the temperature setting is to be 475° F. Company has a number of plants where the Casting are made, Engineer select four as a random sample. He conduct an experiment and measure furnace temperature three times during a work shift for each of four operators from each plant over four different shifts
Plant 1 Plant 2 Plant 3 Plant 4
Operator -1
Batch 1
Operator -2 Operator -3 Operator -4
Shift –11 Shift –12 Shift –13 Shift –14
Batch 2 Batch3
Df = (N-1)=(4-1)=3
Df = (N-1)*Nplnat =(4-1)*4=12
Df = (N-1)*N(ope)r*N(plant)=(4-1)*4*4=48
Df = (N-1)*N(shift)*N(oper)*N(plant)=(3-1)*4*4*4=128
(factors chosen at random from a population)
Nested structure
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Using Minitab to obtain nested ANOVA Use menus:Stat / ANOVA / Fully Nested ANOVA ...
Response Variable
Factors in hierarchical order ( Higher To lower order based on Nested Diagram )
Hypothesis for Nested ANOVA
H0 (the null hypothesis): That the Variances for the factors is Zero
H1 (the alternative hypothesis): That the Variances for the factors is significantly different from Zero
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Minitab Output Degree of Freedom for the each factor
Sequential Sum of square for the factor
P value is based on the following hypothesis test Ha = Variance of the factor is not ZeroHo = Variance of the factor is ZeroIf P value is <0.05 for α =0.05reject H0
Variance of the Component
% of total Variance
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“B” Vs. “C”
OBJECTIVE :
• VALIDATES A BETTER OR A SUPERIOR PRODUCTOR PROCESS (B) OVER CURRENT ONE (C) WITH ADESIRED STATISTICAL CONFIDENCE.(UAUALLY 95%)
• EVALUATES ENGINEERING CHANGES
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PROCEDURE FOR “ B” Vs “ C”
STEP 1 : CHOOSE AN ACCEPTABLE LEVEL OF “ D” RISK(ASSUMING IMPROVEMENT WHEN NO IMPROVEMENT EXISTS)
STEP 2 : DECIDE ON SAMPLE SIZES FOR “ B” AND “ C”TESTS
STEP 3 : RANDOMIZE AND CONDUCT THE TESTSTEP 4 : RANK ORDER THE RESULTSSTEP 5 : DECISION RULESSTEP 6 : SEPARATE THE MEANS - THE “ ß” RISK
(ASSUMING NO IMPROVEMENT WHEN IMPROVEMENT EXISTS)
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B Vs. CDECISION RULES
• No overlap End count techniques:B B B C C C
Best WorstB end count = 3 ; C end count = 3Total end counts = Total no. of rankings ( No overlap )• The Overlap End count techniques:B B B C B C C C C
Best WorstB end count = 3 ; C end count = 4Total end counts = 7Confidence Level End Counts >
90 695 7
99 10• Separate the meansX - bar ( B) - X - bar ( C ) > K V
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SHANIN RANGE METHOD(Previous example evaluated)
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plant1 operatorshift 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4Batch
1 17 18 12 22 11 19 10 20 15 0 17 16 5 17 21 1014.46 2 12 15 15 17 14 22 17 13 12 9 23 20 4 15 17 15
3 21 14 8 14 10 17 23 18 10 12 15 11 11 11 15 14
Range batch 9 4 7 8 4 5 13 7 5 12 8 9 7 6 6 5avg.shift 16.7 15.7 11.7 17.7 11.7 19.3 16.7 17 12.3 7 18.3 15.7 6.67 14.3 17.7 13range of mean of shift w ithin operatoraverage of operatorrange of operator w ithin plant
plant2 1 24 17 19 17 12 12 12 22 25 22 17 19 17 16 15 2117.29 2 17 22 17 10 15 15 17 17 21 23 16 16 15 11 15 10
3 21 21 22 19 15 10 15 23 17 25 21 25 16 12 12 12
Range batch 7 5 5 9 3 5 5 6 8 3 5 9 2 5 3 11avg.shift 20.7 20 19.3 15.3 14 12.3 ‘ 20.7 21 23.3 18 20 16 13 14 14.3range of mean of shift w ithin operatoraverage of operatorrange of operator w ithin plant
Plant3 1 15 17 9 17 10 11 18 17 6 13 11 15 9 13 17 311.98 2 10 11 13 15 6 13 20 11 5 15 9 17 11 15 10 11
3 9 14 8 13 8 16 14 9 11 18 11 12 8 13 9 9
Range batch 6 6 5 4 4 5 6 8 6 5 2 5 3 2 8 8avg.shift 11.3 14 10 15 8 13.3 17.3 12.3 7.33 15.3 10.3 14.7 9.33 13.7 12 7.67range of mean of shift w ithin operatoraverage of operatorrange of operator w ithin plant
plant4 1 24 16 15 21 9 22 17 12 10 15 9 25 9 17 10 2315.79 2 17 15 10 16 15 23 19 16 21 10 17 19 13 13 8 17
3 20 14 9 12 19 19 15 19 21 15 22 14 15 11 14 16
Range batch 7 2 6 9 10 4 4 7 11 5 13 11 6 6 6 7avg.shift 20.3 15 11.3 16.3 14.3 21.3 17 15.7 17.3 13.3 16 19.3 12.3 13.7 10.7 18.7range of mean of shift w ithin operatoraverage of operatorrange of operator w ithin plant
range of plantmax range of operatormax range of shiftmax range of batch
11.3313
3.25
2.08
5.316.25
15.75 17.08 16.50 13.839 7 6 8
12.58 12.75 11.92 10.67
14.336.25
5.00 9.33 8.00 6.00
18.83 15.42 20.58
3.25
5.33 8.33 5.33 3.00
11.0015.42 16.17 13.33 12.926.00 7.67 11.33
1 2 3 4
Ran
ge M
e tho
d o f
Sha
n in
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F test
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� Hypothesis Testing for the ratio of two variances: (F-test)
We assume that both populations are normally distributed.
