Copy Data to MinitabStatistical Tests

Open Excel File > Highlight All Data & Column Headings > Copy

Launch Minitab: Start > Programs > Minitab > Minitab

Paste Data into Minitab

Put the Data into Minitab for AnalysisPut the Data into Minitab for Analysis

Remember to paste in cell above row labeled 1 – This brings column headings and data in without confusing Minitab

Do you have Normal Data?

Ho: Data distribution = Normal Distribution

Ha: Data distribution Normal Distribution

Continuous

Non Normal Normal

If the p-value >.05 …… Then Ho is true and there is no difference in the groups you had me test(Accept Ho) The Data is normal

If the p-value < .05 …... Then Ho is false and there is a statically significant difference (Reject Ho ) The Data is not Normal

Normality Test

91 8971 9279 8386 7994 7288 8580 7495 7389 8777 8195 8587 96101 8582 9083 74

Staffing Cycle time

MINITAB: Stat> Basic Statistics> Display Descriptive statistics>

Normality Test>

If the p-value >.05 …… Then Ho is true and there is no difference in the groups you had me test(Accept Ho) The Data is normal

If p-value < .05 …... Then Ho is false and there is a statically significant difference (Reject Ho ) The Data is not Normal

If the p-value >.05 …… Then Ho is true and there is no difference in the groups you had me test(Accept Ho) The Data is normal

If p-value < .05 …... Then Ho is false and there is a statically significant difference (Reject Ho ) The Data is not Normal

P-value = 0.000 < 0.05 Therefore The data is NOT NormalP-value = 0.000 < 0.05 Therefore The data is NOT Normal

Cycle Time

Perc

ent

120100806040200

99

95

90

80

70

605040

30

20

10

5

1

Mean

<0.005

79.39StDev 15.59N 34AD 5.353P-Value

Probability Plot of Cycle TimeNormal

Normality Test

Capability

10080604020

USLProcess Data

Sample N 34Shape 8.67521Scale 83.60229

LSL *

Target *USL 75.00000Sample Mean 79.39412

Overall CapabilityZ.Bench -0.46

Z.LSL *Z.USL -0.65Ppk -0.22

Observed PerformancePPM < LSL *

PPM > USL 882353PPM Total 882353

Exp. Overall PerformancePPM < LSL *

PPM > USL 677155PPM Total 677155

Process Capability of Cycle TimeCalculations Based on Weibull Distribution Model

MINITAB: Stat> Quality tools Capability analysis Nonnormal

Engineering Finance IT Sourcing

0

10

20

30

40

50

60

70

80

90

Department

Cyc

le T

ime

Kathy Sam Tom

0

10

20

30

40

50

60

70

80

90

Sourcing agent

Cyc

le T

ime

Visualize The Data – Box PlotsStatistical Tests

Minitab: Graph > Box Plots > Y=Cycle Time > X1 = DepartmentX2 = BandX3 = Agent

Which Xs Look they effect the Cycle Time (Mean, Median, Variance)? If they look like they have an impact they probably do and we can prove it with Statistics

Which Xs Look they effect the Cycle Time (Mean, Median, Variance)? If they look like they have an impact they probably do and we can prove it with Statistics

P SP

0

10

20

30

40

50

60

70

80

90

Band

Cyc

le T

ime

We have these tests to choose from because the Data is NOT NORMAL

Statistical Tests

Variances

HOVLevine’s

Correlation

Sign test

Wilcoxon

Mann-Whitney

Mood’s

Friedman

Continuous

Non Normal

Medians

Non Normal DataVariance Tests

Homogeneity of Variance - Levine’s - Compares two or more sample variances.

Medians Tests

Sign Test - Tests if a sample median is equal to a known or target value.

Wilcoxon Test - Tests if a sample median is equal to a known or hypothesized value.

Mann-Whitney Test - Test if two sample medians are equal.

Mood’s Median Test - Test if two or more sample medians are equal.

Friedman Test - Tests if medians from samples classified by two categories are equal.

Correlation - Tests linear relationship between two variables.

