Download Analyze - Hypothesis Testing Normal Data - P2

Analyze Phase
Hypothesis Testing Normal Data
Part 2
Hypothesis Testing Normal Data Part 2
Welcome to Analyze
“X” Sifting
Inferential Statistics
Intro to Hypothesis Testing
Hypothesis Testing ND P1
Calculate Sample Size
Hypothesis Testing ND P2
Variance Testing
Analyze Results
Hypothesis Testing NND P1
Hypothesis Testing NND P2
Wrap Up & Action Items
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Tests of Variance
Tests of Variance are used for both Normal and Non-normal
Data.
Normal Data
– 1 Sample to a target
– 2 Samples: F-Test
– 3 or More Samples: Bartlett’s Test
Non-Normal Data
– 2 or more samples: Levene’s Test
The null hypothesis states there is no difference between the
Standard Deviations or variances.
– Ho: σ1 = σ2 = σ3 …
– Ha: at least one is different
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1-Sample Variance
A 1-sample variance test is used to compare an expected
population variance to a target.
Stat > Basic Statistics > Graphical Summary
If the target variance lies inside the confidence interval then we
fail to reject the null hypothesis.
– Ho: σ2Sample = σ2Target
– Ha: σ2Sample ≠ σ2Target
Use the sample size calculations for a 1 sample t-test.
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1-Sample Variance
1. Practical Problem:
• We are considering changing supplies for a part we currently
purchase from a supplier that charges a premium for the
hardening process and has a large variance in their process.
• The proposed new supplier has provided us with a sample of
their product. They have stated they can maintain a variance of
0.10.
2. Statistical Problem:
Ho: σ2 = 0.10 or
Ha: σ2 ≠ 0.10
Ho: σ = 0.31
Ha: σ ≠ 0.31
3. 1-sample variance:
α = 0.05 β = 0.10
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1-Sample Variance
4. Sample Size:
• Open the MINITABTM worksheet: “Exh_Stat.MTW”
• This is the same file used for the 1 Sample t example.
– We will assume the sample size is adequate.
5. State Statistical Solution
Stat > Basic Statistics > Graphical Summary
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1-Sample Variance
Recall the target Standard Deviation is 0.31.
Summary for Values
A nderson-Darling N ormality Test
4.4
4.6
4.8
A -S quared
P -V alue
0.33
0.442
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
4.7889
0.2472
0.0611
-0.02863
-1.24215
9
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
5.0
4.4000
4.6000
4.7000
5.0500
5.1000
95% C onfidence Interv al for M ean
4.5989
4.9789
95% C onfidence Interv al for M edian
4.6000
5.0772
95% C onfidence Interv al for S tD ev
95% Confidence Intervals
0.1670
0.4736
Mean
Median
4.6
4.7
LSS Green Belt v11.1 MT - Analyze Phase
4.8
4.9
7
5.0
5.1
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Test of Variance Example
1. Practical Problem:
We want to determine the effect of two different storage methods on
the rotting of potatoes. You study conditions conducive to potato rot
by injecting potatoes with bacteria that cause rotting and subjecting
them to different temperature and oxygen regimes. We can test the
data to determine if there is a difference in the Standard Deviation
of the rot time between the two different methods.
2. Statistical Problem:
Ho: σ1 = σ2
Ha: σ1 ≠ σ2
3. Equal Variance test (F-test since there are only 2 factors.)
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Test of Variance Example
4. Sample Size:
α = 0.05 β = 0.10
Stat > Power and Sample Size > s2 2 Variances…
Open Minitab Worksheet “EXH_AOV.MTW”
MINITABTM Session Window
Power and Sample Size
Test for Two Standard Deviations
Testing (StDev 1 / StDev 2) = 1 (versus not =)
Calculating power for (StDev 1 / StDev 2) = ratio
Alpha = 0.05
Method: Levene's Test
Ratio
0.684
Minimum sample size of 89 required
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Sample Target
Size Power
89
0.9
Actual Power
0.900312
The sample size is for each group.
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Normality Test – Follow the Roadmap
5. Statistical Solution:
Stat>Basic Statistics>Normality Test
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Normality Test – Follow the Roadmap
Ho: Data is Normal
Ha: Data is NOT Normal
Stat>Basic Stats> Normality Test (Use Anderson Darling)
Probability Plot of Rot 1
Normal
99.9
Mean
StDev
N
AD
P-Value
99
Percent
95
90
4.871
0.9670
100
0.306
0.559
80
70
60
50
40
30
20
10
5
1
0.1
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3
4
11
5
Rot 1
6
7
8
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Test of Equal Variance – Normal Data
Stat>ANOVA>Test for Equal Variance
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Test of Equal Variance – Normal Data
6. Practical Solution:
The difference between the Standard Deviations from the two samples
is not significant.