Hypotheses H0: σ2= σ20 H0: σ2≤ σ2
0 H0: σ2≥ σ20
H1: σ2≠σ20 H1: σ2>σ2
0 H1: σ2< σ2
0
Rejection Region F0> Fα/2,n1-1,n2-1, F0 > Fα,n1-1,n2-1 F0 < F(1-α),n1-1,n2-1
F0 < F(1-α/2 ,n1-1,n2-1
s21
s22
Test Statistic F0= ( d.o.f.: n1-1,n2-1)
(
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TATA MOTORS LIMITEDExample of F-test :The variabilities of the service times of two bank tellers are of interest. We want to determine whether the variance of service time for the first teller is greater than that for the second. A random sample of 8 observations from the first teller yields a sample average of 3.4 min. with a standard deviation of 1.8 min. A random sample of 10 observations from the second teller yields a sample average of 2.5 min. with a standard deviation of 0.9 min. Can we conclude that the variance of the service time is greater for the first teller than for the second? Use a level of significance of 0.05.
Solution:
The hypotheses are H0: σ21≤ σ2
2 H1: σ21>σ2
2
The test statistic is F =
(1.8)(1.8)2
(0.9)2= 4.00
From table F0.05,7,9= 3.29. The test statistic lies in the rejection region, and so we reject the null hypothesis.
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LEVENES TEST
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In a experimental study conditions conducive to potato rot by injecting potatoes with bacteria that cause rotting and subjecting them to different temperature and oxygen regimes. Check the equal variance assumption.
Rot Temp13 1011 103 1010 104 107 1015 102 107 1026 1619 1624 1615 1622 1618 1620 1624 168 16
Example
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Using Minitab on Variance test Use menus:Stat / ANOVA /Test for equal of variance test ...
Response Variable
Factor Level of data
Null hypothesis (Ho) is that the population variances under consideration (or equivalently, the population standard deviations) are equal.
Alternative hypothesis (H1) is that not all variances are equal.
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Note -If P<0.05 reject null hypothesis and accept alternate hypothesis
This means that in this example variances are not equal to each others
Levene’s test is not sensitive to assumption of normality
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SHANIN DOE
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PAIRED COMPARISON
Are incoming parameters on product affecting the response?
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Why is it required ?
1. To identify what suspect product characteristic or quality parameter is contributing to the problem.
2. This tool can be used only if there are suspect product parameters ( hardness, yield strength, ovality,etc.) that are contributing to the problem.
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. Identify eight Very good parts and eight Very bad parts which are having the problem that is under analysis.
. List as many Product characteristics/parameters which may explain the difference between the Good and Bad parts (This list is based on the engineering judgment of the person/team, but later will be proved using this tool). List it in the descending order or importance to the problem .
. Measure all the Good parts and bad parts for the parameters identified above. There will then be totally 16 values.
. Arrange the 16 values in the rank order (start from the smallest to the largest or the largest to the smallest) irrespective of whether they are good or bad.
Steps followed in paired comparison :
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. Write against each value whether that particular value corresponds to a bad or Good part. If it belongs to the good part write ‘G’ within bracket after the value. If it belongs to the bad part, write ‘B’ within bracket after the value.
. From the top, find out when for the first time, the Good changes to Bad (or) the Bad changes to Good.
. Draw a line at the change point (For e.g.., if the change occurs after the 5th data, draw a line after the fifth data).
. From the bottom, find out when for the first time, the Good changes to Bad (or) the Bad changes to Good.
. Draw a line at the change point (For e.g.., if the change occurs after the 15th value from the bottom, draw a line between 15 & 16).
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. Find out the top count (Count the number of values till the line is drawn).
. Find out the Bottom count (Count the number of values till the line is drawn from the bottom).
. Add both these counts to find the Total count.
. If the Total count <=5, then that quality parameter or productcharacteristic is not the reason for the problem.
. If the Total count is = 6, it can be concluded that at 90% confidence level, it is this quality parameter or product characteristic that is leading to the problem.
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Rules for Count1. If first & last parameter are of same family( Both Bad or both good ) then there is no significance .
2. At the point of transition ( when good converting to bad or Bad converting to good ) if same two values appeared then we subtract ½ from the count for the two value.
3. At the point of transition ( when good converting to bad or Bad converting to good ) if same three or more values appeared then we delete the entire block from count.
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Confidence Level for other counts:
Confidence level90%95%99%99.90%
Number of Total count671013
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The DOE team suspected the hardness of the shell as one of the product characteristics that could lead to crack . Pair comparison was done to find out whether this indeed was the parameter leading to the crack The eight good(Best of Best) & eight bad (Worse of Worse) are selected & it’s results are shown below
Hardness Value Component 55 B54 B
54.5 B49 B51 B50 B52 B
51.5 B46 G42 G42 G49 G
51.5 G50 G44 G
Example
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The results after arranging in the rank order ( Lower To Higher)
Hardness Value Component 42 G42 G44 G46 G49 B49 G50 B50 G51 B
51.5 B51.5 G52 B54 B
54.5 B55 B
Top Count= 4
Bottom Count= 5-12 =4.5
Total Count =8.5 ,explains that the hardness is creating the problem at 95% confidence level
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Process parameter search
Does a process parameter have significant impact on the output?
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/ PROCESS SEARCH
OBJ ECTIVE :
�Process Search separates
important Process parameters from
unimportant ones.
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Met hodology :
1. If process parameters variations are suspected as possible causes for good and bad finished products, make a list of these process parameters, in descending order of likelihood.
2. Determine how each process parameter is to be measured, who is to measure it, and where precisely it is to be measured.
3. Make sure that the accuracy of the measuring instrument is at least 5 times the accuracy ( tolerance) of the process parameter.
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Met hodology : conti….
4. Make sure that the actual process parameters, not just mere settings, measured.
5. If a particular process parameter does not vary at all while being monitored, it can be eliminated from further consideration.
6. Run a 100 % sample of units or a multi-vari sample of units until :¾ A minimum of eight good units and eight bad units are collected
at the end of the process.
¾ The spread between the best unit and the worst unit is a minimum of 80% of the historic variation observed on the product produced by the process.
