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Page 1: Data and Graphical Analysismerck.qualitycampus.com/materials/IM_09442536512604_139.pdfParticipant Workbook My Notes © 2014 The Quality Group All Rights Reserved ver. 5.1 DATA AND

Participant Workbook

My Notes

© 2014 The Quality Group All Rights Reserved ver. 5.1 DATA AND GRAPHICAL ANALYSIS - 1© The Quality Group -Revised 8/20/09

© The Quality Group. All Rights Reserved – Revised 10/6/2014

Participant Workbook

Data and Graphical Analysis

Page 2: Data and Graphical Analysismerck.qualitycampus.com/materials/IM_09442536512604_139.pdfParticipant Workbook My Notes © 2014 The Quality Group All Rights Reserved ver. 5.1 DATA AND

Participant Workbook

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© 2014 The Quality Group All Rights Reserved ver. 5.1 DATA AND GRAPHICAL ANALYSIS - 2

© 1992, 1995, 2008, 2012, 2014 by The Quality Group. All rights reserved.Ver. 5.1 October, 2014

Terms of UseThis guide can only be used by those with a paid license to the corresponding course in the e-Learning curriculum produced and distributed by The Quality Group. No part of this Student Guide may be altered, reproduced, stored, or transmitted in any form by any means without the prior written permission of The Quality Group.

TrademarksAll terms mentioned in this guide that are known to be trademarks or service marks have been appropriately capitalized.

CommentsPlease address any questions or comments to your distributor or to The Quality Group at [email protected].

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Time: 3 minutes

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.

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Time: 2 minutes

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Time: 3 minutes

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Time: 3 minutes

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X

Time: 5 minutes

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+1-2

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Control charts will be covered in detail in a future class. They are a very important tool for assessing stability of a process over time and have application in all phases of a project.

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File:  CASESTUDY_SHIPPING1

One of several issues identified by the project team is customer complaints due to shipping.  The CTQ is on‐time shipment and the KPOV being measured is the number of days late.  

The customers have requested that all shipments be on time and nothing be shipped earlier that 5 days.

Upper Spec:_________________    Lower Spec:___________________

Using the above file, calculate the days late from the SHIP DATE – DUE DATE using the Minitab calculator function    Calc  >> Calculator

Using the tools you learned in this module, analyze the data to determine if the variation in Cycle Time is due to shape, center, spread, stability, or some combination.Be prepared to share your results with the class. 

TOOL USED OBSERVATIONS, CONCLUSIONS

Exercise – Putting It All Together

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The data represents shipping performance compared to due date..  Ship Delta is the difference between the ship date and due date measure in number of days.  A negative value represents an early shipment, positive is late.

File:  SHIPPING PERFORMANCE .xlsStat  >> Basic Statistics  >>  Display Descriptive Statistics

Stat  >> Basic Statistics  >>  Graphical Summary

Exercise 1 – Descriptive Statistics

Minitab 16

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The data represents shipping performance compared to due date..  Ship Delta is the difference between the ship date and due date measure in number of days.  A negative value represents an early shipment, positive is late.

File:  SHIPPING PERFORMANCE .xlsGraphical Tools >> Histogram and Descriptive Statistics

Exercise 1 – Descriptive Statistics

SigmaXL

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The data represents shipping performance compared to due date..  Ship Delta is the difference between the ship date and due date measure in number of days.  A negative value represents an early shipment, positive is late.