Use F-Test for 2 samples
Normally distributed data.
P-value < 0.05 (.002)
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Normality Test
Perform another test using the column Rot. First run the
Normality Test…
Probability Plot of Rot
Normal
99
Mean
StDev
N
AD
P-Value
95
90
13.78
7.712
18
0.285
0.586
Percent
80
70
60
50
40
The p-value is > 0.05
We can assume our data is
Normally Distributed.
30
20
10
5
1
-5
0
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10
15
Rot
14
20
25
30
35
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Test for Equal Variance (Normal Data)
Then run a Test for Equal Variance using using Rot as a
“Response:” and Temp as “Factors:”.
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Test of Equal Variance
Test for Equal Variances for Rot
F-Test
Test Statistic
P-Value
10
0.68
0.598
Temp
Lev ene's Test
Test Statistic
P-Value
16
2
4
6
8
10
95% Bonferroni Confidence Intervals for StDevs
10
0.05
0.824
12
Ho: σ1 = σ2
Ha: σ1≠ σ2
Temp
P-value > 0.05; there is no
statistically significant difference.
16
0
5
10
15
20
25
Rot
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Test of Equal Variance
Use F- Test for 2
samples of Normally
Distributed Data.
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Continuous Data - Normal
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Test For Equal Variances
Stat>ANOVA>Test for Equal Variance
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Test For Equal Variances Graphical Analysis
Test for Equal Variances for Rot
Temp
Oxygen
Bartlett's Test
2
10
Test Statistic
P-Value
6
2.71
0.744
Lev ene's Test
Test Statistic
P-Value
10
0.37
0.858
2
16
6
10
0
20
40
60
80
100
120
140
95% Bonferroni Confidence Intervals for StDevs
P-value > 0.05 shows insignificant
difference between variance
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Test For Equal Variances Statistical Analysis
Test for Equal Variances: Rot versus Temp, Oxygen
95% Bonferroni confidence intervals for Standard Deviations Temp
10
10
10
16
16
16
Oxygen
2
6
10
2
6
10
N
3
3
3
3
3
3
Lower
2.26029
1.28146
2.80104
1.54013
1.50012
3.55677
StDev
5.29150
3.00000
6.55744
3.60555
3.51188
8.32666
`Upper
81.890
46.427
101.481
55.799
54.349
128.862
Bartlett's Test (Normal distribution)
Use this if
data is Normal and
for Factors > or = 2
Test statistic = 2.71 P-value = 0.744
Levene's Test (any continuous distribution)
Use this if
data is Non-normal and
for Factors > or = 2
Test statistic = 0.37 P-value = 0.858
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Tests for Variance Exercise
Exercise objective: Utilize what you have learned to conduct
and analyze a test for Equal Variance using MINITABTM.
1. The quality manager was challenged by the plant director as to why
the VOC levels in the product varied so much. After using a
Process Map some potential sources of variation were identified.
These sources included operating shifts and the raw material
supplier. Of course the quality manager has already clarified the
Gage R&R results were less than 17% study variation so the gage
was acceptable.
2. The quality manager decided to investigate the effect of the raw
material supplier. He wants to see if the variation of the product
quality is different when using supplier A or supplier B. He wants to
be at least 95% confident the variances are similar when using the
two suppliers.
3. Use data ppm VOC and RM Supplier to determine if there is a
difference between suppliers.
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Tests for Variance Exercise: Solution
First we want to do a graphical summary of the two samples from the
two suppliers.
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Tests for Variance Exercise: Solution
In “Variables:” enter ‘ppm
VOC’
In “By variables:” enter
‘RM Supplier’
We want to see if the two
samples are from Normal
populations.
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Tests for Variance Exercise: Solution
The P-value is greater than 0.05 for both Anderson-Darling Normality
Tests so we conclude the samples are from Normally Distributed
populations because we “failed to reject” the null hypothesis that the
data sets are from Normal Distributions.