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Met hodology :conti….
7. Measure all the designated process parameters associated with each unit.
8. As the units go through the process, a determination of whether the units are going to be good or bad cannot be made until the end of the process. This means that many potentially good units will have to be measured, along with many potentially bad units, in order to obtain a minimum of eight eventually bad units.
9. A Paired Comparison of the process parameters associated with the eight good and eight bad units is then run. And a Tukey test is performed on each parameter, with total end count calculation.
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Met hodology :conti…..10. If there are several process parameters identified with
90% or greater confidence, run a “ B verses C” test to verify that the important parameters are truly captured.
11.Next, a variable search or a full factorial should be run to quantify the most important parameters and their interaction effect.
12.Further optimization, through Scatter Plots or Evolutionary Operation, followed by Positrol, Process Certification, and Pre-control should be conducted.
13. The tolerances of the unimportant parameters in step 8 can be expanded to reduce costs, although some experimentation may be necessary to determine how much to expand them.
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Cooling Time Zone Temp. Molding TempSpec 3Sec 415-425 Deg. 100-110Good 3.1 423 106Good 3 415 103Good 3.1 422 105Bad 2.9 422 109Bad 2.8 419 101
Good 3 419 105Bad 2.9 417 102
Good 3 418 105Good 3 420 105Bad 3 420 108Bad 2.8 417 108Bad 2.8 414 104
Good 3 414 103Good 3 420 105Bad 2.8 415 102Bad 2.8 416 103
In a foundry process following process parameters where identified as the suspects for this problem.A/ Cooling timeB/ Zone temperatureC/Mould temperatureThe data collected for 8 good & 8 bad parts as given below.
Identify the parameter that is creating the problem.
Example
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Spec 3SecCooling Time
Bad 2.8Bad 2.8Bad 2.8Bad 2.8Bad 2.9Bad 2.9
Good 3Good 3Good 3Good 3Bad 3
Good 3Good 3Good 3.1Good 3.1
Bad 2.8Bad 2.8Bad 2.9Bad 2.9
Good 3.1Good 3.1
Spec Cooling Time
The results of cooling time after arranging in the rank order ( Lower To Higher)
Based on Rule 4 of paired comparison Total block of data is removed.
Reduce matrix after Block removal
4
2
Total Count =6,explains that the cooling time is suspected parameterfor the problem at 90% confidence level
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Spec 415-425 Deg.Zone Temp.
Bad 415Good 415Bad 416Bad 417Bad 417
Good 418Good 419Good 420Bad 420
Good 420Bad 422
Good 422Good 423
1/2 1
Spec 100-110Molding Temp
Bad 102Good 103Bad 103
Good 105Good 105Good 105Good 105Good 105Good 106Bad 108Bad 108Bad 109
2-1/2
3
The results of Zone Temp after arranging in the rank order ( Lower To Higher)
The results of Molding Temp after arranging in the rank order ( Lower To Higher)
Total Count =2,explains that the Zone temp is not suspected parameter
Total Count =4,explains that the Molding is not suspected parameter
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Regression Analysis
ESTABLISHES THE RELATION SHIP BETWEEN A CONTINUOUS INPUT VARIABLE AND A CONTINUOUS OUTPUT VARIABLE AND ALSO HELPS IN FINDING OUTTHE TOLERANCE OF INPUT VARIABLE REQUIRED TO ACHIEVE OUTPUT SPECIFICATION
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Simple Regression Analysis
The regression analysis is performed when
there is single X(Predictor)
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What is Regression AnalysisRegression analysis is a statistical technique used to model & investigate or predict
the relationship between dependent variable & the independent variable.
Where Y is the response, X is the predictor, and are the regression coefficients,
e the error terms with normal distribution with mean equal to zero & standard deviation
equal to sigma
The regression analysis perform when the X &Y are continuous .
Regression Is a hypothesis test
Ha :The model is significant predictor of the response
Ho : The model is not a significant predictor of the response
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Using Minitab to obtain regression equation
Use menus:
Stat / Regression / Regression ...
Dependent variable
Independent variable
(Mintab labels as ‘Predictor’)
Click Graph
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Select following order
Analysis of residuals ( Difference between the observed value & corresponding fitted value )
Residual analysis investigate adequacy of fitted model & detecting
departures from the model
By residual analysis technique
1. Check the normality assumption through a normal probability plot and /or histogram of the residual
2. Check the correlation between residuals by plotting residuals in time sequence
3. Check the correctness of the model by plotting residual versus fitted values.
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Analysis of Residuals
1. Normality Assessment : The histogram plot of residuals should look like a sample
from normal distribution which is the assumption for regression analysis.
2. Time Sequence :A plot of residuals against the observation order as a graphical
way of detecting a correlation that may exist between random errors.
3. Fitted Value : Residuals plotted against the fits should be randomly scatter &
having no pattern .
Check for outliers ,non constant variance ( Difference in the residual values either
increases or decreases for the increase in the fitted value ) & poor model fit
( Residual value tend to increase and then decrease with an increase in the fitted
value )
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Regression Analysis: Drug_conc versus Height
The regression equation isDrug_conc = 79.3 - 6.30 Height
Predictor Coef SE Coef T PConstant 79.296 6.023 13.17 0.000Height -6.296 1.176 -5.35 0.000
S = 9.751 R-Sq = 68.8% R-Sq(adj) = 66.4%
Analysis of Variance
Source DF SS MS F PRegression 1 2726.0 2726.0 28.67 0.000Residual Error 13 1236.0 95.1Total 14 3962.0
Minitab output
Regression equation
Regression Analysis: Drug_conc versus Height
The regression equation isDrug_conc = 79.3 - 6.30 Height
Predictor Coef SE Coef T PConstant 79.296 6.023 13.17 0.000Height -6.296 1.176 -5.35 0.000
S = 9.751 R-Sq = 68.8% R-Sq(adj) = 66.4%
Analysis of Variance
Source DF SS MS F PRegression 1 2726.0 2726.0 28.67 0.000Residual Error 13 1236.0 95.1Total 14 3962.0
Minitab output
Regression equation
P value of predictor is based on the following hypothesis test Ha = Coefficient is not zero Ho = Coefficient is zero If P value is <0.05 for α =0.05 reject Ho or coefficient of predictor is not zero
P value of regression is based on the following hypothesis test Ha = The model is a significant predictor of response Ho = The model not a significant predictor of response If P value is <0.05 for α =0.05 reject Ho or the model having significant predictor of response
Value represent the proportion of variation in the response data explained by the predictorsIn general , the closer is to 1 ,The better the fit of the model
R2 Adj(R-Sq) is the % of variation explained by the model adjusted for the no. of terms in your model & no. of data points
R2
S represent the std.error of the prediction = Sigma of individual data point
P value of constant is based on the following hypothesis test Ha = Coefficient is not constant Ho = Coefficient is constant If P value is <0.05 for α =0.05reject Ho or coefficient is not constant
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Using Minitab on Regression Analysis :Confidence & prediction Band Use menus:Stat / Regression / Fitted Line Plot ...
