The Certified Six Sigma Green Belt Handbook - ASQ Groups

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The Certified Six Sigma Green Belt Handbook - ASQ Groups

Page 14

2nd paragraph.

The Certified Six Sigma Green Belt Handbook

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“form” should be “from”

Page 23

2nd paragraph.

“Marker” should be “maker”

Page 32

1st paragraph.

“son” should be “so”

Page 56

1st paragraph

“Deming’s’” should be “Deming’s”

Page 57

9th line from bottom.

“reference QFD in Chapter 6” should be “reference QFD in Chapters 3 and 9”

Page 74

Third paragraph, 2nd bullet.

"...Team leader still monitors the goals..”

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Page 89 Figure 8.3.

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Multi Voting- Weighted approach

Venue: Caribou meeting room

Date: 3-Feb-07

Subject: How to improve ASQ attendance of ASQ Section Programs?

Scale 1 to 5- Most Important 5, Least important 1

Member 1 Member 2 Member 3 Member 4 Member 5 Member 6 Member 7 Member 8 Member 9 Total

A 30 20 25 35 20 25 25 35 30 245

B 15 25 20 20 25 20 20 20 15 180

C 10 30 30 25 30 30 30 25 10 220

D 20 15 10 15 15 10 10 15 20 130

E 25 10 15 5 10 15 15 5 25 125

There is NO ranking scale applicable to this approach. The column total should add up to 100 for all individual columns and the relative

importance of A to E to be understood by the points allotted by each member (from that member’s point of view). Overall relative importance is

understood from reviewing the “Total” column. Based on consolidated input from all members, in this example, A is most important, followed by

C, B, D and E.

Page 113 Figure 9.15.

Consistency (not error):

Indicated Δ as 'moderate' relationship, and 'o' as 'weak' relationship.... should be reversed.

'O' as 'Moderate' and Δ as 'Weak' relationship

Legends used in text and figure on pp 139, 141 are general usage for these legends.

- Also refer to figure 9.20 QFD.

Pages 114-115

In the example in Figure 9.16, option A and C are the most desirable (Lowest) cost so it is assigned 3.5 both (A and C are tied). The package with

the next lowest cost is option D and is assigned 3. Option B has the highest cost and is assigned 1.

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Options

Options

Options

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Step1: Each option is ranked against the criteria with the desirable numbers being larger.

(For cost, smaller the desireable).

Determine the most suitable software package

Lowest is most desirable.

Criteria

Compatibility Cost Ease of Use Training Time Total

Package A

1.00 0.45 1.20 0.15 2.80

Package B

0.25 1.20 0.80 0.05 2.30

Package C

0.75 0.45 1.60 0.20 3.00

Package D

0.50 0.90 0.40 0.10 1.90

Figure 9.16 Prioritization matrix example.

Step2: Assign ranking, 4 to 1, 4 being most desirable, 1 least desirable.

Apply averages for tie. Package A and C tied for lowest cost.

Determine the most suitable software package

Criteria

Compatibility Cost Ease of Use Training Time Total

Package A

4 3.5 3 3

Package B

1 1 2 1

Package C

3 3.5 4 4

Package D

2 2 1 2

Figure 9.16 Prioritization matrix example.

Step3: Assign Weights based on relative importance.

Multiply each of the option values by criteria weights at the top of the column and calculate row totals.

Determine the most suitable software package

Criteria

Compatibility

(0.25)

Cost (0.3)

Ease of Use

(0.40)

Training Time

(0.05)

Total

Package A

1.00 1.05 1.20 0.15 3.40

Package B

0.25 0.30 0.80 0.05 1.40

Package C

0.75 1.05 1.60 0.20 3.60

Package D

0.50 0.60 0.40 0.10 1.60

Figure 9.16 Prioritization matrix example.

Step4: Option with the highest total is the one most favored by the prioritization matrix (Highlighted)

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Page 115

Package A Row total.

4.15 should be 2.80.

Page 116

Third line from bottom.

In paragraph describing PERT and CPM: the line should read ACEGH in two places, not ACFGH.

Page 136 Paragraph 5, line 3.

Vilfredo Pareto is stated as an 18th century economist. The correct period is 20th century.