File:  SHIPPING PERFORMANCE .xlsAnalyze >> Distribution

Exercise 1 – Descriptive Statistics

JMP 10

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Above upper spec = 1 ‐

Calc  >>  Probability Distributions  >> Normal

Above upper spec = 1 - .8413 = .1587 or 15.87%

Below upper spec

Below Lower spec = .0228 or 2.28%

Amount out of spec = .1587 + .0228 = .1815 or 18.15%

Exercise 2 – Normal Probability

Minitab 16

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Above upper spec = 1 ‐

Templates and Calculators  >>  Probability Distribution Calculators  >> Normal

Exercise 2 – Normal Probability

SigmaXL

a) Percent of parts below the LSL - .0227 = 2.27%b) Percent of parts above the USL - .1587 = 15.87%c) Percent of parts out-of-spec - .1814 = 18.14%

Enter data

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Above upper spec = 1 ‐

Launch JMP and create a new data table.Compute and display the normal probability.

1. Add one row to the data table2. Double-click the column header, "Column 1."3. In the dialog box that follows, select "New

Property --> Formula." Screenshot:4. Click "Edit Formula."5. In the dialog box that follows, double-click on

the words, "no formula." Enter "Normal Distribution(1.0)" Note: JMP requires that the user enter a z-value rather than x, mean, and standard deviation.

6. Note: When you finish editing the function, JMP will display it in a graphically-editable form. Functions can also be entered graphically using the "Functions" list and the arithmetic buttons. This approach is more intuitive, though more time-consuming. The "Normal Distribution" function is found under "Probability" when the functions are shown grouped.

7. Click "OK" to close the function dialog box. 8. Click "OK" in the column properties dialog box. 9. Repeat Step 5 and enter (-2.0)

JMP computes the probability and displays it in the first

row of Column 1.Above upper spec = 1 - .8413 = .1587 or 15.87%

Below Lower spec = .0228 or 2.28%

Amount out of spec = .1587 + .0228 = .1815 or 18.15%

Exercise 2 – Normal Probability

JMP 10

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File: Pareto.mtwStat >> Quality Tools >> Pareto Chart

A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.

Exercise 3 – Pareto Diagram

Minitab 17

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A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.

File: Pareto.xlsGraphical Tools >> Basic Pareto Chart

Exercise 3 – Pareto Diagram

SigmaXL

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File: Pareto.mtwAnalyze >> Quality and Process >> Pareto Plot

A project is looking at the number of discrepancies in company travel expense reports with the intent of reducing the cycle time to process the expenses.

Exercise 3 – Pareto Diagram

JMP 10

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCE .MTWGraph  >>  Time Series Plot  >>  Simple

Exercise 4 – Time Series Plot

Minitab 17

Recall the histogram from an earlier exercise which shows a distribution which appears to be normal.

What additional information do you now have about this process?

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCE .MTWGraphical Tool >> Run Chart

Exercise 4 – Time Series Plot

SigmaXL

Recall the histogram from an earlier exercise which shows a distribution which appears to be normal.

What additional information do you now have about this process?

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCE .MTWAnalyze >> Quality and Process >>  Control Chart >> Run Chart

Exercise 4 – Time Series Plot

JMP 10

Recall the histogram from an earlier exercise which shows a distribution which appears to be normal.

What additional information do you now have about this process?

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCE .MTWStat >> Control Charts >> Charts for Individuals >> ImR

Exercise 5 – Control Chart – Individual and Moving Range

Minitab 17

Control charts will be covered in detail in a later module

Tests for Special causesCTL‐E to return to dialog boxImR Options Tests

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCEControl Charts >> Individuals & Moving Range

Exercise 5 – Control Chart – Individual and Moving Range

SigmaXL

Control charts will be covered in detail in a later module

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Construct a Time Series Plot (Run Chart) for shipping performance

File:  SHIPPING PERFORMANCE .MTWAnalyze >> Quality and Process >>  Control Chart >> IR

Exercise 5 – Control Chart – Individual and Moving Range

JMP 10

Control charts will be covered in detail in a later module

Tests for Special causes

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Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.

This is typical of data collected in Excel and it will need to be stacked by rows.Create columns Age (yrs) and Car Price

Data >> Stack >> Stack Rows

Graph >> Scatter Plot, Simple

File: carprices.xls

Exercise 6 – Scatter Diagrams

Minitab 17

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Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.