Dependent variable
Independent variable
(Minitab labels as ‘Predictor’)
Click Option
Click both the box
Click Ok
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Minitab Output
A confidence band ( or interval ) is a measure of the certainty of the shape of the fitted line at 95% confidence interval .
A prediction band ( or interval ) is a measure of the certainty of the scatter of individual points about the regression line
The regression line always passes through the average value of the cause variable X and the average value
of the response variable Y.
The prediction interval & confidence interval are at a minimum width at X = X bar.
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Use of Regression plot in Tolerancing The prediction interval is used in statistical tolerancing
USL
LSL
Note1 -In positive co-relation The USL should cut the upper prediction line & on negative co-relation vice versa Note2 – If the line do not cross in a negative relationship or line cross inpositive relationship, you can not use this relationship to control the output
Shows the Band for controllingX to meet tolerance
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Multiple Regression Analysis
The regression analysis is perform , when
there is Multiple X(Predictor)
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yi = β0 + β1x1i + β2x2i + ... + βkxki
where yi = a value of the dependent variable, y
β0 = the y-intercept
x1i, x2i, ... , xki = individual values of the independent variables, x1, x2, ... , xk
β1, β2 ,... , βk = the partial regression coefficients for the independent variables, x1, x2, ... , xk
Regression Is a hypothesis test
Ha :The model is significant predictor of the response
Ho : The model is not a significant parameter.
Multiple linear regression examines the linear relationships between one continuous response(Y) and two or more predictors (X).
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Using Minitab on Multiple Regression Analysis :Using Subset Method
The Regression analysis using subset method is used to select or identify a group of predictor variable (X’s) which are important. This technique is used to funnel group of predictor to important predictors. These important predictors are used further in multiple regression analysis.
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Using Minitab to obtain regression equation by best Sub –Set methodUse menus:
Stat / Regression / Best subset ...
Dependent variable
Independent variables
(Minitab labels as ‘Predictors)
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Highest Sum of square
Lowest C-p(which is used to determine minimum no. of parameter which best fits the model . It measures Sum of square of biases + squares of random error in Y at all n data points)
Using given criteria, the
model (row 5) with four
predictors,( Score-2,
insulation ,East ,South &
north appears to be the best
among all the X’s
Small Sd of error in the model
Note : Start with the single factor & select the variable having highest R-sq(adj) & lowest C-P ( In single factor) . Then take two factor & so on.. . Stop When C-P value become minimum & then increases
Do multiple regression on these Selected Predictor variable
Minitab Output
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Using Minitab to obtain regression equationUse menus:
Stat / Regression / Regression ...
Dependent variable
Independent variables
(Mintab labels as ‘Predictors)
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Value represent the proportion of variation in the response data explained by the predictorsIn general , the closer R- Sq is to 1 ,The better the fit of the model
R2Adj(R-Sq) is the % of
variation explained by the model adjusted for the no. of terms in your model & no. of data points
P value of regression is based on the following hypothesis test Ha = The model is a significant predictor of response Ho = The model is not a significant predictor of response If P value is <0.05 for α =0.05 reject Ho or the model is a significant predictor & response
P value of predictor is based on the following hypothesis test Ha = Coefficient is not zero Ho = Coefficient is zero If P value is <0.05 for α =0.05 reject Ho or coefficient of predictors are not zero
Minitab Output Regression Equation P value of constant is based on the
following hypothesis test Ha = Coefficient is not constant Ho = Coefficient is constant If P value is <0.05 for α =0.05reject Ho or coefficient is not constant
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Total Sum of all Seq SS is SS Regression given in previous table sum
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Select following order
Analysis of residuals ( Difference between the observed value & corresponding fitted value )
Residual analysis investigate adequacy of fitted model & detecting
departures from the model
By residual analysis technique
1. Check the normality assumption through a normal probability plot and /or histogram of the residual
2. Check the correlation between residuals by plotting residuals in time sequence
3. Check the correctness of the model by plotting residual versus fitted values.
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Design Of Experiment
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Y=f(X1,X2,X3,X4)
Find out the optimum combination of levels of x1,x2,x3,x4 Which will optimize Y
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The above table shows the durability values of pins found with combination of factors namely vendor size of clip and heat-treated state of clip.Each has been tried out at two levels.Are the three factors and their interaction Significant?
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Each row in the previous slide is a experimental condition with a combination of various level of the factor coded as -1 and +1.Each experimental condition has a output that is measured during experimental run.The example in the previous slide is a 23 factorial experiment with two replication thus requiring 16 experimental runs.Once the output of all experimental condition using all possible combination of factor levels is obtained these are than analyzed to find out the mean effect of the factor and the interactions.
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What is a Interaction
•If we get a graph like the one shown above, there is an interaction between the engine and the carburetor.
•The engine contributes to an increase in mileage of 5 when the carburetor is bad, whereas it contributes to an increase in 13 when the carburetor is good.
•The carburetor contributes to an increase in mileage of 2 when the engine is bad, whereas it contributes to an increase of 10 when the engine is good.