Page 147

Line 4, summary of key probability rules.

General addition rule to be read as

General multiplication rule:

P(A &B)= P(A) X P(B|A)

Page 149

Example box.

In second equation, change 0.84 to 0.64. (0.8 x 0.8 = 0.64). Answer: 1.6 - 0.64 = 0.96.

Page 150

Example.

The assembly of an electronic board has three major components. Probability of component A working is 0.2, component B is 0.3, component C is

0.5. What is the probability that either B or C is working?”

This problem was written to given some practical application of probability deviating from traditional Red ball, Black ball and Blue ball from an

urn or throwing a die example. Unfortunately the problem statement is not strongly supporting “mutually exclusive” interpretation.

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Rephrase the Problem Statement as:

The assembly of product requires an electronic board. This electronic board is supplied by 3 different suppliers. Probability of the board from

Supplier A working on the product is 0.2, board from Supplier B is 0.3, board from Supplier C is 0.5. What is the probability that either board from

B or C is working on the product?”

Page 151

A secured software program requires creating 5 character alpha numeric passwords. How many passwords of five symbols (letters and digits) can

be generated using at least two digits each? (Assume that all letters are lowercase.)

Let us assume this is one of the Permutations of 5 letter password. abc12

n=5 and r =2

5P 2 = 20

It was intended to create an interesting permutation problem, but although the problem looked simple at the outset it turned out to be very complex

problem.

Alpha has 26 possibilities in each position (a to z), numeric has 10 possibilities (0-9) with this there can be at 20 permutations as per the solution

below many permutations. This is an interesting problem but certainly not at the green belt level problem. This may be a good problem for a

Statistics Major at graduate level.

For a Green belt level, the problem should be rephrased as follows: A secured software program requires creating 5 character alpha numeric

passwords. Let us assume for simplicity the software engineer is given letters a, b, c and numbers1, 2 to set up a password. How many passwords

of five characters can be generated using the two numbers?

n=5 and r =2

5P 2 = 20

(Google: “Permutation generator”)

listing of permutations by positions of the 2 numbers

1 2

2 1

1 3

3 1

1 4

4 1

1 5

5


5 1

2 3

3 2

2 4

4 2

2 5

5 2

3 4

4 3

3 5

5 3

4 5

5 4

Example: 12abc, 21abc, 1a2bc, …

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Page 152

The Counting rule.

Number of

combinations of

nCr

r objects

n!

r!

n r !



from a collection

of

n objects

Note : Another

symbol

for

number of

combinations is

n


r

The numerator n ! was wrongly printed as r !

Page 154 Link under Table 12.1.

Hyperlink has changed. The link should be removed and replaced with something like: http://en.wikipedia.org/wiki/Level_of_measurement

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Page 172

Correlation and Causation

Change “This is called Causation” to “This is an example of correlation does not imply causation”.

Causation by definition happens as an effect of cause. Causation cannot be proved statistically. Variable A can cause Effect B, B can cause A or A

and B can each other. There is also a possibility that a third Variable C can cause both A and B.

In the example, the data of gold price from 1930 to 2000 increases while the infant mortality rate decreases. The question is if the increase in gold

price causes the infant mortality going down from 1930 to 2000? This scenario is complicated since it is driven by economy growth and maturity in

the medical field. One argument could be that affluent countries may very likely invest in medical research and improve the infant mortality. This

is an example of correlation does not imply causation.

See Examples in http://www.stat.tamu.edu/stat30x/notes/node42.html

Page 176 Table 13.1.

Formula for binomial distribution, the numerator should be n!, not r!

Page 177

Figure 13.1, Binomial distribution.

Ignore the (k,k) in the figure and interpret the vertical axis as probability of occurrence and horizontal axis as number of occurrences.

Page 178

Second paragraph.

Beginning with "We can…", the web address is www.stat.tamu... not www/stat.tamu…

Page 180

Figure 13.3 Poisson distribution.

Ignore the (k,k) in the figure and interpret the vertical axis as probability of occurrence and horizontal axis as number of occurrences.

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Page 181

Top of page.

Equation should read: “P(x=5) = (e^-5 * 5^5)/5! = 0.175, ~ 18%,” instead of “0.0017 (0.18%)”

Page 182 2nd paragraph, line 5.

