six sigma black belt certification demo week2

7/30/2019 Six Sigma Black Belt Certification Demo Week2

1/27

Qi2, 1995-2002

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3/27

OVER

PROCESSING

INVENTORY

MOTION

MOVE IT OVER

THERE UNTIL WE

NEED IT

CONVEYANCE

CORRECTION

OVER PRODUCTION

WAITING

The 7 MajorWastes

Source: Ford 6 S Training Course



4/27

EXERCISE

Assembling two parts together

Searching for information

Making initial fixture setup

Transporting materials to next station

Capturing data once at the source

Examining castings for defectsWalking to parts storage for assembly screws

Storing a lot on a movable rack

Readjusting the fixture setup

Trouble-shooting a rejected assembly

QC final inspection

Leaving a form in an in-basket

Attending a benefits meeting

VA NVA

For each task, select whether it is value

added or non value added



5/27

SETUP TIME REDUCTION

The objective is to facilitate small-lot production,

notto reduce setup personnel headcount/labor

The people involved in the setup should do their ownsetup time reduction

- Look at the setup as a process

- Use videotape recording, with a built-in time clock(down to seconds)

The team should follow a 4-step process:Analyze the complete videotaped setup and

classify the time elements into 2 categories:

1. Internaltime

2. Externaltime

Eliminate the external downtime

Reduce the remaining internal setup time withbetter methods; then, practice on them

Eliminate as many adjustments as possibleMore



6/27

USING TESTS OF STATISTICAL SIGNIFICANCE

It is not always possible to draw valid conclusions about a

population just by looking at sample data. We need to

make sure any conclusions we draw about the underlyingpopulation are truly supported by the data.

You may ask yourself, for example,

Is our root-cause hypothesis proven based on the data we have collected?

Are these two groups of sample data from the same or different underlying

populations? = ?

Has the proportion of defectives truly changed from what it used to be? p1 p0?

Do these two processes have the same amount of variation? = ?

Is there any significant difference in the quality characteristics between these

three vendors? p1 = p2= p3?

1

2

2

1 2

2



7/27

TWO - TAIL TESTS

Use a two- tail test if you are testing whether a sample value(statistic) is different (whether bigger or smaller) than anestablished (generally a population) value

a The risk is divided in twoand placed on both ends of adistribution of test statistics:

in bothits previous value or is significantly different than another

of the existing or reference distribution0).the right and left tails (the Rejection Regions, for rejecting H

populations value, the critical region will be divided

To test whether a statistic has changed significantly from

curve.

0

a

H : value 1 = value 2

H : value 1value 2

2

The Acceptance Region,

for accepting H 0

2



8/27

DECISION TREE ON 2 MEANS TESTS

with known variance

or with n > 30

Compare sample

mean to a specified

value

with unknown &

assumed equal

variances, and

n 30

Compare means

from 2 samples

Compare sample

mean to a specifiedvalue

Compare means

of paired sets ofdata (e.g., on mult.

samples from 2

sources)

with unknown &

not assumed equal

variances, & n 30

Compare means

from 2 samples

The Non-Parametric alternatives to the 2 Means tests are the Sign Test on 2 Medians,

the Rank Sum Test (Wilcoxon) on 2 Means, and the Mann-Whitney U Test. p. 68

p. 42

p. 47 p. 51

p. 54

p. 58

Back to Means Test Exercise



9/27

= 9.37

The AcceptanceRegion, for

accepting H0

+ 1.645x

= 10.11

The Rejection Region,for rejecting H0 (5%)

+2.71x

=10.59

Location of thetest statistic

.

Z TEST ON TWO MEANS (Contd.)

The actual p value, i.e., the probability that if H0 is true, we will

observe the statistic x deviating by chance from the parameter being

tested ( = 9.37) by a greater degree than is observed, is .0034.

This is the area under the curve to the right of Z = 2.71.to Z Table



10/27

t TEST ON TWO MEANS

(With Unknown but Assumed Equal Variances, & n 30)

Purpose:

To compare the mean from a sample to the known orspecified mean of a population to see if the difference

between them should be considered significant (at a

selected a level of risk)

Assumptions:

1. Both distributions are normal, and the samples areindependent and randomly taken

2. 1 = 2 (unknown but assumed equal; a test on thetwo variances is often applied first in order to assess

this assumption).

example next



11/27

MINITAB EXERCISE APPLYING TESTS ON 2 MEANS (Contd.)

2. File > Open Worksheet \Data\exh_stat.mtw (no 1-sample tests in Excel)

Test to see if the likely population mean material thickness (Values inCol. 1) is less than the historical average of 5.0 mils. Use a 90%

confidence level. is not known. Also, create a indiv. value plot of the

data as part of the test.

What is the attained significance level (p-value) ?

Does this suggest that the new population mean

is less than 5.0?

What does the plot show?

.017

Yes

The C.I. does not contain H0 (5.0).