0
5
10
15
20
25
30
E- E+
Mile
age
25
10 13
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Before any formal data collection can start planning of experiment has to be done to ensure replication and randomization
• Replication
-- Multiple execution of all or part of the experimental process with the same factor setting.
-- It is done to measure the experimental variability. By doing this we can decide whether the difference in response is due to change in factor levels ( an induced special cause) or common cause variability i.e. noise factor ( which is uncontrollable)
-- to obtain the responses for each set of experimental conditions i.e.. Location & spread
• Properties of Replication
-- It estimates experimental error This estimate becomes the unit of measurement for determining whether observed difference are statistically significant.
-- It Yields more precise estimate of factor effects.
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Randomization
• Randomization -- To assign the order in which the experimental trails will be run using a random
mechanism *It is not the the standard order * It is not running in an order that is convenient * Computer help or random table help is taken in randomization
-- this is done to average out the effect of lurking variables over all the factors of experiment & help to validate statistical conclusions made from the experiment
• Lurking Variable -- A variable that has an important effect & yet is not included among the factors
under consideration because* Its existence is unknown *. Its Influence is thought to be negligible * data on it are unavailable
• Experimental trial are randomized to protect the effect of lurking variable
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.FortunatelyMinitab helps us formulating the structure.In this volume we will be limiting ourselves to creation of factorial design of Experiments with 2 levels through minitab.Any DOE experimentation has to gothrough the following steps of analysis.
•Perform analysis¾Identify factors¾Choose factor levels¾Select design¾Randomize runs¾Collect data¾Analyze data¾Draw conclusions¾Verify results
•Determine solutions•Record results•Standardization
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The project is to increase the durability of clip. The durability of clip is measured by number of bends a paper clip will withstand before breaking .The vendors are claiming improvement in durability due to new heat treatment process.As heat treatment procedure was identified as a important parameter from cause andeffect along with clip size and vendor a DOE is planned with twovendors,two clip sizes for heat treated and non heat treated clips.
Identify factors
example
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Creating factorial design Creating factorial design
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Input factors Input factors
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Random design Random design
Select randomize runsTo create run order and experimental conditions
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Data Sheet
The order in which experiment was run after randomizing
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Analyze data–Is data OK
•Plot the raw data in time order by factors-• Assess process stability over time to see the presence of
manufacturing trends and systematic influences.
•Do box plot to identify outliers.to focus quick improvement efforts on the outliers and reduce variation
•Do one way ANOVA to find out if any experimental level issignificant with respect to residuals
•Plot the residuals•Plot the residuals against time order to ensure the experimentalvariability has only common causes and absence of lurking variable (trends,outliers,or non-random patterns) that might influence conclusions.
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Using Minitab on Time Order Plot Use menus:Stat / Quality Tools /Run Chart ...
Response Variable
Subgroup Size of data
If P<0.05 then reject the Null hypothesisIn this case P value is grater then 0.05 for Mixture Trend & oscillation & clustering
Check for any Non random Patterns
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Y Variable & X category
Display Type in the graph
The Box Plot is graphical tool to look at several aspect of data Such as spread , symmetry & outliers.
The left most edge of the box plot represent the first quartile (25% of points ) . The right most edge represents the third quartile (75% of points) The centre line represent the 50th
percentage or median . A outlier is any points at the distance grater then 1.5times the width of the box from the first or third quartile , plotted with Asterisks (*)
Click Option
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Bend * Exp_Cond
Bend * Size
Bend * Vendor
Bend * HeatBox Plot of Bend
Check the highest variability Exp. Condition
There is difference between the heat treatment & Non
heat treatment
There is slight difference between the vendors
There is no outliers
Minitab Output There is slight difference
between Sizes
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Using Minitab for 1 Way ANOVAUse menus:Stat /ANOVA /1 Way ANOVA ...
Response Variable
Use each condition of the experimental condition column as factors
Click Ok
The null hypothesis Ho for the test is that all means for all experimental condition are the same.
The alternative hypothesis Ha is that one or more experimental condition has different mean.
Hypothesis for 1 way ANOVA
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Minitab Output Note -If P<0.05 reject null
hypothesis and accept alternate hypothesisThere is difference in mean of experimental condition.
If the Sp (pooled Std. Deviation ) is very large any factor effects may
appear insignificant
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•Plot the residuals
•Plot the residuals against time order to ensure the experimental variability has only common causes and absence of lurking variable (trends,outliers,or non-random patterns) that might influence conclusions.
•Plot the residual against average fits on Minitab for each experimental condition-look for presence of trends like megaphone shape.Note:ignore the pattern indicated by symmetry of dots around 0.
•Plot residuals on normal probability scale-any non normal trend indicates the variation in the experiment is non random and assignable causes are present beyond ones identified for experiment.
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Response data
Click Graph
Click four residual Plot
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Check for any non random patterns or
outliers
Check for any correlation between &
residual
Check for normal distribution of
residuals
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Action For Residual Plot
• If there are outliers , try to find an explanation & correct the
data If not able to find an explanation , leave the data as it is.
• If there is trend in the residuals over time ,there may be lurking
variable . Look for the cause & correct the situation.
• If there is a correlation between the residuals & fits ,understand
the cause of relation .
• If the residual are not normally distributed ,there may be some
significant factors which are not accounted . Look for that factor
& include it in the experiment
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Analysis –Make Inferences
The main aim of analysis is to understand which Xs and interaction affect the Y 1.Examine the factor Effect Using Minitab for Pareto Use menus:Stat /Doe /Factorial /Analyze Factorial Design ...
Response data
Click Graph
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Click Pataro
Pareto chart of the effects to compare the relative magnitude and the statistical
significance of both main and interaction effects in decreasing order of the
absolute value of the standardized effects and draws a reference line on the chart.
Any effect that extends past this reference line is significant.
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The null hypothesis Ho Factor has no effect on the results.
The alternative hypothesis Ha is that Factors has an effect on the result.