Should read: “…even if the individual data…”

Page 211 Figure 15.3.

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Test Results for Xbar

Chart of Length:

TEST 1. One point more than

3.00 standard deviations from

center line.

Test Failed at points: 12

TEST 4. 14 points in a row

alternating up and down.

Test Failed at points: 54, 55,

56, 57

TEST 5. 2 out of 3 points more

than 2 standard deviations from

center line (on one side of

CL).Test Failed at points: 24,

45

Test Results for R Chart of

Length

TEST 1. One point more than

3.00 standard deviations from

center line.

Test Failed at points: 8

Above arrow pointed figures: Data points that Fail Test 1 of X bar and R chart identified along side the

Violating point. Currently the identification “1” referring to Test 1 is missing in the print version of the

Hand book.

Test Failed at points:12

Correct “points” to “Point”

Test Failed at points:8

Correct “points” to “Point”

Page 212

1st formula.

0.24 lbs. should be 0.024 lbs.

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The point estimate for is 1.065 lbs. The point estimate for process standard deviation is given by the formula


R

d

2


0.05

2.059

.024lbs

Page 213

Final full paragraph.

0.010 should be 0.1, making the equation read as: C p = 0.1 ÷ 0.144 ≈ 0.69

Page 214 Figure 15.4.

The figure in the lower left corner for Cp = 1.67 is the same as Cp = 1.33. The Cp=1.67 should be narrower than Cp=1.33 but wider than

Cp=2.00.

Page 215 Bullet 6.

Explanations for Abbreviations: CPU and CPL

(CP Upper Limit, CP Lower Limit)

CPL is a capability index defined as the ratio of the interval formed by the process mean and LSL and one-sided spread of the potential process.

CPU is a capability index defined as the ratio of the interval formed by the process mean (m) and USL and one-sided spread of the potential

process.

For two-sided tolerances, Cpk is the lowest of CPU and CPL.

For one-sided tolerance, obviously it is either CPU or CPL – only one of them will be applicable.

(Source: MINITAB Help)

Page 235 Example – data sets, line 2.

Based on data provided in line 1, the data sets should read: (3,2) (3,30) (6,4)

In the 2nd data set, change “30” to “3” as (3,3)

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Page 235

Formula, bottom of page.

In the denominator, replace the “-“ (minus sign) that is in-between the two bracketed segments with an “x” (multiplication sign) […] x […]

Page 237

Example at bottom of page.

Calculating t-statistics, t = 0.972/.01175 = 8.273

Page 240

Example.

Result shows y = 6.785 + 0.52(16) = 15.1, should be y = 6.785 + 0.52(14) = 14.1

The b= Value should be 0.5833 and a = 6.5 (Error in the text as b=0.52 and a= 6.785)

The problem was to predict in another 6 months, which seems to the 8 + 6 = 14th month.

Obtain the least square prediction for the table containing month and number of complaints received in a manufacturing facility.

Month, x i Complaints, y i

2

x i

1

8

1

2

6

4

3

10

9

4

6

16

5

10

25

6

13

36

7

9

49

8

11

64

x i y i

2

y i

8

64

12

36

30

100

24

36

50

100

78

169

63

81

88

121

∑ x i = 36 ∑ y i = 73 ∑ x i 2 = 204 ∑x i y i = 353 ∑y i 2 = 707

x 4.5 y 9.125

11


Complaints

b

n

xy

x

y

2

2

nx


x

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8(353) (36)(73)


0.5833

8(204) (36x36)

a y bx = 9.125 – 0.5833(4.5) = 6.50015

The regression line would be ŷ = 6.50115+0.5833x

Because of the slope of this equation is positive 0.5833, there is evidence that the number of complaints increases over time at an average rate of 1

per month. If some one wants to predict for the complaints in another 6 months at this rate, would be

ŷ = 6.785 + 0.52(16) = 15.1 ≈ 15 complaints There are already 8 data points in Month column. Hence for predicting the complaints in another 6

months, x has to take the value (8+6) i.e. 14. ŷ = 6.50115 + 0.5833 (14) = 14.667 ≈ 15 complaints.