Summary for ppm VOC
Summary for ppm VOC
RM Supplier = B
RM Supplier = A
A nderson-Darling N ormality Test
A nderson-Darling N ormality Test
20
25
30
35
40
45
A -S quared
P -V alue
0.33
0.465
A -S quared
P -V alue
0.49
0.175
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
37.583
7.090
50.265
0.261735
-0.091503
12
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
30.500
6.571
43.182
-0.555911
-0.988688
12
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
50
25.000
33.250
35.500
42.000
50.000
20
25
30
35
40
45
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
50
95% C onfidence Interv al for M ean
95% C onfidence Interv al for M ean
33.079
26.325
42.088
33.263
42.000
5.022
12.038
25.000
95% Confidence Intervals
95% C onfidence Interv al for S tD ev
Mean
34.675
95% C onfidence Interv al for M edian
95% C onfidence Interv al for M edian
95% Confidence Intervals
19.000
25.000
31.500
37.000
38.000
37.000
95% C onfidence Interv al for S tD ev
Mean
4.655
11.157
Median
Median
32
34
36
38
40
25.0
42
27.5
30.0
32.5
35.0
37.5
Are both Data Sets Normal?
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Tests for Variance Exercise: Solution
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Tests for Variance Exercise: Solution
For “Response:” enter ‘ppm VOC’
For “Factors:” enter ‘RM Supplier’
Note MINITABTM defaults to 95% confidence level which is exactly the
level we want to test for this problem.
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Tests for Variance Exercise: Solution
Because both populations were considered to be Normally Distributed the F-test is
used to evaluate whether the variances (Standard Deviation squared) are equal.
The P-value of the F-test was greater than 0.05 so we “fail to reject” the null
hypothesis.
So once again in English: The variances are equal between the results from the two
suppliers on our product’s ppm VOC level.
Test for Equal Variances for ppm VOC
RM Supplier
F-Test
Test Statistic
P-Value
A
Lev ene's Test
Test Statistic
P-Value
B
4
RM Supplier
1.16
0.806
6
8
10
12
95% Bonferroni Confidence Intervals for StDevs
0.02
0.890
14
A
B
20
LSS Green Belt v11.1 MT - Analyze Phase
25
30
35
ppm VOC
28
40
45
50
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Hypothesis Testing Roadmap
Normal
Test of Equal Variance
1 Sample Variance
Variance Equal
Variance Not Equal
Two Samples
Two
Samples
Two
Samples
Two
Samples
2 Sample T
1 Sample t-test
One Way ANOVA
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2 Sample T
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One Way ANOVA
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Purpose of ANOVA
Analysis of Variance (ANOVA) is used to investigate and model
the relationship between a response variable and one or more
independent variables.
Analysis of Variance extends the two sample t-test for testing the
equality of two population Means to a more general null hypothesis
of comparing the equality of more than two Means versus them not
all being equal.
– The classification variable, or factor, usually has three or more
levels (If there are only two levels, a t-test can be used).
– Allows you to examine differences among Means using multiple
comparisons.
– The ANOVA test statistic is:
Avg SS between S2 between
 2
Avg SS within
S within
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What do we want to know?
Is the “between group” variation large enough to be
distinguished from the “within group” variation?
X
delta
(δ)
(Between Group Variation)
Total (Overall) Variation
Within Group Variation
(level of supplier 1)
X
X
X X
X
X X X
μ1
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μ2
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Calculating ANOVA
Where:
G = the number of groups (levels in the study)
xij = the individual in the jth group
nj = the number of individuals in the jth group or level
X = the grand Mean
Xj = the Mean of the jth group or level
Total (Overall) Variation
delta
(δ)
Within Group Variation
(Between Group Variation)
Between Group Variation
g

j1
nj (Xj  X )
Within Group Variation
g
2
LSS Green Belt v11.1 MT - Analyze Phase
nj
 (X
j1 i 1
ij  X)
32
Total Variation
g
2
nj
 (X
j1 i 1
ij
 X) 2
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Alpha Risk and Pair-Wise t-tests
The alpha risk increases as the number of Means increases with a
pair-wise t-test scheme. The formula for testing more than one
pair of Means using a t-test is:
1  1  α 
k
where k  number of pairs of means
so, for 7 pairs of means and an α  0.05 :
1 - 1 - 0.05  0.30
7
or 30% alpha risk
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Three Samples
We have three potential suppliers claiming to have equal levels
of quality. Supplier B provides a considerably lower purchase
price than either of the other two vendors. We would like to
choose the lowest cost supplier but we must ensure we do not
effect the quality of our raw material.
File>Open Worksheet > ANOVA.MTW
We would like test the data to determine if there is
a difference between the three suppliers.