Analyze Factorial Design
Note -If P<0.05 reject null hypothesis and accept alternate hypothesisOnly Heat & vendor *Size Interaction have a significant affect on the number of bends
•Conform impressions with statistical Procedures
The sum of square for the main effects , 2-way interactions , &3-way interactions represent SSBThe residual Error represent SSW.Further exploration of sum of squares is done throughANOVA(GLM)
The Equation is Bend = 15.688-0.437V+0.563S+4.063H-2.563V*S
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Analysis –confirm Inferences (Practical Significance) Practical Significance is evaluated by analyzing the sum of Square (SS) andsubsequently mean square.
Using Minitab for ANOVA GLM Use menus:Stat /ANOVA /General Linear model ...
Response Variable
Model contain the main effect & Interaction (Crossed ) for the factors
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Minitab Output The null hypothesis Ho Factor has no effect on the results.
The alternative hypothesis Ha is that Factors has an effect on the result.
Note -If P<0.05 reject null hypothesis and accept alternate hypothesisOnly Heat & vendor *Size Interaction have a significant affect on the number of bends
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Using Minitab for factorial Plot Use menus:Stat /Doe /Factorial /Factorial Plot ...
Select the type of plot required
The center value for the plot
Select the Factor for the Plots
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influence the response and to compare the relative
strength of the effects
• If the line is horizontal (parallel to the x-axis), there is no main effect present. The response mean does not change depending on the factor level.
• If the line is not horizontal (parallel to the x-axis), there is a main effect present. The greater the degree of departure from being horizontal, the stronger the effect.
Use an interaction plot to determine if two factors interact in their effect on the response and to compare the relative strength of the effects.
• .If the lines are parallel to each other, there is no interaction present. The change in the response mean from the low to the high level of a factor does not depend on the level of a second factor.
• If the lines are not parallel to each other, there is an interaction present. The change in the response mean from the low to the high level of a factor depends on the level of a second factor. The greater the degree of departure from being parallel, the stronger the effect.
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Suppose one finds seven significant factors.With 2 replications this will require 256 experiments.Conducting so many experiments will consume lot of time .In such cases minitab allows fractional factorial.The only disadvantage with fractional factorial is that the higher order interactions are Confounded.The confounding is determined by degree of resolution and fraction applied.
SCREENING
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WITH SEVEN FACTORS 16 TRIALS CAN BE CONDUCTED WITH RESOLUTION IV32 TRIALS CAN BE CONDUCTED WITH RESOLUTION 1V64 TRIALS CAN BE CONDUCTED WITH RESOLUTION SEVEN128 TRIALS FOR FULL FACTORIAL
THE TABLE SHOWS NUMBER OF TRIALS WITH SINGLE REPLICATION
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STAT-DOE-CREATE FACTORIAL DESIGN-DESIGNS gives the followingScreen gives the following screen with all alternatives
32 trials with seven factors corresponds to ¼ fraction as (1/4)*27=25=32
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What is resolution?
R CONFOUNDING
III 1-2
IV 1-3 2-2
V 1-4 2-3
Table shows that for resolution III all main effects are confounded withTwo way interaction.For resolution IV all main effects are confounded with three way interactions and all two way interaction are confounded with each other
Confounding implies that Calculated effects are due toCombinations determinedby degree of resolution.
1-main effect2-2way interaction3-3way interaction4-4way interaction
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Confounding effect due to fractional factorial
When one is choosing a fractional factorial experiment a certain amount of Confounding is introduced with main effect for example in a ¼fraction experiment with 7 factors involving 25 trials (A,B,C,D,E,F,G,H) two of the 5 factors will have Confounding.Two of the factors i.e. G and H will be Confounded with ABCD and ABDE respectively. ALL THE CONFOUNDING DUE TO RESOLUTION AND FRACTIONALFACTORIAL ARE SHOWN IN MINITAB BY ALIAS STRUCTUREAND DESIGN GENERATORS AND BEFORE PROCEEDING ONE MUSTEVALUATE WHETHER KNOWN SIGNIFICANT INTERACTION WILL BE LOST OR NOT.
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Prediction Model for Standard Deviation Step 1Prepare the reduce matrix in standard order. Then transfer the mean & standard deviation calculated earlier by using 1 way ANOVA ( Using std. Order as factor) in this reduce matrix.
Mean & standard Deviation is achieved from –Way ANOVA
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Step 2Use menus:Stat /Doe /Factorial / Define Custom Factorial Design ...
This is done in order for Minitab to recognise the new design .
Select the design
Select the factor
Select the 2 level design
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Select these option for std. Order , run order ,centre point
& block
Now new Design recognized by Minitab
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TATA MOTORS LIMITEDStep 3Run the Factorial Analysis Using St. dev as Response Use menus:Stat /Doe /Factorial /AnalyzeFactorial Design ...
Select Term Menu
Include the model Having up to 3 Level of interaction
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Rule : Look For the coefficients With magnitude that is
roughly greater then half of the constants
Vendor *Heat Interaction is a significant Dispersion Effect
Model with Coefficient value plugged inSt dev = 3.094+0.795*Vendor +0.972*Heat –1.503*(Vendor*Heat)
Step 4Built The prediction model
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Step 5
Explore Prediction model
• Optimize centre & spread
• Explore “ Trade Off” vs. goals (CTQs)
•Improve Setting
•Track The Result
Step6
Determine the factors settings
The six sigma solution should be trade off between target Mean & Variation
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RESPONSE SURFACE •Used to determine how a response is affected by a set of quantitative variables/ factors over some specified region.
•For a given number of variables response surface analysis techniques require more trials than two level fractional factorial design techniques.Hence the number of variables considered in an experiment may first need to be reduced through either technical considerations or fractional factorial experiments.
•Information on response surface is used to optimise the settings of a processto give a maximum or minimum response.
•Knowledge of the response surface analysis techniques can help in choosingsettings for a process so that day to day variations which are typically foundIn a manufacturing environment which will have a minimum effect on Degradation of product quality.
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225Six Sigma Central Coordination Group
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Variable Search Variable Search
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VARIABLE SEARCH
OBJECTIVE :
PINPOINTS “RED X” , “PINK X” etc.. CAPTURES
THE MAGNITUDE OF ALL IMPORTANT MAIN
EFFECTS AND INTERACTION EFFECTS.
