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Six Sigma Six Sigma 1 Basic Statistics GB Module

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Page 1: Six Sigma Basic Stats Module

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1

Basic StatisticsGB Module

Page 2: Six Sigma Basic Stats Module

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2

Continuous Improvement Road Map

Improve

Control

• Define CTQ• Determine Current State

• Determine Key Input / Output Variables

• Perform MSA• Calculate initial process

capabilities

Measure

• Verify Effects of Key inputs with DOE’s

• Determine Optimum Settings

• Update Control Plan• Verify Improvements

Analyze• Evaluate Existing Control Plan• Using statistical methods to determine potential key inputs• Prioritize key input variables

N

Define

Page 3: Six Sigma Basic Stats Module

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Customers and Variation

• Customers complain when they believe the product or service they receive differs from their expectations; there is variation

• Variation has many faces:– Missing functionality/actions– Defects and faults– Delays etc

• All variation is caused• Six Sigma is about reducing and controlling variation• We need to understand variation and the causes of variation

Page 4: Six Sigma Basic Stats Module

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Causes of Variation

YY DependentDependent OutputOutput EffectEffect SymptomSymptom MonitorMonitor

XX11 . . . X . . . XNN

IndependentIndependent Input-ProcessInput-Process CauseCause ProblemProblem ControlControl

Therefore we need to understand the Xs and improve and control the ones with most influence on Y

The variation in Y is caused by variation of the Xs

Output (Ys)Process

(Xs)Input (Xs)

Page 5: Six Sigma Basic Stats Module

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Long Term & Short Term Variation

ProcessResponse

Y

Short-Term includes: common cause variation only

Long-Term includes: common cause & (some) special cause variation

EXAMPLEI drive to work. It takes me 35 +/- 3 minutes. This is the common cause variation. One day it takes me 50 minutes due to roadworks - this is a special cause.

Time

Short-Term Variation due toCommon causes

Long

-Ter

m V

aria

tion

Special Causes

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Pro

cess

Res

pons

e

Time

Examples of Special Causes

Special causes are assignable and can include:

Weather (season, time of day)Lighting ConditionsMachine TypeMachine AgeMaintenanceSupplierOperatoretc…special causes

Page 7: Six Sigma Basic Stats Module

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Exercise – Special Causes

• Consider the process in your project

• Make a list of the potential special causes

• Be prepared to share your list with the rest of the group

• Time: 10 Minutes

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Measuring Variation

• Variation is not simple to measure because it is RANDOM• Random does not mean erratic! While it may not be

possible to predict what an individual process output is, there is usually a pattern if we measure a number of outputs

• Process outputs will group together and we are interested in their central value, the value they group around, and their spread.

• This grouping forms a pattern that is often predictable

Page 9: Six Sigma Basic Stats Module

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Example

• If a coin is tossed we cannot predict whether it will be a head or tail

• If we tossed the coin say 100 times we would expect that on 50 occasions it would be a head and on 50 occasions it would be a tail

• So there is a pattern - but we cannot predict any individual toss

• We relate the expectations to chance (probability), there is a 50% chance it will be a head

• Randomness is about chance

Page 10: Six Sigma Basic Stats Module

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Coin Toss Exercise

• Everyone needs a coin of some type

• Flip the coin 25 times and record the number of “heads”

• Report the total number of “heads” obtained and create a dot plot

• Repeat the experiment• How do the dot plots

compare?

Page 11: Six Sigma Basic Stats Module

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Randomness and Distributions• Outputs group together to form a pattern

• This pattern describes the distribution of the variation

• We cannot predict where an individual value will fall, but we can predict the overall pattern

X

X

X

X X

X X X

X X X X

X X X X X X X X

10 11 12 13 14 15 16 17 18

Time to deliver

Distribution

Page 12: Six Sigma Basic Stats Module

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Real Life and Distributions• Distributions can be modelled mathematically

• If we collect data from a process or product we can “match” it to a distribution and use the properties of the distribution for analysis and predictions

Real situationModelled by adistribution

AnalysisReal Solution

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Probability Distributions• Probability

distributions (normally just called distributions) are a way of being able to make predictions about random events

• There are many “standard” distributions which enable us to model real world variation

