Javier Garcia - Verdugo Sanchez - Six Sigma Training - W2 Simple Variance Analysis

Simple Variancep
Analysis
1 2 3 4
O ANOVAOne-way ANOVA
F - Test
Week 2
Knorr-Bremse Group
Introduction
Th i l i l i id th ibilit tThe simple variance analysis provides the possibility to
determine the effect of one factor on the result (Y).
Based on the variance analysis you can assign the
variation of the result to factors.
You can explain the portion of the of the variance caused
b th f tby the factor.
It is the foundation for further statistical methods likeIt is the foundation for further statistical methods like
measurement system analysis, Multivari…, DOE…
Knorr-Bremse Group 04 BB W2 Simple Variance Analysis 08, D. Szemkus/H. Winkler Page 2/40

Content
O i f th t ti ti l d l ANOVA• Overview of the statistical model one-way ANOVA
• Conducting the ANOVA and the interpretation of the• Conducting the ANOVA and the interpretation of the
results (p - value and F - value)
• Different graphical investigation methods
• Statistical test for comparison of variances
• The diagnosis of experiment and evaluation
• Many exercises with Minitab
Validation of Factors Y = f(x)
Factor X = Input
Discrete / Attributive Continuous / VariableDiscrete / Attributive Continuous / Variable
ut
te
ve
L i ti
=Outpu
Discret
Attributiv
Chi - Square
Logistic
Regression
nseY=
A
s
T Test
Respon
ntinuou
ariable
Regression
T - Test
ANOVA ( F - Test)
R
Con
Va
( )
Median Tests
Statistical techniques for all combination of data types are available
y

Validation of Factors y = f (x)
Factor X has two
settings
Factor X has two or
more settings
Factor X has one
setting
Representation of the Stability
(Investigation if needed)
Investigation of the
distribution type
(Investigation if needed) (Investigation if needed) (Investigation if needed)
distribution type
distribution type
distribution type
Investigation of the process
distribution type distribution type
variabilitycenter (target value) variability
Investigation of the processInvestigation of the process Investigation of the process
center (target value)
T - Test
T - Test
ANOVA
ANOVA
Single Factor Experiments
The mathematical model:
x = µ + α + εxik = µ + αi + εik
xik observed value at trial k with setting iik g
αi the true effect of factor A on level (setting) i
εik experimental error
H0 says that there is no difference between the settings!
Math. hypothesis model Common hypothesis model
Ho: all αi‘s = 0
Ha: one αi is different
Ho: µ1 = µ2 = µ3 = µ4
Ha: at least one µi different
Ha: one αi is different Ha: at least one µi different

The Components of the Variance Analysis
Average: Mean = (MeanGroup1 + MeanGroup2)/ 2
Variance: SSTotal = SSBetween + SSWithin
Variance
within groups
17
18
within groups
lt
14
15
16
Total
variance
Variance
between groups
Resu
12
13
14
1 2
Variance
within groups9
10
11
1 2 within groups
Factor setting
8
Simple Variance Analysis
The ANOVA Evaluation
Degrees of Freedom …
Sum of Squares
DF SS MS F p
Factor A 1 25 25 6 0,022
Unexplained 18 75 4 15Unexplained 18 75 4,15
Total 19 100
Mean Square
P – Value
Probability
F - Value
F Value

The Evaluation with ANOVA: SS Column
Observed total Variation (Variability) = Factor Variation +
SS = Sum of squares...
Observed total Variation (Variability) Factor Variation +
Random Variation (noise or statistical error)
Observed Total VariationObserved Total Variation
= Portion between the groups
SS (Total) = SS (Between) + SS (Within)
+ Portion within the groups
SS (Total) = SS (Between) + SS (Within)
SS (Total) = SS (Factor) + SS (Error)
( ) ( )∑∑∑∑∑ −+−=−
g k
jij
g
j
g k
ij xxxxkxx
222
)(( ) ( )∑∑∑∑∑ = === =
+
j i
jij
j
j
j i
ij xxxxkxx
1 111 1
)(
g = Number of subgroups
g = Number of subgroups
k = Subgroup size
The Evaluation with ANOVA: DF Column
DF = Degrees of Freedomg
The Degrees of Freedom are defined by the
independent number of possibilities to compare
That means:
Total = Total Number (all data points) - 1
Factor = Number of Groups - 1
Error = Total Number – Number of Groups