15.0

12.5

Trend Analysis Plot for Complaints

Linear Trend Model

Yt = 6.50 + 0.583*t

Variable

Actual

Fits

Forecasts

10.0

7.5

5.0

1

2

3

4

5

6

7 8

Month

MINITAB analysis graph.

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The Certified Six Sigma Green Belt Handbook

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Page 242

Example.

S = √4.4319 = 2.105

In handbook, answer is 2.165 since the calculation is using the incorrect values from page 240.

Also, Y^ = a^ + b^ x = 6.5015 + 0.583 (8) = 11.1655, not 10.94 as stated in book.

Also, in calculating 95% confidence level, 7.84 ≤ CI ≤ 14.49 instead of 7.53 ≤ CI ≤ 14.35 as stated in book.

Page 242

Example: t-statistic values in “CI” calculation and bottom paragraph.

There is a discrepancy in the t c values indicated. In CI equation, 2.447 is correctly substituted for t c ; in bottom explanation it is incorrectly noted as

3.707.

For 95% confidence interval, α=0.05 α/2=0.025.

For n-2 deg freedom (=6), the table reads 2.447. The conclusion that t < t c is still accurate.

The bottom paragraph should state:

“With a two-tail test and α=0.10 0.05, t c =3.707 2.447 as per the t-table.”

Page 243 Statement just above Figure 16.14.

States: “…Where se and sxx are as defined previously…”

Se (Standard Error of Mean) and

Sxx is the Standard deviation of variable X.

Formula for Se is in Page 240 (right bottom)

Formula for Sxx is in Page 171 example.

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The Certified Six Sigma Green Belt Handbook

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Page 245 Example box, bullet 5.

Wrong data is used in the formula, with a wrong result as the consequence. Correct formula t= .3/(.71/(sqrt(50))) = .3/.1004 = 2.99 with t=

.3/(.5/(sqrt(50))) = .3/.071 = 4.24 – conclusion on Hypothesis is still valid.

Page 249

First formula.

“” is shown twice; one should be “>”

The alternative hypothesis, denoted by H 1 , represents the opposite of the null hypothesis is found and holds true if the null hypothesis is found to be

false. The alternative hypothesis always states the mean of the population is a specific value. The alternative hypothesis would be stated as

H 1 :μ ≠ 6.0, H 1 :μ < 6.0, H 1 :μ > 6.0

Page 249

The error matrix is erroneously reversed in the book.

False

True

Reject H 0 p = 1 – α ,

Correct outcome

p = β, Type II

error

Do nor reject H 0 p = α, Type I

rror

p = 1 – β,

Correct outcome

Note: p = 1 – β is also called power. Higher the power is

better in a hypothesis test.

Correct arrangement:

False

True

Reject H 0 p = 1 – β,

Correct outcome

p = α, Type I

errr

Do not reject H 0 p = β, Type II p = 1 – α ,

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The Certified Six Sigma Green Belt Handbook

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error

Correct outcome

Note: p = 1 – β is also called power. Higher the power is

better in a hypothesis test.

See also a better arrangement of the error matrix:

In Deciding to Reject or Not, We Could Make One of Two

Decision Errors:

Your Decision

H o True

“Don’t Reject

H o”

Correct

Reject

H o

Type I

Error

(-Risk)

H o False

Type II

Error

(-Risk)

Correct

Page 253

Formula for “Continuous data….”

“=” should be “+/-“

Page 253

Both examples.

The inequality symbols are reversed, and the solutions should be “91.25 ≥ µ ≥ 65.25” and “93.75 ≥ µ ≥ 62.77” (similar to the error on Page 255)

Page 254

Example.

Denominators are 36.42 and 13.85, and should be 48.60 and 21.66, respectively, for n=35

The sample variance for a set of 35 samples was found to be 46. Calculate the 90% confidence interval for the variance

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The Certified Six Sigma Green Belt Handbook

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3446 34

46

2

36.42


42.94 2 112.92

3446 34

46

2

48.6


32.18 2 72.21



21.66



13.85

(Please refer to the Chi square table in the appendix. It should be 48.60 and 21.66, respectively, for n=35)

Page 255

Example.

The answer, currently as: 0.689 < p < 0.569

should be reversed, and written as:

0.569 < p < 0.689

Page 258

Example.