Minitab

to next t test



12/27

ANOVA TEST ON 3 MEANS

Purpose:

To compare the means from 3 or more samples to see if the

difference between them should be considered significant

(at a selected a level of risk)

Assumptions:

We use an F test to see whether the 3 or more sample means

underlying populations are different. This test requires that we

assume homogeneous variances and that the measurements

were taken randomly from normally distributed populations.

If the sample sizes are equal, slight departures from the

assumption of equal variances are acceptable.

Example:

Three molding machines are being studied to see if their average molded

part dimensions are the same, with the following samples:

Machine A: .450, .458, .449, .451, .455; B: .456, .461, .458, .451, .463C: .464, .467, .463, .459, .462 (see next slide for Minitab instructions)



13/27

In Minitab:

Enter the 15 values in a column; Enter the machine ID letters

(A, B, or C) in another column to match up with the response data

Stats\ANOVA\One-way

In the 1-way Analysis of Variance dialog box:

Response: select the column that contains the 15 test data.

Factor: select the column that contains the machine ID letters.Graphs: select boxplot of data; click on OK

What does the output tell you?

Save the worksheet for use later with ANOM & multiple variances test

The boxplot shows machines A and C have different means. The small p value

in the ANOVA shows there is a statistically significant difference between at least2 of the machines. The C.I.s for machines A and C do not overlap.



14/27

DETERMINING SAMPLE SIZE

When estimating the population mean forvariabledata from a

normaldistribution,

n =

where = the table Z value for the desired

level of confidence

= the standard deviation of the normal

population from which we are to

select our sample, and

e = the specified amount of error between

and ; the allowable differencebetween them

a

2

2

Z

e

a2

Z

x

We will be 1 - % confident that our estimate of will

be off by no more than e; there is a % chance that

we think is not between e and it actually is

a

a

x



15/27

DETERMINING SAMPLE SIZE (Contd.)

Example

If an initial sample of 32 product samples yielded a mean ( ) of 24.3 mm

and a standard deviation (S) of 1.0 mm, what total sample size do we

need to be 95% confident that our estimate of will be off by .1 mm orless (i.e., it is within 24.2 and 24.4)? As long as the initial sample size is

30, we can use the samples S as an estimate of. The Z value for a .95

area under the curve (two-tail) is 1.96.

x

Note: The beta risk defaults to 50% when it is not included

in the sample size calculation.

21.96x1.0

n 384.2 385.1

to Z Table



16/27

Column A B CCondition 1 2 3 Data

1 1 1 1

2 1 2 2

3 2 1 2

4 2 2 1

L4 (23)

This design can be used to study 3 factors at 2 levels

each but not all combinations are tested . . .

Qi2, 2002

Experimental

Runs

Factors

Responses

Levels



17/27

CALCULATING MAIN EFFECTS (Contd.)

To begin the analysis of the data, you need to isolate (or separate) the data

that corresponds to each level of each factor:

Column A B C Response

Condition 1 2 3 Data (Yi)

1 1 1 1 3

2 1 2 2 5

3 2 1 2 6

4 2 2 1 4

L4 (23)

The average of A at level 1 ( ) is the average of the data from

experimental run conditions 1 and 2 (Why?):1

A

1A =

Cond.1 Cond.2

n

= = 4.0

3 5

2



18/27

A1 A2 B1 B2 C1 C2 X

Y

6

4

2

PLOTTING OF MAIN EFFECT RESPONSE GRAPHS

Now, in order to visualize our main effect calculations, we plot

them on response graphs:

Plot the response graphs for factors B and C above, too.

If we wanted a smaller-is-better response, which level is

better for each factor? ans.

Which factor is the most significant? The least?

Why?

A1 B1 or B2 C1

C B

Size of the delta between the levels (slope)

4.0

5.0

Note how we plot all 3

responses on one graph

to make the interpretation

more convenient



19/27

low oven temp (B1)

X

slow conveyor spd fast conveyor spd

A1 A2

30

20

10

40

Y

Cure

Quality

An interaction exists when a change in the level of one factor

effects how another factor influences the measured output.

For example, if Conveyor Speed is slower (A1), the effect of Oven Temperature, when

changed from high (B2) to low (B1), has a positive effect on the quality of cure; but when

Conveyor Speed is faster (A2), the effect of Oven Temperature changing from high to low

is negative on the quality of cure.

WHAT IS AN INTERACTION?

Qi2, 2002

+ -

high oven temp (B2)



20/27

1 3 2 5 4 7 6

2 1 6 7 4 5

3 7 6 5 4

4 1 2 3

1 2 3 4 5 6 7

6 1

7

35 2

A

BA X B

Triangular Table for the L4 and L8 2-level OAs:

USE OF TRIANGULAR TABLES

FOR DESIGNING INTERACTIONS

How would you assign C X D in addition to A X B?

Qi2, 2002

Cols

If we want to study or isolate an

interaction, we need to assign it to

a column in an array (instead of

assigning a main effect). Thistable was developed by Dr.