The Evaluation with ANOVA: MS; F; P-value
The MS value = Mean Square value
I h i f h S f S D f F dIs the portion of the Sum of Squares per Degree of Freedom…
MS = SS / DF
F-Value
Is the quotient of MS (Factor) divided by MS (Error), or the quotient of the
factor of the explained variation divided by the not explained variation.
P-Value
This value is an outcome of the Hypothesis Test, the value varies between
0 and 1
High values confirm no change (we don’t reject the Null Hypothesis)
Low values confirm a change (we reject the Null Hypothesis and accept theg ( j yp p
Alternative Hypothesis)
P 0 05 i th i ifi li it i l
P = 0.05 is the significance limit in general
Statistical Assumptions
• The variance of the population of the output is for all levels equal
(H it f i )(Homogeneity of variance).
– We can test this assumption with the Minitab function
Stat>ANOVA> Test for Equal Variances.
• The mean values of the start variables are independent fromThe mean values of the start variables are independent from
each other and normal distributed.
If the samples are randomized and the sample size– If the samples are randomized and the sample size
appropriate this assumption is usually valid.
Att ti At h i l th i k i hi h t h– Attention: At chemical processes the risk is high to have
dependent variables, therefore randomization is a need.
• Residuals (Error of the mathematical model) have to be
independent and normal distributed, with a mean of 0 and a
constant variance
constant variance.

Remarks to Single Factor Experiments
• The input variable will be also called “factor”.
– In single factor designs the factor will be treated as a attribute
variable although it is representing an interval or a ratio.variable although it is representing an interval or a ratio.
• If the factor is a continuous variable, it has to be divided in
subgroupssubgroups.
– E.g. measuring pressure gives us values from high to low.
– We can split the set of values using the median in two ranges
– high and low.
– In our example we have day, time and shifts as factors.
• The output will be measured usually with interval or ratio scales
(Yield, Temperature, Volt etc.)
The Evaluation with Minitab
The known example, file: WATER CONTENT.MTW
The process runs 24hrs at 6 days in a week with 3 shifts This will be checked every 2
Day Time Shift Water Content
1 6 1 5 67
The process runs 24hrs. at 6 days in a week with 3 shifts. This will be checked every 2
hrs which results in 72 measurement values.
1 6 1 5,67
1 8 1 6
1 10 1 6,27
1 12 1 6,33
1 14 2 6 53 6,50
Boxplot of Water Content vs Day
1 14 2 6,53
1 16 2 5,93
1 18 2 6
1 20 2 6,27
1 22 3 6 07
ent
6,50
6,25
1 22 3 6,07
1 0 3 6,33
1 2 3 6,13
1 4 3 6,07
WaterConte
6,00
2 6 1 6,33
2 8 1 6,47
2 10 1 6
5,75
5,50
Day
654321
5,50
?
Are the differences statistical significant?