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Follow the Roadmap…Test for Normality
Probability Plot of Supplier A
Normal
99
Mean
StDev
N
AD
P-Value
95
90
3.664
0.4401
5
0.246
0.568
The samples of all three
suppliers are Normally
Distributed.
Supplier A P-value = 0.568
Supplier B P-value = 0.385
Supplier C P-value = 0.910
70
60
50
40
30
Probability Plot of Supplier B
20
Normal
10
99
Mean
3.968
StDev
0.2051
N
5
AD
0.314
Probability
P-Value
0.385
5
95
1
2.5
3.0
90
3.5
Supplier
A
80
Percent
4.0
4.5
60
50
40
Mean
StDev
N
AD
P-Value
95
90
30
20
4.03
0.4177
5
0.148
0.910
80
10
5
1
Plot of Supplier C
Normal
99
70
Percent
Percent
80
3.50
3.75
4.00
Supplier B
70
60
50
40
30
20
4.25
4.50
3.0
3.5
10
5
1
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4.0
Supplier C
4.5
5.0
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Test for Equal Variance…
Test for Equal Variance (must stack data to create “Response” & “ Factors”):
Test for Equal Variances for Data
Bartlett's Test
Test Statistic
P-Value
Supplier A
2.11
0.348
Lev ene's Test
Suppliers
Test Statistic
P-Value
0.59
0.568
Supplier B
Supplier C
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8
95% Bonferroni Confidence Intervals for StDevs
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ANOVA MINITABTM
Stat>ANOVA>One-Way Unstacked
Enter Stacked Supplier data in
“Responses:”
Click on “Graphs…”,
Check “Boxplots of data”
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ANOVA
What does this graph tell us?
Boxplot of Supplier A, Supplier B, Supplier C
4.6
4.4
4.2
Data
4.0
3.8
3.6
3.4
3.2
3.0
Supplier A
LSS Green Belt v11.1 MT - Analyze Phase
Supplier B
38
Supplier C
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ANOVA Session Window
One-way ANOVA: Supplier A, Supplier B, Supplier C
P-value > .05
No Difference between suppliers
Source DF SS MS
F
P
Factor 2 0.384 0.192 1.40 0.284
Error 12 1.641 0.137
Total 14 2.025
S = 0.3698 R-Sq = 18.95% R-Sq(adj) = 5.44%
Level
N
Supplier A 5
Supplier B 5
Supplier C 5
Mean
3.6640
3.9680
4.0300
StDev
0.4401
0.2051
0.4177
Stat>ANOVA>One Way (unstacked)
Individual 95% CIs For Mean Based on Pooled StDev
Level
+---------+---------+---------+--------Supplier A (-----------*-----------)
Supplier B
(-----------*-----------)
Supplier C
(-----------*-----------)
+---------+---------+---------+--------3.30
3.60
3.90
4.20
Pooled StDev = 0.3698
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ANOVA
One-way ANOVA: Supplier A, Supplier B, Supplier C
Source DF SS MS
F
P
Factor 2 0.384 0.192 1.40 0.284
Error 12 1.641 0.137
Total 14 2.025
S = 0.3698 R-Sq = 18.95% R-Sq(adj) = 5.44%
F-Critical
F-Calc
Level
N
Supplier A 5
Supplier B 5
Supplier C 5
Mean
3.6640
3.9680
4.0300
StDev
0.4401
0.2051
0.4177
Individual 95% CIs For Mean Based on Pooled StDev
Level
+---------+---------+---------+--------Supplier A (-----------*-----------)
Supplier B
(-----------*-----------)
Supplier C
(-----------*-----------)
+---------+---------+---------+--------3.30
3.60
3.90
4.20
Pooled StDev = 0.3698
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D/N
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
161.40
18.51
10.13
7.71
6.61
5.99
5.59
5.32
5.12
4.96
4.84
4.75
4.67
4.60
4.54
2
199.50
19.00
9.55
6.94
5.79
5.14
4.74
4.46
4.26
4.10
3.98
3.89
3.81
3.74
3.68
3
215.70
19.16
9.28
6.59
5.41
4.76
4.35
4.07
3.86
3.71
3.59
3.49
3.41
3.34
3.29
4
224.60
19.25
9.12
6.39
5.19
4.53
4.12
3.84
3.63
3.48
3.36
3.26
3.18
3.11
3.06
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Sample Size
Let’s check how much difference we can see with a sample of 5.