Staffing cycle time is exceeding 75 days 88% of the time

Measurement Variation

Process (or product)Variation

Measurement System Valid

Timeto

Time

Departmentto

Department

Job type to

Job type

Bandto

Band

AgentTo

AgentOther?

Let’s see what we can eliminate from our suspect list

Let’s start with Department to department

•Check the variances

•Check the means

Let’s start with Department to department

•Check the variances

•Check the means

Statistical Tests

HOV

Posting # Department Band Date posted Date closed

Sourcing agent

Cycle Time

1 Finance SP 1/1/1999 3/25/1999 Kathy 83.72 Engineering P 1/7/1999 4/6/1999 Tom 89.73 IT E 1/21/1999 4/19/1999 Kathy 88.54 Finance P 2/2/1999 4/27/1999 Kathy 84.55 Engineering SP 2/2/1999 4/24/1999 Tom 81.46 Finance SP 2/2/1999 4/27/1999 Kathy 84.77 Engineering SP 2/3/1999 4/28/1999 Tom 84.58 Finance P 2/26/1999 5/23/1999 Sam 86.19 Sourcing P 2/28/1999 5/18/1999 Sam 79.4

10 Sourcing P 3/1/1999 5/20/1999 Sam 80.111 Sourcing LP 3/1/1999 5/23/1999 Sam 83.412 Sourcing LP 3/4/1999 5/25/1999 Sam 82.213 Finance SP 3/5/1999 5/27/1999 Sam 83.214 IT SP 3/5/1999 6/3/1999 Kathy 90

Staffing Cycle TimeStaffing cycle time is exceeding 75

days 88% of the time

Process (or product)Variation

Departmentto

Department

Department to Department - Variances

Appropriate Test: HOV-Levine’s

Ho: Variances are the same department to department

Ha: Variances are different

Response = Cycle TimeFactor = Department

Statistical Tests

Variances

HOVLevine’s

Non Normal

MINITAB: Stat> ANOVA Test of equal variances

Remember: Min I.Tab can only give us a P-value, we have to determine what it means

Statistical Tests

HOV

If the p-value >.05 …… Then Ho is true and there is no difference in variances (Accept Ho)

If the p-value < .05 …... Then Ho is false and there is a statically significant difference in variances (Reject Ho )

P-value = 0.000

95% Confident that the variances are different – Validates what we saw on the Box Plot (more scatter in IT Data)

Depart

ment

95% Bonferroni Confidence Intervals for StDevs

Sourcing

IT

Finance

Engineering

9080706050403020100

Bartlett's Test

0.000

Test Statistic 63.50P-Value 0.000

Levene's Test

Test Statistic 13.61P-Value

Test for Equal Variances for Cycle Time

Department to Department - Medians

Appropriate Test: Mood’s Median

Ho: Cycle Time Medians are the same department to department

Ha: Cycle Time Medians are different

Response = Cycle TimeFactor = Department

Staffing cycle time is exceeding 75 days 88%

of the time

Process (or product)Variation

Departmentto

Department

Statistical Tests

Moods Median

Correlation

Sign test

Wilcoxon

Mann-Whitney

Mood’s

Friedman

Non Normal

Medians

MINITAB: Stat> Nonparametrics Mood’s Median test

If the p-value >.05 …… Then Ho is true and there is no difference in Means (Accept Ho)

If the p-value < .05 …... Then Ho is false and there is a statically significant difference in Means (Reject Ho )

P-value = 0.06

The Department Medians are NOT different at 95% Confidence Level – but close

Statistical Tests

Moods Median

Mood Median Test

Mood median test for Cycle Ti

Chi-Square = 7.42 DF = 3 P = 0.060

Individual 95.0% CIs

Departme N<= N> Median Q3-Q1 -------+---------+---------+---------

Engineer 3 5 84.6 5.3 (-+-)

Finance 3 8 84.5 3.1 (+-)

IT 4 3 66.7 43.5 (-------------------+--------------)

Sourcing 7 1 81.4 3.2 (+-)

-------+---------+---------+---------

45 60 75

Overall median = 83.6

Means

t - test1 - sample2 - sample

ANOVAOne wayTwo way

NormalDepartment to Department - Means

Appropriate Test: One-Way ANOVA, (Tests if more than two sample means are equal)

Ho: Cycle Time Means are the same department to department

Ha: Cycle Time Means are different

Staffing cycle time is exceeding 75 days 80%

of the time

Process (or product)Variation

Departmentto

Department

Statistical Tests

ANOVAIf our staffing cycle time data had been normal...