Factors with 2 or more Settings; Y = f (x)
Minitab What do we ask or want to find?
SPC Chart Indications of trends or patterns that data are notSPC Chart
I-MR
Indications of trends or patterns that data are not
from the same process or population?
Descriptive
Data are normal distributed? Small
Factor X has two or
more settings
Descriptive
Stats und
Normality Tests
P-values (<.05) are the evidence of non normal
distribution
Watch for the sample size!
Stack Data &
Homogeneity of
Levene’s TestBartlett Test (F-Test)
Normal distributed Not normal distributedInvestigation of the
distribution type
Homogeneity of
Variance Test Ho: σ2
A= σ2
BHo: σ2
A= σ2
B
Small P-values (<0,05) indicate unequal variances!
variability
Equal Variances
1 Way ANOVA
Ho: µA= µB =µc
1 Way ANOVA
(if N >25 or transformed)
Kruskal-Wallis / Moods Median
(Med A = Med B = Med C)
Unequal Variances
Ask for MBB support
1 Way ANOVA
Small P-values (<0,05) indicates different distribution!
Stability and Distribution
Stat
>Control Charts
lValue
6,50
6,25
UC L=6,638
I-MR Chart of Water Content
>Variable Charts Individuals
>I-MR…
Observation
Individua
70635649423528211471
6,00
5,75
5,50
_
X=6,054
LC L=5,469
Observation
gRange
0,8
0,6
0,4
UC L=0,7179
Summary for Water Content
Observation
Movin
70635649423528211471
0,2
0,0
__
MR=0,2197
LC L=0
A nderson-Darling Normality Test
V ariance 0,0646
Sk 0 124017
A -Squared 0,39
P-V alue 0,380
Mean 6,0537
StDev 0,2541
Summary for Water Content
Stat 6,46,26,05,85,6
Skewness -0,124017
Kurtosis -0,609103
N 72
Minimum 5,5333
1st Q uartile 5,8667
Median 6,0667
3rd Q uartile 6,2500
Maximum 6,5333
>Basic Statistics
>Graphical Summary…
M
,
95% C onfidence Interv al for Mean
5,9940 6,1134
95% C onfidence Interv al for Median
6,0000 6,1333
95% C onfidence Interv al for StDev
0,2183 0,3040
95% Confidence Intervals
Wh i d i i ?
Median
Mean
6,1506,1256,1006,0756,0506,0256,000
What is your decision?

Test for Equal Variances
Stat
>ANOVA
T t f E l V i>Test for Equal Variances…
1
Bartlett's Test
Test for Equal Variances for Water Content
Barlett’s test needs normal
2
1
0,759
Test Statistic 1,10
P-Value 0,954
Levene's Test
Test Statistic 0,52
P-Value
Barlett s test needs normal
distributed data. For 2 or more
groups Minitab displays the F
Day
4
3
– Test.
The Levene’s test is used for
non normal data
6
5
non normal data
95% Bonferroni Confidence Intervals for StDevs
0,50,40,30,20,1
Wh i d i i ?
The ANOVA Table in the Session Window
Stat
>ANOVA
>One way>One way…
One-way ANOVA: Water Content versus Day
Source DF SS MS F P
Day 5 1,1686 0,2337 4,52 0,001
Error 66 3,4148 0,0517Error 66 3,4148 0,0517
Total 71 4,5835
S = 0,2275 R-Sq = 25,50% R-Sq(adj) = 19,85%
Individual 95% CIs For Mean Based onIndividual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ---------+---------+---------+---------+
1 12 6,1333 0,2292 (--------*--------)
2 12 6 1833 0 2241 (--------*--------)2 12 6,1833 0,2241 ( )
3 12 6,1611 0,2386 (--------*-------)
4 12 5,8500 0,2560 (--------*--------)
5 12 5,9111 0,1871 (--------*--------)
6 12 6 0833 0 2241 (--------*-------)6 12 6,0833 0,2241 ( )
---------+---------+---------+---------+
5,85 6,00 6,15 6,30
Wh i d i i ?

R2 und R2 adj.: Practical Significance
• R² is a method within the statistics, to show the practical significance of
an effect.
255,0
5835,4
1686,1Re2
===
Total
gression
SS
SS
R
Explained variation (SS Regression) divided by the
total variation (SS Total).
Approximately 26% of the variation of the water
content is explained by daily fluctuations.content is explained by daily fluctuations.
• R² adj. is a similar method to explain the practical significance of anj p p g
effect. It is helpful, if we use several factors in a model. E.g. R2 adj. gets
smaller, if an additional factor is added in the model, because every
reduction of SS can be balanced by the loss of degrees of freedomreduction of SS error can be balanced by the loss of degrees of freedom.
The values for R² adj. are always a little bit smaller than for R².
19850
0517,0
112 ErrorMS
djR 1985,0
71
5835,4
,
112
=−=−=
Total
Total
Error
DF
SS
adjR
• S is the pooled standard deviation (averaged within group variation) The
square root of S is the MS Error
square root of S is the MS Error.
The Graphical Evaluation in the ANOVA Menu
Stat
>ANOVA
>O>One way…
Boxplot of Water Content by DayIndividual Value Plot of Water Content vs Day
t
6,50
6,25
t
6,50
6,25
WaterContent
6,00
WaterContent
6,00
D
654321
5,75
5,50
D
654321
5,75
5,50
DayDay