• Standard distributions– Attribute data

• Binomial• Poisson

– Variable Data• Normal (Gaussian)• Lognormal (skewed)• Student t• F-distribution• Exponential

Page 14: Six Sigma Basic Stats Module

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Key Properties of DistributionsCentral Tendency: the value the data

groups around

Spread or dispersion of the values

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Measures of Central Tendency– Mean (mu)

– Median

- middle value of ranked data

– Mode

- most frequently occurring value

x

n i

If the distribution is symmetric

The mean, medianand mode have the same value

If the distribution is NOT symmetric

The mean, medianand mode have different values

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Measures of Spread• Range R = Biggest value – smallest value

• The range is susceptible to outlying values, as a result we need a better measure

• One approach is to calculate the average deviation from the mean:

(X - )n

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• The average of the deviations squared is called the variance and is a measure of spread

• It suffers from having units the same as the mean2. To overcome this we take the square root to give the standard deviation which has the symbol

• We use standard deviation as a measure of spread

Variance and Standard Deviation

n

xi2

i 21 xn

V

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Descriptive Statistics• Can be calculated using Minitab or Excel

• Gives information about a data set’s central tendency, spread and shape

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Descriptive StatisticsDescriptive Statistics: Data1

Variable N Mean Median TrMean StDev SE MeanData1 500 56.421 56.355 56.486 5.563 0.249

Variable Minimum Maximum Q1 Q3Data1 38.260 69.801 52.693 60.227

Central Tendency

Shape

Spread

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Computer Exercise!• Open the file 3L54

Stone.mpj

• Calculate the Descriptive Statistics for the data set

• Additionally, create a graphical descriptive statistics output

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Descriptive Statistics Result

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Descriptive Statistics – Graphical Summary

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Descriptive Statistics – Graphical Summary

15 25 35 45 55 65

95% Confidence Interval for Mu

41 42 43 44

95% Confidence Interval for Median

Variable: MPH

A-Squared:P-Value:

MeanStDevVarianceSkewnessKurtosisN

Minimum1st QuartileMedian3rd QuartileMaximum

40.5931

8.4792

40.5544

0.3990.360

42.2730 9.5282

90.78578.53E-02-4.0E-02

126

16.900036.675042.450048.050065.2000

43.9530

10.8755

43.5000

Anderson-Darling Normality Test

95% Confidence Interval for Mu

95% Confidence Interval for Sigma

95% Confidence Interval for Median

Descriptive Statistics

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XX

XXXXX

XX

X

X

X

On Target Reduce theSpread

X

XXXX

XX

XXXX

XX X

Mean and standard deviation tell us a great deal about a process

Off-Target Measured by the mean

Spread Measured by the Standard Deviation

X

X

X

X

XX

X

X

XXX

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Off-Target Spread

On Target Reduce the Spread

Distributions and Variation …..

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Normal Distribution

Mean

• Frequently occurs in practice • Models random behaviour• Shows that the variation groups around the mean and tails off• Symmetric about the mean with a 50% chance of falling either side of the mean• Is the basis for Six Sigma and many Six Sigma tools

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Area and ProbabilityArea under Normal curve = probability or chance of

being in that region

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

Area = 0.5 probability = 0.5 or 50%

Area = 0.159 probability = 0.159 or15.9%

-4 -3 -2 -1 0 1 2 3 4

Area = 1.0 probability = 1.0 or 100%

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Area, Probability & Standard Deviation

Mean

Total Area100%

-134

.13%

34.1

3%

-2-3 1 2 3

13.5

9%

13.5

9%

2.14

%

2.14

%

0.135% 0.135%

99.73% between -3 and + 3

0.27% lies outside -3 and + 3

The area under theNormal distributionrelates to probability

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A Common Situation

Mean

X

What is the probabilitythat the characteristic

is greater orequal to X?