Step 4 Based on a significance level of 0.10 (α=10), and a left-tail test (as stated in step 2), the text should read:

The positive critical value is in the 19th row of the t 0.05 t 0.10 column of the t-table. This value is 1.729 1.328. Since this is a left-tail test, the critical

value is -1.729 -1.328. The reject region consists of the area to the right of 1.729 and the area to the left of -1.729 -1.328.”

Read as:

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The Certified Six Sigma Green Belt Handbook

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The positive critical value is in the 19th row of the t 0.10 column of the t-table. This value is 1.328. Since this is a left-tail test, the critical value is -

1.328. The reject region consists of the area to the left of -1.328.”

Page 258

Verification using MINITAB software.

Test of mu = 4.125 vs < 4.125

90% Upper

N Mean StDev SE Mean Bound T P

20 4.12300 0.00800 0.00179 4.12538 -1.12 0.139

A “P” value of > 0.10 indicates that there is not enough statistical evidence to say that mean length has decreased after the new blade is installed.

Page 259

Step in the step by step explanation and in the Example.

Formulas should read as follows:

p'

p0

Z

p (1 p )

0

n

0

Z


0.03 0.02

0.02(0.98)

500

1.597

Page 259

Example.

The first line of the problem statement says “at most two percent of a shipment of parts is defective.” H 0 should be “p ≤ 0.02” and that H 1 should be

“p > 0.02”.

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Page 259

Procedure and example.

Point 5 in procedure: In denominator, the “division sign” under the square root sign should be “multiplied by”

5. Calculate the test statistic using

Z

The above Formula should read as:


Z

p'

np

0

0

np (1 p


0

)

x np

0

0

np (1 p

Where p’ = sample proportion = x/n

p = population proportion

n = number of samples

x = number of items in the sample with the defined attribute

p 0 = the hypothesized proportion

0

)

Example, Step 5:

z

0.03 0.02

0.020.98

500

Above error should be corrected as:

1.597

z

15 10

0.02

0.98500

1.597

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The Certified Six Sigma Green Belt Handbook

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Page 261

Middle of page.

The printed formula is missing square on both sides of the denominator notations.

The rounded off Degrees of Freedom in the bottom example is correct. The calculation comes to 7.627.

Page 261

Bottom example.

The subscript for t be “0.025, 8” and not “0.025, 5” in the last line. The t-value is for 8 degrees of freedom.

Page 263

F test definitions.

Where s 2 1 and s 2 2 = sample variances = sample standard deviation for population 1 & 2.

2 1 and 2 2 = standard deviation for population 1 & 2

Should read as:

Where s 2 1 and s 2 2 = sample variances of the two samples 1 & 2 under comparison.

2 1 and 2 2 are population variances from which the sample may have been drawn.

Page 272

Example box (top).

Wrong heading in column 5 of the table. Replace column title (O-E) 2 with (O-E)

Page 288 Table 18.4.

In the Example, the last column of A x B x C should be all '+' (plus), not '-' for Runs 1-3.

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The Certified Six Sigma Green Belt Handbook

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Page 292 Table 18.7.

A 2 2 full-factorial completely randomized experiment with results.

Average values from this

column should have been

used to compute the main

and interaction effects.

Erroneously this column

was used to compute main

and interaction effects.

The main effect of factor B

(32.9 +37) ÷ 2 – (28.2 +24.8) ÷ 2 = 8.45

should be corrected to (33.0+37.3) ÷ 2 – (28.4+24.7) ÷ 2 = 8.6

The interaction effect of A x B

(28.4 +37.3) ÷ 2 – (33.0+24.7) ÷ 2 = 4.0

should be corrected to (28.4+24.7) ÷ 2 – (33.0+37.3) ÷ 2 = - 8.6

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Page 312

Final paragraph.

Formulas for control limits are given in Appendix C (not Appendix G)

Page 317

Chart.

The Moving R data above the graphs show the difference between the 7th and 8th samples as being 8 (hence, R = 8), but the range should be 6

(290-284).

Page 321

Example.

The equation for p-bar should end “ = 177/1496 = .118”. That is, it should be an equals sign prior to 177/1496, not a minus sign.