OPENS UP TOLERANCES OF ALL UNIMPORTANT
VARIABLES TO REDUCE COST.
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THE FOUR STAGES OF COMPONENTS/VARIABLESEARCH
To quantity the magnitudes of the important main causes and their interaction effect.
4Factorial Analysis
To verify that the important causes are truly important and that the unimportant causes are truly unimportant.
3Capping Run
To eliminate all unimportant causes and their associated interaction effects.
2Elimination
To determine if Red “X”, Pink “X” are among the causes being considered
1Ball Park
OBJECTIVESTAGE
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Procedure for Variable SearchStage I )
¾ List the most important input variables or factors or causes :
A,B,C,D,E,F,G,H and so on in descending order of importance of each
factor’s ability to influence the output. The selection can be made by using
prior DOE techniques like Paired comparison or Multivari etc.
¾ Assign two levels to each factor : a high (+) which is most likely to
contributed to good results and a low (-) which represents a most likely
deviation from the high level in day to day production.
¾ Run two experiments, one with all factors at their high levels, the other
with all factors at their low levels. Repeat these two experiments twice more
so that there are three runs with all factors at high levels and three runs
with all factors at low levels.
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Proc edure for Var iab le Searc h
Apply the D : d bar > 1.25 : 1 rule
where D is the difference between the median of the high readings and the median of
the low readings and d bar is the average of the differences in repeatability within
each set of high and low readings.
Stage I is successful when all three high levels are better than all three low levels and
where the above mentioned ratio is met. If these conditions are not met then look for
any cancellation effect or an important factor that has been left out.
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Proc edure for Var iab le Searc hSt age I I
Run a pair of tests with the low level of the most important factor, A, I.e. AL
along with the high levels of all the remaining factors labeled RH and with the
high level of the most important factor, A, I.e. AH, along with the low levels of all
the remaining factors RL. Calculate the high side and low side control limits. The
formula is Median + 2.776 ( d bar/ 1.81 ).
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Procedure for Variable SearchPossible Results :
I) If both pairs of tests AL RH and AHRL falls within the control limits for all
high level combination and all low level combination respectively, factor A,
along with all of its associated interaction effects, is unimportant and can be
eliminated from further study.
II) If there is a complete reversal, I.e. if ALRH falls within the control limits of all
low level combination and AHRL falls within the control limits of all high level
combination, A is a main effect Red X. the rest of the factors – B,C,D etc.
are all unimportant and can be eliminated.
III ) If either of both pairs of tests show results outside the low side and high side
control limits respectively, but not a complete reversal, factors A along with
its associated interaction effects can not be eliminated. A plus some other
factor or factors must be considered.
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Proc edure for Var iab le Searc hStage3
If the results are I)and III)in Stage 1 repeat steps with next factor B .
I)If the result is complete reversal B is red x and no further search is required.
II)If results is partial swing then a capping run with A is made at high and low levels of
A and B i.e. with ALBLRH and AHBHRL if in stage1 A had shown partial swing.
complete reversal indicates end of experiment and no further search is required.
Partial swing indicates search is not complete and third factor is to be selected and
and the steps I and ii) repeated until a reversal is Obtained with the new factor or a
capping run of the new factor with the old factors which had caused partial swing in
earlier stages.
Normally search does not go beyond capping run with three factors.a three factor
capping run. ( Generally four factor capping run is rare ).
Finally draw up a factorial analysis using the factors identified to quantify the main
effects and interaction effects of the important factors.
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233Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY : THE PRESS BRAKE
IN A METAL STAMPING FORMING OPERATION,PARTS PRODUCED ON THE
PRESS BRAKES COULD NOT BE HELD TO A + 0.005 INCHES TOLERANCE
(OR PROCESS WIDTH OF 0.010 INCHES). VARIATION AS HIGH AS + 0.010
INCHES WERE BEING MEASURED
A VARIABLE SEARCH EXPERIMENT WAS THEN CONDUCTED. THE OBJECTIVE
WAS TO BRING THE PROCESS UNDER CONTROL, CONSISTENTLY, TO
+ 0.005 INCHES OR CLOSER.
SIX FACTORS WERE SELECTED, IN DESCENDING ORDER OF PERCEIVED
IMPORTANCE, AND THE HIGH AND LOW LEVELS
FOR EACH FACTOR DETERMINED.
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234Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY: THE PRESS BRAK E
NOT ALIGNEDTHINSOFT
BOWEDAIR FORMAT ANGLE
ALIGNEDTHICKHARDFLAT
COIN FORMLEVEL
A. PUNCH & DIE ALIGNMENTB. METAL THICKNESSC. METAL HARDNESSD. METAL BOWE. RAM STROKEF. HOLDING MATERIAL
LOWHIGHFACTORS
RESULTS : NUMBER BELOW ARE EXPRESSED IN DEVIATION FROMNOMINAL IN MULTIPLES OF 0.001 INCHES ( LOWER
TOLERANCE ARE BETTER AND VICE VERSA ).STAGE 1 ALL HIGH LEVELS ALL LOW LEVELS INITIAL 4 471st REPLICATION 4 612nd REPLICATION 3 68D = ( 61 – 4 ) = 57 d = ( 1 + 21 ) / 2 = 11 SO D/d = 5.2
(STAGE 1 SUCCESSFUL)CONTROL LIMITS = MEDIAN + 2.776 d/ 1.81 = MEDIAN + 16.87SO CONTROL LIMITS ( LOW LEVEL ) = 44.13 TO 77.87SO CONTROL LIMITS (HIGH LEVEL) = -12.87 TO 28.87
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235Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY: THE PRESS BRAK E
STAGE - 2
F IMPORTANT WITH ANOTHER FACTOR.