Power and Sample Size
One-way ANOVA
Alpha = 0.05 Assumed Standard Deviation = 1 Number of
Levels = 3
Sample
Size
5
Power
0.9
SS Means
3.29659
Maximum
Difference
2.56772
The sample size is for each level.
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ANOVA Assumptions
1. Observations are adequately described by the model.
2. Errors are Normally and independently distributed.
3. Homogeneity of variance among factor levels.
In one-way ANOVA model adequacy can be checked by either of
the following:
1. Check the data for Normality at each level and for homogeneity
of variance across all levels.
2. Examine the residuals (a residual is the difference in what the
model predicts and the true observation).

Normal plot of the residuals

Residuals versus fits

Residuals versus order
If the model is adequate the residual plots will be structureless.
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Residual Plots
Stat>ANOVA>One-Way Unstacked>Graphs
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Histogram of Residuals
Histogram of the Residuals
(responses are Supplier A, Supplier B, Supplier C)
5
Frequency
4
3
2
1
0
-0.6
-0.4
-0.2
0.0
Residual
0.2
0.4
0.6
The Histogram of Residuals should
show a bell-shaped curve.
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Normal Probability Plot of Residuals
Normality plot of the Residuals should follow a straight line.
Results of our example look good.
The Normality assumption is satisfied.
Normal Probability Plot of the Residuals
(responses are Supplier A, Supplier B, Supplier C)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-1.0
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-0.5
0.0
Residual
45
0.5
1.0
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Residuals versus Fitted Values
The plot of Residuals versus fits examines constant variance.
The plot should be structureless with no outliers present.
Our example does not indicate a problem.
Residuals Versus the Fitted Values
(responses are Supplier A, Supplier B, Supplier C)
0.75
0.50
Residual
0.25
0.00
-0.25
-0.50
3.65
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3.70
3.75
3.80
46
3.85
3.90
Fitted Value
3.95
4.00
4.05
© Open Source Six Sigma, LLC
ANOVA Exercise
Exercise objective: Utilize what you have learned to
conduct an analysis of a one way ANOVA using
MINITABTM.
1. The quality manager was challenged by the plant director as
to why the VOC levels in the product varied so much. The
quality manager now wants to find if the product quality is
different because of how the shifts work with the product.
2. The quality manager wants to know if the average is
different for the ppm VOC of the product among the
production shifts.
3. Use Data in columns “ppm VOC” and “Shift” in “hypotest
stud.mtw” to determine the answer for the quality manager
at a 95% confidence level.
LSS Green Belt v11.1 MT - Analyze Phase
47
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
First we need to
do a graphical
summary of the
samples from the
3 shifts.
LSS Green Belt v11.1 MT - Analyze Phase
Stat>Basic Stat>Graphical Summary
48
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
We want to see if the 3
samples are from Normal
populations.
In “Variables:” enter ‘ppm
VOC’
In “By variables:” enter
‘Shift’
LSS Green Belt v11.1 MT - Analyze Phase
49
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
The P-value is greater than 0.05
for both Anderson-Darling
Normality Tests so we conclude
the samples are from Normally
Distributed populations because
we “failed to reject” the null
hypothesis that the data sets are
from Normal Distributions.
Summary for ppm VOC
Summary for ppm VOC
A nderson-Darling N ormality Test
20
25
30
35
40
45
25
30
35
40
45
50
32.000
33.500
38.000
46.500
50.000
33.847
32.936
45.153
95% Confidence Intervals
48.129
95% C onfidence Interv al for S tD ev
Mean
4.470
13.761
Median
35
40
45
50
Summary for ppm VOC
P-Value 0.658
Shift = 3
A nderson-Darling N ormality Test
0.37
0.334
A -S quared
P -V alue
0.24
0.658
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
34.625
5.041
25.411
-0.74123
1.37039
8
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
28.000
6.525
42.571
0.06172
-1.10012
8
30.411
25.000
31.750
35.500
37.000
42.000
20
25
30
35
40
45
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
50
30.614
22.545
37.322
20.871
95% Confidence Intervals
33.322
95% C onfidence Interv al for S tD ev
Mean
10.260
33.455
95% C onfidence Interv al for M edian
95% C onfidence Interv al for S tD ev
3.333
19.000
22.000
28.000
32.750
38.000
95% C onfidence Interv al for M ean
38.839
95% C onfidence Interv al for M edian
Mean
39.500
6.761
45.714
0.58976
-1.13911
8
95% C onfidence Interv al for M edian
95% C onfidence Interv al for M ean
95% Confidence Intervals
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
A -S quared
P -V alue
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
50
0.32
0.446
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
A nderson-Darling N ormality Test
20
A -S quared
P -V alue
95% C onfidence Interv al for M ean
P-Value 0.334
Shift = 2
P-Value 0.446
Shift = 1
4.314
13.279
Median
Median
30
32
34
36
38
LSS Green Belt v11.1 MT - Analyze Phase
20.0
40
50
22.5
25.0
27.5
30.0
32.5
35.0
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
First we need to determine if our
data has Equal Variances.