Page 325

1st group of formulas.

Four of the square root signs should be division.

Last group of formulas: Show value of 4.1 that should be 16.7. 4.1 is the square root of 16.7

Page 325

Start of the page.

The calculations for u Control chart should read as:

u = 66 1496 0.044

n = 1496 12 124.7

3 u

UCL u = 0.044 + 3 (0.210 11.167) 0.100

n

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Sample Count Per Unit

The Certified Six Sigma Green Belt Handbook

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3 u

LCL u = 0.044 - 3 (0.210 11.167) - 0.012 0

n

If LCL < 0, LCL = 0.

MINITAB Analysis output equivalent to Figure 19.14

Float Plate

0.10

UCL=0.0978

0.08

0.06

0.04

_

U=0.0441

Process In control

No Violations

0.02

0.00

LCL=0

1

2

3

4

5

6 7

Sample

8

9

10

11

12

Tests performed with unequal sample sizes

Page 325

Bottom of the page.

The calculations for c Control chart should read as:

c – = 217 ÷ 13 ≈ 16.7.

UCL= 16.7 + (3 × 4.1) = 29

LCL= 16.7 – (3 × 4.1) = 4.4

MINITAB Analysis output equivalent to Figure 19.15

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Sample Count

The Certified Six Sigma Green Belt Handbook

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30

Imprint

UCL=28.95

25

20

15

_

C=16.69

Process In control

No Violations

10

5

LCL=4.44

1

2

3

4

5

6

7

Sample

8

9

10

11

12

13

Page 350

Report.

The bottom left hand box for total variation should say “TV” and not “PV”.

Page 352

Fifth paragraph, second line.

The sentence should end “not responded to” instead of “not responded too”.

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Pages 358-359 Appendix A.

The ASQ Code of Ethics should appear as follows:

ASQ Code of Ethics

Fundamental Principles

ASQ requires its members and certification holders to conduct themselves ethically by:

I. Being honest and impartial in serving the public, their employers, customers, and clients.

II.

III.

Striving to increase the competence and prestige of the quality profession, and

Using their knowledge and skill for the enhancement of human welfare.

Members and certification holders are required to observe the tenets set forth below:

Relations With the Public

Article 1 – Hold paramount the safety, health, and welfare of the public in the performance of their professional duties.

Relations With Employers and Clients

Article 2 – Perform services only in their areas of competence.

Article 3 – Continue their professional development throughout their careers and provide opportunities for the professional and ethical development

of others.

Article 4 – Act in a professional manner in dealings with ASQ staff and each employer, customer or client.

Article 5 – Act as faithful agents or trustees and avoid conflict of interest and the appearance of conflicts of interest.

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Relations With Peers

The Certified Six Sigma Green Belt Handbook

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Article 6 – Build their professional reputation on the merit of their services and not compete unfairly with others.

Article 7 – Assure that credit for the work of others is given to those to whom it is due.

Ref: http://www.asq.org/about-asq/who-we-are/ethics.html

Page 368 Appendix C.

Attribute Chart formulas including 'n' should read nbar for p chart and u chart, not just 'n'.

Page 368 Appendix C.

Placement of UCL/LCL formulas for c and u charts: locations should be swapped. c-charts are for constant sample size; u-charts are for variable.

Also flip-flopped on the Appendix C document on the CD

CD-ROM

CSSGB Question Base One.pdf

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41. On the basis of the control chart above, which of the

following statements is true?

(A) Components 1, 2, 3, and 5 should be reinspected

because they are below the mean.

(B) Only component 4 should be investigated because it

is closest to the upper control limit.

(C) Components 4, 6, 7, 8 and 9 should be investigated

because they are above the mean.

(D) No action is required; all data points are within

acceptable statistical variance.

Ref: CSQE pdf Study Guide Q31

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The Certified Six Sigma Green Belt Handbook

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The answer is D, not C. It is a typo in the answer key.

As for the SPC technicality, the only rule that comes close to the figure is 4 out of 5 points > 1 standard deviation from center line (same side).

We don’t have evidence to prove this in the given diagram. So the data in the diagram is basically a random process variation (i.e., no action is

required; all data points are within acceptable statistical variance).

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