Given productcharacteristic that isnormally distributedwith a mean and

standard deviation

If we calculate the number of standard deviations betweenthe mean and X, we can use

the area results to determine theprobability

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Standard Normal Distribution

Tables exist for the probability vs. number of standard deviationsfor the case of a “Standard Normal distribution” which has

a mean = 0 and a standard deviation = 1

Z value or number of standard deviations

z .00 .01 .02 .03 .04 .05

0.0 .5000 .4960 .4920 .4880 .4840 .48010.1 .4602 .4562 .4522 .4483 .4443 .44040.2 .4207 .4168 .4129 .4090 .4052 .40130.3 .3821 .3783 .3745 .3707 .3669 .36320.4 .3446 .3409 .3372 .3336 .3300 .32640.5 .3085 .3050 .3015 .2981 .2946 .29120.6 .2743 .2709 .2676 .2643 .2611 .25780.7 .2420 .2389 .2358 .2327 .2296 .22660.8 .2119 .2090 .2061 .2033 .2005 .19770.9 .1841 .1814 .1788 .1762 .1736 .17111.0 .1587 .1562 .1539 .1515 .1492 .14691.1 .1357 .1335 .1314 .1292 .1271 .1251

Mean =0

= 1

Tables exist that give the probability of a pointof interest X being

greater or equal to Z

Z

X

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• The probability values are often called p-values– p-value = tail area– Area under curve beyond point or value of interest– Probability of being at value of interest or beyond

• A small p-value (0 to 0.05) indicates– The probability is small that the value of interest comes from that distribution

by chance– Something else is going on

P-values are Probabilities of Interest

-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4

Value ofInterest

Value ofInterest

Value ofInterest

Value ofInterest

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Z Values

Mean 10

3

Mean 45

6.1

Mean 7.98

1.2

Mean =0

= 1

Z is calculated using the equation

P-value

Z

XZ =

X -

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Using Z Values

Mean = 1.735

= 0.0855

1.65 1.85

Total area in tails 0.1611 + 0.0885 = 0.2496Area of interest = 1- 0.2496 = 0.75 = yield

345.10855.0

735.185.1 Z994.00855.0

735.165.1 Z

NoteThe sign of the Z value

simply indicates direction from the mean.

A customer demands that a product’s specification is 1.75±0.1 We collect datafrom the manufacturing process and find the mean to be 1.735 and = 0.0855

From Standard Normal tables Area 0.1611

From Standard Normal tables Area 0.0885

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Mini Exercise

Mean = 1.5

= 0.25

1.75

?ZProbability that an item

is greater than 1.75?

From Standard Normal tables Area/Probability?

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Variation and 6-Sigma

Customer Critical RequirementsLSL

Target

USL1 2 3 4 5 6

99.9999998%99.9999998%99.9999998%99.9999998%

Z= 6.0Tail Area

No of Defects1 in a billion

Tail AreaNo of Defects1 in a billion

This is a 6-Sigma (Process or Product)

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Long-Term & Short-Term Variation

The presence of special causes will act to increase the variationseen by the customer. A gross assumption is a 1.5 sigma shift.

Cha

ract

eris

tics

or R

espo

nse

Time

Short-Term Variation

Long-Term Variation

What the Customer

Sees

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1.5 Sigma Shift

Customer Critical RequirementsLSL

Target

USL

This is a Six Sigma (Process or Product)

99.99966%99.99966%99.99966%99.99966%

Tail AreaNo of Defects

0

Tail AreaNo of Defects3.4 in a million

Z= 4.5

1 2 3 4 5 6

1.5

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1.5 Sigma Shift Demonstration

• Open Minitab file Glass Strength.mpj• Calculate the subgroup standard

deviations using Descriptive Statistics• Calculate the average standard

deviation across the subgroups• Stack the subgroups• Calculate the combined Standard

Deviation of the stacked data• Divide the standard deviation of the

stacked data by the average standard deviation of the subgroups.

What did you find?

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Summary

• In the short term we need a Zst = 6.0 to “guarantee” a long term Zlt = 4.5

• Note to achieve 3.4 defects per million requires Zlt = 4.5 - we should not strive to achieve Zst = 6.0 if

Zshift = Zst - Zlt < 1.5

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Z values and Sigma levels

• Zst values are related to Sigma levels• In 6-Sigma we look at short and long term values: Zst and Zlt • In 6-Sigma if we cannot calculate long term variation we assume a 1.5

sigma shift

Zlt = Zst - 1.5

• Note Z-tables generally do not have the 1.5 sigma shift and give Zlt

• Sigma/DPMO tables do have the 1.5 sigma shift and give Zst

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Testing for Normality• The Normal distribution is important to 6-Sigma since many of the tools

and techniques are affected by Non Normal data

Tool Consequence

Process sigma Incorrect process sigma

Individuals control chart False detection of special causes

Hypothesis testing Incorrect conclusions

Regression False identification of important factors poor predictive properties

DOE Incorrect conclusions about important factors poor prediction abilities

Page 42: Six Sigma Basic Stats Module

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U S L

a v e r a g e

W h a t p e r c e n t a g e f a l l s h e r e ?