Taguchi to help assign interactions

to the correct columns in an L4

and in an L8. This table shows the

interacting column relationships.

The bold numbers along the top

and along the diagonal are columnnumbers to which the main effects

(e.g., A and B) would be assigned.

The numbers in the inside of the

table are the columns to which the

interaction (AXB, or AB) would be

assigned.

Once you assign the main effects to

their columns (5 and 6 for example),

then their interaction can only go in

one place (column 3), according to

where they intersect in this table.

Note: you cannot find another set of 3 interacting columns that do not use

columns 3, 5, or 6. Try it.

back to interaction effect proof 22Brought To you by sixsigmatutorial.com


21/27

Column B C C A D E F

Condition 1 2 3 4 5 6 7 Data

1 1 1 1 1 1 1 1

2 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 26 2 1 2 2 1 2 1

7 2 2 1 1 2 2 1

8 2 2 1 2 1 1 2

B

X

ANALYSIS OF AN INTERACTION EXAMPLE (Contd.)

Next, we average the response data for each of the four factor-level

combinations of the interaction.

38

42

23

17

21

24

15

19

B1C2: B2C1: B2C2:(23+17)/2 =20.0 (21+24)/2=22.5 (15 + 19)/2 = 17.0

The level average ofB1C1 is (38 + 42) 2 = 40.0.

What are the level averages (& formulas) forB1C2, B2C1, and B2C2?

B1C1 bar is calculated

by averaging together

the data for conditions 1

and 2, where, and only

where, both factors B

and C were tested at

level 1.

Next we will graph these 4 plot points for the interaction.



22/27

DESIGN LAYOUT L8 (27) X F2

Column A B C D E F G Response Data

Condition 1 2 3 4 5 6 7 R/M Lot 1 R/M Lot 2

1 1 1 1 1 1 1 1 Y1, Y2, Y3 Y4, Y5, Y62 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 2

6 2 1 2 2 1 2 1

7 2 2 1 1 2 2 1

8 2 2 1 2 1 1 2 Y43, Y44, Y45 Y46, Y47, Y48

Is this is a 1-, 2-, or 3-way design?

Why?

How might it be used?

2-way

2 groups 1 inside OA and 1 outside factor

parameter design



23/27

Column

Condition 1 2 3 4 5 6 7

1 1 1 1 1 1 1 1

2 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 2

6 2 1 2 2 1 2 1

7 2 2 1 1 2 2 18 2 2 1 2 1 1 2

RANDOMIZATION

It is generally recommended to run your

conditions in random order. The reasonwhy will now be demonstrated. Lets say

you decide to run in standard order, which

is 1, 2, 3, 4, and so on. And lets further

say that you can only run the first 4

conditions in the cool, dry morning and the

last 4 in the hot, humid afternoon. Next,

you collect data and analyze it, to find outthat the first column was the most

significant. Can you be sure it was the

factor assigned to that column or the

ambient conditions that caused the

changes in the measured response?

Running in a random order would minimize(but not eliminate) that risk of an unwanted

factor coinciding with the levels of a

column. Hopefully, the noise would be

scattered among the levels of the factors.



24/27

GRAND, EXPECTED AND LEVEL AVERAGES

32.0

29.0

28.5

25.5

27.5

30.0

B2

B2

A1 C1A1

C1

A1 A2 B1 B2 C1 C2

= 33.5

T

n= 28.75

1. Graphical Solution:

2. Calculated Solution:

(e.g., AOPT. = A1 - )T

n

= 28.75 + .25 + 3.25 + 1.25 = 33.5

= + (AOPT. - ) + (BOPT. - ) + (COPT. - ) + T

n

T

n

T

n

T

n



25/27

A 2 2.00 1.00

B 2 .67 .33

C 2 50.67 25.33

D 2 8.67 4.33

(e) 0

e2 0

Total 8 62.00

ANOVA BEFORE POOLING

Sum of Mean F Percent

Factor df Squares Squares Ratio Contrib.

S V F

4 2.67

At this point we have no way to estimate the denominator of the F ratio, the within

variation. For this we will have to pool one or more of the least significant columns

into error to come up with an error estimate. Note that experimental error refers tounexplained variation, not any kind of a mistake. For this analysis, lets pool factors

A and B into pooled error, (e). We will pool both the degrees of freedom and the

sum of squares for the insignificant column(s).



26/27

USING STATIC SIGNAL-TO-NOISE RATIOS TO MEASURE VARIATION

Among many available Static S/N ratios are three of the most common:

1. Nominal is Best (Case I), the most common S/N ratio

N = +10 log10 (MSD)N

where (MSD)N =m e

e

(S V )1

n V



27/27

INTRODUCTION TO DESIGN OF EXPERIMENTS

Lesson 6 Intermediate Analysis of Results II

Objectives

Attendees will be able to

- Calculate ANOVA values for a simple design

- Design and analyze an experiment using parameter design

- Calculate static S/N ratios for a parameter design

- Identify and use tuning factors in a 2-step analysis

- Use response tables


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