Data was collected for 4 cars at each age. This is typical of data collected in Excel and it will need to be stacked by rows.

Data Manipulation >> Stack Subgroups Across Rows

File: carprices.xls

Exercise 6 – Scatter Diagrams

SigmaXL

Graphical Tools >> Scatter Plots

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Selling price for used cars of a particular model and option package were captured. Construct a scatter plot to see the relationship between the 2 variables.

This is typical of data collected in Excel and it will need to be transposed by rows.Create columns Age (yrs) and Car PriceTables >> Transpose

File: carprices.xls

Exercise 6 – Scatter Diagrams

JMP 10

Change to Car Price

Analyze >> Fit Y by X

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Exercise 7 – Putting It All Together - Solution

The histogram shows a distribution which appears “normal”’ or bell=shaped. However, the P-Value for the normality test (P less than .05 means non-normal) shows a possible bi-modal distribution. Since the sample size is large (>30) there is not much tolerance for minor departures from normality. This passes the “eyeball” test so assume normality and go on, but look for reasons for this distribution

Percent above upper spec = 1 - .802 = .198

Percent below lower spec = .250

Total out of spec = .198 + .250 = .448or 44.8% defective

Minitab 17

NOTE – the non-statistical thinker would look at the data and note that the mean is less than zero, and might “observe” that the average shipping time is actually early – so what’s the problem?

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Exercise 7 – Putting It All Together - Solution

A box plot using ship day of the week as a grouping variable shows a possible source of the clustering. Shipments on Thursday and Friday tend to be later than the rest of the week.

Minitab 17

The run chart shows a cyclic variation. The P-values (less than .05 is significant) show that the data is clustered and shows a lack of randomness. This could account for the abnormality with the distribution.

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Exercise 7 – Putting It All Together - Solution

Looking at the Order Entry data, a Pareto chart shows that “No Problem” has the largest number –nothing interesting here.

A Box Plot using status as the grouping variable shows that pricing is the biggest contributor to the data entry time. Credit and new part also contribute.

Minitab 17

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Exercise 7 – Putting It All Together - Solution

The histogram shows adistribution which appears“normal”’ or bell=shaped.However, the P-Value forthe normality test (P lessthan .05 means non-normal) shows a possiblebi-modal distribution. Sincethe sample size is large(>30) there is not muchtolerance for minordepartures from normality.This passes the “eyeball”test so assume normalityand go on, but look forreasons for this distribution

The histogramshows a distributionwhich appears“normal”’ orbell=shaped.However, the P-Value for thenormality test (P lessthan .05 means non-normal) shows apossible bi-modaldistribution. Sincethe sample size islarge (>30) there isnot much tolerancefor minor departuresfrom normality.

SigmaXL

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Exercise 7 – Putting It All Together - Solution

Fri Mon Thurs Tues Wed‐13

‐11

‐9

‐7

‐5

‐3

‐1

1

3

5

7

*Day

s L

ate

Ship Day

The run chart shows a cyclic variation. The P-values (less than .05 is significant) show that the data is clustered and shows a lack of randomness.

A box plot using ship day of the week as a grouping variable shows a possible source of the clustering. Shipments on Thursday and Friday tend to be later than the rest of the week.

SigmaXL

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Exercise 7 – Putting It All Together - Solution

no problem new part credit pricing

250 25 18 14

Cumulative Sum 0.814332248 0.895765472 0.954397394 1

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0

50

100

150

200

250

300

Count

status

credit new part no problem pricing0

2

4

6

8

10

12

Dat

a E

ntr

y T

ime

(hrs

)

status

Looking at the OrderEntry data, a Paretochart shows that “NoProblem” has thelargest number –nothing interestinghere.

A Box Plot using status asthe grouping variableshows that pricing is thebiggest contributor to thedata entry time. Credit andnew part also contribute.

SigmaXL