Samples MUST BE NORMAL

Staffing cycle time is exceeding 75 days 88% of the time

Measurement Variation

Process (or product)Variation

Measurement System Valid

AgentTo

Agent

Departmentto

Department

Bandto

Band

Let’s see what we can eliminate from our suspect list

Moods p-value = 0.060HOV p-value = 0.000

Continue to work through until you find the Red X or X’s

Continue to work through until you find the Red X or X’s

Statistical Tests

Staffing Cycle Time

Staffing cycle time is exceeding 75 days 88% of the time

Measurement Variation

Process (or product)Variation

Measurement System Valid

AgentTo

Agent

Departmentto

Department

BandTo

Band

Staffing Cycle TimeStatistical Tests

Department to Department is the Red XDepartment to Department is the Red X

Moods p-value = 0.060HOV p-value = 0.000

Moods p-value = 1.000HOV p-value = 0.0166

Moods p-value = 0.211HOV p-value = 0.067

Statistical Tests

What if you have Attribute data? 75 days

Anything less than 75 daysis Good

Anything greater than 75 daysis a Defect

WhoStaffing

Cycle timeDate posted Start date Department Band

Sourcing agent Month

Fred 71 1/7/1999 3/19/1999 Engineering SP Tom JanDoug 72 3/22/1999 6/2/1999 Sourcing P Kathy MarchUrsla 73 4/14/1999 6/26/1999 Finance SP Bin AprilZack 74 3/28/1999 6/10/1999 Sourcing SP Kathy MarchBen 74 5/31/1999 8/13/1999 Engineering SP Tom MayTim 77 2/11/1999 4/29/1999 Engineering SP Tom FebWalt 79 1/13/1999 4/2/1999 Engineering SP Tom JanKathy 79 3/17/1999 6/4/1999 Engineering SP Tom MarchAmy 80 2/1/1999 4/22/1999 Finance SP Bin Feb

God / Defect

DefectGoodDefectDefectDefectDefectDefectDefect

Hypothesis Testing

Continuous Non-Continuous

Chi-sq

Correlation

Staffing cycle time is exceeding 75 days 88% of the time

Measurement Variation

Process (or product)Variation

Timeto

Time

Departmentto

Department

Bandto

Band

Operatorto

Operator

Attribute Data

Non-Continuous

Chi-sq

Correlation

Ho: All Departments are the sameHa: Departments are different

If the p-value >.05 …… Then Ho is true and there is no difference in the groups you had me test(Accept Ho) Departments are the same

If the p-value < .05 …... Then Ho is false and there is a statically significant difference (Reject Ho ) Departments are different

Staffing cycle time is exceeding 75 days 88% of the time

Process (or product)Variation

Departmentto

Department

Pass Fail

Engineering

Finance

Sourcing

Department to Department

Counts of p

ass / f

ail

Chi 2

Statistical Tests

75 days

Anything less than 75 daysis Good

Anything greater than 75 daysis a Defect

Statistical Tests

If the p-value >.05 …… Then Ho is true and there is no difference in the groups you had me test(Accept Ho) Departments are the same

If the p-value < .05 …... Then Ho is false and there is a statically significant difference (Reject Ho ) Departments are different

P-value > .05There is no difference between departments

Chi 2

MINITAB: Stat> Tables Chi Square test

Chi-Square Test: Defects, Good

Expected counts are printed below observed countsChi-Square contributions are printed below expected counts

Defects Good Total 1 7 1 8 6.12 1.88 0.127 0.414

2 9 2 11 8.41 2.59 0.041 0.134

3 3 4 7 5.35 1.65 1.034 3.361

4 7 1 8 6.12 1.88 0.127 0.414

Total 26 8 34

Chi-Sq = 5.652, DF = 3, P-Value = 0.1304 cells with expected counts less than 5.