Multiple Comparison of Means
• A significant F-test indicates that at least one meang
value deviates with a high probability from the others.
f ff f• The question is which of the mean values differ from
the others.
• We detect that using multiple comparisons.
• We apply that to our water content example.
• The F-test indicates that at least one mean value
deviates, the probability that this is by chance is pretty
low. But which mean value is that?
You will see the following options: Stat
• Tukey’s
>ANOVA
>One-way…
• Fishers’s
D tt’
>Comparison
• Dunnett’s
• Hsu’s MCB
Select one of this options. I recommend Fisher’s, this
t t i b d With th d ill fi dtest is broader. With other words, you will find a
difference easier, at the same time the risk is higher to
k d i i Y ld diffmake a wrong decision. You would see a difference
where no exists.

The matrix shows the comparison of the mean values. The
both numbers in each field are the 95% CI for the difference of
the mean values. If both are positive or negative than a mean
deviation exists.
Is there a zero within the two numbers, the data of this group
b l t th di t ib ti
Day = 1 subtracted from:
Day Lower Center Upper -------+---------+---------+---------+--
belongs more to the same distribution.
y pp
2 -0,1354 0,0500 0,2354 (------*-----)
3 -0,1576 0,0278 0,2132 (-----*-----)
4 -0,4687 -0,2833 -0,0979 (------*-----)
5 0 4076 0 2222 0 0368 ( * )5 -0,4076 -0,2222 -0,0368 (------*-----)
6 -0,2354 -0,0500 0,1354 (-----*------)
-------+---------+---------+---------+--
-0,30 0,00 0,30 0,600,30 0,00 0,30 0,60
Day = 2 subtracted from:
3 -0,2076 -0,0222 0,1632 (-----*-----)
4 -0,5187 -0,3333 -0,1479 (-----*-----)
5 -0,4576 -0,2722 -0,0868 (-----*-----)
6 -0,2854 -0,1000 0,0854 (------*-----)
+ + + +
-------+---------+---------+---------+--
-0,30 0,00 0,30 0,60
Day = 3 subtracted from:Day = 3 subtracted from:
4 -0,4965 -0,3111 -0,1257 (------*-----)
5 -0,4354 -0,2500 -0,0646 (------*-----), , , ( )
6 -0,2632 -0,0778 0,1076 (-----*------)
-------+---------+---------+---------+--
-0,30 0,00 0,30 0,60
Day = 4 subtracted from
5 -0,1243 0,0611 0,2465 (-----*-----)
6 0 0479 0 2333 0 4187 ( * )6 0,0479 0,2333 0,4187 (-----*-----)
-------+---------+---------+---------+--
-0,30 0,00 0,30 0,60
Day = 5 subtracted from:Day = 5 subtracted from:
6 -0,0132 0,1722 0,3576 (-----*-----)
-------+---------+---------+---------+--
-0,30 0,00 0,30 0,60