-12,87 TO 20.8744.13 TO 77.87
461
7318
FLRH
FHRL
1112
E NOT IMPORTAN-12,87 TO 20.8744.13 TO 77.87
461
750
ELRH
EHRL
910
D IMPORTANT WITH ANOTHER FACTOR
-12,87 TO 20.8744.13 TO 77.87
461
2330
DLRH
DHRL
78
C NOT IMPORTANT-12,87 TO 20.8744.13 TO 77.87
461
772
CLRH
CHRL
56
B NOT IMPORTANT-12,87 TO 20.8744.13 TO 77.87
461
547
BLRH
BHRL
34
A NOT IMPORTANT-12,87 TO 20.8744.13 TO 77.87
461
372
AL RH
AL RH
12
CONCLUSIONCONTROL LIMITS
MEDIANRESULTSCOMBINATIONTEST
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VARIABLE SEARCH CASE STUDY: THE PRESS BRAK E
STAGE – 3 : CAPPING RUN
•COMPLETE REVERSAL•END OF TEST
44.13 TO 77.876170DLFLRH#
•DF INTERACTION MAY BE PRESENT•R UNIMPORTANT
-12,87 TO 20.8744DHFHRL#
CONCLUSIONCONTROL LIMITS
MEDIANRESULTSCOMBINATIONTEST
# : DENOTES CAPPING RUN
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237Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY: THE PRESS BRAK ESTAGE 4 : FACTORIAL ANALYSIS
47 7261 4768 7270 50
MEDIAN = 64.5
7330
MEDIAN = 51.5
2318
MEDIAN = 20.5
4 34 53 74 7
MEDIAN = 4
DLOWDHIGH
FHIGH
FLOW
72 55.585
24.5
116
68.5
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238Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY: THE PRESS BRAK E
MAIN EFFECT : D = (20.5+64.5) – (4+51.5) = (85-55.5 ) = 14.75 PINK ‘X ’2 2
MAIN EFFECT : F = (51.5+64.5) – (4+20.5) = (116-24.5) = 45.75 RED ‘X ’2 2
DF INTERACTION = (51.5+20.5) – (4+64.5) = (72-68.5) = 1 .752 2
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DESIGN OF EXPERIMENTS
GREEN ‘Y’ : BURNING MARKS ON BALL TRACK OF STEERING NUT(407 / 207)PROBLEM HISTORY : (87 Rej. 19570 Prod.) i.e. 15% OF TOTAL REJN.FOR APR TO OCT.COST OF REJECTION : Rs 138 /- APPROX. PER PIECE.COST OF REJN. FOR BURNING MARKS : 1715 /- PER MONTHDOE TECHNIQUE USED : VARIABLE SEARCHLIKERT SCALE : TO CONVERT ATTRIBUTE DATA INTO VARIABLE DATA.
4HEAVY BURNING MARKS
3MEDIUM BURNING MARKS
2LIGHT BURNING MARKS
1NO BURNING MARKS
VARIABLE SCALEATTRIBUTE SCALE
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VARIABLE SEARCH CASE STUDY : STEERING NUT BURNING MARK S PROBLEM
FIXED FACTORS : QUILL R/OUT, DIAMOND ROLL FACE R/OUT, BACKLASH OFLEAD SCREW & BOX NUT, COOLENT FLOW WERE OPTIMISED
90150000.061817GRADE:RA8017VF8
80120000.0226S17 (SUPLHUR TREATED)GRADE : RA8017VF8
A. WORK HEAD FEEDB. WHEEL R.P.MC. DRESSING DEPTH OF CUTD. NO. OF PASSESE. GRINDING WHEEL
WORSTBESTFACTORS
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RESULTS : NUMBERS BELOW ARE CONVERSION OF ATTRIBUTE DATA INTOVARIABLE ( LIKERT SCALE )
STAGE I ALL BEST LEVELS ALL WORST LEVELINTIAL 2 31st REPLICATION 2 32nd REPLICATION 2 2
D = (3 – 2) = 1 d + (0+1) /2 = 0.5 So, D/d = 2.0STAGE I SUCCESSFUL
CONTROL LIMITS = MEDIAN + 2.776 d / 1.81 = MEDIAN + 0.767
SO CONTROL LIMITS ( WORST LEVEL ) = 2.233 TO 3.767
SO CONTROL LIMITS ( BEST LEVEL ) = 1.233 TO 2.767
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242Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY : STEERING NUT BURNING MARK S PROBLEM
STAGE 2
E IS NOT IMPORTANT
1.233 TO 2.7672.233 TO 3.767
23
1.53
EWRB
EBRW
910
D IS IMPORTANT1.233 TO 2.7672.233 TO 3.767
23
31
DWRB
DBRW
78
C IS NOT IMPORTANT
1.233 TO 2.7672.233 TO 3.767
23
1.53
CWRB
CBRW
56
B IS IMPORTANT1.233 TO 2.7672.233 TO 3.767
23
31.5
BWRB
BBRW
34
A IS IMPORTANT1.233 TO 2.7672.233 TO 3.767
23
31
AWRB
ABRW
12
CONCLUSIONCONTROL LIMITSMEDIANRESULTSCOMBINATIONTEST
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243Six Sigma Central Coordination Group
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VARIABLE SEARCH CASE STUDY : STEERING NUT BURNING MARK S PROBLEM
STAGE 4 : FACTORIAL ANALYSIS
3.00 2.00 3.003.00 3.00 3.00
2.001.003.00
1.503.002.001.50 2.00 2.001.00 1.50 2.00Media =1.75
DWORSTDBESTDWORSTDBEST
AWROSTABEST
BBEST
BWORST
Media =3
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RE
D ‘X
’
E:- GRINDING WHEEL
D – NO. OF PASSES
C:- DRESSING DEPTH OF CUT
B:- WHEEL R.P.M
A:- FEED
A B D A X B A X D B X D AX BXD-1 -1 -1 1 1 1 -1 31 -1 -1 -1 -1 1 1 1-1 1 -1 -1 1 -1 1 1.51 1 -1 1 -1 -1 -1 2-1 -1 1 1 -1 -1 1 21 -1 1 -1 1 -1 -1 3-1 1 1 -1 -1 1 -1 31 1 1 1 1 1 1 1.75
-1.75 -0.75 2.25 0.25 1.25 0.25 -4.75Effect
Pin
k X