Stat > ANOVA > Test for Equal Variances…
Now we need to test the variances.
For “Response:” enter ‘ppm VOC’
For “Factors:” enter ‘Shift’
LSS Green Belt v11.1 MT - Analyze Phase
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© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
The P-value of the F-test was greater than 0.05 so we “fail to
reject” the null hypothesis.
Test for Equal Variances for ppm VOC
Bartlett's Test
Test Statistic
P-Value
1
0.63
0.729
Lev ene's Test
Shift
Test Statistic
P-Value
0.85
0.440
2
3
2
4
6
8
10
12
14
16
95% Bonferroni Confidence Intervals for StDevs
18
Are the variances are equal…Yes!
LSS Green Belt v11.1 MT - Analyze Phase
52
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
We need to use the One-Way ANOVA to
determine if the Means are equal of
product quality when being produced by
the 3 shifts. Again we want to put 95.0 for
the confidence level.
Stat > ANOVA > One-Way…
For “Response:” enter ‘ppm VOC’
For “Factor:” enter ‘Shift’
Also be sure to click “Graphs…” to select “Four
in one” under residual plots.
Also, remember to click “Assume equal
variances” because we determined the
variances were equal between the 2 samples.
LSS Green Belt v11.1 MT - Analyze Phase
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© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
We must look at the Residual Plots to be sure our ANOVA analysis
is valid.
Since our residuals look Normally Distributed and randomly
patterned we will assume our analysis is correct.
Residual Plots for ppm VOC
Normal Probability Plot
Residuals Versus the Fitted Values
99
N
24
AD
0.255
P-Value 0.698
10
Residual
Percent
90
50
10
1
0
-5
-10
-10
0
Residual
10
30
Histogram of the Residuals
4.8
10
3.6
5
2.4
40
0
-5
1.2
0.0
35
Fitted Value
Residuals Versus the Order of the Data
Residual
Frequency
5
-10
-10
-5
LSS Green Belt v11.1 MT - Analyze Phase
0
Residual
5
10
54
2
4
6
8 10 12 14 16 18 20 22 24
Observation Order
© Open Source Six Sigma, LLC
ANOVA Exercise: Solution
Since the P-value of the ANOVA test is less than 0.05 we “reject”
the null hypothesis that the Mean product quality as measured
in ppm VOC is the same from all shifts.
We “accept” the alternate hypothesis that the Mean product
quality is different from at least one shift.
Don’t miss that
shift!
Since the confidence intervals
of the Means do not overlap
between Shift 1 and Shift 3 we
see one of the shifts is
delivering a product quality with
a higher level of ppm VOC.
LSS Green Belt v11.1 MT - Analyze Phase
55
© Open Source Six Sigma, LLC
Summary
At this point you should be able to:
• Be able to conduct Hypothesis Testing of Variances
• Understand how to Analyze Hypothesis Testing Results
LSS Green Belt v11.1 MT - Analyze Phase
56
© Open Source Six Sigma, LLC
IASSC Certified Lean Six Sigma Green Belt (ICGB)
The International Association for Six Sigma Certification (IASSC) is a Professional
Association dedicated to growing and enhancing the standards within the Lean Six Sigma
Community. IASSC is the only independent third-party certification body within the Lean Six
Sigma Industry that does not provide training, mentoring and coaching or consulting
services. IASSC exclusively facilitates and delivers centralized universal Lean Six Sigma
Certification Standards testing and organizational Accreditations.
The IASSC Certified Lean Six Sigma Green
Belt (ICGB) is an internationally recognized
professional who is well versed in the Lean
Six Sigma Methodology who both leads or
supports improvement projects. The Certified
Green Belt Exam, is a 3 hour 100 question
proctored exam.
Learn about IASSC Certifications and Exam options at…
http://www.iassc.org/six-sigma-certification/
LSS Green Belt v11.1 MT - Analyze Phase
© Open Source Six Sigma, LLC