How sure are we about our decisions?
After the evaluation of the water content there is still the open questions how
sure we are feeling us with the decision.
With Minitab we have the possibility to establish a quick diagnosis.
We evaluate the so called residuals.
Minitab sees at the ANOVA the mean values of each group as theoretical
values (Fitted Values).values (Fitted Values).
The difference of these theoretical to the actual found values result in the
residuals.residuals.
With an optimal experiment the residuals have the following behavior:
• Normal distributed
• Gives a symmetric histogram with 0 as a mean valueGives a symmetric histogram with 0 as a mean value
• Don’t show increasing or decreasing pattern over time
• Don’t show pattern against the theoretical values
The Analysis of the Residuals
ANOVA allows us to save the residuals
and conduct a graphical evaluation.
Normal Probability Plot of the Residuals Residuals Versus the Fitted Values
Residual Plots for Water Content
rcent
99,9
99
90
50
sidual
0,50
0,25
0,00
Normal Probability Plot of the Residuals Residuals Versus the Fitted Values
Residual
Per
0,80,40,0-0,4-0,8
10
1
0,1
Fitted Value
Res
6,26,16,05,9
-0,25
-0,50
ncy
16
12
al
0,50
0,25
Histogram of the Residuals Residuals Versus the Order of the Data
Frequen
0 40 20 0-0 2-0 4
8
4
0
Residua
7065605550454035302520151051
0,00
-0,25
-0,50
Residual
0,40,20,00,20,4
Observation Order
7065605550454035302520151051

Analysis of Single Factor Experiments
1 Th lt h t b i l i Mi it b A dditi l l i1. The results have to be in one column in Minitab. An additional column is
needed for the input variable (factor) where the factor settings are noted.
2 Start Stat>ANOVA>One way2. Start Stat>ANOVA>One-way .
3. Interpret the F- relation. Reject H0 if the F-value is high and the p-value
smaller than 5 - 10%smaller than 5 10%.
4. If you reject H0 use Stat>ANOVA>Main Effects Plot or Graph>Interval
Plot, to show the differences of the means graphically., g p y
5. Conduct diagnosis of the residuals using "Residual Plot" or ANOVA
function of Minitab.
6. Calculate the epsilon square of the effect to proof the practical
significance.
7. Test for homogeneity of variances with Stat>ANOVA>Test for Equal
Variances function.
8. Formulate the conclusions and recommendations.
9. Replicate the optimal setup.
10. Implement the improvements.
Another Example
• A technician in the product development area evaluates the effect
of 5 different currents on the welding strength for a resistance
welding systemwelding system.
• The range is 15 - 19 A
– The settings of the input variable (factor) are 15A, 16A, 17A, 18A
and 19A
– For every level he takes 5 samples.
O t t W ldi t th (G l i hi h t th)• Output: Welding strength (Goal is a high strength)
• Input: Current
• This is an example for an one single factor (current) experiment
with 5 levels.with 5 levels.
The current procedure for this process steps recommends not to
exceed 16A
exceed 16A.

Example
A Design-Matrix with response data as shown below:
Exercise: Open the file: Welding strength.mtw and present the
results in a graphical way. Subsequent investigate the results forg p y q g
significant differences. Follow the validation plan.
Current (Ampere)
W ldi t thTrial 15A 16A 17A 18A 19A
Welding strengthTrial 15A 16A 17A 18A 19A
1 7 12 14 19 7
2 7 17 18 25 10
3 15 12 18 22 113 15 12 18 22 11
4 11 18 19 19 15
5 9 18 19 23 11
Catapult Exercise
• The goal is to determine if a certain setting of the catapult results in a
wider range (distance) than others.
• The output variable is the difference.
• The factor is the position .
– Select four different angles for shooting of the Ping-Pong-balls.
– Shoot minimum 5 balls for each angleg
• Every team member shall:
– Enter the results in Minitab and store the data (will be used later– Enter the results in Minitab and store the data (will be used later
again!)
– Perform a hypothesis test, create graphics for the main effects,yp , g p ,
residual diagnosis, check for equal variances.
– Define how much variation can be explained by the angle (R²)
• Present your results on a flip chart in line with the procedure for
hypothesis testing.