T h e p e r c e n t a g e i s d i f f e r e n t f o r t h e N o r m a l c u r v e

Effect of not checking Normality• Example: Effect of skewed

distribution on calculating the process Sigma Level

– Process Sigma Level is determined by finding the area beyond the specification limits using Z-tables

– If the data is not Normal, the area will be incorrectly estimated from the Z-tables and therefore give a misleading Process Sigma Level

Page 43: Six Sigma Basic Stats Module

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Exercise – Normal?

– Look at each histogram on the following pages and decide which data sets come from a Normal distribution

• Circle or mark the ones you think are Normal.

– Work in pairs to confirm your answers

– Be prepared to share your answers with the whole group

– Time 10 minutes

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Assess Data for Normality•25 Data Points

Mark the histograms that you think come from a Normal distribution

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Assess Data for Normality, cont.50 Data Points

Mark the histograms that you think come from a Normal distribution

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Assess Data for Normality, cont.100 Data Points

Mark the histograms that you think come from a Normal distribution

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Exercise : Answers– Just looking at histograms can be deceptive

• Each of the Histograms on the previous pages were randomly generated in Minitab as a Normal distribution with a mean = 50, and a standard deviation = 10: they are all Normal.

– It is difficult to tell if data is Normal by looking at histograms of n = 25, n = 50, and sometimes even n = 100

– Plotting the data is very good practice, but do not be misled by small amounts of data

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Other DistributionsExponential Poisson Uniform

25

50

100

Sam

ple

Siz

e

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Data Not Normal• If the data is not Normal there may be

reasons which can be corrected– Extreme values, Typographical errors

- correct them

– Multiple modes - separate them

– Data rounded - increase precision

– Not enough data – collect more

– Special causes present – remove them

– Underlying distribution is not normal

Always check these first!

before concluding

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Check for Normal Distribution

• Both variable and discrete data (if there is enough) can often be modelled by a Normal distribution

• Many Statistical tools are based upon a normal distribution. However, many of the statistical tools will produce outputs even if the data is not normal. These outputs could be misleading

• Hence one of the first steps having collected data is to check for normality

• There is a test in Minitab for this

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Normality TestIn a normality test the datais plotted on “normal probability paper” - if thedata follows a straight lineit is normal.

The test also includes a hypothesis test (seeWeek 2). This test providesa quantitative value as towhether the data is normalthrough the p-value.

If p>0.05 we can say thatthe data is normallydistributed

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P-values• The p-value is the risk of making the wrong decision - in this case

concluding that the data is not normally distributed when it is.• In this case the p-value is 54% - a 54% risk of making the wrong

decision - in this case concluding that the data is not normally distributed when it is.

• This risk is too high so we conclude the data is normally distributed• We can never be risk free or 100% certain. Hence we need to set a

decision level. Experience shows that this is 5% (or 95% confidence)

• Hence we test to see if p > 0.05 if it is the data is normally distributed

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Using Minitab to Check for Normality

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Normality Test Exercise

• Using the data from Glass Strength.mpj, check for normality for each of the subgroups

• Then check for normality on the combined data

• Be prepared to report your findings

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Non Normal Distributions• Having checked the data for typographical errors etc

and concluded that the data is not normally distributed progress can still be made– In some cases of non-normal distributions (typically

Skewed Distributions) it is possible to transform the data to make it normal

– In some cases the data may be “close enough” to a normal distribution to use the statistical tools with care

– In some cases it does not matter that the data is not normally distributed

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Summary• This ppt has covered

– Variation• Common and Special causes, Long and short term data

– Distributions• Central Value (tendency): mean, median and mode

• Spread or dispersion: range, variance and standard deviation

– Normal Distribution• Z-values and p-values, Six Sigma, 1.5 Sigma shift and Z-values

• Checking for normality and dealing with Non normal data