Factors of the Catapult
Ball type
R bb b d
yp
6Rubber band -
fix point Rubber band -
tension adjustment
3
4 4
Rubber band5
Stop -Position3
2
3
1
2
1
2
345
1Tension angle
6
1
1
Summary
O i f th t ti ti l d l ANOVA• Overview of the statistical model one-way ANOVA
• Conducting the ANOVA and the interpretation of the• Conducting the ANOVA and the interpretation of the
results (p - value and F - value)
• Different graphical investigation methods
• Statistical test for comparison of variances
• The diagnosis of experiment and evaluation
• Many exercises with Minitab

ANOVA
Appendix
Critical F-Values
Critical F – values (rounded) for α error of 0,05
1 2 3 4 5 6 7 8 9 10
Degrees of freedom, numerator
1 2 3 4 5 6 7 8 9 10
1 161,4 199,5 215,7 224,6 230,2 234 236,8 238,9 240,5 241,9
2 18,5 19 19,2 19,3 19,3 19,4 19,4 19,4 19,4 19,4
3 10 1 9 6 9 3 9 1 9 8 9 8 9 8 9 8 8 8 8
nator
3 10,1 9,6 9,3 9,1 9 8,9 8,9 8,9 8,8 8,8
4 7,7 6,9 6,6 6,4 6,3 6,2 6,1 6 6 6
5 6,6 5,8 5,4 5,2 5,1 5 4,9 4,8 4,8 4,7
6 6 5,1 4,8 4,5 4,4 4,3 4,2 4,2 4,1 4,1
denomin
6 6 5,1 4,8 4,5 4,4 4,3 4,2 4,2 4,1 4,1
7 5,6 4,7 4,4 4,1 4 3,9 3,8 3,7 3,7 3,6
8 5,3 4,5 4,1 3,8 3,7 3,6 3,5 3,4 3,4 3,4
9 5,1 4,3 3,9 3,6 3,5 3,4 3,3 3,2 3,2 3,1
reedom,
10 5 4,1 3,7 3,5 3,3 3,2 3,1 3,1 3 3
15 4,5 3,7 3,3 3,1 2,9 2,8 2,7 2,6 2,6 2,5
20 4,4 3,5 3,1 2,9 2,7 2,6 2,5 2,5 2,4 2,4
30 4 2 3 3 2 9 2 7 2 5 2 4 2 3 2 3 2 2 2 2
reesoffr
30 4,2 3,3 2,9 2,7 2,5 2,4 2,3 2,3 2,2 2,2
60 4 3,2 2,8 2,5 2,4 2,3 2,3 2,2 2,1 2
120 3,9 3,1 2,7 2,5 2,3 2,2 2,1 2 2 1,9
Deg
F - value: the quotient of MS factor and MS error or the
quotient factor of the explainable variation by the not
l i bl i ti
explainable variation

Evaluation of the Welding Strength
A nderson-Darling Normality Test
V ariance 26,540
Sk 0 008799
A -Squared 0,49
P-V alue 0,205
Mean 15,040
StDev 5,152
Summary for StrengthStat
>Basic Statistics
>Graphical Summary
25201510
Skewness 0,008799
Kurtosis -0,908100
N 25
Minimum 7,000
1st Q uartile 11,000
Median 15,000
3rd Q uartile 19,000
Maximum 25,000
95% C onfidence Interv al for Mean
>Graphical Summary…
Median
Mean
12,913 17,167
95% C onfidence Interv al for Median
11,198 18,000
95% C onfidence Interv al for StDev
4,023 7,167
95% Confidence Intervals
Data normal distributed,
Stat
>Control Charts
>Variable Charts for Individuals
18,016,515,013,512,0
Data normal distributed,
P-value = 0,205
>Variable Charts for Individuals
>I-MR…
Increase of the welding strength
until 18 A followed by a strong drop
35
30
15A 16A 17A 18A 19A
I Chart of Strength by Current
25 UCL=24,57
1
I Chart of Strength
until 18 A followed by a strong drop
ividualValue
25
20
15
UCL=18,78
1
dividualValue
20
15
_
X=15,04
Ind
24222018161412108642
10
5
0
_
X=10,8
LCL=2,82
1
Ind
24222018161412108642
10
5
LCL=5,51
Observation
24222018161412108642
Observation
24222018161412108642
Stat
>ANOVA
>Test for Equal Variances…
15A Test Statistic 0,93
P-Value 0,920
Bartlett's Test
Test for Equal Variances for Strength
17A
16A
rent
,
Test Statistic 0,32
P-Value 0,863
Levene's Test
18A
17A
Curr
Because of the high P-values H0
will be not rejected!
No significant differences of the
19A
1614121086420
95% Bonferroni Confidence Intervals for StDevs
variation between the factor levels!
Bartlett’s Test to be used for normal
distributed data, see also the
session window!
session window!

Stat
>ANOVA
>One way
One-way ANOVA: Strength versus Current
>One way…
Source DF SS MS F P
Current 4 475,76 118,94 14,76 0,000
Error 20 161,20 8,06
Total 24 636,96
S = 2,839 R-Sq = 74,69% R-Sq(adj) = 69,63%
Individual 95% CIs For Mean Based on
Pooled StDevPooled StDev
Level N Mean StDev ------+---------+---------+---------+---
15A 5 9,800 3,347 (-----*----)
16A 5 15,400 3,130 (----*----)
17A 5 17 600 2 074 (----*----)17A 5 17,600 2,074 ( )
18A 5 21,600 2,608 (----*----)
19A 5 10,800 2,864 (-----*----)
------+---------+---------+---------+---
10,0 15,0 20,0 25,010,0 15,0 20,0 25,0
The current change explains about 70 to 75 % of the welding strength variation. The P-value of 0
indicates that H0 has to be rejected with high certainty. The F-value shows that the explained
variation is about 15 times higher than the unexplained variation. The graphic of the confidence
intervals shows significant differences between 15, 16 and 18 Ampere. The area around 18
Ampere has to be investigated further in detail to understand when the strength decrease starts.
Stat
>ANOVA
>One-way…
>[Graphs]
Boxplot of Strength Individual Value Plot of Strength vs Current
25
20
25
20
15
Strength
15
Strength
19A18A17A16A15A
10
5
C ent
19A18A17A16A15A
10
5
C ent
Current Current

Normal Probability Plot Versus Fits
Residual Plots for Strength
99
90
50
ercent
N 25
AD 0,519
P-Value 0,170
5,0
2,5
0,0
esidual
5,02,50,0-2,5-5,0
10
1
Residual
P
20,017,515,012,510,0
-2,5
-5,0
Fitted Value
Re
6,0
4,5
cy
5,0
2,5
l
Histogram Versus Order
3,0
1,5
0 0
Frequenc
0,0
-2,5
5 0
Residua
420-2-4
0,0
Residual
24222018161412108642
-5,0
Observation Order
The residuals are normal distributed. The residuals versus fitted values and also versus order of
data don’t show obvious trends. The histogram shows no clear symmetry. For 25 values Minitab
uses to many intervals. At this point the number of intervals could be reduced to 5 (> Edit X
Scale > Binning) The histogram shows then a better symmetry
Scale… > Binning). The histogram shows then a better symmetry.
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons
Simultaneous confidence level = 73,57%
The groups are not significant different
in the pair wise comparison, if a 0 is
within the 95% confidence interval
15A subtracted from:
Lower Center Upper --------+---------+---------+---------+-
16A 1,855 5,600 9,345 (----*----)
within the 95% confidence interval.
, , , ( )
17A 4,055 7,800 11,545 (----*---)
18A 8,055 11,800 15,545 (----*---)
19A -2,745 1,000 4,745 (---*----)
--------+---------+---------+---------+-
-8 0 0 0 8 0 16 0-8,0 0,0 8,0 16,0
17A -1,545 2,200 5,945 (----*---)
18A 2 455 6 200 9 945 ( * )18A 2,455 6,200 9,945 (----*---)
19A -8,345 -4,600 -0,855 (---*----)
--------+---------+---------+---------+-
-8,0 0,0 8,0 16,0
18A 0,255 4,000 7,745 (----*----)
19A -10,545 -6,800 -3,055 (----*---)
--------+---------+---------+---------+-
-8,0 0,0 8,0 16,0
19A -14 545 -10 800 -7 055 (----*---)
19A 14,545 10,800 7,055 ( * )
--------+---------+---------+---------+-
-8,0 0,0 8,0 16,0

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